$\begin{array}[]{lcl}\Delta(s_{\mathfrak{g}}(f))&=&\Delta(s_{\mathfrak{a}}(a)+t^{2}a_{0}+t{\mathscr{D}}(a))\\\\[5.69054pt]
&=&\tilde{\Delta}(s_{\mathfrak{a}}(a)+t^{2}a_{0}+t{\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a)+t^{2}a_{0}+t{\mathscr{D}}(a),b_{0})x\\\\[5.69054pt]
&=&\tilde{\Delta}(s_{\mathfrak{a}}(a)+t^{2}a_{0}+t{\mathscr{D}}(a)).\end{array}$

The RHS (for $f=rx+a+tx^{*}$):

$\begin{array}[]{lcl}[\Delta(f),f]_{\mathfrak{g}}&=&[r\lambda
x+\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x+t(\lambda
x^{*}+b_{0}+\mu x),rx+a+tx^{*}]_{\mathfrak{g}}\\\\[5.69054pt]
&=&[\tilde{\Delta}(a)+t(\lambda
x^{*}+b_{0}),a+tx^{*}]_{\mathfrak{g}}\\\\[5.69054pt]
&=&[\tilde{\Delta}(a),a]_{\mathfrak{a}}+t{\mathscr{D}}(\tilde{\Delta}(a))+t\lambda{\mathscr{D}}(a)+t[b_{0},a]_{\mathfrak{a}}+t^{2}{\mathscr{D}}(b_{0}).\end{array}$

Taking into account Eq.(56), we see that

(59) $\displaystyle\tilde{\Delta}(s_{\mathfrak{a}}(a))$ $\displaystyle=$
$\displaystyle[a,\tilde{\Delta}(a)]_{\mathfrak{a}},$ (60)
$\displaystyle\tilde{\Delta}(a_{0})$ $\displaystyle=$
$\displaystyle{\mathscr{D}}(b_{0}).$

Eqs. (57), (59) imply that $\tilde{\Delta}$ is a derivation of
${\mathfrak{a}}$; obviously $\tilde{\Delta}$ is invertible. Define

$\omega_{\mathfrak{a}}(\cdot,\cdot):={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(\cdot),\cdot).$

The form $\omega_{\mathfrak{a}}$ on ${\mathfrak{a}}$ is closed,
${\bar{1}}$-antisymmetric and ortho-orthogonal.∎

###  5.3. Double extensions (DEs) of odd NIS-Lie superalgebras with closed
non-degenerate anti-symmetric bilinear forms

Here we continue the study we carried out in Section 5.2 in the case where NIS
is odd. The first two theorems below have been proved in [BeBou].

######  Theorem (${\mathscr{D}}_{\bar{1}}$-extension – NIS odd).

Let $({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$ be a NIS-Lie superalgebra
in characteristic $2$ such that ${\mathscr{B}}_{\mathfrak{a}}$ is odd. Let
${\mathscr{D}}\in{\mathfrak{der}}_{\bar{1}}({\mathfrak{a}})$ and
$a_{0}\in{\mathfrak{a}}_{\bar{0}}$ satisfy the following conditions:

(61)
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)+{\mathscr{B}}_{\mathfrak{a}}(a,{\mathscr{D}}(b))$
$\displaystyle=$ $\displaystyle 0\quad\text{for any }a,b\in{\mathfrak{a}};$
(62) $\displaystyle{\mathscr{B}}_{\mathfrak{a}}(a,{\mathscr{D}}(a))$
$\displaystyle=$ $\displaystyle 0\quad\text{for any
$a\in{\mathfrak{a}}_{\bar{0}}$};$ (63) $\displaystyle{\mathscr{D}}^{2}$
$\displaystyle=$ $\displaystyle\mathop{\mathrm{ad}}\nolimits_{a_{0}};$ (64)
$\displaystyle{\mathscr{D}}(a_{0})$ $\displaystyle=$ $\displaystyle 0.$

Let $\alpha$ be a quadratic form on ${\mathfrak{a}}_{\bar{1}}$ such that
${\mathscr{B}}_{\alpha}(a,b)={\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)$.
Then, there exists a NIS-Lie superalgebra structure on
${\mathfrak{g}}:=\mathscr{K}\oplus{\mathfrak{a}}\oplus\mathscr{E}$ , where
$\mathscr{K}:=\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$ and $x$ is even,
$\mathscr{E}:=\mathop{\mathrm{{Span}}}\nolimits\\{e\\}$ and $e$ is odd,
defined as follows. The squaring is given by (where $\mu\in{\mathbb{K}}$):

$s_{\mathfrak{g}}(a+\mu
e):=s_{\mathfrak{a}}(a)+(\mu^{2}m+\alpha(a))x+\mu^{2}a_{0}+\mu{\mathscr{D}}(a)\qquad\text{
for any }\;a+\mu e\in{\mathfrak{g}}_{\bar{1}}.$

The bracket is given by

$\begin{array}[]{l}[x,{\mathfrak{g}}]_{\mathfrak{g}}:=0;\qquad[a,b]_{\mathfrak{g}}:=[a,b]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)x\quad\text{~{}~{}for
any $a,b\in{\mathfrak{a}}$};\\\\[5.69054pt]
[e,a]_{\mathfrak{g}}:={\mathscr{D}}(a)\quad\text{ for any
}a~{}\in{\mathfrak{a}}.\end{array}$

The bilinear form $B_{\mathfrak{g}}$ on ${\mathfrak{g}}$ defined by:

$\displaystyle{{\mathscr{B}}_{\mathfrak{g}}}|_{{\mathfrak{a}}\times{\mathfrak{a}}}:={\mathscr{B}}_{\mathfrak{a}},\quad{\mathscr{B}}_{\mathfrak{g}}({\mathfrak{a}},\mathscr{K}):=0,\quad{\mathscr{B}}_{\mathfrak{g}}({\mathfrak{a}},\mathscr{E}):=0,$
$\displaystyle{\mathscr{B}}_{\mathfrak{g}}(x,e):=1,\quad{\mathscr{B}}_{\mathfrak{g}}(x,x):={\mathscr{B}}_{\mathfrak{g}}(e,e):=0,$

is odd, non-degenerate, invariant, and ${\bar{0}}$-antisymmetric. Therefore,
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ is a NIS-Lie superalgebra.

