# The signed random-to-top operator on tensor space (draft)

Darij Grinberg

##  1 Introduction

The purpose of this note is to answer a question I asked in 2010 in
[Grinbe10]. It concerns the kernel of a certain operator on the tensor algebra
$T\left(L\right)$ of a free module $L$ over a commutative ring $\mathbf{k}$
(an operator that picks out a factor from a tensor and moves it to the front,
and takes an alternating sum of the results ranging over all factors – an
algebraic version of what probabilists call the “random-to-top shuffle”,
albeit with signs). Originating in pure curiosity, this question has been
tempting me with its apparent connections to the random-to-top and random-to-
random shuffling operators as studied in [ReSaWe14] and [Schock02]. I have not
(yet?) grown any wiser from these connections, but I was able to answer the
question (with some help from a 1950 paper by Specht [Specht50]), and the
answer seems (to me) to be interesting enough to warrant some publicity.

We shall not use the notations of [Grinbe10] (indeed, our notations in the
following will be incompatible with those in [Grinbe10]).

###  1.1 Outline

Let me outline what will be proven in this note. (Everything mentioned here
will be defined again in more detail later on.)

We fix a commutative ring $\mathbf{k}$ and a $\mathbf{k}$-module $L$, and we
consider the tensor algebra $T\left(L\right)$. We define a $\mathbf{k}$-linear
map $\mathbf{t}:T\left(L\right)\rightarrow T\left(L\right)$ by setting

$\mathbf{t}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{k}\right)=\sum\limits_{i=1}^{k}\left(-1\right)^{i-1}u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{k}$

for all pure tensors $u_{1}\otimes u_{2}\otimes\cdots\otimes u_{k}\in
T\left(L\right)$. 111This map $\mathbf{t}$ is the map $-T$ defined in
[Grinbe10].

Roughly speaking, what the map $\mathbf{t}$ does to a pure tensor can be
described as picking out the $i$-th tensorand and moving it to the front of
the tensor, multiplying the new tensor with $\left(-1\right)^{i-1}$, and
summing the result over all $i$’s. Thus, the map $\mathbf{t}$ is a signed
multilinear analogue of the “random-to-top shuffling operator” known from
combinatorics (essentially the element
$\sum\limits_{i=1}^{k}\left(-1\right)^{i-1}\left(1,2,\ldots,i\right)$ of the
group algebra $\mathbf{k}S_{k}$, acting on $L^{\otimes k}$). Alternatively, we
can view the restriction of $\mathbf{t}$ to $L^{\otimes k}$ as the action of
the “random-to-top shuffling element”
$\sum\limits_{i=1}^{k}\left(1,2,\ldots,i\right)\in\mathbf{k}S_{k}$ (this is
the antipode of the $\Xi_{n,1}$ of [Schock02]) on $L^{\otimes k}$ via the
$\mathbf{k}S_{k}$-module structure on $L^{\otimes k}$ which is given by
permuting the $k$ tensorands, twisted with the sign representation. For $L$ a
free $\mathbf{k}$-module of rank $\geq k$, this $\mathbf{k}S_{k}$-module
structure is faithful, and so from the behavior of $\mathbf{t}$ one can draw
conclusions about the random-top-shuffling operator.

Our main goal in the first few sections is to describe the kernel of the map
$\mathbf{t}$. One of our first observations (Proposition 3.4) is that if $L$
is a free $\mathbf{k}$-module, then this kernel is the set of all tensors
$U\in T\left(L\right)$ which are annihilated by $\partial_{g}^{\prime}$ for
all $g\in L$, where the maps $\partial_{g}^{\prime}$ are certain “interior
product” operators (see Definition 3.1 for a precise definition). This rather
simple fact will come out useful in understanding
$\operatorname*{Ker}\mathbf{t}$. 222This is very close to what I wrote about
bilinear forms in [Grinbe10], but the use of bilinearity instead of linearity
was a red herring.

Once this is proven, we will come to the actual description of
$\operatorname*{Ker}\mathbf{t}$. The tensor algebra $T\left(L\right)$ is
$\mathbb{Z}_{2}$-graded, and thus a superalgebra. Thus, any two elements $U$
and $V$ of $T\left(L\right)$ have a supercommutator
$\left[U,V\right]_{\operatorname*{s}}$ (which equals
$UV-\left(-1\right)^{nm}VU$ if $U$ and $V$ are homogeneous of degrees $n$ and
$m$; otherwise it is determined by $\mathbf{k}$-bilinearity). Define

  * •

a sequence $\left(L_{1},L_{2},L_{3},\ldots\right)$ of $\mathbf{k}$-submodules
of $T\left(L\right)$ recursively by $L_{1}=L$ and
$L_{i+1}=\left[L,L_{i}\right]_{\operatorname*{s}}$;

  * •

a $\mathbf{k}$-submodule $\overline{\mathfrak{g}}$ of $T\left(L\right)$ by
$\overline{\mathfrak{g}}=L_{2}+L_{3}+L_{4}+\cdots$;

  * •

a $\mathbf{k}$-submodule $P$ of $T\left(L\right)$ as the $\mathbf{k}$-linear
span of all $xx$ for $x\in L$.

(Notice that if $2$ is invertible in the ground ring $\mathbf{k}$, then
$P\subseteq L_{2}\subseteq\overline{\mathfrak{g}}$.)

Then, $\operatorname*{Ker}\mathbf{t}$ is the $\mathbf{k}$-subalgebra of
$T\left(L\right)$ generated by $\overline{\mathfrak{g}}+P$, at least when $L$
is a free $\mathbf{k}$-module. This result (Theorem 6.10 below) will be proven
after several auxiliary observations. Our proof will rely on ideas of Wilhelm
Specht in his 1950 paper [Specht50] on (what would now be called) PI-algebras
(specifically, Sections V and VI of said paper). Specht characterized
“properly $n$-linear forms”333In the original: “eigentlich $n$-fach lineare
Formen”., which, in our notations, would correspond to multilinear elements of
$\operatorname*{Ker}\mathbf{t}$ when $L$ is the free $\mathbf{k}$-module
$\mathbf{k}^{n}$. (The correspondence is not immediate – Specht’s analogue of
the map $\mathbf{t}$ has no $\left(-1\right)^{i-1}$ signs.) The fact that we
consider arbitrary, not just multilinear, elements of $T\left(L\right)$
somewhat complicates our arguments (and prevents us from going as deep as
Specht did – e.g., we shall not find a basis for
$\operatorname*{Ker}\mathbf{t}$, although this appears to be doable using
Lyndon methods).

Then, we will study an “unsigned” analogue of the map $\mathbf{t}$. Namely, we
will define a $\mathbf{k}$-linear map
$\mathbf{t}^{\prime}:T\left(L\right)\rightarrow T\left(L\right)$ by setting

$\mathbf{t}^{\prime}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{k}\right)=\sum\limits_{i=1}^{k}u_{i}\otimes u_{1}\otimes
u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes u_{k}$

for all pure tensors $u_{1}\otimes u_{2}\otimes\cdots\otimes u_{k}\in
T\left(L\right)$. From a superalgebraic viewpoint, $\mathbf{t}$ and
$\mathbf{t}^{\prime}$ are particular cases of a common general construction,
but we will witness their kernels behaving differently when the additive group
of $\mathbf{k}$ is not torsionfree. I am not able to describe
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$ in the same generality
as $\operatorname*{Ker}\mathbf{t}$ (for arbitrary $\mathbf{k}$), but we will
see separate descriptions of
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$

  * •

in the case when the additive group of $\mathbf{k}$ is torsionfree (Theorem
7.22), and

  * •

in the case when $\mathbf{k}$ is a commutative $\mathbb{F}_{p}$-algebra for
some prime $p$ (Theorem 8.15).

Much of our reasoning related to $\operatorname*{Ker}\mathbf{t}$ will apply to
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$ as long as some changes
are made; supercommutators are replaced by commutators, the
$\mathbf{k}$-submodule $P$ is replaced by either $0$ (when $\mathbf{k}$ is
torsionfree) or the $\mathbf{k}$-submodule of $T\left(L\right)$ spanned by
$x^{p}$ for all $x\in L$ (when $\mathbf{k}$ is an $\mathbb{F}_{p}$-algebra).

It has come to my attention that the description of
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$ (Theorem 7.22) in the
case when $\mathbf{k}$ is a $\mathbb{Q}$-algebra is a consequence of Amy
Pang’s [Pang15, Theorem 5.1] (applied to $\mathcal{H}=T\left(L\right)$, $q=1$
and $j=0$). (Actually, when $\mathbf{k}$ is a $\mathbb{Q}$-algebra, [Pang15,
Theorem 5.1] gives a basis of each eigenspace of $\mathbf{t}^{\prime}$, thus
in particular a basis of
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$, but this latter basis
is what one would obtain using the symmetrization map and the Poincaré-
Birkhoff-Witt theorem from Theorem 7.22. Conversely, Pang’s [Pang15, Theorem
5.1] immediately yields Theorem 7.22 when $\mathbf{k}$ is a
$\mathbb{Q}$-algebra.)

###  1.2 Acknowledgments

Communications with Franco Saliola (who is studying the random-to-top shuffle
as an element of the group algebra of the symmetric group) have helped
rekindle my interest in this question. Parts of what comes below might be
related (or even equivalent) to some of his recent unpublished work. The
SageMath computer algebra system [sage] was used to verify some of the results
below (in small degrees and for small ranks of $L$) before a general proof was
found.

##  2 The map $\mathbf{t}$

######  Convention 2.1.

For the rest of this note, we fix a commutative ring $\mathbf{k}$. All
unadorned tensor signs (i.e., signs $\otimes$ without a subscript) in the
following are understood to mean $\otimes_{\mathbf{k}}$.

We also fix a $\mathbf{k}$-module $L$.

######  Definition 2.1.

Let $T\left(L\right)$ be the tensor algebra of $L$ (over $\mathbf{k}$). Notice
that $T\left(L\right)=L^{\otimes 0}\oplus L^{\otimes 1}\oplus L^{\otimes
2}\oplus\cdots$ as $\mathbf{k}$-module.

The tensor algebra $T\left(L\right)$ is $\mathbb{Z}$-graded (an element of
$L^{\otimes n}$ has degree $n$) and $\mathbb{Z}_{2}$-graded (here an element
of $L^{\otimes n}$ has degree $n\operatorname{mod}2$).

(Here, I use $\mathbb{Z}_{2}$ to denote the quotient ring
$\mathbb{Z}/2\mathbb{Z}$, and I use the notation $n\operatorname{mod}2$ to
denote the remainder class of $n$ modulo $2$.)

######  Definition 2.2.

Let $\mathbf{t}:T\left(L\right)\rightarrow T\left(L\right)$ be the
$\mathbf{k}$-linear map which acts on pure tensors according to the formula

$\mathbf{t}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{k}\right)=\sum\limits_{i=1}^{k}\left(-1\right)^{i-1}u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{k}$

(for all $k\in\mathbb{N}$ and $u_{1},u_{2},\ldots,u_{k}\in L$), where the
$\widehat{u_{i}}$ is not an actual tensorand but rather a symbol that means
that the factor $u_{i}$ is removed from the place where it would usually occur
in the tensor product. (This is clearly well-defined.) Thus, $\mathbf{t}$ is a
graded $\mathbf{k}$-module endomorphism of $T\left(L\right)$.

##  3 $\operatorname*{Ker}\mathbf{t}$ is the joint kernel of the
superderivations $\partial_{g}$

######  Definition 3.1.

Let $L^{\ast}$ denote the dual $\mathbf{k}$-module
$\operatorname*{Hom}\left(L,\mathbf{k}\right)$ of $L$. If $g\in L^{\ast}$,
then we define a $\mathbf{k}$-linear map
$\partial_{g}:T\left(L\right)\rightarrow T\left(L\right)$ by

$\partial_{g}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{k}\right)=\sum\limits_{i=1}^{k}\left(-1\right)^{i-1}g\left(u_{i}\right)\cdot
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{k}$

for all $k\in\mathbb{N}$ and $u_{1},u_{2},\ldots,u_{k}\in L$. (Again, it is
easy to check that this is well-defined.)

The map $\partial_{g}$ is a lift to $T\left(L\right)$ of what is called the
“interior product by $g$” on the Clifford algebra of $L$ endowed with any
quadratic form. This observation provides a motivation for studying
$\partial_{g}$; it will not be used below.

For any $g\in L^{\ast}$, the map $\partial_{g}$ is a superderivation of the
superalgebra $T\left(L\right)$. Rather than explaining these notions, let us
state explicitly what the previous sentence means:

######  Proposition 3.2.

Let $g\in L^{\ast}$.

(a) Then, $\partial_{g}\left(1\right)=0$.

(b) Also, if $n\in\mathbb{N}$, $a\in L^{\otimes n}$ and $b\in
T\left(L\right)$, then
$\partial_{g}\left(ab\right)=\partial_{g}\left(a\right)b+\left(-1\right)^{n}a\partial_{g}\left(b\right)$.

######  Proof 3.3 (Proof of Proposition 3.2.).

We give this straightforward proof purely for the sake of completeness.

(a) The unity $1$ of the ring $T\left(L\right)$ is the empty tensor product.
The definition of $\partial_{g}$ thus shows that $\partial_{g}\left(1\right)$
is an empty sum, and therefore equal to $0$. This proves Proposition 3.2 (a).

(b) Let $n\in\mathbb{N}$, $a\in L^{\otimes n}$ and $b\in T\left(L\right)$. We
need to prove the equality
$\partial_{g}\left(ab\right)=\partial_{g}\left(a\right)b+\left(-1\right)^{n}a\partial_{g}\left(b\right)$.
Since this equality is $\mathbf{k}$-linear in each of $a$ and $b$, we can WLOG
assume that both $a$ and $b$ are pure tensors. Assume this. Since $a\in
L^{\otimes n}$ is a pure tensor, we have $a=a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}$ for some $a_{1},a_{2},\ldots,a_{n}\in L$.
Consider these $a_{1},a_{2},\ldots,a_{n}$. Since $b$ is a pure tensor, we have
$b=b_{1}\otimes b_{2}\otimes\cdots\otimes b_{m}$ for some $m\in\mathbb{N}$ and
$b_{1},b_{2},\ldots,b_{m}\in L$. Consider this $m$ and these
$b_{1},b_{2},\ldots,b_{m}$. Multiplying the equalities $a=a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}$ and $b=b_{1}\otimes b_{2}\otimes\cdots\otimes
b_{m}$, we obtain

$\displaystyle ab$ $\displaystyle=\left(a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\right)\left(b_{1}\otimes b_{2}\otimes\cdots\otimes b_{m}\right)$
$\displaystyle=a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\otimes
b_{1}\otimes b_{2}\otimes\cdots\otimes b_{m}.$

Hence,

$\displaystyle\partial_{g}\left(ab\right)$
$\displaystyle=\partial_{g}\left(a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\otimes b_{1}\otimes b_{2}\otimes\cdots\otimes b_{m}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot\underbrace{a_{1}\otimes
a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes a_{n}\otimes
b_{1}\otimes b_{2}\otimes\cdots\otimes b_{m}}_{=\left(a_{1}\otimes
a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes
a_{n}\right)\cdot\left(b_{1}\otimes b_{2}\otimes\cdots\otimes b_{m}\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=n+1}^{n+m}\left(-1\right)^{i-1}g\left(b_{i-n}\right)\cdot\underbrace{a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}\otimes b_{1}\otimes
b_{2}\otimes\cdots\otimes\widehat{b_{i-n}}\otimes\cdots\otimes
b_{m}}_{=\left(a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\right)\cdot\left(b_{1}\otimes
b_{2}\otimes\cdots\otimes\widehat{b_{i-n}}\otimes\cdots\otimes b_{m}\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of
}\partial_{g}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot\left(a_{1}\otimes
a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes
a_{n}\right)\cdot\underbrace{\left(b_{1}\otimes b_{2}\otimes\cdots\otimes
b_{m}\right)}_{=b}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=n+1}^{n+m}\left(-1\right)^{i-1}g\left(b_{i-n}\right)\cdot\underbrace{\left(a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}\right)}_{=a}\cdot\left(b_{1}\otimes
b_{2}\otimes\cdots\otimes\widehat{b_{i-n}}\otimes\cdots\otimes b_{m}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot\left(a_{1}\otimes
a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes a_{n}\right)\cdot
b$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=n+1}^{n+m}\left(-1\right)^{i-1}g\left(b_{i-n}\right)\cdot
a\cdot\left(b_{1}\otimes
b_{2}\otimes\cdots\otimes\widehat{b_{i-n}}\otimes\cdots\otimes b_{m}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot\left(a_{1}\otimes
a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes a_{n}\right)\cdot
b$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=1}^{m}\left(-1\right)^{i+n-1}g\left(b_{i}\right)\cdot
a\cdot\left(b_{1}\otimes
b_{2}\otimes\cdots\otimes\widehat{b_{i}}\otimes\cdots\otimes b_{m}\right)$ (1)
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{here, we have substituted
}i\text{ for }i-n\text{ in the second sum}\right).$

But $a=a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}$ shows that

$\displaystyle\partial_{g}\left(a\right)$
$\displaystyle=\partial_{g}\left(a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot
a_{1}\otimes a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes
a_{n}$

(by the definition of $\partial_{g}$). Similarly,

$\partial_{g}\left(b\right)=\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}g\left(b_{i}\right)\cdot
b_{1}\otimes b_{2}\otimes\cdots\otimes\widehat{b_{i}}\otimes\cdots\otimes
b_{m}.$

Hence,

$\displaystyle\underbrace{\partial_{g}\left(a\right)}_{=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot
a_{1}\otimes a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes
a_{n}}b+\left(-1\right)^{n}a\underbrace{\partial_{g}\left(b\right)}_{=\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}g\left(b_{i}\right)\cdot
b_{1}\otimes b_{2}\otimes\cdots\otimes\widehat{b_{i}}\otimes\cdots\otimes
b_{m}}$
$\displaystyle=\underbrace{\left(\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot
a_{1}\otimes a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes
a_{n}\right)b}_{=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot\left(a_{1}\otimes
a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes a_{n}\right)\cdot
b}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\underbrace{\left(-1\right)^{n}a\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}g\left(b_{i}\right)\cdot
b_{1}\otimes b_{2}\otimes\cdots\otimes\widehat{b_{i}}\otimes\cdots\otimes
b_{m}}_{=\sum\limits_{i=1}^{m}\left(-1\right)^{i+n-1}g\left(b_{i}\right)\cdot
a\cdot\left(b_{1}\otimes
b_{2}\otimes\cdots\otimes\widehat{b_{i}}\otimes\cdots\otimes b_{m}\right)}$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}g\left(a_{i}\right)\cdot\left(a_{1}\otimes
a_{2}\otimes\cdots\otimes\widehat{a_{i}}\otimes\cdots\otimes a_{n}\right)\cdot
b$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=1}^{m}\left(-1\right)^{i+n-1}g\left(b_{i}\right)\cdot
a\cdot\left(b_{1}\otimes
b_{2}\otimes\cdots\otimes\widehat{b_{i}}\otimes\cdots\otimes b_{m}\right)$
$\displaystyle=\partial_{g}\left(ab\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by
(\ref{pf.prop.1.ab1})}\right).$

This proves Proposition 3.2 (b).

The maps $\partial_{g}$ relate to $\operatorname*{Ker}\mathbf{t}$ as follows:

######  Proposition 3.4.

(a) We have $\partial_{g}\left(\operatorname{Ker}\mathbf{t}\right)=0$ for
every $g\in L^{\ast}$.

(b) Assume that $L$ is a free $\mathbf{k}$-module. Then,

$\operatorname*{Ker}\mathbf{t}=\left\\{U\in T\left(L\right)\ \mid\
\partial_{g}\left(U\right)=0\text{ for every }g\in L^{\ast}\right\\}.$

######  Proof 3.5 (Proof of Proposition 3.4.).

For every $g\in L^{\ast}$, we define a $\mathbf{k}$-module homomorphism
$\mathbf{c}_{g}:T\left(L\right)\rightarrow T\left(L\right)$ by the formula

$\mathbf{c}_{g}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{k}\right)=\left\\{\begin{array}[c]{c}0,\text{ if }k=0;\\\
g\left(u_{1}\right)u_{2}\otimes u_{3}\otimes\cdots\otimes u_{k},\text{ if
}k>0\end{array}\right.$

for all $k\in\mathbb{N}$ and $u_{1},u_{2},\ldots,u_{k}\in L$. (Again, this is
well-defined for rather obvious reasons.)