######  Theorem (${\mathscr{D}}_{\bar{0}}$-extension – NIS odd).

Let $({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$ be a NIS-Lie superalgebra
in characteristic $2$ such that ${\mathscr{B}}_{\mathfrak{a}}$ is odd. Let
${\mathscr{D}}\in{\mathfrak{der}}_{\bar{0}}({\mathfrak{a}})$ be such that:

(65)
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)+{\mathscr{B}}_{\mathfrak{a}}(a,{\mathscr{D}}(b))$
$\displaystyle=$ $\displaystyle 0\text{ for any }a,b\in{\mathfrak{a}}.$

Then, there exists a NIS-Lie superalgebra structure on
${\mathfrak{g}}:=\mathscr{K}\oplus{\mathfrak{a}}\oplus\mathscr{E}$, where
$\mathscr{K}:=\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$ and $x$ is odd, and
where $\mathscr{E}:=\mathop{\mathrm{{Span}}}\nolimits\\{e\\}$ and $e$ is even,
defined as follows. The squaring is given by (where $\mu\in{\mathbb{K}}$):

$s_{\mathfrak{g}}(a+\mu x):=s_{\mathfrak{a}}(a)\qquad\text{ for any }\;a+\mu
x\in{\mathfrak{g}}_{\bar{1}}.$

The bracket is given by:

$\begin{array}[]{l}[x,{\mathfrak{g}}]_{\mathfrak{g}}:=0;\qquad[a,b]_{\mathfrak{g}}:={\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)x+[a,b]_{\mathfrak{a}}\quad\text{~{}~{}for
any $a,b\in{\mathfrak{a}}$};\\\\[5.69054pt]
[a,e]_{\mathfrak{g}}:={\mathscr{D}}(a)\quad\text{ for any
}a~{}\in{\mathfrak{a}}.\end{array}$

The bilinear form ${\mathscr{B}}_{\mathfrak{g}}$ on ${\mathfrak{g}}$ defined
by:

$\displaystyle{{\mathscr{B}}_{\mathfrak{g}}}|_{{\mathfrak{a}}\times{\mathfrak{a}}}:={\mathscr{B}}_{\mathfrak{a}},\quad{\mathscr{B}}_{\mathfrak{g}}({\mathfrak{a}},\mathscr{K}):=0,\quad{\mathscr{B}}_{\mathfrak{g}}({\mathfrak{a}},\mathscr{E}):=0,$
$\displaystyle{\mathscr{B}}_{\mathfrak{g}}(x,e):=1,\quad{\mathscr{B}}_{\mathfrak{g}}(x,x):={\mathscr{B}}_{\mathfrak{g}}(e,e):=0,$

is odd, non-degenerate, invariant, and ${\bar{0}}$-antisymmetric. Therefore,
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ is a NIS-Lie superalgebra.

######  Theorem (Periplectic forms for ${\mathscr{D}}_{\bar{1}}$-extensions).

Let $({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}},\omega_{\mathfrak{a}})$ be a
NIS Lie superalgebra equipped with a closed ${\bar{1}}$-antisymmetric
periplectic form $\omega_{\mathfrak{a}}$, and let
$\tilde{\Delta}\in{\mathfrak{der}}({\mathfrak{a}})_{\bar{0}}$ be the unique
and invertible derivation such that
$\omega_{\mathfrak{a}}(\cdot,\cdot)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(\cdot),\cdot)$.
Let $({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ be the double extension of
$({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$ by means of a derivation
${\mathscr{D}}\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{1}}$ as in Theorem
5.3.1. Theorem. Suppose there exist $\lambda\in{\mathbb{K}}\backslash\\{0\\}$
and $a_{1}\in{\mathfrak{a}}_{\bar{1}}$ such that

(66) $\displaystyle[\tilde{\Delta},{\mathscr{D}}]$ $\displaystyle=$
$\displaystyle\lambda{\mathscr{D}}+\mathop{\mathrm{ad}}\nolimits_{a_{1}},$
(67)
$\displaystyle\lambda\alpha(a)+{\mathscr{B}}_{\alpha}(a,{\mathscr{D}}\circ\tilde{\Delta}(a))$
$\displaystyle=$
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a),a)\text{ for
all $a\in{\mathfrak{a}}_{\bar{1}}$},$ (68) $\displaystyle m$ $\displaystyle=$
$\displaystyle\lambda^{-1}{\mathscr{B}}_{\mathfrak{a}}(a_{0},a_{1}),$ (69)
$\displaystyle{\mathscr{D}}(a_{1})$ $\displaystyle=$
$\displaystyle\tilde{\Delta}(a_{0}).$

Then, the linear map $\Delta:{\mathfrak{g}}\rightarrow{\mathfrak{g}}$ defined
by

$\begin{array}[]{lcl}\Delta(x)&=&\lambda x,\\\ \Delta(e)&=&\lambda e+a_{1},\\\
\Delta(a)&=&\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{1})x.\end{array}$

is an invertible derivation on ${\mathfrak{g}}$ and
${\mathscr{B}}_{\mathfrak{g}}$ is $\Delta$-invariant. Consequently,
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ is a
NIS-Lie superalgebra with a closed ${\bar{1}}$-antisymmetric periplectic form
$\omega_{\mathfrak{g}}:={\mathscr{B}}_{\mathfrak{g}}(\Delta(\cdot),\cdot)$.

###### Proof.

First, observe that $\Delta$ is obviously invertible since $\tilde{\Delta}$ is
invertible. Let us prove that $\Delta$ is a derivation. We will check the
condition on the squaring. We have (where $a\in{\mathfrak{a}}_{\bar{1}}$)

$\Delta(s_{\mathfrak{g}}(a))=\Delta(s_{\mathfrak{a}}(a)+\alpha(a)x)=\Delta(s_{\mathfrak{a}}(a))+\lambda\alpha(a)x=\tilde{\Delta}(s_{\mathfrak{a}}(a))+{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a),a_{1})x+\lambda\alpha(a)x.$

Now,

$\Delta(s_{\mathfrak{g}}(e))=\Delta(mx+a_{0})=m\lambda
x+\tilde{\Delta}(a_{0})+{\mathscr{B}}_{\mathfrak{a}}(a_{0},a_{1})x,$

while

$[\Delta(e),e]_{\mathfrak{g}}=[\lambda
e+a_{1},e]_{\mathfrak{g}}=[a_{1},e]_{\mathfrak{g}}={\mathscr{D}}(a_{1}).$

It follows that $\Delta(s_{\mathfrak{g}}(e))=[\Delta(e),e]_{\mathfrak{g}}$ is
equivalent to Eqn. (68), (69).

On the other hand, we have (where $a\in{\mathfrak{a}}_{\bar{1}}$)

$[\Delta(a),a]_{\mathfrak{g}}=[\tilde{\Delta}(a),a]_{\mathfrak{g}}=[\tilde{\Delta}(a),a]_{\mathfrak{a}}+{\mathscr{B}}_{\alpha}({\mathscr{D}}\circ\tilde{\Delta}(a),a)x.$

The fact that $\tilde{\Delta}\in{\mathfrak{der}}({\mathfrak{a}})_{\bar{0}}$
and Eq. (67) implies that
$\Delta(s_{\mathfrak{g}}(a))=[\Delta(a),a]_{\mathfrak{g}}$.

Now let us check the condition

$\Delta([f,g]_{\mathfrak{g}})=[\Delta(f),g]_{\mathfrak{g}}+[f,\Delta(g)]_{\mathfrak{g}}\text{
for all $f,g\in{\mathfrak{g}}$}.$

We will only check the case where $(f,g)=(a,e)$; checking the other cases is a
routine. Indeed,

$\begin{array}[]{lcl}\Delta([a,e]_{\mathfrak{g}})&=&\Delta({\mathscr{D}}(a))=\tilde{\Delta}({\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),a_{1})x.\end{array}$

On the other hand,

$\begin{array}[]{lcl}[\Delta(a),e]_{\mathfrak{g}}+[a,\Delta(e)]_{\mathfrak{g}}&=&[\tilde{\Delta}(a),e]_{\mathfrak{g}}+[a,\lambda
e+a_{1}]_{\mathfrak{g}}\\\\[5.69054pt]
&=&{\mathscr{D}}(\tilde{\Delta}(a))+\lambda{\mathscr{D}}(a)+[a,a_{1}]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),a_{1})x.\end{array}$

Therefore,
$\Delta([a,e]_{\mathfrak{g}})=[\Delta(a),e]_{\mathfrak{g}}+[a,\Delta(e)]_{\mathfrak{g}}$
is equivalent to Eq. (66).

Let us show that ${\mathscr{B}}_{\mathfrak{g}}$ is $\Delta$-invariant. For all
$a,b\in{\mathfrak{a}}$, we have

${\mathscr{B}}_{\mathfrak{g}}(\Delta(a),b)={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{1})x,b)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(a),b)={\mathscr{B}}_{\mathfrak{a}}(a,\tilde{\Delta}(b))={\mathscr{B}}_{\mathfrak{g}}(a,\Delta(b)).$

Moreover,

$\begin{array}[]{lcl}{\mathscr{B}}_{\mathfrak{g}}(\Delta(e),a)+{\mathscr{B}}_{\mathfrak{g}}(e,\Delta(a))&=&{\mathscr{B}}_{\mathfrak{g}}(\lambda
e+a_{1},a)+{\mathscr{B}}_{\mathfrak{g}}(e,\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{1})x.)\\\\[5.69054pt]
&=&{\mathscr{B}}_{\mathfrak{a}}(a_{1},a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{1})=0.\end{array}$

The $\Delta$-invariant of ${\mathscr{B}}_{\mathfrak{g}}$ can easily be checked
for the remaining elements of the basis of ${\mathfrak{g}}$.∎

######  Theorem (Converse of Theorem 5.3.3. Theorem).

Let $({\mathfrak{g}},\mathscr{B}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ be an
irreducible NIS-Lie superalgebra such that
$p({\mathscr{B}}_{\mathfrak{g}})={\bar{1}}$,. Let it be equipped with a closed
${\bar{1}}$-antisymmetric periplectic form $\omega_{\mathfrak{g}}$. Suppose
that ${\mathfrak{z}}({\mathfrak{g}})_{\bar{0}}\not=\\{0\\}.$ Then,
$({\mathfrak{g}},\mathscr{B}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ is
obtained as an $\mathscr{D}$-extension from a NIS-Lie superalgebra
$({\mathfrak{a}},\mathscr{B}_{\mathfrak{a}},\omega_{\mathfrak{a}})$ with a
closed ${\bar{1}}$-antisymmetric periplectic form $\omega_{\mathfrak{a}}$.

###### Proof.

Let $\Delta\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{0}}$ be the unique
invertible derivation such that
$\omega_{\mathfrak{g}}(\cdot,\cdot)=\mathscr{B}_{\mathfrak{g}}(\Delta(\cdot),\cdot)$.
It follows that
$\Delta({\mathfrak{z}}({\mathfrak{g}})_{\bar{0}})={\mathfrak{z}}({\mathfrak{g}})_{\bar{0}}$.
Since ${\mathbb{K}}$ is algebraically closed, let us choose
$x\in{\mathfrak{z}}({\mathfrak{g}})_{\bar{0}}$ such that $\Delta(x)=\lambda x$
for some $\lambda\in{\mathbb{K}}\backslash\\{0\\}$. It has been shown in
[BeBou, Prop. 4.1.2], that the subspaces
$\mathscr{K}:=\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$ and
$\mathscr{K}^{\perp}$ are ideals in
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$. Besides,
$\Delta(\mathscr{K}^{\perp})\subset\mathscr{K}^{\perp}$. Indeed, (for all
$f\in\mathscr{K}^{\perp}$)

${\mathscr{B}}_{\mathfrak{g}}(\Delta(f),x)={\mathscr{B}}_{\mathfrak{g}}(f,\Delta(x))={\mathscr{B}}_{\mathfrak{g}}(f,\lambda
x)=0.$

Now, since $\mathscr{K}$ is 1-dimensional, then either
$\mathscr{K}\cap\mathscr{K}^{\perp}=\\{0\\}$ or
$\mathscr{K}\cap\mathscr{K}^{\perp}=\mathscr{K}$. The first case is to be
disregarded because otherwise
${\mathfrak{g}}=\mathscr{K}\oplus\mathscr{K}^{\perp}$ and the Lie superalgebra
${\mathfrak{g}}$ will not be irreducible. Hence,
$\mathscr{K}\cap\mathscr{K}^{\perp}=\mathscr{K}$. It follows that
$\mathscr{K}\subset\mathscr{K}^{\perp}$ and
$\dim(\mathscr{K}^{\perp})=\dim({\mathfrak{g}})-1$. Therefore, there exists
$e\in{\mathfrak{g}}_{\bar{1}}$ such that (since
${\mathscr{K}}^{\perp}_{\bar{0}}={\mathfrak{g}}_{\bar{0}}$ and
$\omega_{\mathfrak{g}}$ is odd ):

${\mathfrak{g}}=\mathscr{K}^{\perp}\oplus\mathscr{E},\quad\text{ where
$\mathscr{E}:=\mathop{\mathrm{{Span}}}\nolimits\\{e\\}$}.$

This $e$ can be normalized to have ${\mathscr{B}}_{\mathfrak{g}}(x,e)=1$.
Besides, ${\mathscr{B}}_{\mathfrak{g}}(x,x)=0$ since
$\mathscr{K}\cap\mathscr{K}^{\perp}=\mathscr{K}$.