It is now easy to prove that $\partial_{g}=\mathbf{c}_{g}\circ\mathbf{t}$ for
every $g\in L^{\ast}$ 444Indeed, $\partial_{g}$ and
$\mathbf{c}_{g}\circ\mathbf{t}$ are two $\mathbf{k}$-linear maps which equal
each other on each pure tensor (this can be checked readily).. Thus, every
$g\in L^{\ast}$ satisfies
$\operatorname{Ker}\mathbf{t}\subseteq\operatorname{Ker}\left(\partial_{g}\right)$,
so that $\partial_{g}\left(\operatorname{Ker}\mathbf{t}\right)=0$. This proves
Proposition 3.4 (a).

(b) The definition of $\mathbf{c}_{g}$ easily yields

$\mathbf{c}_{g}\left(vU\right)=g\left(v\right)U\ \ \ \ \ \ \ \ \ \ \text{for
any }v\in L\text{ and }U\in T\left(L\right).$ (2)

We denote by $\overline{T\left(L\right)}$ the $\mathbf{k}$-submodule
$L^{\otimes 1}\oplus L^{\otimes 2}\oplus L^{\otimes 3}\oplus\cdots$ of
$T\left(L\right)$. We notice that
$\mathbf{t}\left(T\left(L\right)\right)\subseteq\overline{T\left(L\right)}$
(since $\mathbf{t}$ is a graded map which equals $0$ in degree $0$).

Now, let $V\in\left\\{U\in T\left(L\right)\ \mid\
\partial_{g}\left(U\right)=0\text{ for every }g\in L^{\ast}\right\\}$. We are
going to show that $V\in\operatorname*{Ker}\mathbf{t}$.

We have $V\in\left\\{U\in T\left(L\right)\ \mid\
\partial_{g}\left(U\right)=0\text{ for every }g\in L^{\ast}\right\\}$. Hence,
$V\in T\left(L\right)$, and we have $\partial_{g}\left(V\right)=0$ for every
$g\in L^{\ast}$.

We fix a basis $\left(e_{i}\right)_{i\in I}$ of the $\mathbf{k}$-module $L$
(this exists since $L$ is free). For every $i\in I$, let $e_{i}^{\ast}\in
L^{\ast}$ be the $\mathbf{k}$-linear map $L\rightarrow\mathbf{k}$ which sends
$e_{i}$ to $1$ and sends all other $e_{j}$ to $0$. In other words,
$e_{i}^{\ast}\in L^{\ast}$ satisfies
$e_{i}^{\ast}\left(e_{j}\right)=\delta_{j,i}$ for all $j\in I$. (If $I$ is
finite, then $\left(e_{i}^{\ast}\right)_{i\in I}$ is thus the basis of
$L^{\ast}$ dual to the basis $\left(e_{i}\right)_{i\in I}$ of $L$. For
arbitrary $I$, it might not be a basis of $L^{\ast}$, but we don’t care.)

We have
$\mathbf{t}\left(V\right)\in\mathbf{t}\left(T\left(L\right)\right)\subseteq\overline{T\left(L\right)}$.
Thus, we can write the tensor $\mathbf{t}\left(V\right)$ in the form
$\mathbf{t}\left(V\right)=\sum\limits_{i\in I}e_{i}V_{i}$ for some tensors
$V_{i}\in T\left(L\right)$ (all but finitely many of which are zero). Consider
these $V_{i}$. For every $j\in I$, we can apply the map
$\mathbf{c}_{e_{j}^{\ast}}$ to both sides of the equality
$\mathbf{t}\left(V\right)=\sum\limits_{i\in I}e_{i}V_{i}$ and obtain

$\mathbf{c}_{e_{j}^{\ast}}\left(\mathbf{t}\left(V\right)\right)=\mathbf{c}_{e_{j}^{\ast}}\left(\sum\limits_{i\in
I}e_{i}V_{i}\right)=\sum\limits_{i\in
I}\underbrace{\mathbf{c}_{e_{j}^{\ast}}\left(e_{i}V_{i}\right)}_{\begin{subarray}{c}=e_{j}^{\ast}\left(e_{i}\right)V_{i}\\\
\text{(by (\ref{pf.prop.2.1}))}\end{subarray}}=\sum\limits_{i\in
I}\underbrace{e_{j}^{\ast}\left(e_{i}\right)}_{=\delta_{i,j}}V_{i}=\sum\limits_{i\in
I}\delta_{i,j}V_{i}=V_{j}.$

But every $g\in L^{\ast}$ satisfies $\partial_{g}\left(V\right)=0$. Since
$\underbrace{\partial_{g}}_{=\mathbf{c}_{g}\circ\mathbf{t}}\left(V\right)=\mathbf{c}_{g}\left(\mathbf{t}\left(V\right)\right)$,
this rewrites as follows: Every $g\in L^{\ast}$ satisfies
$\mathbf{c}_{g}\left(\mathbf{t}\left(V\right)\right)=0$. Applied to
$g=e_{j}^{\ast}$, this yields
$\mathbf{c}_{e_{j}^{\ast}}\left(\mathbf{t}\left(V\right)\right)=0$ for every
$j\in I$. Compared with
$\mathbf{c}_{e_{j}^{\ast}}\left(\mathbf{t}\left(V\right)\right)=V_{j}$, this
yields $V_{j}=0$. This holds for each $j\in I$. Thus, $V_{i}=0$ for each $i\in
I$. Thus, $\mathbf{t}\left(V\right)=\sum\limits_{i\in
I}e_{i}\underbrace{V_{i}}_{=0}=0$, so that
$V\in\operatorname*{Ker}\mathbf{t}$.

Now, let us forget that we fixed $V$. We thus have shown that
$V\in\operatorname*{Ker}\mathbf{t}$ for every $V\in\left\\{U\in
T\left(L\right)\ \mid\ \partial_{g}\left(U\right)=0\text{ for every }g\in
L^{\ast}\right\\}$. In other words, we have shown the inclusion

$\left\\{U\in T\left(L\right)\ \mid\ \partial_{g}\left(U\right)=0\text{ for
every }g\in L^{\ast}\right\\}\subseteq\operatorname*{Ker}\mathbf{t}.$

The reverse inclusion also holds (its proof is a trivial application of
$\partial_{g}=\mathbf{c}_{g}\circ\mathbf{t}$). Combined, the two inclusions
yield

$\operatorname*{Ker}\mathbf{t}=\left\\{U\in T\left(L\right)\ \mid\
\partial_{g}\left(U\right)=0\text{ for every }g\in L^{\ast}\right\\}.$

Proposition 3.4 (b) is thus proven.

##  4 $\operatorname*{Ker}\mathbf{t}$ is a subalgebra of $T\left(L\right)$

We now want to describe $\operatorname*{Ker}\mathbf{t}$. Clearly,
$\operatorname*{Ker}\mathbf{t}$ is a graded $\mathbf{k}$-submodule of
$T\left(L\right)$ (since $\mathbf{t}$ is a graded map). We first introduce
some notations:

######  Definition 4.1.

We define a $\mathbf{k}$-bilinear map
$\operatorname*{scomm}:T\left(L\right)\times T\left(L\right)\rightarrow
T\left(L\right)$ as follows: We set

$\operatorname*{scomm}\left(U,V\right)=UV-\left(-1\right)^{nm}VU$

for any $n\in\mathbb{N}$, $m\in\mathbb{N}$, $U\in L^{\otimes n}$ and $V\in
L^{\otimes m}$. (This is easily seen to be well-defined.)

If $U\in T\left(L\right)$ and $V\in T\left(L\right)$, then we denote the
tensor $\operatorname*{scomm}\left(U,V\right)$ by
$\left[U,V\right]_{\operatorname*{s}}$, and we call it the supercommutator of
$U$ and $V$. Thus, $\left[U,V\right]_{\operatorname*{s}}$ depends
$\mathbf{k}$-linearly on each of $U$ and $V$, and satisfies

$\left[U,V\right]_{\operatorname*{s}}=UV-\left(-1\right)^{nm}VU$

for any $n\in\mathbb{N}$, $m\in\mathbb{N}$, $U\in L^{\otimes n}$ and $V\in
L^{\otimes m}$.

This definition should not surprise anyone familiar with superalgebras. In
fact, recall that the $\mathbf{k}$-algebra $T\left(L\right)$ is
$\mathbb{Z}_{2}$-graded; thus, $T\left(L\right)$ is a
$\mathbf{k}$-superalgebra. Consequently, it has a supercommutator. This
supercommutator is precisely the map $\operatorname*{scomm}$ that we have just
defined. We just preferred not to use the language of superalgebras.

Clearly,
$\left[U,V\right]_{\operatorname*{s}}=-\left(-1\right)^{nm}\left[V,U\right]_{\operatorname*{s}}$
for any $n\in\mathbb{N}$, $m\in\mathbb{N}$, $U\in L^{\otimes n}$ and $V\in
L^{\otimes m}$. The supercommutator $\operatorname*{scomm}$ (or, differently
written, $\left[\cdot,\cdot\right]_{\operatorname*{s}}$) furthermore satisfies
the following analogue of the Leibniz and Jacobi identities:

######  Proposition 4.2.

Let $n\in\mathbb{N}$, $m\in\mathbb{N}$, $U\in L^{\otimes n}$, $V\in L^{\otimes
m}$ and $W\in T\left(L\right)$. Then:

(a) We have
$\left[U,VW\right]_{\operatorname*{s}}=\left[U,V\right]_{\operatorname*{s}}W+\left(-1\right)^{nm}V\left[U,W\right]_{\operatorname*{s}}$.

(b) We have
$\left[U,\left[V,W\right]_{\operatorname*{s}}\right]_{\operatorname*{s}}=\left[\left[U,V\right]_{\operatorname*{s}},W\right]_{\operatorname*{s}}+\left(-1\right)^{nm}\left[V,\left[U,W\right]_{\operatorname*{s}}\right]_{\operatorname*{s}}$.

######  Proof 4.3 (Proof of Proposition 4.2.).

This is straightforward and left to the reader.

######  Definition 4.4.

We will apply the same notations to supercommutators that are usually applied
to commutators. For instance, when $P$ and $Q$ are two $\mathbf{k}$-submodules
of $T\left(L\right)$, we will use the notation
$\left[P,Q\right]_{\operatorname*{s}}$ for the $\mathbf{k}$-linear span of the
supercommutators $\left[U,V\right]_{\operatorname*{s}}$ for $U\in P$ and $V\in
Q$ (just as one commonly writes $\left[P,Q\right]$ for the $\mathbf{k}$-linear
span of the commutators $\left[U,V\right]$ for $U\in P$ and $V\in Q$).

######  Proposition 4.5.

(a) We have $L^{\otimes 0}\subseteq\operatorname*{Ker}\mathbf{t}$.

(b) We have
$\operatorname*{Ker}\mathbf{t}\cdot\operatorname*{Ker}\mathbf{t}\subseteq\operatorname*{Ker}\mathbf{t}$.

(c) We have
$\left[L,L\right]_{\operatorname*{s}}\subseteq\operatorname*{Ker}\mathbf{t}$.

(d) We have
$\left[L,\operatorname*{Ker}\mathbf{t}\right]_{\operatorname*{s}}\subseteq\operatorname*{Ker}\mathbf{t}$.

(e) We have $xx\in\operatorname*{Ker}\mathbf{t}$ for each $x\in L$.

######  Proof 4.6 (Proof of Proposition 4.5.).

(a) This is obvious.

(b) For every $U\in T\left(L\right)$, we define a $\mathbf{k}$-linear map
$\mathbf{i}_{U}:T\left(L\right)\rightarrow T\left(L\right)$ by the formula

$\mathbf{i}_{U}\left(v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{k}\right)=\left\\{\begin{array}[c]{c}0,\text{ if }k=0;\\\ v_{1}\cdot
U\cdot\left(v_{2}\otimes v_{3}\otimes\cdots\otimes v_{k}\right),\text{ if
}k>0\end{array}\right.$

for all $k\in\mathbb{N}$ and $v_{1},v_{2},\ldots,v_{k}\in L$. Then, it is
straightforward to see that

$\mathbf{t}\left(UV\right)=\mathbf{t}\left(U\right)V+\left(-1\right)^{n}\mathbf{i}_{U}\left(\mathbf{t}\left(V\right)\right)$
(3)

for all $n\in\mathbb{N}$, $U\in L^{\otimes n}$ and $V\in T\left(L\right)$
555Proof of (3): Let $n\in\mathbb{N}$, $U\in L^{\otimes n}$ and $V\in
T\left(L\right)$. We need to prove the equality (3). Since this equality is
$\mathbf{k}$-linear in each of $U$ and $V$ (because $\mathbf{i}_{U}$ is
$\mathbf{k}$-linear in $U$), we can WLOG assume that both $U$ and $V$ are pure
tensors. Assume this. Since $U\in L^{\otimes n}$ is a pure tensor, we have
$U=u_{1}\otimes u_{2}\otimes\cdots\otimes u_{n}$ for some
$u_{1},u_{2},\ldots,u_{n}\in L$. Consider these $u_{1},u_{2},\ldots,u_{n}$.
Since $V$ is a pure tensor, we have $V=v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{m}$ for some $m\in\mathbb{N}$ and $v_{1},v_{2},\ldots,v_{m}\in L$. Consider
this $m$ and these $v_{1},v_{2},\ldots,v_{m}$. Multiplying the equalities
$U=u_{1}\otimes u_{2}\otimes\cdots\otimes u_{n}$ and $V=v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{m}$, we obtain  $\displaystyle UV$
$\displaystyle=\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{n}\right)\left(v_{1}\otimes v_{2}\otimes\cdots\otimes v_{m}\right)$
$\displaystyle=u_{1}\otimes u_{2}\otimes\cdots\otimes u_{n}\otimes
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{m}.$ Hence,
$\displaystyle\mathbf{t}\left(UV\right)$
$\displaystyle=\mathbf{t}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{n}\otimes v_{1}\otimes v_{2}\otimes\cdots\otimes v_{m}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}\underbrace{u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}\otimes v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{m}}_{=\left(u_{i}\otimes u_{1}\otimes
u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}\right)\cdot\left(v_{1}\otimes v_{2}\otimes\cdots\otimes v_{m}\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=n+1}^{n+m}\left(-1\right)^{i-1}\underbrace{v_{i-n}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes u_{n}\otimes v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i-n}}\otimes\cdots\otimes
v_{m}}_{=v_{i-n}\cdot\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{n}\right)\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i-n}}\otimes\cdots\otimes v_{m}\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of
}\mathbf{t}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}\left(u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}\right)\cdot\underbrace{\left(v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{m}\right)}_{=V}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=n+1}^{n+m}\left(-1\right)^{i-1}v_{i-n}\cdot\underbrace{\left(u_{1}\otimes
u_{2}\otimes\cdots\otimes u_{n}\right)}_{=U}\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i-n}}\otimes\cdots\otimes v_{m}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}\left(u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}\right)\cdot V$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=n+1}^{n+m}\left(-1\right)^{i-1}v_{i-n}\cdot
U\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i-n}}\otimes\cdots\otimes v_{m}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}\left(u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}\right)\cdot V$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=1}^{m}\left(-1\right)^{i+n-1}v_{i}\cdot
U\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes v_{m}\right)$ (4)
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{here, we have substituted
}i\text{ for }i-n\text{ in the second sum}\right).$ But $U=u_{1}\otimes
u_{2}\otimes\cdots\otimes u_{n}$ shows that
$\displaystyle\mathbf{t}\left(U\right)$
$\displaystyle=\mathbf{t}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{n}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}$ (by the definition of $\mathbf{t}$). Similarly,
$\mathbf{t}\left(V\right)=\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}v_{i}\otimes
v_{1}\otimes v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes
v_{m}.$ Applying the map $\mathbf{i}_{U}$ to both sides of this equality, we
obtain  $\displaystyle\mathbf{i}_{U}\left(\mathbf{t}\left(V\right)\right)$
$\displaystyle=\mathbf{i}_{U}\left(\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}v_{i}\otimes
v_{1}\otimes v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes
v_{m}\right)$
$\displaystyle=\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}\underbrace{\mathbf{i}_{U}\left(v_{i}\otimes
v_{1}\otimes v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes
v_{m}\right)}_{\begin{subarray}{c}=v_{i}\cdot U\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes v_{m}\right)\\\
\text{(by the definition of }\mathbf{i}_{U}\text{, since }m>0\text{ (because
}i\in\left\\{1,2,\ldots,m\right\\}\text{))}\end{subarray}}$
$\displaystyle=\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}v_{i}\cdot
U\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes v_{m}\right).$
Hence,
$\displaystyle\underbrace{\mathbf{t}\left(U\right)}_{=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}}V+\left(-1\right)^{n}\underbrace{\mathbf{i}_{U}\left(\mathbf{t}\left(V\right)\right)}_{=\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}v_{i}\cdot
U\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes v_{m}\right)}$
$\displaystyle=\underbrace{\left(\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}\right)V}_{=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}\left(u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}\right)\cdot V}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\underbrace{\left(-1\right)^{n}\sum\limits_{i=1}^{m}\left(-1\right)^{i-1}v_{i}\cdot
U\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes
v_{m}\right)}_{=\sum\limits_{i=1}^{m}\left(-1\right)^{i+n-1}v_{i}\cdot
U\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes v_{m}\right)}$
$\displaystyle=\sum\limits_{i=1}^{n}\left(-1\right)^{i-1}\left(u_{i}\otimes
u_{1}\otimes u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes
u_{n}\right)\cdot V$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\sum\limits_{i=1}^{m}\left(-1\right)^{i+n-1}v_{i}\cdot
U\cdot\left(v_{1}\otimes
v_{2}\otimes\cdots\otimes\widehat{v_{i}}\otimes\cdots\otimes v_{m}\right)$
$\displaystyle=\mathbf{t}\left(UV\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by
(\ref{pf.prop.3.b.tUV.pf.1})}\right).$ This proves (3)..

Now, we need to prove
$\operatorname*{Ker}\mathbf{t}\cdot\operatorname*{Ker}\mathbf{t}\subseteq\operatorname*{Ker}\mathbf{t}$.
In other words, we need to prove that $UV\in\operatorname*{Ker}\mathbf{t}$ for
all $U\in\operatorname*{Ker}\mathbf{t}$ and
$V\in\operatorname*{Ker}\mathbf{t}$. So let us fix
$U\in\operatorname*{Ker}\mathbf{t}$ and $V\in\operatorname*{Ker}\mathbf{t}$.
Since $\operatorname*{Ker}\mathbf{t}$ is a graded $\mathbf{k}$-submodule of
$T\left(L\right)$ (because $\mathbf{t}$ is a graded map), we can WLOG assume
that $U$ is homogeneous, i.e., that $U\in L^{\otimes n}$ for some
$n\in\mathbb{N}$. Assume this, and consider this $n$. From (3), we thus obtain
$\mathbf{t}\left(UV\right)=\underbrace{\mathbf{t}\left(U\right)}_{\begin{subarray}{c}=0\\\
\text{(since
}U\in\operatorname*{Ker}\mathbf{t}\text{)}\end{subarray}}V+\left(-1\right)^{n}\mathbf{i}_{U}\left(\underbrace{\mathbf{t}\left(V\right)}_{\begin{subarray}{c}=0\\\
\text{(since
}V\in\operatorname*{Ker}\mathbf{t}\text{)}\end{subarray}}\right)=0$, so that
$UV\in\operatorname*{Ker}\mathbf{t}$, just as we wished to prove. Proposition
4.5 (b) is thus proven.

(c) This is straightforward: For all $x,y\in L$, we have
$\left[x,y\right]_{\operatorname*{s}}=xy-\left(-1\right)^{1\cdot 1}yx=xy+yx$
and $\mathbf{t}\left(xy\right)=xy-yx$ and $\mathbf{t}\left(yx\right)=yx-xy$.
Thus, for all $x,y\in L$, we have

$\mathbf{t}\left(\underbrace{\left[x,y\right]_{\operatorname*{s}}}_{=xy+yx}\right)=\underbrace{\mathbf{t}\left(xy\right)}_{=xy-
yx}+\underbrace{\mathbf{t}\left(yx\right)}_{=yx-xy}=\left(xy-
yx\right)+\left(yx-xy\right)=0,$

and thus
$\left[x,y\right]_{\operatorname*{s}}\in\operatorname*{Ker}\mathbf{t}$. In
other words,
$\left[L,L\right]_{\operatorname*{s}}\subseteq\operatorname*{Ker}\mathbf{t}$.

(d) It is enough to show that
$\left[u,V\right]_{\operatorname*{s}}\in\operatorname*{Ker}\mathbf{t}$ for
every $u\in L$ and $V\in\operatorname*{Ker}\mathbf{t}$. So let $u\in L$ and
$V\in\operatorname*{Ker}\mathbf{t}$. Then, $\mathbf{t}\left(V\right)=0$.

Since $\operatorname*{Ker}\mathbf{t}$ is a graded $\mathbf{k}$-submodule of
$T\left(L\right)$, we WLOG assume that $V$ is homogeneous. That is, $V\in
L^{\otimes m}$ for some $m\in\mathbb{N}$. Consider this $m$. Applying (3) to
$n=1$ and $U=u$, we obtain

$\mathbf{t}\left(uV\right)=\underbrace{\mathbf{t}\left(u\right)}_{=u}V+\left(-1\right)^{1}\mathbf{i}_{u}\left(\underbrace{\mathbf{t}\left(V\right)}_{=0}\right)=uV.$

On the other hand, we can apply (3) to $m$, $V$ and $u$ instead of $n$, $U$
and $V$. As a result, we obtain

$\mathbf{t}\left(Vu\right)=\underbrace{\mathbf{t}\left(V\right)}_{=0}u+\left(-1\right)^{m}\mathbf{i}_{V}\left(\underbrace{\mathbf{t}\left(u\right)}_{=u}\right)=\left(-1\right)^{m}\underbrace{\mathbf{i}_{V}\left(u\right)}_{\begin{subarray}{c}=uV\\\
\text{(by the definition of
}\mathbf{i}_{V}\text{)}\end{subarray}}=\left(-1\right)^{m}uV.$

Now,

$\mathbf{t}\left(\underbrace{\left[u,V\right]_{\operatorname*{s}}}_{=uV-\left(-1\right)^{1\cdot
m}Vu}\right)=\underbrace{\mathbf{t}\left(uV\right)}_{=uV}-\left(-1\right)^{1\cdot
m}\underbrace{\mathbf{t}\left(Vu\right)}_{=\left(-1\right)^{m}uV}=uV-\underbrace{\left(-1\right)^{1\cdot
m}\left(-1\right)^{m}}_{=1}uV=0,$

so that
$\left[u,V\right]_{\operatorname*{s}}\in\operatorname*{Ker}\mathbf{t}$. This
completes our proof of Proposition 4.5 (d).

(e) This is also straightforward.

##  5 The submodules $\overline{\mathfrak{g}}$, $P$ and $\mathfrak{h}$

Parts (a) and (b) of Proposition 4.5 show that $\operatorname*{Ker}\mathbf{t}$
is a $\mathbf{k}$-subalgebra of $T\left(L\right)$. Parts (c) and (d) show that
nontrivial iterated supercommutators of elements of $L$ (that is, tensors in
$\left[L,L\right]_{\operatorname*{s}}$ or
$\left[L,\left[L,L\right]_{\operatorname*{s}}\right]_{\operatorname*{s}}$,
etc.) belong to $\operatorname*{Ker}\mathbf{t}$, and (by parts (a) and (b)) so
do their products. Part (e) shows that elements of the form $xx$ with $x\in L$
are in $\operatorname*{Ker}\mathbf{t}$ as well. Of course, $\mathbf{k}$-linear
combinations of elements of $\operatorname*{Ker}\mathbf{t}$ are also elements
of $\operatorname*{Ker}\mathbf{t}$. Our goal is to show that all elements of
$\operatorname*{Ker}\mathbf{t}$ are obtained in these ways. We shall, however,
first formalize and somewhat improve this goal.

######  Definition 5.1.

We recursively define a sequence $\left(L_{1},L_{2},L_{3},\ldots\right)$ of
$\mathbf{k}$-submodules of $T\left(L\right)$ as follows: We set $L_{1}=L$, and
$L_{i+1}=\left[L,L_{i}\right]_{\operatorname*{s}}$ for every positive integer
$i$.

For instance, $L_{2}=\left[L,L\right]_{\operatorname*{s}}$ and
$L_{3}=\left[L,L_{2}\right]_{\operatorname*{s}}=\left[L,\left[L,L\right]_{\operatorname*{s}}\right]_{\operatorname*{s}}$.

If you are familiar with Lie superalgebras, you will recognize
$L_{1}+L_{2}+L_{3}+\cdots$ as the Lie subsuperalgebra of $T\left(L\right)$
generated by $L$. I suspect that it is the free Lie superalgebra over $L$
(though I am not sure if this is unconditionally true).

By induction, it is clear that $L_{i}\subseteq L^{\otimes i}$ for every
positive integer $i$.

######  Definition 5.2.

Let $\overline{\mathfrak{g}}$ denote the $\mathbf{k}$-submodule
$L_{2}+L_{3}+L_{4}+\cdots$ of $T\left(L\right)$.

It is easy to see (but unnecessary for us) that $\overline{\mathfrak{g}}$ is a
Lie superalgebra under the supercommutator
$\left[\cdot,\cdot\right]_{\operatorname*{s}}$. (See Proposition 5.9 below.)

######  Definition 5.3.

Let $P$ denote the $\mathbf{k}$-submodule of $L^{\otimes 2}$ spanned by
elements of the form $x\otimes x$ with $x\in L$. Notice that $x\otimes x=xx$
in the $\mathbf{k}$-algebra $T\left(L\right)$ for every $x\in L$.

######  Proposition 5.4.

If $2$ is invertible in $\mathbf{k}$, then $P\subseteq
L_{2}\subseteq\overline{\mathfrak{g}}$.

######  Proof 5.5 (Proof of Proposition 5.4.).

Assume that $2$ is invertible in $\mathbf{k}$. Let $x\in L$. Then,
$xx=x\otimes x$ in $T\left(L\right)$, and since $x$ has degree $1$, we have
$\left[x,x\right]_{\operatorname*{s}}=xx-\left(-1\right)^{1\cdot
1}xx=xx+xx=2xx=2x\otimes x$. Hence, $x\otimes
x=\dfrac{1}{2}\left[\underbrace{x}_{\in L},\underbrace{x}_{\in
L}\right]_{\operatorname*{s}}\in\dfrac{1}{2}\underbrace{\left[L,L\right]_{\operatorname*{s}}}_{=L_{2}}\subseteq
L_{2}$. Since we have proven this for each $x\in L$, we thus obtain
$P\subseteq L_{2}$ (since $P$ is spanned by the $x\otimes x$ with $x\in L$).
Combined with $L_{2}\subseteq\overline{\mathfrak{g}}$, this proves Proposition
5.4.

######  Definition 5.6.

Let $\mathfrak{h}=\overline{\mathfrak{g}}+P$.

Proposition 5.4 shows that $\mathfrak{h}=\overline{\mathfrak{g}}$ if $2$ is
invertible in $\mathbf{k}$. (But in general, $\mathfrak{h}$ can be larger than
$\overline{\mathfrak{g}}$.) Obviously,
$\overline{\mathfrak{g}}\subseteq\mathfrak{h}$ and $P\subseteq\mathfrak{h}$.

######  Proposition 5.7.

(a) We have
$\left[L,\overline{\mathfrak{g}}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$.

(b) Furthermore,
$\left[L,P\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$ and
$\left[L,\mathfrak{h}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}\subseteq\mathfrak{h}$.

######  Proof 5.8 (Proof of Proposition 5.7.).

(a) Since $\overline{\mathfrak{g}}=L_{2}+L_{3}+L_{4}+\cdots=\sum\limits_{i\geq
2}L_{i}$, we have

$\left[L,\overline{\mathfrak{g}}\right]_{\operatorname*{s}}=\left[L,\sum\limits_{i\geq
2}L_{i}\right]_{\operatorname*{s}}=\sum\limits_{i\geq
2}\underbrace{\left[L,L_{i}\right]_{\operatorname*{s}}}_{=L_{i+1}}=\sum\limits_{i\geq
2}L_{i+1}=\sum\limits_{i\geq 3}L_{i}\subseteq\sum\limits_{i\geq
2}L_{i}=\overline{\mathfrak{g}}.$

Thus, Proposition 5.7 (a) is proven.

Also, we have
$\left[\overline{\mathfrak{g}},L\right]_{\operatorname*{s}}=\left[L,\overline{\mathfrak{g}}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$.

(b) It is easy to check that

$\left[U,xx\right]_{\operatorname*{s}}=\left[\left[U,x\right]_{\operatorname*{s}},x\right]_{\operatorname*{s}}$
(5)

for every $U\in T\left(L\right)$ and every $x\in L$. Hence, for every $U\in L$
and $x\in L$, we have

$\left[U,xx\right]_{\operatorname*{s}}=\left[\left[\underbrace{U}_{\in
L},\underbrace{x}_{\in L}\right]_{\operatorname*{s}},\underbrace{x}_{\in
L}\right]_{\operatorname*{s}}\in\left[\underbrace{\left[L,L\right]_{\operatorname*{s}}}_{=L_{2}\subseteq\overline{\mathfrak{g}}},L\right]_{\operatorname*{s}}\subseteq\left[\overline{\mathfrak{g}},L\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}.$

Thus, $\left[L,P\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$
(since $P$ is the $\mathbf{k}$-linear span of all elements of
$T\left(L\right)$ of the form $x\otimes x=xx$ with $x\in L$).

Since $\mathfrak{h}=\overline{\mathfrak{g}}+P$, we have
$\left[L,\mathfrak{h}\right]_{\operatorname*{s}}=\underbrace{\left[L,\overline{\mathfrak{g}}\right]_{\operatorname*{s}}}_{\subseteq\overline{\mathfrak{g}}}+\underbrace{\left[L,P\right]_{\operatorname*{s}}}_{\subseteq\overline{\mathfrak{g}}}\subseteq\overline{\mathfrak{g}}+\overline{\mathfrak{g}}\subseteq\overline{\mathfrak{g}}\subseteq\mathfrak{h}$.
This finishes the proof of Proposition 5.7 (b).

The following proposition will not be used in the following, but provides an
interesting aside (and explains why we are using Fraktur letters for
$\overline{\mathfrak{g}}$ and $\mathfrak{h}$):

######  Proposition 5.9.

(a) We have
$\left[\mathfrak{h},\mathfrak{h}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}\subseteq\mathfrak{h}$.

(b) The four $\mathbf{k}$-submodules $\overline{\mathfrak{g}}$,
$\mathfrak{h}$, $\overline{\mathfrak{g}}+L$ and $\mathfrak{h}+L$ of
$T\left(L\right)$ are invariant under the supercommutator
$\left[\cdot,\cdot\right]_{\operatorname*{s}}$. (In superalgebraic terms, they
are Lie subsuperalgebras of $T\left(L\right)$ (with the supercommutator
$\left[\cdot,\cdot\right]_{\operatorname*{s}}$ as the Lie bracket).)

######  Proof 5.10 (Proof of Proposition 5.9.).

(a) First, we notice that

$\left[P,\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}.$ (6)
666Proof of (6): It is clearly enough to show that
$\left[P,L_{i}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$ for
all positive integers $i$. So let us do this. Let $i$ be a positive integer.
We need to show that
$\left[P,L_{i}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$. In
other words, we need to show that
$\left[xx,L_{i}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$
for every $x\in L$ (because the $\mathbf{k}$-module $P$ is spanned by elements
of the form $x\otimes x=xx$ with $x\in L$). So fix $x\in L$. Then,
$\displaystyle\left[xx,L_{i}\right]_{\operatorname*{s}}$
$\displaystyle=\left[L_{i},xx\right]_{\operatorname*{s}}=\left[\left[L_{i},\underbrace{x}_{\in
L}\right]_{\operatorname*{s}},\underbrace{x}_{\in
L}\right]_{\operatorname*{s}}\ \ \ \ \ \ \ \ \ \ \left(\text{by
(\ref{pf.prop.5.b.Uxx})}\right)$
$\displaystyle\subseteq\left[\underbrace{\left[L_{i},L\right]_{\operatorname*{s}}}_{=\left[L,L_{i}\right]_{\operatorname*{s}}=L_{i+1}},L\right]_{\operatorname*{s}}=\left[L_{i+1},L\right]_{\operatorname*{s}}=\left[L,L_{i+1}\right]_{\operatorname*{s}}=L_{i+2}$
$\displaystyle\subseteq L_{2}+L_{3}+L_{4}+\cdots=\overline{\mathfrak{g}}.$
This completes our proof of (6).

Next, we notice that any two positive integers $i$ and $j$ satisfy

$\left[L_{i},L_{j}\right]_{\operatorname*{s}}\subseteq L_{i+j}.$ (7) 777Proof
of (7): We shall prove (7) by induction over $j$: Induction base: For every
positive integer $i$, we have
$\left[L_{i},\underbrace{L_{1}}_{=L}\right]_{\operatorname*{s}}=\left[L_{i},L\right]_{\operatorname*{s}}=\left[L,L_{i}\right]_{\operatorname*{s}}=L_{i+1}$.
Thus, (7) holds for $j=1$. This completes the induction base. Induction step:
Let $J$ be a positive integer. Assume that (7) is proven for $j=J$. We now
need to prove (7) for $j=J+1$. We have assumed that (7) is proven for $j=J$.
In other words,  $\left[L_{i},L_{J}\right]_{\operatorname*{s}}\subseteq
L_{i+J}\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }i.$ (8) Now, let
$i$ be a positive integer. Then, $L_{i}$, $L_{J}$ and $L$ are graded
$\mathbf{k}$-submodules of $T\left(L\right)$, and we have
$\displaystyle\left[L_{i},\underbrace{L_{J+1}}_{=\left[L,L_{J}\right]_{\operatorname*{s}}}\right]_{\operatorname*{s}}$
$\displaystyle=\left[L_{i},\left[L,L_{J}\right]_{\operatorname*{s}}\right]_{\operatorname*{s}}\subseteq\left[\underbrace{\left[L_{i},L\right]_{\operatorname*{s}}}_{=\left[L,L_{i}\right]_{\operatorname*{s}}=L_{i+1}},L_{J}\right]_{\operatorname*{s}}+\left[L,\underbrace{\left[L_{i},L_{J}\right]_{\operatorname*{s}}}_{\begin{subarray}{c}\subseteq
L_{i+J}\\\ \text{(by
(\ref{pf.prop.6.a.LiLj.pf.1}))}\end{subarray}}\right]_{\operatorname*{s}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{by Proposition \ref{prop.jacobi}
{(b)}}\right)$
$\displaystyle\subseteq\underbrace{\left[L_{i+1},L_{J}\right]_{\operatorname*{s}}}_{\begin{subarray}{c}\subseteq
L_{\left(i+1\right)+J}\\\ \text{(by (\ref{pf.prop.6.a.LiLj.pf.1}), applied
to}\\\ i+1\text{ instead of
}i\text{)}\end{subarray}}+\underbrace{\left[L,L_{i+J}\right]_{\operatorname*{s}}}_{=L_{i+J+1}=L_{i+\left(J+1\right)}}$
$\displaystyle\subseteq\underbrace{L_{\left(i+1\right)+J}}_{=L_{i+\left(J+1\right)}}+L_{i+\left(J+1\right)}=L_{i+\left(J+1\right)}+L_{i+\left(J+1\right)}=L_{i+\left(J+1\right)}.$
In other words, (7) holds for $j=J+1$. This completes the induction step.
Thus, (7) is proven by induction.

Now,

$\displaystyle\left[\sum\limits_{i\geq 1}L_{i},\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}$ $\displaystyle=\left[\sum\limits_{i\geq
1}L_{i},\sum\limits_{j\geq
1}L_{j}\right]_{\operatorname*{s}}=\sum\limits_{i\geq 1}\sum\limits_{j\geq
1}\underbrace{\left[L_{i},L_{j}\right]_{\operatorname*{s}}}_{\begin{subarray}{c}\subseteq
L_{i+j}\\\ \text{(by
(\ref{pf.prop.6.a.LiLj}))}\end{subarray}}\subseteq\sum\limits_{i\geq
1}\sum\limits_{j\geq 1}L_{i+j}$ $\displaystyle\subseteq\sum\limits_{k\geq
2}L_{k}\ \ \ \ \ \ \ \ \ \ \left(\text{since }i+j\geq 2\text{ for any }i\geq
1\text{ and }j\geq 1\right)$
$\displaystyle=L_{2}+L_{3}+L_{4}+\cdots=\overline{\mathfrak{g}}.$

Recall now that
$\overline{\mathfrak{g}}=L_{2}+L_{3}+L_{4}+\cdots=\sum\limits_{i\geq
2}L_{i}\subseteq\sum\limits_{i\geq 1}L_{i}$. Thus,

$\left[\overline{\mathfrak{g}},\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}\subseteq\left[\sum\limits_{i\geq
1}L_{i},\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}.$

Since $\mathfrak{h}=\overline{\mathfrak{g}}+P$, we have

$\left[\mathfrak{h},\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}=\underbrace{\left[\overline{\mathfrak{g}},\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}}_{\subseteq\overline{\mathfrak{g}}}+\underbrace{\left[P,\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}\\\
\text{(by
(\ref{pf.prop.6.a.PLi}))}\end{subarray}}\subseteq\overline{\mathfrak{g}}+\overline{\mathfrak{g}}=\overline{\mathfrak{g}}.$

Since $\overline{\mathfrak{g}}\subseteq\sum\limits_{i\geq 1}L_{i}$, we now
have

$\left[\mathfrak{h},\overline{\mathfrak{g}}\right]_{\operatorname*{s}}\subseteq\left[\mathfrak{h},\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}.$ (9)

But we also have $L=L_{1}\subseteq\sum\limits_{i\geq 1}L_{i}$ and thus

$\left[\mathfrak{h},L\right]_{\operatorname*{s}}\subseteq\left[\mathfrak{h},\sum\limits_{i\geq
1}L_{i}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}.$ (10)

From this, we easily obtain

$\left[\mathfrak{h},P\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$
888Proof. It is clearly enough to show that
$\left[\mathfrak{h},xx\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$
for every $x\in L$ (since the $\mathbf{k}$-module $P$ is spanned by elements
of the form $x\otimes x=xx$ for $x\in L$). So let $x\in L$. Then,
$\displaystyle\left[\mathfrak{h},xx\right]_{\operatorname*{s}}$
$\displaystyle\subseteq\left[\left[\mathfrak{h},\underbrace{x}_{\in
L}\right]_{\operatorname*{s}},\underbrace{x}_{\in
L}\right]_{\operatorname*{s}}\ \ \ \ \ \ \ \ \ \ \left(\text{by
(\ref{pf.prop.5.b.Uxx})}\right)$
$\displaystyle\subseteq\left[\underbrace{\left[\mathfrak{h},L\right]_{\operatorname*{s}}}_{\subseteq\overline{\mathfrak{g}}},L\right]_{\operatorname*{s}}\subseteq\left[\overline{\mathfrak{g}},L\right]_{\operatorname*{s}}=\left[L,\overline{\mathfrak{g}}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}\
\ \ \ \ \ \ \ \ \ \left(\text{by Proposition \ref{prop.5} {(a)}}\right),$ qed.

.

Now, using $\mathfrak{h}=\overline{\mathfrak{g}}+P$ again, we obtain

$\left[\mathfrak{h},\mathfrak{h}\right]_{\operatorname*{s}}=\underbrace{\left[\mathfrak{h},\overline{\mathfrak{g}}\right]_{\operatorname*{s}}}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}\\\
\text{(by
(\ref{pf.prop.6.a.4}))}\end{subarray}}+\underbrace{\left[\mathfrak{h},P\right]_{\operatorname*{s}}}_{\subseteq\overline{\mathfrak{g}}}\subseteq\overline{\mathfrak{g}}+\overline{\mathfrak{g}}=\overline{\mathfrak{g}}.$

This proves Proposition 5.9 (a).