Let us define ${\mathfrak{a}}:=(\mathscr{K}+\mathscr{E})^{\perp}$. We then
have a decomposition
${\mathfrak{g}}=\mathscr{K}\oplus{\mathfrak{a}}\oplus\mathscr{E}$.

It has been shown in [BeBou, Prop. 4.1.2] that
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ is a double extension of
$({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$, where
${\mathscr{B}}_{\mathfrak{a}}={\mathscr{B}}_{{\mathfrak{g}}}|_{{\mathfrak{a}}\times{\mathfrak{a}}}$,
by means of a derivation $\mathscr{D}$ satisfying (61), (62) and a quadratic
form $\alpha$.

Let us write $\Delta$ as (where
$a_{1}\in{\mathfrak{a}}_{\bar{1}},\beta(a),\mu,\nu\in{\mathbb{K}}$):

$\begin{array}[]{lcl}\Delta(x)&=&\lambda x,\\\ \Delta(e)&=&\nu e+a_{1},\\\
\Delta(a)&=&\tilde{\Delta}(a)+\beta(a)x+\gamma(a)x^{*}.\end{array}$

Since
${\mathscr{B}}_{\mathfrak{g}}(\Delta(e),x)={\mathscr{B}}_{\mathfrak{g}}(e,\Delta(x))$,
then $\nu=\lambda$. Similarly, since
${\mathscr{B}}_{\mathfrak{a}}(a,\Delta(x))={\mathscr{B}}_{\mathfrak{a}}(a,\Delta(x))=0$,
then $\gamma(a)=0$ for all $a\in{\mathfrak{a}}$.

In addition, the conditions

$\begin{array}[]{l}{\mathscr{B}}_{\mathfrak{g}}(\Delta(a),e)={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+\beta(a)x,e)=\beta(a),\\\\[5.69054pt]
{\mathscr{B}}_{\mathfrak{g}}(a,\Delta(e))={\mathscr{B}}_{\mathfrak{g}}(a,\lambda
e+a_{1}+\mu x)={\mathscr{B}}_{\mathfrak{a}}(a,a_{1}),\end{array}$

imply that

(70) $\beta(a)={\mathscr{B}}_{\mathfrak{a}}(a,a_{1})\text{ for all
$a\in{\mathfrak{a}}$}.$

Now, we have

$\begin{array}[]{lcl}\Delta([e,a]_{\mathfrak{g}})&=&\Delta({\mathscr{D}}(a))=\tilde{\Delta}({\mathscr{D}}(a))+\beta({\mathscr{D}}(a))x\;\text{
for all $a\in{\mathfrak{a}}$}.\end{array}$

On the other hand, we have

$[\Delta(e),a]_{\mathfrak{g}}+[e,\Delta(a)]_{\mathfrak{g}}=\lambda{\mathscr{D}}(a)+[a_{1},a]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a_{1}),a)x+{\mathscr{D}}(\tilde{\Delta}(a))\text{
for all $a\in{\mathfrak{a}}$}.$

Since $\Delta$ is a derivation on ${\mathfrak{g}}$ and thanks to Eq. (70), we
see that

(71)
$[\tilde{\Delta},{\mathscr{D}}]=\lambda{\mathscr{D}}+\mathop{\mathrm{ad}}\nolimits_{a_{1}}.$

Following the same computation as in [BaB], the condition

$\Delta([a,b]_{\mathfrak{g}})=[\Delta(a),b]_{\mathfrak{g}}+[a,\Delta(b)]_{\mathfrak{g}}\;\text{
for all $a,b\in{\mathfrak{a}}$,}$

implies that

(72) $\displaystyle\tilde{\Delta}([a,b]_{\mathfrak{a}})$ $\displaystyle=$
$\displaystyle[\tilde{\Delta}(a),b]_{\mathfrak{a}}+[a,\tilde{\Delta}(b)]_{\mathfrak{a}}$
(73) $\displaystyle\beta([a,b]_{\mathfrak{a}})$ $\displaystyle=$
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(\tilde{\Delta}(a)),b)+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),\tilde{\Delta}(b)).$

Eq. (73) is certainly satisfied, thanks to Eqs. (70), (71) and the fact that
is ${\mathscr{B}}_{\mathfrak{a}}$ is $\tilde{\Delta}$-invariant.

For every $a\in{\mathfrak{a}}_{\bar{1}}$, let us study the condition

$\Delta(s_{\mathfrak{g}}(a))=[a,\Delta(a)]_{\mathfrak{g}}.$

Its LHS reads

$\begin{array}[]{l}\Delta(s_{\mathfrak{g}}(a))=\Delta(s_{\mathfrak{a}}(a)+\alpha(a)x)=\Delta(s_{\mathfrak{a}}(a))+\lambda\alpha(a)x=\tilde{\Delta}(s_{\mathfrak{a}}(a))+\beta(s_{\mathfrak{a}}(a))x+\lambda\alpha(a)x,\end{array}$

while the RHS reads

$\begin{array}[]{l}[a,\Delta(a)]_{\mathfrak{g}}=[\tilde{\Delta}(a)+\beta(a)x,a]_{\mathfrak{g}}=[\tilde{\Delta}(a),a]_{\mathfrak{g}}=[\tilde{\Delta}(a),a]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}\circ\tilde{\Delta}(a),a).\end{array}$

It follows that

(74) $\displaystyle\tilde{\Delta}(s_{\mathfrak{a}}(a))$ $\displaystyle=$
$\displaystyle[a,\tilde{\Delta}(a)]_{\mathfrak{a}},$ (75)
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a),a_{0})+\lambda\alpha(a)$
$\displaystyle=$
$\displaystyle{\mathscr{B}}_{\alpha}(a,{\mathscr{D}}\circ\tilde{\Delta}(a)).$

Eqs. (72), (74) imply that $\tilde{\Delta}$ is a derivation on
${\mathfrak{a}}$. On the other hand, since
$\Delta(s_{\mathfrak{g}}(e))=[\Delta(e),e]_{\mathfrak{g}}$, then that