(b) We need to show that
$\left[\overline{\mathfrak{g}},\overline{\mathfrak{g}}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$,
$\left[\mathfrak{h},\mathfrak{h}\right]_{\operatorname*{s}}\subseteq\mathfrak{h}$,
$\left[\overline{\mathfrak{g}}+L,\overline{\mathfrak{g}}+L\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}+L$
and
$\left[\mathfrak{h}+L,\mathfrak{h}+L\right]_{\operatorname*{s}}\subseteq\mathfrak{h}+L$.

The relation
$\left[\overline{\mathfrak{g}},\overline{\mathfrak{g}}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$
follows immediately from
$\left[\underbrace{\overline{\mathfrak{g}}}_{\subseteq\mathfrak{h}},\underbrace{\overline{\mathfrak{g}}}_{\subseteq\mathfrak{h}}\right]_{\operatorname*{s}}\subseteq\left[\mathfrak{h},\mathfrak{h}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}$.
The relation
$\left[\mathfrak{h},\mathfrak{h}\right]_{\operatorname*{s}}\subseteq\mathfrak{h}$
follows from
$\left[\mathfrak{h},\mathfrak{h}\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}\subseteq\mathfrak{h}$.

We have

$\displaystyle\left[\mathfrak{h}+L,\mathfrak{h}+L\right]_{\operatorname*{s}}$
$\displaystyle=\underbrace{\left[\mathfrak{h},\mathfrak{h}\right]_{\operatorname*{s}}}_{\subseteq\overline{\mathfrak{g}}}+\underbrace{\left[\mathfrak{h},L\right]_{\operatorname*{s}}}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}\\\
\text{(by
(\ref{pf.prop.6.a.6}))}\end{subarray}}+\underbrace{\left[L,\mathfrak{h}\right]_{\operatorname*{s}}}_{\begin{subarray}{c}=\left[\mathfrak{h},L\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}\\\
\text{(by
(\ref{pf.prop.6.a.6}))}\end{subarray}}+\underbrace{\left[L,L\right]_{\operatorname*{s}}}_{\begin{subarray}{c}=L_{2}\subseteq
L_{2}+L_{3}+L_{4}+\cdots\\\ =\overline{\mathfrak{g}}\end{subarray}}$
$\displaystyle\subseteq\overline{\mathfrak{g}}+\overline{\mathfrak{g}}+\overline{\mathfrak{g}}+\overline{\mathfrak{g}}=\overline{\mathfrak{g}}.$

Thus,
$\left[\underbrace{\overline{\mathfrak{g}}}_{\subseteq\mathfrak{h}}+L,\underbrace{\overline{\mathfrak{g}}}_{\subseteq\mathfrak{h}}+L\right]_{\operatorname*{s}}\subseteq\left[\mathfrak{h}+L,\mathfrak{h}+L\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}\subseteq\overline{\mathfrak{g}}+L$
and
$\left[\mathfrak{h}+L,\mathfrak{h}+L\right]_{\operatorname*{s}}\subseteq\overline{\mathfrak{g}}\subseteq\mathfrak{h}\subseteq\mathfrak{h}+L$.
This proves everything we needed to show. Proposition 5.9 (b) is thus shown.

##  6 The kernel of $\mathbf{t}$

######  Definition 6.1.

If $U$ is any $\mathbf{k}$-submodule of a $\mathbf{k}$-algebra $A$, then we
define $U^{i}$ to be the $\mathbf{k}$-submodule $\underbrace{UU\cdots
U}_{i\text{ times}}$ of $A$ for every $i\in\mathbb{N}$. When $i=0$, this
$\mathbf{k}$-submodule means the copy of $\mathbf{k}$ in $A$ (that is, the
$\mathbf{k}$-linear span of $1_{A}$).

If $U$ is any $\mathbf{k}$-submodule of a $\mathbf{k}$-algebra $A$, then we
let $U^{\star}$ denote the $\mathbf{k}$-submodule $U^{0}+U^{1}+U^{2}+\cdots$
of $A$. This is the $\mathbf{k}$-subalgebra of $A$ generated by $U$.

The reader should keep in mind that $U^{\star}$ (the $\mathbf{k}$-subalgebra
of $A$ generated by $U$) and $U^{\ast}$ (the dual $\mathbf{k}$-module of $U$)
are two different things; they are to be distinguished by the shape of the
asterisk/star.

######  Proposition 6.2.

We have $\mathfrak{h}^{\star}\subseteq\operatorname*{Ker}\mathbf{t}$.

######  Proof 6.3 (Proof of Proposition 6.2.).

We have $P\subseteq\operatorname*{Ker}\mathbf{t}$ due to Proposition 4.5 (e).
Also,
$L_{2}=\left[L,L\right]_{\operatorname*{s}}\subseteq\operatorname*{Ker}\mathbf{t}$
by Proposition 4.5 (c). Using this and Proposition 4.5 (d), we can show that
$L_{i}\subseteq\operatorname*{Ker}\mathbf{t}$ for each $i\geq 2$ (by induction
over $i$). Thus,
$\overline{\mathfrak{g}}\subseteq\operatorname*{Ker}\mathbf{t}$ (since
$\overline{\mathfrak{g}}=L_{2}+L_{3}+L_{4}+\cdots$). Combined with
$P\subseteq\operatorname*{Ker}\mathbf{t}$, this yields
$\mathfrak{h}\subseteq\operatorname*{Ker}\mathbf{t}$ (since
$\mathfrak{h}=\overline{\mathfrak{g}}+P$). Since
$\operatorname*{Ker}\mathbf{t}$ is a $\mathbf{k}$-subalgebra of
$T\left(L\right)$, this yields that
$\mathfrak{h}^{\star}\subseteq\operatorname*{Ker}\mathbf{t}$. This proves
Proposition 6.2.

Our main goal is to prove that the inclusion in Proposition 6.2 actually
becomes an equality if the $\mathbf{k}$-module $L$ is free. First, we show
three simple lemmas:

######  Lemma 6.4.

Let $u\in L$. Let $S$ be a graded $\mathbf{k}$-submodule of $T\left(L\right)$
such that $uu\in S^{\star}$ and $\left[u,S\right]_{\operatorname*{s}}\subseteq
S^{\star}$. Then, $S^{\star}+S^{\star}u$ is a $\mathbf{k}$-subalgebra of
$T\left(L\right)$.

######  Proof 6.5 (Proof of Lemma 6.4.).

Clearly, $1\in S^{\star}\subseteq S^{\star}+S^{\star}u$. Hence, it only
remains to show that
$\left(S^{\star}+S^{\star}u\right)\left(S^{\star}+S^{\star}u\right)\subseteq
S^{\star}+S^{\star}u$.

We know that $S^{\star}$ is a $\mathbf{k}$-subalgebra of $T\left(L\right)$;
thus, $S^{\star}S^{\star}\subseteq S^{\star}$. Hence,
$S^{\star}\underbrace{uu}_{\in S^{\star}}\subseteq S^{\star}S^{\star}\subseteq
S^{\star}$.

We have

$uS\subseteq S^{\star}+Su$ (11) 999Proof of (11): It clearly suffices to show
that $us\in Su+S^{\star}$ for every $s\in S$. So let us fix some $s\in S$. We
can WLOG assume that $s$ is homogeneous (since $S$ is graded). Assume this.
Then, $u\in L^{\otimes n}$ for some $n\in\mathbb{N}$. Consider this $n$. Thus,
$\left[u,s\right]_{\operatorname*{s}}=us-\left(-1\right)^{1\cdot n}su$, so
that  $us=\left[u,\underbrace{s}_{\in
S}\right]_{\operatorname*{s}}+\underbrace{\left(-1\right)^{1\cdot n}s}_{\in
S}u\in\underbrace{\left[u,S\right]_{\operatorname*{s}}}_{\subseteq
S^{\star}}+Su\subseteq S^{\star}+Su.$ This proves (11).

. Now, for every $i\in\mathbb{N}$, we have

$uS^{i}\subseteq S^{\star}+S^{i}u$ (12) 101010Proof of (12): We will prove
(12) by induction over $i$: Induction base: We have
$u\underbrace{S^{0}}_{=\mathbf{k}}=u\mathbf{k}=\underbrace{\mathbf{k}}_{=S^{0}}u=S^{0}u\subseteq
S^{\star}+S^{0}u$. In other words, (12) holds for $i=0$. This completes the
induction base. Induction step: Let $I\in\mathbb{N}$. Assume that (12) is
proven for $i=I$. We need to prove (12) for $i=I+1$. We know that (12) is
proven for $i=I$. In other words, $uS^{I}\subseteq S^{\star}+S^{I}u$. Thus,
$\displaystyle u\underbrace{S^{I+1}}_{=S^{I}S}$
$\displaystyle=\underbrace{uS^{I}}_{\subseteq
S^{\star}+S^{I}u}S\subseteq\left(S^{\star}+S^{I}u\right)S\subseteq\underbrace{S^{\star}S}_{\subseteq
S^{\star}}+S^{I}\underbrace{uS}_{\begin{subarray}{c}\subseteq S^{\star}+Su\\\
\text{(by (\ref{pf.lem.Kert.1.1}))}\end{subarray}}$ $\displaystyle\subseteq
S^{\star}+S^{I}\left(S^{\star}+Su\right)\subseteq
S^{\star}+\underbrace{S^{I}S^{\star}}_{\subseteq
S^{\star}}+\underbrace{S^{I}S}_{=S^{I+1}}u\subseteq\underbrace{S^{\star}+S^{\star}}_{=S^{\star}}+S^{I+1}u=S^{\star}+S^{I+1}u.$
In other words, (12) holds for $i=I+1$. This completes the induction step, and
thus (12) is proven.

. Hence,

$uS^{\star}\subseteq S^{\star}+S^{\star}u$ 111111since
$S^{\star}=S^{0}+S^{1}+S^{2}+\cdots=\sum\limits_{i\in\mathbb{N}}S^{i}$ and
thus  $\displaystyle
u\underbrace{S^{\star}}_{=\sum\limits_{i\in\mathbb{N}}S^{i}}$
$\displaystyle=u\left(\sum\limits_{i\in\mathbb{N}}S^{i}\right)=\sum\limits_{i\in\mathbb{N}}\underbrace{uS^{i}}_{\begin{subarray}{c}\subseteq
S^{\star}+S^{i}u\\\ \text{(by
(\ref{pf.lem.Kert.1.2}))}\end{subarray}}\subseteq\sum\limits_{i\in\mathbb{N}}\left(S^{\star}+S^{i}u\right)$
$\displaystyle\subseteq
S^{\star}+\sum\limits_{i\in\mathbb{N}}S^{i}u=S^{\star}+\underbrace{\left(\sum\limits_{i\in\mathbb{N}}S^{i}\right)}_{=S^{\star}}u=S^{\star}+S^{\star}u$

. Thus,

$\displaystyle u\left(S^{\star}+S^{\star}u\right)$
$\displaystyle=\underbrace{uS^{\star}}_{\subseteq
S^{\star}+S^{\star}u}+\underbrace{uS^{\star}}_{\subseteq
S^{\star}+S^{\star}u}u$
$\displaystyle\subseteq\left(S^{\star}+S^{\star}u\right)+\left(S^{\star}+S^{\star}u\right)u=S^{\star}+S^{\star}u+S^{\star}u+\underbrace{S^{\star}uu}_{\subseteq
S^{\star}}$ $\displaystyle\subseteq
S^{\star}+S^{\star}u+S^{\star}u+S^{\star}=S^{\star}+S^{\star}u,$

so that

$\displaystyle\left(S^{\star}+S^{\star}u\right)\left(S^{\star}+S^{\star}u\right)$
$\displaystyle=S^{\star}\left(S^{\star}+S^{\star}u\right)+S^{\star}\underbrace{u\left(S^{\star}+S^{\star}u\right)}_{\subseteq
S^{\star}+S^{\star}u}$ $\displaystyle\subseteq
S^{\star}\left(S^{\star}+S^{\star}u\right)+S^{\star}\left(S^{\star}+S^{\star}u\right)=S^{\star}\left(S^{\star}+S^{\star}u\right)=\underbrace{S^{\star}S^{\star}}_{\subseteq
S^{\star}}+\underbrace{S^{\star}S^{\star}}_{\subseteq S^{\star}}u\subseteq
S^{\star}+S^{\star}u,$

and this completes our proof of Lemma 6.4.

######  Lemma 6.6.

Let $N$ be a graded $\mathbf{k}$-submodule of $T\left(L\right)$. Let $g\in
L^{\ast}$ be such that $\partial_{g}\left(N\right)=0$. Then,
$\partial_{g}\left(N^{\star}\right)=0$.

######  Proof 6.7 (Proof of Lemma 6.6.).

First, we claim that

$\partial_{g}\left(N^{i}\right)=0$ (13)

for all $i\in\mathbb{N}$.

Proof of (13). We shall prove (13) by induction over $i$:

Induction base: Proposition 3.2 (a) yields $\partial_{g}\left(1\right)=0$.
Thus, $\partial_{g}\left(N^{0}\right)=0$ (since the $\mathbf{k}$-module
$N^{0}$ is spanned by $1$). In other words, (13) holds for $i=0$. This
completes the induction base.

Induction step: Let $I\in\mathbb{N}$. Assume that (13) holds for $i=I$. We now
need to show that (13) holds for $i=I+1$.

In other words, we need to show that $\partial_{g}\left(N^{I+1}\right)=0$. For
this, it is clearly enough to prove that $\partial_{g}\left(UV\right)=0$ for
all $U\in N$ and $V\in N^{I}$ (since the $\mathbf{k}$-module $N^{I+1}=NN^{I}$
is spanned by elements of the form $UV$ with $U\in N$ and $V\in N^{I}$). So
let $U\in N$ and $V\in N^{I}$. Since both $N$ and $N^{I}$ are graded
$\mathbf{k}$-modules (because $N$ is graded), we can WLOG assume that $U$ and
$V$ are homogeneous elements of $T\left(L\right)$. Assume this, and let
$n\in\mathbb{N}$ and $m\in\mathbb{N}$ be such that $U\in L^{\otimes n}$ and
$V\in L^{\otimes m}$. Proposition 3.2 (b) (applied to $a=U$ and $b=V$) then
yields
$\partial_{g}\left(UV\right)=\partial_{g}\left(U\right)V+\left(-1\right)^{n}U\partial_{g}\left(V\right)$.
But we assumed that (13) holds for $i=I$. In other words,
$\partial_{g}\left(N^{I}\right)=0$, so that $\partial_{g}\left(V\right)=0$
(since $V\in N^{I}$). Also, $\partial_{g}\left(N\right)=0$ and thus
$\partial_{g}\left(U\right)=0$ (since $U\in N$). Now,
$\partial_{g}\left(UV\right)=\underbrace{\partial_{g}\left(U\right)}_{=0}V+\left(-1\right)^{n}U\underbrace{\partial_{g}\left(V\right)}_{=0}=0$,
which is exactly what we wanted to prove. Thus, the induction step is
complete, and (13) is proven.

Now, the definition of $N^{\star}$ yields
$N^{\star}=N^{0}+N^{1}+N^{2}+\cdots=\sum\limits_{i\in\mathbb{N}}N^{i}$, so
that
$\partial_{g}\left(N^{\star}\right)=\partial_{g}\left(\sum\limits_{i\in\mathbb{N}}N^{i}\right)=\sum\limits_{i\in\mathbb{N}}\underbrace{\partial_{g}\left(N^{i}\right)}_{\begin{subarray}{c}=0\\\
\text{(by (\ref{pf.lem.Kert.deriv0.gen.i}))}\end{subarray}}=0$. This proves
Lemma 6.6.

######  Lemma 6.8.

Let $M$ be a $\mathbf{k}$-submodule of $L$. Let $g\in L^{\ast}$ be such that
$g\left(M\right)=0$. Then:

(a) We have $\partial_{g}\left(\left(M+\mathfrak{h}\right)^{\star}\right)=0$.

(b) Let $q\in L$ be such that $g\left(q\right)=1$. Let $U_{0}$ and $U_{1}$ be
any two elements of $\left(M+\mathfrak{h}\right)^{\star}$. If
$\partial_{g}\left(U_{0}+U_{1}q\right)=0$, then $U_{1}=0$.

######  Proof 6.9 (Proof of Lemma 6.8.).

The $\mathbf{k}$-submodules $M$ and $\mathfrak{h}$ of $T\left(L\right)$ are
graded (for $M$, this is because $M\subseteq L\subseteq L^{\otimes 1}$). Thus,
$M+\mathfrak{h}$ is graded.

Proposition 6.2 yields
$\mathfrak{h}^{\star}\subseteq\operatorname*{Ker}\mathbf{t}$, so that
$\mathfrak{h}\subseteq\mathfrak{h}^{\star}\subseteq\operatorname*{Ker}\mathbf{t}$
and therefore
$\partial_{g}\left(\underbrace{\mathfrak{h}}_{\subseteq\operatorname*{Ker}\mathbf{t}}\right)\subseteq\partial_{g}\left(\operatorname*{Ker}\mathbf{t}\right)=0$
(by Proposition 3.4 (a)). Thus, $\partial_{g}\left(\mathfrak{h}\right)=0$.
Also, $\partial_{g}\left(v\right)=g\left(v\right)$ holds for every $v\in L$
(by the definition of $\partial_{g}$), and thus, in particular, for every
$v\in M$. Hence, $\partial_{g}\left(M\right)=g\left(M\right)=0$. Now,

$\partial_{g}\left(M+\mathfrak{h}\right)=\underbrace{\partial_{g}\left(M\right)}_{=0}+\underbrace{\partial_{g}\left(\mathfrak{h}\right)}_{=0}=0.$

Hence, Lemma 6.6 (applied to $N=M+\mathfrak{h}$) yields
$\partial_{g}\left(\left(M+\mathfrak{h}\right)^{\star}\right)=0$. This proves
Lemma 6.8 (a).

(b) We have $\partial_{g}\left(v\right)=g\left(v\right)$ for every $v\in L$
(by the definition of $\partial_{g}$). Applied to $v=q$, this yields
$\partial_{g}\left(q\right)=g\left(q\right)=1$.

We have $U_{0}\in\left(M+\mathfrak{h}\right)^{\star}$ and thus
$\partial_{g}\left(U_{0}\right)=0$ (since
$\partial_{g}\left(\left(M+\mathfrak{h}\right)^{\star}\right)=0$). Similarly,
$\partial_{g}\left(U_{1}\right)=0$.

Let $\mathbf{s}$ be the $\mathbf{k}$-module homomorphism $T\left(L\right)\to
T\left(L\right)$ which is given by

$\mathbf{s}\left(U\right)=\left(-1\right)^{n}U\ \ \ \ \ \ \ \ \ \ \text{for
any }n\in\mathbb{N}\text{ and any }U\in L^{\otimes n}.$

Clearly, this map $\mathbf{s}$ is an isomorphism of graded
$\mathbf{k}$-modules. (It is actually an involution and an isomorphism of
graded $\mathbf{k}$-algebras.)

It is easy to see that any $a\in T\left(L\right)$ and $b\in T\left(L\right)$
satisfy
$\partial_{g}\left(ab\right)=\partial_{g}\left(a\right)b+\mathbf{s}\left(a\right)\partial_{g}\left(b\right)$.
121212Indeed, it is sufficient to prove this when $a$ is homogeneous, but in
this case it follows from Proposition 3.2 (b). Applying this to $a=U_{1}$ and
$b=q$, we obtain
$\partial_{g}\left(U_{1}q\right)=\underbrace{\partial_{g}\left(U_{1}\right)}_{=0}q+\mathbf{s}\left(U_{1}\right)\underbrace{\partial_{g}\left(q\right)}_{=1}=0+\mathbf{s}\left(U_{1}\right)=\mathbf{s}\left(U_{1}\right)$.

Now, assume that $\partial_{g}\left(U_{0}+U_{1}q\right)=0$. Thus,
$0=\partial_{g}\left(U_{0}+U_{1}q\right)=\underbrace{\partial_{g}\left(U_{0}\right)}_{=0}+\underbrace{\partial_{g}\left(U_{1}q\right)}_{=\mathbf{s}\left(U_{1}\right)}=\mathbf{s}\left(U_{1}\right)$.
Since $\mathbf{s}$ is an isomorphism, this yields $0=U_{1}$. This proves Lemma
6.8 (b).

The following is our main result:

######  Theorem 6.10.

Assume that the $\mathbf{k}$-module $L$ is free. Then,
$\mathfrak{h}^{\star}=\operatorname*{Ker}\mathbf{t}$.

######  Proof 6.11 (Proof of Theorem 6.10.).

Proposition 6.2 shows that
$\mathfrak{h}^{\star}\subseteq\operatorname*{Ker}\mathbf{t}$. We thus only
need to verify that
$\operatorname*{Ker}\mathbf{t}\subseteq\mathfrak{h}^{\star}$. This means
proving that every $U\in\operatorname*{Ker}\mathbf{t}$ satisfies
$U\in\mathfrak{h}^{\star}$. So let us fix $U\in\operatorname*{Ker}\mathbf{t}$.

We know that the $\mathbf{k}$-module $L$ is free; it thus has a basis. Since
the tensor $U\in T\left(L\right)$ can be constructed using only finitely many
elements of this basis, we can thus WLOG assume that the basis of $L$ is
finite. Let us assume this, and let us denote this basis by
$\left(e_{1},e_{2},\ldots,e_{n}\right)$.

For every $i\in\left\\{1,2,\ldots,n\right\\}$, let
$e_{i}^{\ast}:L\rightarrow\mathbf{k}$ be the $\mathbf{k}$-linear map which
sends $e_{i}$ to $1$ and sends every other $e_{j}$ to $0$.

For every $k\in\left\\{0,1,\ldots,n\right\\}$, we let $M_{k}$ denote the
$\mathbf{k}$-submodule of $L$ spanned by $e_{1},e_{2},\ldots,e_{k}$. Thus,
$M_{0}=0$ and $M_{n}=L$. Clearly, every $k\in\left\\{1,2,\ldots,n\right\\}$
satisfies

$M_{k}=M_{k-1}+\mathbf{k}e_{k}.$ (14)

For every $k\in\left\\{0,1,\ldots,n\right\\}$, we set
$\mathfrak{h}_{k}=M_{k}+\mathfrak{h}$ and $H_{k}=\mathfrak{h}_{k}^{\star}$.

Notice that $\mathfrak{h}_{n}=M_{n}+\mathfrak{h}\supseteq M_{n}=L$ and thus
$H_{n}=\mathfrak{h}_{n}^{\star}\supseteq L^{\star}=T\left(L\right)$. Hence,
$H_{n}=T\left(L\right)$. Now, $U\in T\left(L\right)=H_{n}$.

On the other hand, the definition of $\mathfrak{h}_{0}$ yields
$\mathfrak{h}_{0}=\underbrace{M_{0}}_{=0}+\mathfrak{h}=\mathfrak{h}$ and thus
$H_{0}=\mathfrak{h}_{0}^{\star}=\mathfrak{h}^{\star}$.

We shall now prove that every $k\in\left\\{1,2,\ldots,n\right\\}$ satisfies
the following implication:

$\text{if }U\in H_{k}\text{, then }U\in H_{k-1}.$ (15)

Once this is proven, we will be able to argue that $U\in H_{n}$ (as we know),
thus $U\in H_{n-1}$ (by (15)), thus $U\in H_{n-2}$ (by (15) again), and so on
– until we finally arrive at $U\in H_{0}$. Since $H_{0}=\mathfrak{h}^{\star}$,
this rewrites as $U\in\mathfrak{h}^{\star}$, and thus we are done.

Therefore, it only remains to prove (15). So let us fix
$k\in\left\\{1,2,\ldots,n\right\\}$, and assume that $U\in H_{k}$. We now need
to show that $U\in H_{k-1}$.

We first notice that $\mathfrak{h}_{k-1}$ is a graded $\mathbf{k}$-submodule
of $T\left(L\right)$ (by its definition, since $M_{k}$ and $\mathfrak{h}$ are
graded $\mathbf{k}$-submodules). Furthermore, $e_{k}e_{k}=e_{k}\otimes
e_{k}\in P$ (by the definition of $P$), and thus $e_{k}e_{k}\in
P\subseteq\mathfrak{h}\subseteq
M_{k-1}+\mathfrak{h}=\mathfrak{h}_{k-1}\subseteq\mathfrak{h}_{k-1}^{\star}$.
Moreover,

$\displaystyle\left[e_{k},\mathfrak{h}_{k-1}\right]_{\operatorname*{s}}$
$\displaystyle=\left[\underbrace{e_{k}}_{\in
L},\underbrace{M_{k-1}}_{\subseteq
L}\right]_{\operatorname*{s}}+\left[\underbrace{e_{k}}_{\in
L},\mathfrak{h}\right]_{\operatorname*{s}}\ \ \ \ \ \ \ \ \ \
\left(\text{since }\mathfrak{h}_{k-1}=M_{k-1}+\mathfrak{h}\right)$
$\displaystyle\subseteq\underbrace{\left[L,L\right]_{\operatorname*{s}}}_{\begin{subarray}{c}=L_{2}\subseteq
L_{2}+L_{3}+L_{4}+\cdots\\\
=\overline{\mathfrak{g}}\end{subarray}}+\underbrace{\left[L,\mathfrak{h}\right]_{\operatorname*{s}}}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}\\\
\text{(by Proposition \ref{prop.5}
{(b)})}\end{subarray}}\subseteq\overline{\mathfrak{g}}+\overline{\mathfrak{g}}=\overline{\mathfrak{g}}\subseteq\mathfrak{h}$
$\displaystyle\subseteq
M_{k-1}+\mathfrak{h}=\mathfrak{h}_{k-1}\subseteq\mathfrak{h}_{k-1}^{\star}.$

Thus, Lemma 6.4 (applied to $u=e_{k}$ and $S=\mathfrak{h}_{k-1}$) yields that
$\mathfrak{h}_{k-1}^{\star}+\mathfrak{h}_{k-1}^{\star}e_{k}$ is a
$\mathbf{k}$-subalgebra of $T\left(L\right)$. In other words,
$H_{k-1}+H_{k-1}e_{k}$ is a $\mathbf{k}$-subalgebra of $T\left(L\right)$
(since $H_{k-1}=\mathfrak{h}_{k-1}^{\star}$). This $\mathbf{k}$-subalgebra
contains $\mathfrak{h}_{k}$ as a subset131313Proof. We have
$\mathfrak{h}_{k}=M_{k}+\mathfrak{h}$ and similarly
$\mathfrak{h}_{k-1}=M_{k-1}+\mathfrak{h}$. Thus,
$\displaystyle\mathfrak{h}_{k}$
$\displaystyle=\underbrace{M_{k}}_{\begin{subarray}{c}=M_{k-1}+\mathbf{k}e_{k}\\\
\text{(by
(\ref{pf.thm.Kert.Mk}))}\end{subarray}}+\mathfrak{h}=M_{k-1}+\mathbf{k}e_{k}+\mathfrak{h}=\underbrace{M_{k-1}+\mathfrak{h}}_{\begin{subarray}{c}=\mathfrak{h}_{k-1}\subseteq\mathfrak{h}_{k-1}^{\star}=H_{k-1}\\\
\text{(since }H_{k-1}\text{ was}\\\ \text{defined as
}\mathfrak{h}_{k-1}^{\star}\text{)}\end{subarray}}+\underbrace{\mathbf{k}}_{\subseteq
H_{k-1}}e_{k}$ $\displaystyle\subseteq H_{k-1}+H_{k-1}e_{k},$ qed., and thus
we have $H_{k}\subseteq H_{k-1}+H_{k-1}e_{k}$ 141414Proof. The
$\mathbf{k}$-subalgebra $H_{k-1}+H_{k-1}e_{k}$ of $T\left(L\right)$ contains
$\mathfrak{h}_{k}$ as a subset. Hence, it also contains
$\mathfrak{h}_{k}^{\star}$ as a subset (since $\mathfrak{h}_{k}^{\star}$ is
the $\mathbf{k}$-subalgebra of $T\left(L\right)$ generated by
$\mathfrak{h}_{k}$). In other words,
$H_{k-1}+H_{k-1}e_{k}\supseteq\mathfrak{h}_{k}^{\star}=H_{k}$, qed..
(Actually, $H_{k}=H_{k-1}+H_{k-1}e_{k}$, but we don’t need this.)

Now, $U\in H_{k}\subseteq H_{k-1}+H_{k-1}e_{k}$. Therefore, there exist two
elements $U_{0}$ and $U_{1}$ of $H_{k-1}$ such that $U=U_{0}+U_{1}e_{k}$.
Consider these $U_{0}$ and $U_{1}$. We have

$\partial_{e_{k}^{\ast}}\left(\underbrace{U_{0}+U_{1}e_{k}}_{=U}\right)=\partial_{e_{k}^{\ast}}\left(\underbrace{U}_{\in\operatorname*{Ker}\mathbf{t}}\right)\in\partial_{e_{k}^{\ast}}\left(\operatorname*{Ker}\mathbf{t}\right)=0$

(by Proposition 3.4 (a), applied to $g=e_{k}^{\ast}$), so that
$\partial_{e_{k}^{\ast}}\left(U_{0}+U_{1}e_{k}\right)=0$.

The elements $U_{0}$ and $U_{1}$ belong to
$H_{k-1}=\mathfrak{h}_{k-1}^{\star}=\left(M_{k-1}+\mathfrak{h}\right)^{\star}$
(since $\mathfrak{h}_{k-1}=M_{k-1}+\mathfrak{h}$). We can thus apply Lemma 6.8
(b) to $M=M_{k-1}$, $g=e_{k}^{\ast}$ and $q=e_{k}$ (since
$e_{k}^{\ast}\left(M_{k-1}\right)=0$ (because $M_{k-1}$ is spanned by
$e_{1},e_{2},\ldots,e_{k-1}$) and $e_{k}^{\ast}\left(e_{k}\right)=1$ and
$\partial_{e_{k}^{\ast}}\left(U_{0}+U_{1}e_{k}\right)=0$). As a result, we see
that $U_{1}=0$. Thus, $U=U_{0}+\underbrace{U_{1}}_{=0}e_{k}=U_{0}\in H_{k-1}$.
This completes the proof of (15). As we already mentioned, this finishes the
proof of Theorem 6.10.

Our idea to prove (15) goes back to Specht; it is, in some sense, an analogue
of the argument [Specht50, VI, Zweiter Schritt] where he gradually moves the
variables $x_{1},x_{2},\ldots,x_{n}$ (which can be roughly seen as
corresponding to our basis vectors $e_{1},e_{2},\ldots,e_{n}$) inside
commutators.

######  Remark 6.12.

We suspect that Theorem 6.10 still holds if we replace the word “free” by
“flat”. (We furthermore suspect that Lazard’s theorem might help prove this.)
However, Theorem 6.10 does not hold if we completely remove the condition that
$L$ be free. A counterexample (one in which $L^{\otimes
2}\cap\operatorname*{Ker}\mathbf{t}\not\subseteq\mathfrak{h}^{\star}$) can be
obtained from [Lundkv08, Example 4.6]. (Indeed, it is not hard to see that
$L^{\otimes 2}\cap\operatorname*{Ker}\mathbf{t}\subseteq\mathfrak{h}^{\star}$
holds if and only if, using the notations of [Lundkv08], the canonical map
$\Gamma_{\mathbf{k}}^{2}\left(L\right)\rightarrow\operatorname*{TS}\nolimits_{\mathbf{k}}^{2}\left(L\right)$
is surjective. And [Lundkv08, Example 4.6] shows that the latter does not
always hold.)

##  7 The even analogue

The map $\mathbf{t}$ we introduced in Definition 2.2 has a natural analogue,
which is obtained by removing the $\left(-1\right)^{i-1}$ signs from its
definition. (One might even argue that this analogue is more natural than
$\mathbf{t}$; at any rate, it is more directly related to both the random-to-
top shuffling operator and Specht’s construction.) We shall denote this
analogue by $\mathbf{t}^{\prime}$; here is its precise definition:

######  Definition 7.1.

Let $\mathbf{t}^{\prime}:T\left(L\right)\rightarrow T\left(L\right)$ be the
$\mathbf{k}$-linear map which acts on pure tensors according to the formula

$\mathbf{t}^{\prime}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{k}\right)=\sum\limits_{i=1}^{k}u_{i}\otimes u_{1}\otimes
u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes u_{k}$

(for all $k\in\mathbb{N}$ and $u_{1},u_{2},\ldots,u_{k}\in L$). (This is
clearly well-defined.) Thus, $\mathbf{t}^{\prime}$ is a graded
$\mathbf{k}$-module endomorphism of $T\left(L\right)$.

Those familiar with superalgebras will immediately notice that the maps
$\mathbf{t}$ and $\mathbf{t}^{\prime}$ can be seen as two particular cases of
one single unifying construction (a map defined on the tensor algebra of a
$\mathbf{k}$-supermodule, which, roughly speaking, differs from $\mathbf{t}$
in that the sign $\left(-1\right)^{i-1}$ is replaced by
$\left(-1\right)^{\left(\deg u_{1}+\deg u_{2}+\cdots+\deg
u_{i-1}\right)\left(\deg u_{i}\right)}$). We shall not follow this lead, but
rather study the map $\mathbf{t}^{\prime}$ separately. Unlike for the map
$\mathbf{t}$, I am not aware of a single general description of
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$ that works with no
restrictions on $\mathbf{k}$ (whenever $L$ is a free $\mathbf{k}$-module).
However, I can describe $\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$
when the additive group $\mathbf{k}$ is torsionfree and when $\mathbf{k}$ is
an $\mathbb{F}_{p}$-algebra for some prime $p$ (of course, in both cases, $L$
still has to be a free $\mathbf{k}$-module). The answers in these two cases
are different, and there does not seem to be an obvious way to extend the
argument to cases such as $\mathbf{k}=\mathbb{Z}/6\mathbb{Z}$.

Let us first present analogues of some objects we constructed earlier in our
study of $\mathbf{t}$. First, here is an analogue of Definition 3.1:

######  Definition 7.2.

Let $L^{\ast}$ denote the dual $\mathbf{k}$-module
$\operatorname*{Hom}\left(L,\mathbf{k}\right)$ of $L$. If $g\in L^{\ast}$,
then we define a $\mathbf{k}$-linear map
$\partial_{g}^{\prime}:T\left(L\right)\rightarrow T\left(L\right)$ by

$\partial_{g}^{\prime}\left(u_{1}\otimes u_{2}\otimes\cdots\otimes
u_{k}\right)=\sum\limits_{i=1}^{k}g\left(u_{i}\right)\cdot u_{1}\otimes
u_{2}\otimes\cdots\otimes\widehat{u_{i}}\otimes\cdots\otimes u_{k}$

for all $k\in\mathbb{N}$ and $u_{1},u_{2},\ldots,u_{k}\in L$. (Again, it is
easy to check that this is well-defined.)

The following proposition is an analogue of Proposition 3.2. (However, unlike
Proposition 3.2, it does not require $a$ to be homogeneous, since there are no
more signs that could change depending on its degree.)

######  Proposition 7.3.

Let $g\in L^{\ast}$.

(a) Then, $\partial_{g}^{\prime}\left(1\right)=0$.

(b) Also, if $a\in T\left(L\right)$ and $b\in T\left(L\right)$, then
$\partial_{g}^{\prime}\left(ab\right)=\partial_{g}^{\prime}\left(a\right)b+a\partial_{g}^{\prime}\left(b\right)$.

Of course, Proposition 7.3 (b) says precisely that $\partial_{g}^{\prime}$ is
a derivation $T\left(L\right)\rightarrow T\left(L\right)$.

Next comes the analogue of Proposition 3.4:

######  Proposition 7.4.

(a) We have
$\partial_{g}^{\prime}\left(\operatorname{Ker}\left(\mathbf{t}^{\prime}\right)\right)=0$
for every $g\in L^{\ast}$.

(b) Assume that $L$ is a free $\mathbf{k}$-module. Then,

$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)=\left\\{U\in
T\left(L\right)\ \mid\ \partial_{g}^{\prime}\left(U\right)=0\text{ for every
}g\in L^{\ast}\right\\}.$

######  Proof 7.5 (Proof of Proposition 7.4.).

The proof of Proposition 7.4 is analogous to that of Proposition 3.4.

The analogue of the supercommutator
$\left[\cdot,\cdot\right]_{\operatorname*{s}}$ is the plain commutator
$\left[\cdot,\cdot\right]$, which is defined by $\left[U,V\right]=UV-VU$ for
any $U\in T\left(L\right)$ and $V\in T\left(L\right)$. Again, this analogue is
less troublesome to work with than the supercommutator
$\left[\cdot,\cdot\right]_{\operatorname*{s}}$ because there is no dependence
on the degrees of $U$ and $V$ in its definition.

For the sake of completeness, we state an analogue of Proposition 4.2 (which
is really well-known):

######  Proposition 7.6.

Let $U\in T\left(L\right)$, $V\in T\left(L\right)$ and $W\in T\left(L\right)$.
Then:

(a) We have $\left[U,VW\right]=\left[U,V\right]W+V\left[U,W\right]$.

(b) We have
$\left[U,\left[V,W\right]\right]=\left[\left[U,V\right],W\right]+\left[V,\left[U,W\right]\right]$.

Next, we formulate an analogue to Proposition 4.5:

######  Proposition 7.7.

(a) We have $L^{\otimes
0}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.

(b) We have
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)\cdot\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.

(c) We have
$\left[L,L\right]\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.

(d) We have
$\left[L,\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)\right]\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.

Notice that Proposition 7.7 has no part (e), unlike Proposition 4.5. Indeed,
there is no analogue to Proposition 4.5 (e) for the map $\mathbf{t}^{\prime}$
in the general case. (We will later see something that can be regarded as an
analogue in the positive-characteristic case.)

We next define an analogue to the $\mathbf{k}$-submodules
$L_{1},L_{2},L_{3},\ldots$:

######  Definition 7.8.

We recursively define a sequence
$\left(L_{1}^{\prime},L_{2}^{\prime},L_{3}^{\prime},\ldots\right)$ of
$\mathbf{k}$-submodules of $T\left(L\right)$ as follows: We set
$L_{1}^{\prime}=L$, and $L_{i+1}^{\prime}=\left[L,L_{i}^{\prime}\right]$ for
every positive integer $i$.

For instance, $L_{2}^{\prime}=\left[L,L\right]$ and
$L_{3}^{\prime}=\left[L,L_{2}^{\prime}\right]=\left[L,\left[L,L\right]\right]$.

The $\mathbf{k}$-submodule
$L_{1}^{\prime}+L_{2}^{\prime}+L_{3}^{\prime}+\cdots$ of $T\left(L\right)$ is
a Lie subalgebra of $T\left(L\right)$. When $L$ is a free $\mathbf{k}$-module,
this Lie subalgebra is isomorphic to the free Lie algebra on $L$.

Of course, $L_{i}^{\prime}\subseteq L^{\otimes i}$ for every positive integer
$i$.

The analogue to $\overline{\mathfrak{g}}$ is what you would expect:

######  Definition 7.9.

Let $\overline{\mathfrak{g}}^{\prime}$ denote the $\mathbf{k}$-submodule
$L_{2}^{\prime}+L_{3}^{\prime}+L_{4}^{\prime}+\cdots$ of $T\left(L\right)$.

The analogue of $P$ is more interesting – in that it is the zero module
$0\subseteq T\left(L\right)$. At least if we don’t make any assumptions on
$\mathbf{k}$, this is the most reasonable choice we could make for the
analogue of $P$. (Later, in the positive-characteristic case, we shall
encounter a more interesting $\mathbf{k}$-submodule similar to $P$.)

The analogue of $\mathfrak{h}$, so far, has to be
$\overline{\mathfrak{g}}^{\prime}$ (since the analogue of $P$ is $0$). There
is no analogue of Proposition 5.4. We have an analogue of Proposition 5.7 (a),
however:

######  Proposition 7.10.

We have
$\left[L,\overline{\mathfrak{g}}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$.

######  Proof 7.11 (Proof of Proposition 7.10.).

This is proven in the same way as Proposition 5.7 (a).

Here is an analogue of parts of Proposition 5.9:

######  Proposition 7.12.

(a) We have
$\left[\overline{\mathfrak{g}}^{\prime},\overline{\mathfrak{g}}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$.

(b) The two $\mathbf{k}$-submodules $\overline{\mathfrak{g}}^{\prime}$ and
$\overline{\mathfrak{g}}^{\prime}+L$ of $T\left(L\right)$ are invariant under
the commutator $\left[\cdot,\cdot\right]$. (In other words, they are Lie
subalgebras of $T\left(L\right)$ (with the commutator
$\left[\cdot,\cdot\right]$ as the Lie bracket).)