$\begin{array}[]{ccl}{\mathscr{D}}(a_{1})&=&\tilde{\Delta}(a_{0}),\\\
m\lambda&=&{\mathscr{B}}_{\mathfrak{a}}(a_{0},a_{1}).\end{array}$

Since $\Delta$ is invertible, then $\tilde{\Delta}$ is invertible as well.
Define

$\omega_{\mathfrak{a}}(a,b):={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(a),b)\;\text{
for all $a,b\in{\mathfrak{a}}$}.$

The form $\omega_{\mathfrak{a}}$ on ${\mathfrak{a}}$ is closed,
${\bar{1}}$-antisymmetric and periplectic.∎

######  Theorem ((Periplectic forms for
${\mathscr{D}}_{\bar{0}}$-extensions)).

Let $({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}},\omega_{\mathfrak{a}})$ be a
NIS Lie superalgebra equipped with a closed ${\bar{1}}$-antisymmetric
periplectic form $\omega_{\mathfrak{a}}$, and let
$\tilde{\Delta}\in{\mathfrak{der}}_{\bar{0}}({\mathfrak{a}})$ be the unique
and invertible derivation such that
$\omega_{\mathfrak{a}}(\cdot,\cdot)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(\cdot),\cdot)$.
Let $({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ be the double extension of
$({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$ by means of a derivation
${\mathscr{D}}\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{0}}$ as in Theorem
5.3.2. Theorem. Suppose there exist $\lambda\in{\mathbb{K}}\backslash\\{0\\}$
and $b_{0}\in{\mathfrak{a}}_{\bar{0}}$ such that

(76) $\displaystyle[\tilde{\Delta},{\mathscr{D}}]$ $\displaystyle=$
$\displaystyle\lambda{\mathscr{D}}+\mathop{\mathrm{ad}}\nolimits_{b_{0}},$

The linear map $\Delta:{\mathfrak{g}}\rightarrow{\mathfrak{g}}$ defined by

$\begin{array}[]{lcl}\Delta(x)&=&\lambda x,\\\ \Delta(e)&=&\lambda e+b_{0},\\\
\Delta(a)&=&\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x.\end{array}$

is an invertible derivation on ${\mathfrak{g}}$ and is
${\mathscr{B}}_{\mathfrak{g}}$ is $\Delta$-invariant. Consequently,
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ is a
NIS-Lie superalgebra, with a closed ${\bar{1}}$-antisymmetric periplectic form
$\omega_{\mathfrak{g}}:={\mathscr{B}}_{\mathfrak{g}}(\Delta(\cdot),\cdot)$.

###### Proof.

First, observe that $\Delta$ is obviously invertible since $\tilde{\Delta}$ is
invertible. Let us prove that $\Delta$ is a derivation. We will check the
condition on the squaring. Indeed, (where $a+\mu
x\in{\mathfrak{a}}_{\bar{1}}$)

$\Delta(s_{\mathfrak{g}}(a+\mu
x))=\Delta(s_{\mathfrak{a}}(a))=\tilde{\Delta}(s_{\mathfrak{a}}(a))+{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a),b_{0})x=\tilde{\Delta}(s_{\mathfrak{a}}(a)).$

On the other hand, we have

$[\Delta(a+\mu x),a+\mu
x]_{\mathfrak{g}}=[\Delta(a),a]_{\mathfrak{g}}=[\tilde{\Delta}(a),a]_{\mathfrak{a}}+{\mathscr{B}}_{\alpha}({\mathscr{D}}\circ\tilde{\Delta}(a),a)x=[\tilde{\Delta}(a),a]_{\mathfrak{a}}.$

The fact that $\tilde{\Delta}\in{\mathfrak{der}}_{\bar{0}}({\mathfrak{a}})$
implies that $\Delta(s_{\mathfrak{g}}(f))=[\Delta(f),f]_{\mathfrak{g}}$ for
all $f\in{\mathfrak{a}}_{\bar{1}}$.

Now let us check the condition

$\Delta([f,g]_{\mathfrak{g}})=[\Delta(f),g]_{\mathfrak{g}}+[f,\Delta(g)]_{\mathfrak{g}}\text{
for all $f,g\in{\mathfrak{g}}$}.$

We will only check the case where $(f,g)=(a,e)$; checking the other cases is a
routine. Indeed,

$\begin{array}[]{lcl}\Delta([a,e]_{\mathfrak{g}})&=&\Delta({\mathscr{D}}(a))=\tilde{\Delta}({\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b_{0})x.\end{array}$

On the other hand,

$\begin{array}[]{lcl}[\Delta(a),e]_{\mathfrak{g}}+[a,\Delta(e)]_{\mathfrak{g}}&=&[\tilde{\Delta}(a),e]_{\mathfrak{g}}+[a,\lambda
e+b_{0}]_{\mathfrak{g}}\\\\[5.69054pt]
&=&{\mathscr{D}}(\tilde{\Delta}(a))+\lambda{\mathscr{D}}(a)+[a,b_{0}]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b_{0})x.\end{array}$

Therefore,
$\Delta([a,e]_{\mathfrak{g}})=[\Delta(a),e]_{\mathfrak{g}}+[a,\Delta(e)]_{\mathfrak{g}}$
is equivalent to Eq. (76).

Let us show that ${\mathscr{B}}_{\mathfrak{g}}$ is $\Delta$-invariant. Indeed,
for all $a,b\in{\mathfrak{a}}$, we have

${\mathscr{B}}_{\mathfrak{g}}(\Delta(a),b)={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x,b)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(a),b)={\mathscr{B}}_{\mathfrak{a}}(a,\tilde{\Delta}(b))={\mathscr{B}}_{\mathfrak{g}}(a,\Delta(b)).$

Moreover,

$\begin{array}[]{lcl}{\mathscr{B}}_{\mathfrak{g}}(\Delta(e),a)+{\mathscr{B}}_{\mathfrak{g}}(e,\Delta(a))&=&{\mathscr{B}}_{\mathfrak{g}}(\lambda
e+b_{0},a)+{\mathscr{B}}_{\mathfrak{g}}(e,\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x.)\\\\[5.69054pt]
&=&{\mathscr{B}}_{\mathfrak{a}}(b_{0},a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})=0.\end{array}$

The $\Delta$-invariance of ${\mathscr{B}}_{\mathfrak{g}}$ can easily be
checked for the remaining elements of the basis of ${\mathfrak{g}}$.∎

In [BeBou], while proving the converse of Theorem 5.3.2. Theorem, we had to
assume the condition ${\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}\cap
s_{\mathfrak{g}}({\mathfrak{g}}_{\bar{1}})^{\perp}\not=\\{0\\}$.222There was a
misprint in [BeBou]. There should be
${\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}\cap
s_{\mathfrak{g}}({\mathfrak{g}}_{\bar{1}})^{\perp}\not=\\{0\\}$ instead of
$s_{\mathfrak{g}}({\mathfrak{z}}({\mathfrak{g}})_{\bar{1}})\cap
s_{\mathfrak{g}}({\mathfrak{g}}_{\bar{1}})^{\perp}\not=\\{0\\}$.