######  Proof 7.13 (Proof of Proposition 7.12.).

This is analogous to the relevant parts of the proof of Proposition 5.9. (The
$\mathbf{k}$-submodule $\overline{\mathfrak{g}}^{\prime}$ takes the roles of
both $\overline{\mathfrak{g}}$ and $\mathfrak{h}$, and the zero module $0$
takes the role of $P$.)

We can now state the analogue of Proposition 6.2:

######  Proposition 7.14.

We have
$\left(\overline{\mathfrak{g}}^{\prime}\right)^{\star}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.

######  Proof 7.15 (Proof of Proposition 7.14.).

Unsurprisingly, this is analogous to the proof of Proposition 6.2.

What is not straightforward is finding the right analogue of Lemma 6.4. We
must no longer assume $uu\in S^{\star}$ (since this won’t be satisfied in the
situation we are going to apply this lemma to). Thus, a two-term sum like the
$S^{\star}+S^{\star}u$ in Lemma 6.4 won’t work anymore; instead we need an
infinite sum
$S^{\star}+S^{\star}u+S^{\star}u^{2}+\cdots=\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}$.
151515This is similar to the difference between the standard basis vectors of
the exterior algebra and the symmetric algebra of a free $\mathbf{k}$-module:
the former have every $e_{i}$ appear at most once, while the latter can have
it multiple times. Here is the exact statement:

######  Lemma 7.16.

Let $u\in L$. Let $S$ be a $\mathbf{k}$-submodule of $T\left(L\right)$ such
that $\left[u,S\right]\subseteq S^{\star}$. Then,
$\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}$ is a $\mathbf{k}$-subalgebra of
$T\left(L\right)$.

######  Proof 7.17 (Proof of Lemma 7.16.).

We have $uS^{\star}\subseteq S^{\star}+S^{\star}u$. (This can be proven just
as in the proof of Lemma 6.4, mutatis mutandis.) Now,

$u^{j}S^{\star}\subseteq\sum\limits_{k=0}^{j}S^{\star}u^{k}\ \ \ \ \ \ \ \ \ \
\text{for every }j\in\mathbb{N}$ (16) 161616Proof of (16): We shall prove (16)
by induction over $j$. Induction base: When $j=0$, the relation (16) rewrites
as $u^{0}S^{\star}\subseteq\sum\limits_{k=0}^{0}S^{\star}u^{k}$. But this is
obvious, since $\underbrace{u^{0}}_{=1}S^{\star}=S^{\star}$ and
$\sum\limits_{k=0}^{0}S^{\star}u^{k}=S^{\star}\underbrace{u^{0}}_{=1}=S^{\star}$.
Thus, the relation (16) holds for $j=0$. The induction base is thus complete.
Induction step: Let $J\in\mathbb{N}$. Assume that (16) holds for $j=J$. We now
need to prove that (16) holds for $j=J+1$. We know that (16) holds for $j=J$.
In other words, $u^{J}S^{\star}\subseteq\sum\limits_{k=0}^{J}S^{\star}u^{k}$.
Now,  $\displaystyle\underbrace{u^{J+1}}_{=uu^{J}}S^{\star}$
$\displaystyle=u\underbrace{u^{J}S^{\star}}_{\subseteq\sum\limits_{k=0}^{J}S^{\star}u^{k}}\subseteq
u\sum\limits_{k=0}^{J}S^{\star}u^{k}=\sum\limits_{k=0}^{J}\underbrace{uS^{\star}}_{\subseteq
S^{\star}+S^{\star}u}u^{k}$
$\displaystyle\subseteq\sum\limits_{k=0}^{J}\underbrace{\left(S^{\star}+S^{\star}u\right)u^{k}}_{=S^{\star}u^{k}+S^{\star}uu^{k}}=\sum\limits_{k=0}^{J}\left(S^{\star}u^{k}+S^{\star}uu^{k}\right)$
$\displaystyle=\sum\limits_{k=0}^{J}S^{\star}u^{k}+\sum\limits_{k=0}^{J}S^{\star}\underbrace{uu^{k}}_{=u^{k+1}}=\sum\limits_{k=0}^{J}S^{\star}u^{k}+\sum\limits_{k=0}^{J}S^{\star}u^{k+1}$
$\displaystyle=\sum\limits_{k=0}^{J}S^{\star}u^{k}+\sum\limits_{k=1}^{J+1}S^{\star}u^{k}=\sum\limits_{k=0}^{J+1}S^{\star}u^{k}.$
In other words, (16) holds for $j=J+1$. This completes the induction step.
Thus, (16) is proven.

. Now,

$\displaystyle\left(\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}\right)\left(\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}\right)$
$\displaystyle=\left(\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}\right)\left(\sum\limits_{i\in\mathbb{N}}S^{\star}u^{i}\right)=\sum\limits_{j\in\mathbb{N}}\sum\limits_{i\in\mathbb{N}}S^{\star}\underbrace{u^{j}S^{\star}}_{\begin{subarray}{c}\subseteq\sum\limits_{k=0}^{j}S^{\star}u^{k}\\\
\text{(by (\ref{pf.lem.Kert'.1.1}))}\end{subarray}}u^{i}$
$\displaystyle\subseteq\sum\limits_{j\in\mathbb{N}}\sum\limits_{i\in\mathbb{N}}S^{\star}\left(\sum\limits_{k=0}^{j}S^{\star}u^{k}\right)u^{i}=\sum\limits_{j\in\mathbb{N}}\sum\limits_{i\in\mathbb{N}}\sum\limits_{k=0}^{j}\underbrace{S^{\star}S^{\star}}_{\begin{subarray}{c}\subseteq
S^{\star}\\\ \text{(since }S^{\star}\text{ is a}\\\
\mathbf{k}\text{-subalgebra)}\end{subarray}}\underbrace{u^{k}u^{i}}_{=u^{k+i}}$
$\displaystyle\subseteq\sum\limits_{j\in\mathbb{N}}\sum\limits_{i\in\mathbb{N}}\sum\limits_{k=0}^{j}S^{\star}u^{k+i}\subseteq\sum\limits_{\ell\in\mathbb{N}}S^{\star}u^{\ell}\
\ \ \ \ \ \ \ \ \ \left(\text{since }k+i\in\mathbb{N}\text{ for all
}i\in\mathbb{N}\text{ and }k\in\left\\{0,1,\ldots,j\right\\}\right)$
$\displaystyle=\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}.$

Combined with $1\in
S^{\star}=S^{\star}\underbrace{1}_{=u^{0}}=S^{\star}u^{0}\subseteq\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}$,
this yields that $\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}$ is a
$\mathbf{k}$-subalgebra of $T\left(L\right)$. This proves Lemma 7.16.

Our next lemma is a straightforward analogue of Lemma 6.6:

######  Lemma 7.18.

Let $N$ be a $\mathbf{k}$-submodule of $T\left(L\right)$. Let $g\in L^{\ast}$
be such that $\partial_{g}^{\prime}\left(N\right)=0$. Then,
$\partial_{g}^{\prime}\left(N^{\star}\right)=0$.

######  Proof 7.19 (Proof of Lemma 7.18.).

The proof is analogous to that of Lemma 6.6.

Next, we state an analogue of Lemma 6.8:

######  Lemma 7.20.

Let $M$ be a $\mathbf{k}$-submodule of $L$. Let $g\in L^{\ast}$ be such that
$g\left(M\right)=0$. Then:

(a) We have
$\partial_{g}^{\prime}\left(\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}\right)=0$.

(b) Assume that the additive group $T\left(L\right)$ is torsionfree. Let $q\in
L$ be such that $g\left(q\right)=1$. Let
$\left(U_{0},U_{1},U_{2},\ldots\right)$ be a sequence of elements of
$\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}$ such that all but
finitely many $i\in\mathbb{N}$ satisfy $U_{i}=0$. If
$\partial_{g}^{\prime}\left(\sum\limits_{i\in\mathbb{N}}U_{i}q^{i}\right)=0$,
then every positive integer $i$ satisfies $U_{i}=0$.

######  Proof 7.21 (Proof of Lemma 7.20.).

The proof of Lemma 7.20 (a) is analogous to that of Lemma 6.8 (a). It remains
to prove part (b).

We assume that the additive group $T\left(L\right)$ is torsionfree. Let $q\in
L$ be such that $g\left(q\right)=1$.

In order to prepare for this proof, we shall make a definition. Given an
$N\in\mathbb{N}$, we say that the sequence
$\left(U_{0},U_{1},U_{2},\ldots\right)$ of elements of
$\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}$ is $N$-supported if
every integer $i\geq N$ satisfies $U_{i}=0$. Of course, the sequence
$\left(U_{0},U_{1},U_{2},\ldots\right)$ of elements of
$\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}$ must be
$N$-supported for some $N\in\mathbb{N}$ (since all but finitely many
$i\in\mathbb{N}$ satisfy $U_{i}=0$). Hence, in order to prove Lemma 7.20 (b),
it suffices to show that, for every $N\in\mathbb{N}$,

$\left(\text{Lemma \ref{lem.Kert'.deriv0} {(b)} holds whenever the sequence
}\left(U_{0},U_{1},U_{2},\ldots\right)\text{ is }N\text{-supported}\right).$
(17)

We shall now prove (17) by induction over $N$:

Induction base: The only $0$-supported sequence
$\left(U_{0},U_{1},U_{2},\ldots\right)$ is $\left(0,0,0,\ldots\right)$. Lemma
7.20 (b) clearly holds for this sequence. Thus, (17) holds for $N=0$. This
completes the induction base.

Induction step: Fix $n\in\mathbb{N}$. Assume that (17) is proven for $N=n$. We
now need to prove (17) for $N=n+1$.

We assumed that (17) is proven for $N=n$. In other words,

$\left(\text{Lemma \ref{lem.Kert'.deriv0} {(b)} holds whenever the sequence
}\left(U_{0},U_{1},U_{2},\ldots\right)\text{ is }n\text{-supported}\right).$
(18)

Let $\left(U_{0},U_{1},U_{2},\ldots\right)$ be a sequence of elements of
$\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}$ such that all but
finitely many $i\in\mathbb{N}$ satisfy $U_{i}=0$. Assume that this sequence
$\left(U_{0},U_{1},U_{2},\ldots\right)$ is $\left(n+1\right)$-supported.
Assume that
$\partial_{g}^{\prime}\left(\sum\limits_{i\in\mathbb{N}}U_{i}q^{i}\right)=0$.
Our goal now is to prove that every positive integer $i$ satisfies $U_{i}=0$.
Once this is shown, it will follow that (17) holds for $N=n+1$, and so the
induction step will be complete.

It is easy to show (using $g\left(q\right)=1$) that

$\partial_{g}^{\prime}\left(q^{i}\right)=iq^{i-1}\ \ \ \ \ \ \ \ \ \ \text{for
every }i\in\mathbb{N}$ (19)

(where $iq^{i-1}$ is to be understood as $0$ when $i=0$). Hence, every
$i\in\mathbb{N}$ satisfies

$\displaystyle\partial_{g}^{\prime}\left(U_{i}q^{i}\right)$
$\displaystyle=\underbrace{\partial_{g}^{\prime}\left(U_{i}\right)}_{\begin{subarray}{c}=0\\\
\text{(since
}U_{i}\in\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}\\\ \text{and
}\partial_{g}^{\prime}\left(\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}\right)=0\text{)}\end{subarray}}q^{i}+U_{i}\partial_{g}^{\prime}\left(q^{i}\right)\
\ \ \ \ \ \ \ \ \ \left(\text{by Proposition \ref{prop.1'} {(b)}}\right)$
$\displaystyle=U_{i}\underbrace{\partial_{g}^{\prime}\left(q^{i}\right)}_{=iq^{i-1}}=U_{i}\cdot
iq^{i-1}.$

Now,

$\displaystyle\partial_{g}^{\prime}\left(\sum\limits_{i\in\mathbb{N}}U_{i}q^{i}\right)$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\underbrace{\partial_{g}^{\prime}\left(U_{i}q^{i}\right)}_{=U_{i}\cdot
iq^{i-1}}=\sum\limits_{i\in\mathbb{N}}U_{i}\cdot
iq^{i-1}=U_{0}\cdot\underbrace{0q^{0-1}}_{\begin{subarray}{c}=0\\\ \text{(by
definition)}\end{subarray}}+\sum\limits_{\begin{subarray}{c}i\in\mathbb{N};\\\
i\text{ is positive}\end{subarray}}U_{i}\cdot iq^{i-1}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}i\in\mathbb{N};\\\ i\text{ is
positive}\end{subarray}}U_{i}\cdot
iq^{i-1}=\sum\limits_{i\in\mathbb{N}}U_{i+1}\cdot\left(i+1\right)q^{i}=\sum\limits_{i\in\mathbb{N}}\left(i+1\right)U_{i+1}q^{i}.$

Hence,
$\sum\limits_{i\in\mathbb{N}}\left(i+1\right)U_{i+1}q^{i}=\partial_{g}^{\prime}\left(\sum\limits_{i\in\mathbb{N}}U_{i}q^{i}\right)=0$,
so that
$\partial_{g}^{\prime}\left(\underbrace{\sum\limits_{i\in\mathbb{N}}\left(i+1\right)U_{i+1}q^{i}}_{=0}\right)=0$.

But $\left(1U_{1},2U_{2},3U_{3},\ldots\right)$ is a sequence of elements of
$\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}$ (because
$U_{1},U_{2},U_{3},\ldots$ are elements of
$\left(M+\overline{\mathfrak{g}}^{\prime}\right)^{\star}$), and is
$n$-supported (since the sequence $\left(U_{0},U_{1},U_{2},\ldots\right)$ is
$\left(n+1\right)$-supported). Hence, we can apply Lemma 7.20 (b) to
$\left(1U_{1},2U_{2},3U_{3},\ldots\right)$ instead of
$\left(U_{0},U_{1},U_{2},\ldots\right)$ (because of (18)). As a result, we
conclude that every positive integer $i$ satisfies
$\left(i+1\right)U_{i+1}=0$. Therefore, every positive integer $i$ satisfies
$U_{i+1}=0$ (since the additive group $T\left(L\right)$ is torsionfree). In
other words, every integer $i>2$ satisfies $U_{i}=0$.

But now, recall that
$\sum\limits_{i\in\mathbb{N}}\left(i+1\right)U_{i+1}q^{i}=0$. Hence,

$\displaystyle 0$
$\displaystyle=\sum\limits_{i\in\mathbb{N}}\left(i+1\right)U_{i+1}q^{i}=\underbrace{\left(0+1\right)}_{=1}\underbrace{U_{0+1}}_{=U_{1}}\underbrace{q^{0}}_{=1}+\sum\limits_{\begin{subarray}{c}i\in\mathbb{N};\\\
i\text{ is
positive}\end{subarray}}\left(i+1\right)\underbrace{U_{i+1}}_{\begin{subarray}{c}=0\\\
\text{(since }i\text{ is}\\\ \text{positive)}\end{subarray}}q^{i}$
$\displaystyle=U_{1}+\underbrace{\sum\limits_{\begin{subarray}{c}i\in\mathbb{N};\\\
i\text{ is positive}\end{subarray}}\left(i+1\right)0q^{i}}_{=0}=U_{1}.$

Hence, $U_{1}=0$. This (combined with the fact that every integer $i>2$
satisfies $U_{i}=0$) shows that every positive integer $i$ satisfies
$U_{i}=0$. Thus, (17) is proven for $N=n+1$. The induction step will be
complete.

We have now proven (17) by induction. Hence, Lemma 7.20 (b) is proven.

We can finally state our characteristic-zero analogue of Theorem 6.10:

######  Theorem 7.22.

Assume that the $\mathbf{k}$-module $L$ is free. Assume that the additive
group $\mathbf{k}$ is torsionfree. Then,
$\left(\overline{\mathfrak{g}}^{\prime}\right)^{\star}=\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.

The proof of this theorem is similar to that of Theorem 6.10, but differs just
enough that we show it in detail.

We notice that the requirement in Theorem 7.22 that the additive group
$\mathbf{k}$ is torsionfree is satisfied whenever $\mathbf{k}$ is a
commutative $\mathbb{Q}$-algebra, but also in cases such as
$\mathbf{k}=\mathbb{Z}$. So Theorem 7.22 is actually a fairly general result.

######  Proof 7.23 (Proof of Theorem 7.22.).

Proposition 7.14 shows that
$\left(\overline{\mathfrak{g}}^{\prime}\right)^{\star}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.
We thus only need to verify that
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)\subseteq\left(\overline{\mathfrak{g}}^{\prime}\right)^{\star}$.
This means proving that every
$U\in\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$ satisfies
$U\in\left(\overline{\mathfrak{g}}^{\prime}\right)^{\star}$. So let us fix
$U\in\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.

The $\mathbf{k}$-module $L$ is free. Hence, the $\mathbf{k}$-module
$T\left(L\right)$ is free as well. Therefore, the additive group
$T\left(L\right)$ is a direct sum of many copies of the additive group
$\mathbf{k}$. Thus, the additive group $T\left(L\right)$ is torsionfree
(because the additive group $\mathbf{k}$ is torsionfree).

We know that the $\mathbf{k}$-module $L$ is free; it thus has a basis. Since
the tensor $U\in T\left(L\right)$ can be constructed using only finitely many
elements of this basis, we can thus WLOG assume that the basis of $L$ is
finite. Let us assume this, and let us denote this basis by
$\left(e_{1},e_{2},\ldots,e_{n}\right)$.

For every $i\in\left\\{1,2,\ldots,n\right\\}$, let
$e_{i}^{\ast}:L\rightarrow\mathbf{k}$ be the $\mathbf{k}$-linear map which
sends $e_{i}$ to $1$ and sends every other $e_{j}$ to $0$.

For every $k\in\left\\{0,1,\ldots,n\right\\}$, we let $M_{k}$ denote the
$\mathbf{k}$-submodule of $L$ spanned by $e_{1},e_{2},\ldots,e_{k}$. Thus,
$M_{0}=0$ and $M_{n}=L$. Clearly, every $k\in\left\\{1,2,\ldots,n\right\\}$
satisfies

$M_{k}=M_{k-1}+\mathbf{k}e_{k}.$ (20)

For every $k\in\left\\{0,1,\ldots,n\right\\}$, we set
$\mathfrak{h}_{k}=M_{k}+\overline{\mathfrak{g}}^{\prime}$ and
$H_{k}=\mathfrak{h}_{k}^{\star}$.

Notice that $\mathfrak{h}_{n}=M_{n}+\overline{\mathfrak{g}}^{\prime}\supseteq
M_{n}=L$ and thus $H_{n}=\mathfrak{h}_{n}^{\star}\supseteq
L^{\star}=T\left(L\right)$. Hence, $H_{n}=T\left(L\right)$. Now, $U\in
T\left(L\right)=H_{n}$.

On the other hand, the definition of $\mathfrak{h}_{0}$ yields
$\mathfrak{h}_{0}=\underbrace{M_{0}}_{=0}+\overline{\mathfrak{g}}^{\prime}=\overline{\mathfrak{g}}^{\prime}$
and thus
$H_{0}=\mathfrak{h}_{0}^{\star}=\left(\overline{\mathfrak{g}}^{\prime}\right)^{\star}$.

We shall now prove that every $k\in\left\\{1,2,\ldots,n\right\\}$ satisfies
the following implication:

$\text{if }U\in H_{k}\text{, then }U\in H_{k-1}.$ (21)

Once this is proven, we will be able to argue that $U\in H_{n}$ (as we know),
thus $U\in H_{n-1}$ (by (21)), thus $U\in H_{n-2}$ (by (21) again), and so on
– until we finally arrive at $U\in H_{0}$. Since
$H_{0}=\left(\overline{\mathfrak{g}}^{\prime}\right)^{\star}$, this rewrites
as $U\in\left(\overline{\mathfrak{g}}^{\prime}\right)^{\star}$, and thus we
are done.

Therefore, it only remains to prove (21). So let us fix
$k\in\left\\{1,2,\ldots,n\right\\}$, and assume that $U\in H_{k}$. We now need
to show that $U\in H_{k-1}$.