######  Lemma (Technical).

If the Lie superalgebra $({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ admits
an invertible derivation $\Delta$ such that ${\mathscr{B}}_{\mathfrak{g}}$ is
$\Delta$-invariant, then

${\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}\subseteq
s_{\mathfrak{g}}({\mathfrak{g}}_{\bar{1}})^{\perp}.$

###### Proof.

Let $y\in{\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}$ and let us write
$\Delta(t)=y$ for $t\in{\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}$. Now, let
$u\in s_{\mathfrak{g}}({\mathfrak{g}}_{\bar{1}})$ and let us write
$u=s_{\mathfrak{g}}(v)$ for $v\in{\mathfrak{g}}_{\bar{1}}$. We have

${\mathscr{B}}_{\mathfrak{g}}(y,u)={\mathscr{B}}_{\mathfrak{g}}(\Delta(t),s_{\mathfrak{g}}(v))={\mathscr{B}}_{\mathfrak{g}}(t,\Delta(s_{\mathfrak{g}}(v)))={\mathscr{B}}_{\mathfrak{g}}(t,[\Delta(v),v])={\mathscr{B}}_{\mathfrak{g}}([t,v],\Delta(v))=0.\qed$

######  Theorem (Converse of Theorem 5.3.5. Theorem).

Let $({\mathfrak{g}},\mathscr{B}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ be a
irreducible NIS-Lie superalgebra equipped with a closed
${\bar{1}}$-antisymmetric periplectic form $\omega_{\mathfrak{g}}$ such that
$p({\mathscr{B}})={\bar{1}}$. Suppose that
${\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}\not=\\{0\\}$. Then
$({\mathfrak{g}},\mathscr{B}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ is
obtained as an $\mathscr{D}$-extension from a NIS-Lie superalgebra
$({\mathfrak{a}},\mathscr{B}_{\mathfrak{a}},\omega_{\mathfrak{a}})$ with a
closed ${\bar{1}}$-antisymmetric periplectic form $\omega_{\mathfrak{a}}$.

###### Proof.

Let $\Delta\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{0}}$ be the unique
invertible derivation such that
$\omega_{\mathfrak{g}}(\cdot,\cdot)=\mathscr{B}_{\mathfrak{g}}(\Delta(\cdot),\cdot)$.
It follows that
$\Delta({\mathfrak{z}}({\mathfrak{g}})_{\bar{1}})={\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}$.
Since ${\mathbb{K}}$ is algebraically closed, let us choose
$x\in{\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}$ such that $\Delta(x)=\lambda x$
for some $\lambda\in{\mathbb{K}}\backslash\\{0\\}$. We assume that
$s_{\mathfrak{g}}(x)=0$; otherwise, $s_{\mathfrak{g}}(x)$ is a non-zero
element of ${\mathfrak{z}}_{\bar{0}}({\mathfrak{g}})$ and hence Theorem 5.3.4.
Theorem can be recalled. Besides,
$\Delta(\mathscr{K}^{\perp})\subset\mathscr{K}^{\perp}$. Indeed, (for all
$f\in\mathscr{K}^{\perp}$)

${\mathscr{B}}_{\mathfrak{g}}(\Delta(f),x)={\mathscr{B}}_{\mathfrak{g}}(f,\Delta(x))={\mathscr{B}}_{\mathfrak{g}}(f,\lambda
x)=0.$

It has been shown in [BeBou, Prop. 4.2.2] that the subspace
$\mathscr{K}:=\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$ is an ideal in
${\mathfrak{g}}$; moreover, the subspace $\mathscr{K}^{\perp}$ is also an
ideal containing $\mathscr{K}$. Same arguments as in Theorem 5.2.5. Theorem
can be used to construct the ideal
${\mathfrak{a}}=(\mathscr{K}\oplus\mathscr{E})^{\perp},$ and a decomposition
${\mathfrak{g}}=\mathscr{K}\oplus{\mathfrak{a}}\oplus\mathscr{E}$, where the
generator $e$ of $\mathscr{E}$ can be normalized so that
${\mathscr{B}}_{\mathfrak{g}}(x,e)=1$.

It has been shown in [BeBou, Prop. 4.2.2] that
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ is a double extension of
$({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$, where
${\mathscr{B}}_{\mathfrak{a}}={\mathscr{B}}_{{\mathfrak{g}}}|_{{\mathfrak{a}}\times{\mathfrak{a}}}$,
by means of a derivation
$\mathscr{D}\in{\mathfrak{der}}({\mathfrak{a}})_{\bar{0}}$.

Let us define $\Delta$ by the formulas (where
$b_{0}\in{\mathfrak{a}}_{\bar{0}}$, and
$\beta(a),\nu,\lambda\in{\mathbb{K}}$):

$\begin{array}[]{lcl}\Delta(x)&=&\lambda x,\\\ \Delta(e)&=&\nu e+b_{0},\\\
\Delta(a)&=&\tilde{\Delta}(a)+\beta(a)x+\gamma(a)x^{*}.\end{array}$

Since
${\mathscr{B}}_{\mathfrak{g}}(\Delta(e),x)={\mathscr{B}}_{\mathfrak{g}}(e,\Delta(x))$,
then $\nu=\lambda$. Similarly, since
${\mathscr{B}}_{\mathfrak{a}}(a,\Delta(x))={\mathscr{B}}_{\mathfrak{a}}(a,\Delta(x))=0$,
then $\gamma(a)=0$ for all $a\in{\mathfrak{a}}$.