We have

$\displaystyle\left[e_{k},\mathfrak{h}_{k-1}\right]$
$\displaystyle=\left[\underbrace{e_{k}}_{\in
L},\underbrace{M_{k-1}}_{\subseteq L}\right]+\left[\underbrace{e_{k}}_{\in
L},\overline{\mathfrak{g}}^{\prime}\right]\ \ \ \ \ \ \ \ \ \
\left(\text{since
}\mathfrak{h}_{k-1}=M_{k-1}+\overline{\mathfrak{g}}^{\prime}\right)$
$\displaystyle\subseteq\underbrace{\left[L,L\right]}_{\begin{subarray}{c}=L_{2}^{\prime}\subseteq
L_{2}^{\prime}+L_{3}^{\prime}+L_{4}^{\prime}+\cdots\\\
=\overline{\mathfrak{g}}^{\prime}\end{subarray}}+\underbrace{\left[L,\overline{\mathfrak{g}}^{\prime}\right]}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}^{\prime}\\\
\text{(by Proposition
\ref{prop.5a'})}\end{subarray}}\subseteq\overline{\mathfrak{g}}^{\prime}+\overline{\mathfrak{g}}^{\prime}=\overline{\mathfrak{g}}^{\prime}$
$\displaystyle\subseteq
M_{k-1}+\overline{\mathfrak{g}}^{\prime}=\mathfrak{h}_{k-1}\subseteq\mathfrak{h}_{k-1}^{\star}.$

Thus, Lemma 7.16 (applied to $u=e_{k}$ and $S=\mathfrak{h}_{k-1}$) yields that
$\sum\limits_{j\in\mathbb{N}}\mathfrak{h}_{k-1}^{\star}e_{k}^{j}$ is a
$\mathbf{k}$-subalgebra of $T\left(L\right)$. In other words,
$\sum\limits_{j\in\mathbb{N}}H_{k-1}e_{k}^{j}$ is a $\mathbf{k}$-subalgebra of
$T\left(L\right)$ (since $H_{k-1}=\mathfrak{h}_{k-1}^{\star}$). This
$\mathbf{k}$-subalgebra contains $\mathfrak{h}_{k}$ as a subset171717Proof. We
have $\mathfrak{h}_{k}=M_{k}+\overline{\mathfrak{g}}^{\prime}$ and similarly
$\mathfrak{h}_{k-1}=M_{k-1}+\overline{\mathfrak{g}}^{\prime}$. Thus,
$\displaystyle\mathfrak{h}_{k}$
$\displaystyle=\underbrace{M_{k}}_{\begin{subarray}{c}=M_{k-1}+\mathbf{k}e_{k}\\\
\text{(by
(\ref{pf.thm.Kert'.Mk}))}\end{subarray}}+\overline{\mathfrak{g}}^{\prime}=M_{k-1}+\mathbf{k}e_{k}+\overline{\mathfrak{g}}^{\prime}=\underbrace{M_{k-1}+\overline{\mathfrak{g}}^{\prime}}_{\begin{subarray}{c}=\mathfrak{h}_{k-1}\subseteq\mathfrak{h}_{k-1}^{\star}=H_{k-1}\\\
\text{(since }H_{k-1}\text{ was}\\\ \text{defined as
}\mathfrak{h}_{k-1}^{\star}\text{)}\end{subarray}}+\underbrace{\mathbf{k}}_{\subseteq
H_{k-1}}e_{k}$ $\displaystyle\subseteq
H_{k-1}+H_{k-1}e_{k}\subseteq\sum\limits_{j\in\mathbb{N}}H_{k-1}e_{k}^{j}$
(since $H_{k-1}$ and $H_{k-1}e_{k}$ are the first two addends of the sum
$\sum\limits_{j\in\mathbb{N}}H_{k-1}e_{k}^{j}$), qed., and thus we have
$H_{k}\subseteq\sum\limits_{j\in\mathbb{N}}H_{k-1}e_{k}^{j}$ 181818Proof. The
$\mathbf{k}$-subalgebra $\sum\limits_{j\in\mathbb{N}}H_{k-1}e_{k}^{j}$ of
$T\left(L\right)$ contains $\mathfrak{h}_{k}$ as a subset. Hence, it also
contains $\mathfrak{h}_{k}^{\star}$ as a subset (since
$\mathfrak{h}_{k}^{\star}$ is the $\mathbf{k}$-subalgebra of $T\left(L\right)$
generated by $\mathfrak{h}_{k}$). In other words,
$\sum\limits_{j\in\mathbb{N}}H_{k-1}e_{k}^{j}\supseteq\mathfrak{h}_{k}^{\star}=H_{k}$,
qed.. (Actually, $H_{k}=\sum\limits_{j\in\mathbb{N}}H_{k-1}e_{k}^{j}$, but we
don’t need this.)

Now, $U\in
H_{k}\subseteq\sum\limits_{j\in\mathbb{N}}H_{k-1}e_{k}^{j}=\sum\limits_{i\in\mathbb{N}}H_{k-1}e_{k}^{i}$.
Therefore, there exists a sequence $\left(U_{0},U_{1},U_{2},\ldots\right)$ of
elements of $H_{k-1}$ such that all but finitely many $i\in\mathbb{N}$ satisfy
$U_{i}=0$ and such that we have
$U=\sum\limits_{i\in\mathbb{N}}U_{i}e_{k}^{i}$. Consider this sequence
$\left(U_{0},U_{1},U_{2},\ldots\right)$. We have

$\partial_{e_{k}^{\ast}}^{\prime}\left(\underbrace{\sum\limits_{i\in\mathbb{N}}U_{i}e_{k}^{i}}_{=U}\right)=\partial_{e_{k}^{\ast}}^{\prime}\left(\underbrace{U}_{\in\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)}\right)\in\partial_{e_{k}^{\ast}}^{\prime}\left(\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)\right)=0$

(by Proposition 7.4 (a), applied to $g=e_{k}^{\ast}$), so that
$\partial_{e_{k}^{\ast}}^{\prime}\left(\sum\limits_{i\in\mathbb{N}}U_{i}e_{k}^{i}\right)=0$.

The entries $U_{0},U_{1},U_{2},\ldots$ of the sequence
$\left(U_{0},U_{1},U_{2},\ldots\right)$ belong to
$H_{k-1}=\mathfrak{h}_{k-1}^{\star}=\left(M_{k-1}+\overline{\mathfrak{g}}^{\prime}\right)^{\star}$
(since $\mathfrak{h}_{k-1}=M_{k-1}+\overline{\mathfrak{g}}^{\prime}$). We can
thus apply Lemma 7.20 (b) to $M=M_{k-1}$, $g=e_{k}^{\ast}$ and $q=e_{k}$
(since $e_{k}^{\ast}\left(M_{k-1}\right)=0$ (because $M_{k-1}$ is spanned by
$e_{1},e_{2},\ldots,e_{k-1}$) and $e_{k}^{\ast}\left(e_{k}\right)=1$ and
$\partial_{e_{k}^{\ast}}^{\prime}\left(\sum\limits_{i\in\mathbb{N}}U_{i}e_{k}^{i}\right)=0$).
As a result, we see that every positive integer $i$ satisfies $U_{i}=0$. Thus,

$U=\sum\limits_{i\in\mathbb{N}}U_{i}e_{k}^{i}=U_{0}\underbrace{e_{k}^{0}}_{=1}+\sum\limits_{\begin{subarray}{c}i\in\mathbb{N};\\\
i\text{ is
positive}\end{subarray}}\underbrace{U_{i}}_{\begin{subarray}{c}=0\\\
\text{(since }i\text{ is}\\\
\text{positive)}\end{subarray}}e_{k}^{i}=U_{0}+\underbrace{\sum\limits_{\begin{subarray}{c}i\in\mathbb{N};\\\
i\text{ is positive}\end{subarray}}0e_{k}^{i}}_{=0}=U_{0}\in H_{k-1}.$

This completes the proof of (21). As we already mentioned, this finishes the
proof of Theorem 7.22.

##  8 The even analogue in positive characteristic

We now come to the question of determining
$\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$ when the ground ring
$\mathbf{k}$ is an $\mathbb{F}_{p}$-algebra for a prime number $p$. In this
case, as we will see, a $\mathbf{k}$-submodule similar to the $P$ of
Definition 5.3 will become relevant once again.

######  Convention 8.1.

For this whole section, we fix a prime number $p$, and we assume that
$\mathbf{k}$ is a commutative $\mathbb{F}_{p}$-algebra.

This assumption yields that $p=0$ in $\mathbf{k}$. Let us immediately put this
to use by stating an analogue of Proposition 4.5 (e) (which, as we recall, had
no analogue in the case of arbitrary $\mathbf{k}$):

######  Proposition 8.1.

We have $x^{p}\in\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$ for each
$x\in L$.

######  Proof 8.2 (Proof of Proposition 8.1.).

Let $x\in L$. It is easy to see that
$\mathbf{t}^{\prime}\left(x^{i}\right)=ix^{i}$ for every positive integer $i$.
Applying this to $i=p$, we obtain
$\mathbf{t}^{\prime}\left(x^{p}\right)=px^{p}=0$ (since $p=0$ in
$\mathbf{k}$). Thus,
$x^{p}\in\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$. This proves
Proposition 8.1.

######  Definition 8.3.

Let $P_{p}$ denote the $\mathbf{k}$-submodule of $L^{\otimes p}$ spanned by
elements of the form $\underbrace{x\otimes x\otimes\cdots\otimes x}_{p\text{
times}}$ with $x\in L$. Notice that $\underbrace{x\otimes
x\otimes\cdots\otimes x}_{p\text{ times}}=x^{p}$ in the $\mathbf{k}$-algebra
$T\left(L\right)$ for every $x\in L$.

Of course, when $p=2$, we have $P_{p}=P_{2}=P$.

######  Definition 8.4.

Let $\mathfrak{h}_{p}^{\prime}=\overline{\mathfrak{g}}^{\prime}+P_{p}$.

Recall that Proposition 7.10 was an analogue of part of Proposition 5.7. Now
that we have $\mathbf{k}$-submodules $P_{p}$ and $\mathfrak{h}_{p}^{\prime}$
similar to our formerly defined $P$ and $\mathfrak{h}$, we can state an
analogue of the remainder of that proposition:

######  Proposition 8.5.

We have $\left[L,P_{p}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$ and
$\left[L,\mathfrak{h}_{p}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}\subseteq\mathfrak{h}_{p}^{\prime}$.

######  Proof 8.6 (Proof of Proposition 8.5.).

Let us first make some general observations on commutators in
$\mathbf{k}$-algebras.

Let $A$ be any $\mathbf{k}$-algebra. Clearly, there is a commutator
$\left[\cdot,\cdot\right]$ on $A$, defined by $\left[U,V\right]=UV-VU$ for all
$U\in A$ and $V\in A$.

For every $a\in A$, we define a $\mathbf{k}$-linear map
$\operatorname*{ad}\nolimits_{a}:A\rightarrow A$ by setting

$\operatorname*{ad}\nolimits_{a}\left(c\right)=\left[a,c\right]\ \ \ \ \ \ \ \
\ \ \text{for all }c\in A.$

Then, every $a\in A$ satisfies

$\operatorname*{ad}\nolimits_{a^{p}}=\left(\operatorname*{ad}\nolimits_{a}\right)^{p}$
(22) 191919Proof of (22): We first notice that $\left(-1\right)^{p}=-1$ in
$\mathbf{k}$. (This is because Fermat’s little theorem yields
$\left(-1\right)^{p}\equiv-1\operatorname{mod}p$.) For every $a\in A$, we
define a $\mathbf{k}$-linear map $\mathcal{L}_{a}:A\rightarrow A$ by setting
$\mathcal{L}_{a}\left(c\right)=ac\ \ \ \ \ \ \ \ \ \ \text{for all }c\in A.$
For every $a\in A$, we define a $\mathbf{k}$-linear map
$\mathcal{R}_{a}:A\rightarrow A$ by setting
$\mathcal{R}_{a}\left(c\right)=ca\ \ \ \ \ \ \ \ \ \ \text{for all }c\in A.$
It is easy to see that $\mathcal{L}_{0}=0$ and
$\mathcal{L}_{1}=\operatorname*{id}$, and that any $a\in A$ and $b\in A$
satisfy $\mathcal{L}_{ab}=\mathcal{L}_{a}\circ\mathcal{L}_{b}$ and
$\mathcal{L}_{a+b}=\mathcal{L}_{a}+\mathcal{L}_{b}$. Hence, the map
$A\rightarrow\operatorname*{End}A,\ \ \ \ \ \ \ \ \ \ a\mapsto\mathcal{L}_{a}$
is a $\mathbf{k}$-algebra homomorphism. Consequently,
$\mathcal{L}_{a^{p}}=\left(\mathcal{L}_{a}\right)^{p}$ for every $a\in A$.
Similarly, $\mathcal{R}_{a^{p}}=\left(\mathcal{R}_{a}\right)^{p}$ for every
$a\in A$. Now, fix $a\in A$. Then, the maps $\mathcal{L}_{a}$ and
$\mathcal{R}_{a}$ commute. (In fact, more generally, the maps
$\mathcal{L}_{a}$ and $\mathcal{R}_{b}$ commute for every $b\in A$. This is
straightforward to check.) Moreover,
$\operatorname*{ad}\nolimits_{a}=\mathcal{L}_{a}-\mathcal{R}_{a}$ (since every
$c\in A$ satisfies
$\operatorname*{ad}\nolimits_{a}\left(c\right)=\left[a,c\right]=\underbrace{ac}_{=\mathcal{L}_{a}\left(c\right)}-\underbrace{ca}_{=\mathcal{R}_{a}\left(c\right)}=\mathcal{L}_{a}\left(c\right)-\mathcal{R}_{a}\left(c\right)=\left(\mathcal{L}_{a}-\mathcal{R}_{a}\right)\left(c\right)$).
The same argument (applied to $a^{p}$ instead of $a$) shows that
$\operatorname*{ad}\nolimits_{a^{p}}=\mathcal{L}_{a^{p}}-\mathcal{R}_{a^{p}}$.
Now, the maps $\mathcal{L}_{a}$ and $\mathcal{R}_{a}$ commute; thus, we can
apply the binomial formula to them. We thus obtain
$\displaystyle\left(\mathcal{L}_{a}-\mathcal{R}_{a}\right)^{p}$
$\displaystyle=\sum\limits_{k=0}^{p}\dbinom{p}{k}\left(\mathcal{L}_{a}\right)^{k}\left(-\mathcal{R}_{a}\right)^{p-k}$
$\displaystyle=\underbrace{\dbinom{p}{0}}_{=1}\underbrace{\left(\mathcal{L}_{a}\right)^{0}}_{=1}\underbrace{\left(-\mathcal{R}_{a}\right)^{p-0}}_{=\left(-\mathcal{R}_{a}\right)^{p}}+\sum\limits_{k=1}^{p-1}\underbrace{\dbinom{p}{k}}_{\begin{subarray}{c}=0\text{
in }\mathbf{k}\\\ \text{(since }p\mid\dbinom{p}{k}\\\ \text{(because }p\text{
is prime and
}0<k<p\text{))}\end{subarray}}\left(\mathcal{L}_{a}\right)^{k}\left(-\mathcal{R}_{a}\right)^{p-k}+\underbrace{\dbinom{p}{p}}_{=1}\left(\mathcal{L}_{a}\right)^{p}\underbrace{\left(-\mathcal{R}_{a}\right)^{p-p}}_{=\left(-\mathcal{R}_{a}\right)^{0}=1}$
$\displaystyle=\underbrace{\left(-\mathcal{R}_{a}\right)^{p}}_{\begin{subarray}{c}=\left(-1\right)^{p}\left(\mathcal{R}_{a}\right)^{p}=-\left(\mathcal{R}_{a}\right)^{p}\\\
\text{(since }\left(-1\right)^{p}=-1\text{ in
}\mathbf{k}\text{)}\end{subarray}}+\underbrace{\sum\limits_{k=1}^{p-1}0\left(\mathcal{L}_{a}\right)^{k}\left(-\mathcal{R}_{a}\right)^{p-k}}_{=0}+\left(\mathcal{L}_{a}\right)^{p}$
$\displaystyle=-\underbrace{\left(\mathcal{R}_{a}\right)^{p}}_{=\mathcal{R}_{a^{p}}}+\underbrace{\left(\mathcal{L}_{a}\right)^{p}}_{=\mathcal{L}_{a^{p}}}=-\mathcal{R}_{a^{p}}+\mathcal{L}_{a^{p}}=\mathcal{L}_{a^{p}}-\mathcal{R}_{a^{p}}=\operatorname*{ad}\nolimits_{a^{p}}.$
Since $\operatorname*{ad}\nolimits_{a}=\mathcal{L}_{a}-\mathcal{R}_{a}$, this
rewrites as
$\left(\operatorname*{ad}\nolimits_{a}\right)^{p}=\operatorname*{ad}\nolimits_{a^{p}}$.
This proves (22).

.

Now, let $A=T\left(L\right)$. Fix $x\in L$. Then,

$\displaystyle\operatorname*{ad}\nolimits_{x}\left(\overline{\mathfrak{g}}^{\prime}\right)$
$\displaystyle=\left[\underbrace{x}_{\in
L},\overline{\mathfrak{g}}^{\prime}\right]\ \ \ \ \ \ \ \ \ \
\left(\text{since
}\operatorname*{ad}\nolimits_{x}\left(c\right)=\left[x,c\right]\text{ for all
}c\in\overline{\mathfrak{g}}^{\prime}\right)$
$\displaystyle\subseteq\left[L,\overline{\mathfrak{g}}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}\
\ \ \ \ \ \ \ \ \ \left(\text{by Proposition \ref{prop.5a'}}\right).$

Thus, the map $\operatorname*{ad}\nolimits_{x}:T\left(L\right)\rightarrow
T\left(L\right)$ restricts to a map
$\operatorname*{ad}\nolimits_{x}:\overline{\mathfrak{g}}^{\prime}\rightarrow\overline{\mathfrak{g}}^{\prime}$.
Consequently,

$\left(\operatorname*{ad}\nolimits_{x}\right)^{n}\left(\overline{\mathfrak{g}}^{\prime}\right)\subseteq\overline{\mathfrak{g}}^{\prime}\
\ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}.$ (23)

Now, if $m$ is a positive integer, then

$\displaystyle\underbrace{\left(\operatorname*{ad}\nolimits_{x}\right)^{m}}_{=\left(\operatorname*{ad}\nolimits_{x}\right)^{m-1}\circ\operatorname*{ad}\nolimits_{x}}\left(L\right)$
$\displaystyle=\left(\left(\operatorname*{ad}\nolimits_{x}\right)^{m-1}\circ\operatorname*{ad}\nolimits_{x}\right)\left(L\right)=\left(\operatorname*{ad}\nolimits_{x}\right)^{m-1}\underbrace{\left(\operatorname*{ad}\nolimits_{x}\left(L\right)\right)}_{=\left[x,L\right]}$
$\displaystyle=\left(\operatorname*{ad}\nolimits_{x}\right)^{m-1}\left(\left[\underbrace{x}_{\in
L},L\right]\right)\in\left(\operatorname*{ad}\nolimits_{x}\right)^{m-1}\left(\underbrace{\left[L,L\right]}_{\begin{subarray}{c}=L_{2}^{\prime}\\\
\subseteq L_{2}^{\prime}+L_{3}^{\prime}+L_{4}^{\prime}+\cdots\\\
=\overline{\mathfrak{g}}^{\prime}\end{subarray}}\right)$
$\displaystyle\in\left(\operatorname*{ad}\nolimits_{x}\right)^{m-1}\left(\overline{\mathfrak{g}}^{\prime}\right)\subseteq\overline{\mathfrak{g}}^{\prime}\
\ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.prop.5b'.sub}), applied to
}n=m-1\right).$ (24)

Hence, every $U\in L$ satisfies

$\left[U,x^{p}\right]=-\underbrace{\left[x^{p},U\right]}_{=\operatorname*{ad}\nolimits_{x^{p}}\left(U\right)}=-\underbrace{\operatorname*{ad}\nolimits_{x^{p}}}_{\begin{subarray}{c}=\left(\operatorname*{ad}\nolimits_{x}\right)^{p}\\\
\text{(by (\ref{pf.prop.5b'.adp}))}\end{subarray}}\left(\underbrace{U}_{\in
L}\right)\in-\underbrace{\left(\operatorname*{ad}\nolimits_{x}\right)^{p}\left(L\right)}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}^{\prime}\\\
\text{(by
(\ref{pf.prop.5b'.sub2}))}\end{subarray}}\subseteq-\overline{\mathfrak{g}}^{\prime}\subseteq\overline{\mathfrak{g}}^{\prime}.$

Let us now forget that we fixed $x$ and $U$. We thus have shown that
$\left[U,x^{p}\right]\in\overline{\mathfrak{g}}^{\prime}$ for every $U\in L$
and every $x\in L$. This shows that
$\left[L,P_{p}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$ (because the
$\mathbf{k}$-module $P_{p}$ is spanned by elements of the form $x^{p}$ with
$x\in L$).