In addition, the conditions

$\begin{array}[]{l}{\mathscr{B}}_{\mathfrak{g}}(\Delta(a),e)={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+\beta(a)x,e)=\beta(a)\;\text{
for all $a\in{\mathfrak{a}}_{\bar{1}}$},\\\\[2.84526pt]
{\mathscr{B}}_{\mathfrak{g}}(a,\Delta(e))={\mathscr{B}}_{\mathfrak{g}}(a,\lambda
e+b_{0})={\mathscr{B}}_{\mathfrak{a}}(a,b_{0})\;\text{ for all
$a\in{\mathfrak{a}}_{\bar{1}}$}\end{array}$

imply that

(77) $\beta(a)={\mathscr{B}}_{\mathfrak{a}}(a,b_{0})\text{ for all
$a\in{\mathfrak{a}}_{\bar{1}}$}.$

Now, we have

$\begin{array}[]{lcl}\Delta([e,a]_{\mathfrak{g}})&=&\Delta({\mathscr{D}}(a))=\tilde{\Delta}({\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b_{0})x=\tilde{\Delta}({\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}(a,{\mathscr{D}}(b_{0}))x.\end{array}$

On the other hand, we have

$\begin{array}[]{lcl}[\Delta(e),a]_{\mathfrak{g}}+[e,\Delta(a)]_{\mathfrak{g}}&=&\lambda{\mathscr{D}}(a)+[b_{0},a]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(b_{0}),a)x+{\mathscr{D}}(\tilde{\Delta}(a)).\end{array}$

Since $\Delta$ is a derivation, it follows that

(78) $\displaystyle[\tilde{\Delta},{\mathscr{D}}]$ $\displaystyle=$
$\displaystyle\lambda{\mathscr{D}}+\mathop{\mathrm{ad}}\nolimits_{b_{0}}.$

Following the same computation of [BaB], the condition

$\Delta([a,b]_{\mathfrak{g}})=[\Delta(a),b]_{\mathfrak{g}}+[a,\Delta(b)]_{\mathfrak{g}}$

implies that

(79) $\displaystyle\tilde{\Delta}([a,b]_{\mathfrak{a}})$ $\displaystyle=$
$\displaystyle[\tilde{\Delta}(a),b]_{\mathfrak{a}}+[a,\tilde{\Delta}(b)]_{\mathfrak{a}},$
(80) $\displaystyle\beta([a,b]_{\mathfrak{a}})$ $\displaystyle=$
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(\tilde{\Delta}(a)),b)+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),\tilde{\Delta}(b)).$

Eq. (80) is certainly satisfied, thanks to Eqs. (77), (78) and the fact that
${\mathscr{B}}_{\mathfrak{a}}$ is $\tilde{\Delta}$-invariant.

Let us study the condition

$\Delta(s_{\mathfrak{g}}(f))=[f,\Delta(f)]_{\mathfrak{g}}\text{ for all
$f\in{\mathfrak{g}}_{\bar{1}}$}.$

Its LHS (for $f=rx+a$) reads as

$\begin{array}[]{lcl}\Delta(s_{\mathfrak{g}}(rx+a))&=&\Delta(s_{\mathfrak{a}}(a))=\tilde{\Delta}(s_{\mathfrak{a}}(a))+{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a),b_{0})x=\tilde{\Delta}(s_{\mathfrak{a}}(a)).\end{array}$

The RHS (for $f=rx+a$) reads as

$\begin{array}[]{lcl}[\Delta(f),f]_{\mathfrak{g}}&=&[r\lambda
x+\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x,rx+a]_{\mathfrak{g}}\\\\[5.69054pt]
&=&[\tilde{\Delta}(a),a]_{\mathfrak{g}}=[\tilde{\Delta}(a),a]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}\circ\tilde{\Delta}(a),a)x=[\tilde{\Delta}(a),a]_{\mathfrak{a}}.\end{array}$

It follows that

(81) $\displaystyle\tilde{\Delta}(s_{\mathfrak{a}}(a))$ $\displaystyle=$
$\displaystyle[a,\tilde{\Delta}(a)]_{\mathfrak{a}},$

Eqs. (79), (81) imply that $\tilde{\Delta}$ is a derivation on
${\mathfrak{a}}$, and obviously invertible. Define

$\omega_{\mathfrak{a}}(\cdot,\cdot):={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(\cdot),\cdot).$

The form $\omega_{\mathfrak{a}}$ on ${\mathfrak{a}}$ is closed,
${\bar{1}}$-antisymmetric and periplectic. ∎

###  5.4. Double extensions of Manin triples in chararcteristic $2$

The notion of double extension of Manin triples over a field of characteristic
zero was introduced in [MR2]. It has been used in [BBM] to describe the
structure of NIS Lie algebras with a closed symplectic form. Later the
construction has been superized in [BaB]. Here this notion is introduced and
studied in the case of Lie superalgebras in characteristic $2$, featuring a
condition on both the squaring and the derivation. This peculiarity has
already been observed in [BeBou] when dealing with double extensions for Lie
superalgebras in characteristic $2$.

######  Theorem (${\mathscr{D}}_{\bar{0}}$-extension of Manin triples).

Let $({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{k}})$ be a Manin triple of
charac-teristic $2$ with a NIS ${\mathscr{B}}_{\mathfrak{h}}$. Let
${\mathscr{D}}\in{\mathfrak{der}}({\mathfrak{h}})_{\bar{0}}$ satisfy Eq. (39),
${\mathscr{D}}({\mathfrak{k}}_{\bar{0}})\subseteq{\mathfrak{k}}_{\bar{0}}$,
and let $\alpha:{\mathfrak{h}}_{\bar{1}}\rightarrow{\mathbb{K}}$ be a
quadratic form satisfying Eq. (40) and $\alpha({\mathfrak{k}}_{\bar{1}})=0$.
Denote by
$(\tilde{\mathfrak{h}}={\mathscr{K}}\oplus{\mathfrak{h}}\oplus{\mathscr{K}}^{*},\tilde{\mathscr{B}}_{\tilde{\mathfrak{h}}})$
the double extension of $({\mathfrak{h}},{\mathscr{B}}_{\mathfrak{h}})$ by the
one-dimensional Lie algebra ${\mathscr{K}}$ (by means of ${\mathscr{D}}$ and
$\alpha$). If ${\mathscr{B}}_{\tilde{\mathfrak{h}}}(x^{*},x^{*})=0$, then the
triple $(\tilde{\mathfrak{h}},\tilde{\mathfrak{g}},\tilde{\mathfrak{k}})$ is
also a Manin triple with a NIS $\tilde{\mathscr{B}}_{\tilde{\mathfrak{h}}}$,
where $\tilde{\mathfrak{g}}={\mathscr{K}}\oplus{\mathfrak{g}}$ and
$\tilde{\mathfrak{k}}={\mathscr{K}}^{*}\ltimes{\mathfrak{k}}$.