It remains to prove that
$\left[L,\mathfrak{h}_{p}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}\subseteq\mathfrak{h}_{p}^{\prime}$.
But this is easy: We have
$\mathfrak{h}_{p}^{\prime}=\overline{\mathfrak{g}}^{\prime}+P_{p}$, and thus

$\left[L,\mathfrak{h}_{p}^{\prime}\right]=\underbrace{\left[L,\overline{\mathfrak{g}}^{\prime}\right]}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}^{\prime}\\\
\text{(by Proposition
\ref{prop.5a'})}\end{subarray}}+\underbrace{\left[L,P_{p}\right]}_{\subseteq\overline{\mathfrak{g}}^{\prime}}\subseteq\overline{\mathfrak{g}}^{\prime}+\overline{\mathfrak{g}}^{\prime}\subseteq\overline{\mathfrak{g}}^{\prime}\subseteq\overline{\mathfrak{g}}^{\prime}+P_{p}=\mathfrak{h}_{p}^{\prime}.$

This proves Proposition 8.5.

Next, we state an analogue of the parts of Proposition 5.9 not covered by
Proposition 7.12:

######  Proposition 8.7.

(a) We have
$\left[\mathfrak{h}_{p}^{\prime},\mathfrak{h}_{p}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}\subseteq\mathfrak{h}_{p}^{\prime}$.

(b) The two $\mathbf{k}$-submodules $\mathfrak{h}_{p}^{\prime}$ and
$\mathfrak{h}_{p}^{\prime}+L$ of $T\left(L\right)$ are invariant under the
commutator $\left[\cdot,\cdot\right]$. (In other words, they are Lie
subalgebras of $T\left(L\right)$ (with the commutator
$\left[\cdot,\cdot\right]$ as the Lie bracket).)

######  Proof 8.8 (Proof of Proposition 8.7.).

(a) First, we notice that

$\left[P_{p},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}.$ (25)
202020Proof of (25): It is clearly enough to show that
$\left[P_{p},L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$
for all positive integers $i$. So let us do this. Let $i$ be a positive
integer. We need to show that
$\left[P_{p},L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$.
In other words, we need to show that
$\left[x^{p},L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$
for every $x\in L$ (because the $\mathbf{k}$-module $P_{p}$ is spanned by
elements of the form $x^{p}$ with $x\in L$). So let us fix $x\in L$. We need
to prove
$\left[x^{p},L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$.
We shall use the notations introduced in the proof of Proposition 8.5. Let
$U\in L_{i}^{\prime}$. Then, the definition of
$\operatorname*{ad}\nolimits_{x}$ yields
$\displaystyle\operatorname*{ad}\nolimits_{x}\left(U\right)$
$\displaystyle=\left[\underbrace{x}_{\in L},\underbrace{U}_{\in
L_{i}^{\prime}}\right]\in\left[L,L_{i}^{\prime}\right]=L_{i+1}^{\prime}\subseteq
L_{2}^{\prime}+L_{3}^{\prime}+L_{4}^{\prime}+\cdots\ \ \ \ \ \ \ \ \ \
\left(\text{since }i+1\geq 2\right)$
$\displaystyle=\overline{\mathfrak{g}}^{\prime}.$ But the definition of
$\operatorname*{ad}\nolimits_{x^{p}}$ yields
$\operatorname*{ad}\nolimits_{x^{p}}\left(U\right)=\left[x^{p},U\right]$, so
that  $\displaystyle\left[x^{p},U\right]$
$\displaystyle=\underbrace{\operatorname*{ad}\nolimits_{x^{p}}}_{\begin{subarray}{c}=\left(\operatorname*{ad}\nolimits_{x}\right)^{p}\\\
\text{(by
(\ref{pf.prop.5b'.adp}))}\end{subarray}}\left(U\right)=\underbrace{\left(\operatorname*{ad}\nolimits_{x}\right)^{p}}_{=\left(\operatorname*{ad}\nolimits_{x}\right)^{p-1}\circ\operatorname*{ad}\nolimits_{x}}\left(U\right)=\left(\left(\operatorname*{ad}\nolimits_{x}\right)^{p-1}\circ\operatorname*{ad}\nolimits_{x}\right)\left(U\right)$
$\displaystyle=\left(\operatorname*{ad}\nolimits_{x}\right)^{p-1}\left(\underbrace{\operatorname*{ad}\nolimits_{x}\left(U\right)}_{\in\overline{\mathfrak{g}}^{\prime}}\right)\in\left(\operatorname*{ad}\nolimits_{x}\right)^{p-1}\left(\overline{\mathfrak{g}}^{\prime}\right)\subseteq\overline{\mathfrak{g}}^{\prime}\
\ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.prop.5b'.sub})}\right).$ Let us now
forget that we fixed $U$. We thus have shown that
$\left[x^{p},U\right]\in\overline{\mathfrak{g}}^{\prime}$ for every $U\in
L_{i}^{\prime}$. In other words,
$\left[x^{p},L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$.
This completes our proof of (25).

Next, we notice that any two positive integers $i$ and $j$ satisfy

$\left[L_{i}^{\prime},L_{j}^{\prime}\right]\subseteq L_{i+j}^{\prime}.$ (26)
212121The proof of (26) is analogous to the proof of (7) given earlier in this
note.

Now,

$\displaystyle\left[\sum\limits_{i\geq 1}L_{i}^{\prime},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]$ $\displaystyle=\left[\sum\limits_{i\geq
1}L_{i}^{\prime},\sum\limits_{j\geq 1}L_{j}^{\prime}\right]=\sum\limits_{i\geq
1}\sum\limits_{j\geq
1}\underbrace{\left[L_{i}^{\prime},L_{j}^{\prime}\right]}_{\begin{subarray}{c}\subseteq
L_{i+j}^{\prime}\\\ \text{(by
(\ref{pf.prop.6b'.a.LiLj}))}\end{subarray}}\subseteq\sum\limits_{i\geq
1}\sum\limits_{j\geq 1}L_{i+j}^{\prime}$
$\displaystyle\subseteq\sum\limits_{k\geq 2}L_{k}^{\prime}\ \ \ \ \ \ \ \ \ \
\left(\text{since }i+j\geq 2\text{ for any }i\geq 1\text{ and }j\geq 1\right)$
$\displaystyle=L_{2}^{\prime}+L_{3}^{\prime}+L_{4}^{\prime}+\cdots=\overline{\mathfrak{g}}^{\prime}.$

Recall now that
$\overline{\mathfrak{g}}^{\prime}=L_{2}^{\prime}+L_{3}^{\prime}+L_{4}^{\prime}+\cdots=\sum\limits_{i\geq
2}L_{i}^{\prime}\subseteq\sum\limits_{i\geq 1}L_{i}^{\prime}$. Thus,

$\left[\overline{\mathfrak{g}}^{\prime},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]\subseteq\left[\sum\limits_{i\geq
1}L_{i}^{\prime},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}.$

Since $\mathfrak{h}_{p}^{\prime}=\overline{\mathfrak{g}}^{\prime}+P_{p}$, we
have

$\left[\mathfrak{h}_{p}^{\prime},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]=\underbrace{\left[\overline{\mathfrak{g}}^{\prime},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]}_{\subseteq\overline{\mathfrak{g}}^{\prime}}+\underbrace{\left[P_{p},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}^{\prime}\\\
\text{(by
(\ref{pf.prop.6b'.a.PLi}))}\end{subarray}}\subseteq\overline{\mathfrak{g}}^{\prime}+\overline{\mathfrak{g}}^{\prime}=\overline{\mathfrak{g}}^{\prime}.$

Since $\overline{\mathfrak{g}}^{\prime}\subseteq\sum\limits_{i\geq
1}L_{i}^{\prime}$, we now have

$\left[\mathfrak{h}_{p}^{\prime},\overline{\mathfrak{g}}^{\prime}\right]\subseteq\left[\mathfrak{h}_{p}^{\prime},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}.$ (27)

But we also have $L=L_{1}^{\prime}\subseteq\sum\limits_{i\geq
1}L_{i}^{\prime}$ and thus

$\left[\mathfrak{h}_{p}^{\prime},L\right]\subseteq\left[\mathfrak{h}_{p}^{\prime},\sum\limits_{i\geq
1}L_{i}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}.$ (28)

From this, we easily obtain

$\left[\mathfrak{h}_{p}^{\prime},P_{p}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$
222222Proof. It is clearly enough to show that
$\left[\mathfrak{h}_{p}^{\prime},x^{p}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$
for every $x\in L$ (since the $\mathbf{k}$-module $P_{p}$ is spanned by
elements of the form $x^{p}$ for $x\in L$). So let $x\in L$. We shall use the
notations introduced in the proof of Proposition 8.5. Let
$U\in\mathfrak{h}_{p}^{\prime}$. Then, the definition of
$\operatorname*{ad}\nolimits_{x}$ yields
$\operatorname*{ad}\nolimits_{x}\left(U\right)=\left[\underbrace{x}_{\in
L},\underbrace{U}_{\in\mathfrak{h}_{p}^{\prime}}\right]\in\left[L,\mathfrak{h}_{p}^{\prime}\right]=\left[\mathfrak{h}_{p}^{\prime},L\right]\subseteq\overline{\mathfrak{g}}^{\prime}\
\ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.prop.6b'.a.6})}\right).$ But the
definition of $\operatorname*{ad}\nolimits_{x^{p}}$ yields
$\operatorname*{ad}\nolimits_{x^{p}}\left(U\right)=\left[x^{p},U\right]$, so
that  $\displaystyle\left[x^{p},U\right]$
$\displaystyle=\underbrace{\operatorname*{ad}\nolimits_{x^{p}}}_{\begin{subarray}{c}=\left(\operatorname*{ad}\nolimits_{x}\right)^{p}\\\
\text{(by
(\ref{pf.prop.5b'.adp}))}\end{subarray}}\left(U\right)=\underbrace{\left(\operatorname*{ad}\nolimits_{x}\right)^{p}}_{=\left(\operatorname*{ad}\nolimits_{x}\right)^{p-1}\circ\operatorname*{ad}\nolimits_{x}}\left(U\right)=\left(\left(\operatorname*{ad}\nolimits_{x}\right)^{p-1}\circ\operatorname*{ad}\nolimits_{x}\right)\left(U\right)$
$\displaystyle=\left(\operatorname*{ad}\nolimits_{x}\right)^{p-1}\left(\underbrace{\operatorname*{ad}\nolimits_{x}\left(U\right)}_{\in\overline{\mathfrak{g}}^{\prime}}\right)\in\left(\operatorname*{ad}\nolimits_{x}\right)^{p-1}\left(\overline{\mathfrak{g}}^{\prime}\right)\subseteq\overline{\mathfrak{g}}^{\prime}\
\ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.prop.5b'.sub})}\right).$ Hence,
$\left[U,x^{p}\right]=-\underbrace{\left[x^{p},U\right]}_{\in\overline{\mathfrak{g}}^{\prime}}\in\overline{\mathfrak{g}}^{\prime}$.
Let us now forget that we fixed $U$. We thus have shown that
$\left[U,x^{p}\right]\in\overline{\mathfrak{g}}^{\prime}$ for every
$U\in\mathfrak{h}_{p}^{\prime}$. In other words,
$\left[\mathfrak{h}_{p}^{\prime},x^{p}\right]\subseteq\overline{\mathfrak{g}}^{\prime}$,
qed.

.

Now, using $\mathfrak{h}_{p}^{\prime}=\overline{\mathfrak{g}}^{\prime}+P_{p}$
again, we obtain

$\left[\mathfrak{h}_{p}^{\prime},\mathfrak{h}_{p}^{\prime}\right]=\underbrace{\left[\mathfrak{h}_{p}^{\prime},\overline{\mathfrak{g}}^{\prime}\right]}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}^{\prime}\\\
\text{(by
(\ref{pf.prop.6b'.a.4}))}\end{subarray}}+\underbrace{\left[\mathfrak{h}_{p}^{\prime},P_{p}\right]}_{\subseteq\overline{\mathfrak{g}}^{\prime}}\subseteq\overline{\mathfrak{g}}^{\prime}+\overline{\mathfrak{g}}^{\prime}=\overline{\mathfrak{g}}^{\prime}.$

This proves Proposition 8.7 (a).

(b) We need to show that
$\left[\mathfrak{h}_{p}^{\prime},\mathfrak{h}_{p}^{\prime}\right]\subseteq\mathfrak{h}_{p}^{\prime}$
and
$\left[\mathfrak{h}_{p}^{\prime}+L,\mathfrak{h}_{p}^{\prime}+L\right]\subseteq\mathfrak{h}_{p}^{\prime}+L$.

The relation
$\left[\mathfrak{h}_{p}^{\prime},\mathfrak{h}_{p}^{\prime}\right]\subseteq\mathfrak{h}_{p}^{\prime}$
follows from
$\left[\mathfrak{h}_{p}^{\prime},\mathfrak{h}_{p}^{\prime}\right]\subseteq\overline{\mathfrak{g}}^{\prime}\subseteq\mathfrak{h}_{p}^{\prime}$.

We have

$\displaystyle\left[\mathfrak{h}_{p}^{\prime}+L,\mathfrak{h}_{p}^{\prime}+L\right]$
$\displaystyle=\underbrace{\left[\mathfrak{h}_{p}^{\prime},\mathfrak{h}_{p}^{\prime}\right]}_{\subseteq\overline{\mathfrak{g}}^{\prime}}+\underbrace{\left[\mathfrak{h}_{p}^{\prime},L\right]}_{\begin{subarray}{c}\subseteq\overline{\mathfrak{g}}^{\prime}\\\
\text{(by
(\ref{pf.prop.6b'.a.6}))}\end{subarray}}+\underbrace{\left[L,\mathfrak{h}_{p}^{\prime}\right]}_{\begin{subarray}{c}=\left[\mathfrak{h}_{p}^{\prime},L\right]\subseteq\overline{\mathfrak{g}}^{\prime}\\\
\text{(by
(\ref{pf.prop.6b'.a.6}))}\end{subarray}}+\underbrace{\left[L,L\right]}_{\begin{subarray}{c}=L_{2}^{\prime}\subseteq
L_{2}^{\prime}+L_{3}^{\prime}+L_{4}^{\prime}+\cdots\\\
=\overline{\mathfrak{g}}^{\prime}\end{subarray}}$
$\displaystyle\subseteq\overline{\mathfrak{g}}^{\prime}+\overline{\mathfrak{g}}^{\prime}+\overline{\mathfrak{g}}^{\prime}+\overline{\mathfrak{g}}^{\prime}=\overline{\mathfrak{g}}^{\prime}\subseteq\mathfrak{h}_{p}^{\prime}\subseteq\mathfrak{h}_{p}^{\prime}+L.$

Proposition 8.7 (b) is thus shown.

Next, we state an analogue of Proposition 6.2 which is stronger than
Proposition 7.14 (of course under the assumption that $\mathbf{k}$ is an
$\mathbb{F}_{p}$-algebra):

######  Proposition 8.9.

We have
$\left(\mathfrak{h}_{p}^{\prime}\right)^{\star}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.

######  Proof 8.10 (Proof of Proposition 8.9.).

We have $P_{p}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$
due to Proposition 8.1. Also,
$L_{2}^{\prime}=\left[L,L\right]\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$
by Proposition 7.7 (c). Using this and Proposition 7.7 (d), we can show that
$L_{i}^{\prime}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$
for each $i\geq 2$ (by induction over $i$). Thus,
$\overline{\mathfrak{g}}^{\prime}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$
(since
$\overline{\mathfrak{g}}^{\prime}=L_{2}^{\prime}+L_{3}^{\prime}+L_{4}^{\prime}+\cdots$).
Combined with
$P_{p}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$, this
yields
$\mathfrak{h}_{p}^{\prime}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$
(since $\mathfrak{h}_{p}^{\prime}=\overline{\mathfrak{g}}^{\prime}+P_{p}$).
Since $\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$ is a
$\mathbf{k}$-subalgebra of $T\left(L\right)$ (because of Proposition 7.7 (a)
and Proposition 7.7 (b)), this yields that
$\left(\mathfrak{h}_{p}^{\prime}\right)^{\star}\subseteq\operatorname*{Ker}\left(\mathbf{t}^{\prime}\right)$.
This proves Proposition 8.9.

Next, we state some lemmas. The first lemma is an analogue of Lemma 6.4 again:

######  Lemma 8.11.

Let $u\in L$. Let $S$ be a $\mathbf{k}$-submodule of $T\left(L\right)$ such
that $u^{p}\in S^{\star}$ and such that $\left[u,S\right]\subseteq S^{\star}$.
Then, $\sum\limits_{j=0}^{p-1}S^{\star}u^{j}$ is a $\mathbf{k}$-subalgebra of
$T\left(L\right)$.

######  Proof 8.12 (Proof of Lemma 8.11.).

Lemma 7.16 yields that $\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}$ is a
$\mathbf{k}$-subalgebra of $T\left(L\right)$. We shall now show that
$\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}=\sum\limits_{j=0}^{p-1}S^{\star}u^{j}$.

We have $S^{\star}u^{p}\subseteq S^{\star}u^{0}$ 232323Proof. We have
$S^{\star}\underbrace{u^{p}}_{\in S^{\star}}=S^{\star}S^{\star}\subseteq
S^{\star}$ (since $S^{\star}$ is a $\mathbf{k}$-algebra). But
$S^{\star}\underbrace{u^{0}}_{=1}=S^{\star}$, so that $S^{\star}u^{p}\subseteq
S^{\star}=S^{\star}u^{0}$, qed.. Thus,

$S^{\star}u^{k}\subseteq\sum\limits_{j=0}^{p-1}S^{\star}u^{j}\ \ \ \ \ \ \ \ \
\ \text{for every }k\in\mathbb{N}.$ (29) 242424Proof of (29): We shall prove
(29) by induction over $k$. Induction base: We have
$S^{\star}u^{0}\subseteq\sum\limits_{j=0}^{p-1}S^{\star}u^{j}$ (since
$S^{\star}u^{0}$ is an addend of the sum
$\sum\limits_{j=0}^{p-1}S^{\star}u^{j}$). In other words, (29) holds for
$k=0$. This completes the induction base. Induction step: Let
$K\in\mathbb{N}$. Assume that (29) holds for $k=K$. We must prove that (29)
holds for $k=K+1$. We have
$S^{\star}u^{K}\subseteq\sum\limits_{j=0}^{p-1}S^{\star}u^{j}$ (since (29)
holds for $k=K$). Hence,  $\displaystyle
S^{\star}\underbrace{u^{K+1}}_{=u^{K}u}$
$\displaystyle=\underbrace{S^{\star}u^{K}}_{\subseteq\sum\limits_{j=0}^{p-1}S^{\star}u^{j}}u\subseteq\left(\sum\limits_{j=0}^{p-1}S^{\star}u^{j}\right)u$
$\displaystyle=\sum\limits_{j=0}^{p-1}S^{\star}\underbrace{u^{j}u}_{=u^{j+1}}=\sum\limits_{j=0}^{p-1}S^{\star}u^{j+1}=\sum\limits_{j=1}^{p}S^{\star}u^{j}$
$\displaystyle=\sum\limits_{j=1}^{p-1}S^{\star}u^{j}+\underbrace{S^{\star}u^{p}}_{\subseteq
S^{\star}u^{0}}\subseteq\sum\limits_{j=1}^{p-1}S^{\star}u^{j}+S^{\star}u^{0}=\sum\limits_{j=0}^{p-1}S^{\star}u^{j}.$
In other words, (29) holds for $k=K+1$. This completes the induction step.
Thus, (29) is proven.

Now,

$\sum\limits_{j\in\mathbb{N}}S^{\star}u^{j}=\sum\limits_{k\in\mathbb{N}}\underbrace{S^{\star}u^{k}}_{\begin{subarray}{c}\subseteq\sum\limits_{j=0}^{p-1}S^{\star}u^{j}\\\
\text{(by