We call the Manin triple
$(\tilde{\mathfrak{h}},\tilde{\mathfrak{g}},\tilde{\mathfrak{k}})$ with NIS
$\tilde{\mathscr{B}}_{\tilde{\mathfrak{h}}}$ the double extension of the Manin
triple $({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{k}})$ with NIS
${\mathscr{B}}_{{\mathfrak{h}}}$ by the one-dimensional Lie algebra
${\mathscr{K}}$.

###### Proof.

Let us first show that if $\alpha({\mathfrak{k}}_{\bar{1}})=0$, then
${\mathscr{D}}({\mathfrak{k}}_{\bar{1}})\subseteq{\mathfrak{k}}_{\bar{1}}$.
Indeed, let ${\mathscr{D}}(k)\in{\mathscr{D}}({\mathfrak{k}}_{\bar{1}})$ and
write
${\mathscr{D}}(k)=\mathrm{proj}_{{\mathfrak{g}}}({\mathscr{D}}(k))+\mathrm{proj}_{{\mathfrak{k}}}({\mathscr{D}}(k))$.
For all $l\in{\mathfrak{k}}_{\bar{1}}$, we have 333Recall that
${\mathscr{B}}_{\alpha}$ is the polar form of the quadratic form $\alpha$.

$0={\mathscr{B}}_{\alpha}(k,l)={\mathscr{B}}_{\mathfrak{h}}({\mathscr{D}}(k),l)={\mathscr{B}}_{\mathfrak{h}}(\mathrm{proj}_{{\mathfrak{g}}}({\mathscr{D}}(k)),l).$

On the other hand, since ${\mathscr{B}}_{\mathfrak{h}}$ is even and
${\mathscr{B}}_{\mathfrak{h}}({\mathfrak{g}},{\mathfrak{g}})=0$, it follows
that

${\mathscr{B}}_{\mathfrak{h}}(\mathrm{proj}_{{\mathfrak{g}}}({\mathscr{D}}(k)),l)=0\;\text{
for all $l\in{\mathfrak{h}}$}.$

It follows that $\mathrm{proj}_{{\mathfrak{g}}}({\mathscr{D}}(k))=0$ since
${\mathscr{B}}_{\mathfrak{h}}$ is non-degenerate.

Let us show that $\tilde{\mathfrak{g}}={\mathfrak{g}}\oplus{\mathscr{K}}$ is
Lie sub-superalgebra of $\tilde{\mathfrak{h}}$. Indeed, for every $\lambda
x+f,\mu x+g\in\tilde{\mathfrak{g}}$, where $\lambda,\mu\in{\mathbb{K}}$ and
$f,g\in{\mathfrak{g}}$, we have

$[\lambda x+f,\lambda
x+g]_{\tilde{\mathfrak{h}}}=[f,g]_{\tilde{\mathfrak{h}}}=[f,g]_{\mathfrak{h}}+{\mathscr{B}}_{\mathfrak{h}}(f,g)x\in\tilde{\mathfrak{g}},$

since ${\mathfrak{g}}$ is a Lie sub-superalgebra of ${\mathfrak{h}}$. Now the
squaring; for all $f\in\tilde{\mathfrak{g}}_{\bar{1}}$, we have

$s_{\tilde{\mathfrak{h}}}(f)=s_{\mathfrak{h}}(f)+\alpha(f)x\in\tilde{\mathfrak{g}}_{\bar{0}},$

since ${\mathfrak{g}}$ is a Lie sub-superalgebra of ${\mathfrak{h}}$.

Let us show that $\tilde{\mathfrak{k}}={\mathscr{K}}^{*}\ltimes{\mathfrak{k}}$
is a Lie sub-superalgebras of ${\mathfrak{h}}$. For every ${\lambda
x^{*}+f,\mu x^{*}+g\in\tilde{\mathfrak{k}}}$, where
$\lambda,\mu\in{\mathbb{K}}$ and $f,g\in{\mathfrak{k}}$, we have

$\begin{array}[]{lcl}[\lambda x^{*}+f,\mu
x^{*}+g]_{\tilde{\mathfrak{h}}}&=&[f,g]_{\mathfrak{h}}+{\mathscr{B}}_{{\mathfrak{h}}}({\mathscr{D}}(f),g)x+\mu{\mathscr{D}}(f)+\lambda{\mathscr{D}}(g)\in\tilde{\mathfrak{k}}\\\\[5.69054pt]
&=&[f,g]_{\mathfrak{h}}+\mu{\mathscr{D}}(f)+\lambda{\mathscr{D}}(g)\in\tilde{\mathfrak{k}}.\end{array}$

since ${\mathfrak{k}}$ is a Lie subsuperalgebra of ${\mathfrak{h}}$ and
${\mathscr{B}}_{\mathfrak{h}}({\mathscr{D}}(f),g)=0$ since
${\mathscr{D}}({\mathfrak{k}})\subseteq{\mathfrak{k}}$. Now the squaring; for
all $f\in\tilde{\mathfrak{k}}_{\bar{1}}$, we have

$s_{\tilde{\mathfrak{h}}}(f)=s_{\mathfrak{h}}(f)+\alpha(f)x=s_{\mathfrak{h}}(f)\in\tilde{\mathfrak{k}}_{\bar{0}},$

since ${\mathfrak{k}}$ is a Lie subsuperalgebra of ${\mathfrak{h}}$ and
$\alpha({\mathfrak{k}}_{\bar{1}})=0$.

The condition
${\mathscr{B}}_{\tilde{\mathfrak{h}}}(\tilde{\mathfrak{g}},\tilde{\mathfrak{g}})={\mathscr{B}}_{\tilde{\mathfrak{h}}}(\tilde{\mathfrak{k}},\tilde{\mathfrak{k}})=0$
follows from the explicit construction of
${\mathscr{B}}_{\tilde{\mathfrak{h}}}$ as in Theorem 5.2.1. Theorem and
${\mathscr{B}}_{\tilde{\mathfrak{h}}}(x^{*},x^{*})=0$. ∎

######  Theorem (Converse of Theorem 5.4.1. Theorem).

Let $({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{k}})$ be a Manin triple such
that $\dim({\mathfrak{h}})>1$. If
$({\mathfrak{z}}_{s}({\mathfrak{h}})\cap{\mathfrak{g}})_{\bar{0}}\not=0$ or
$({\mathfrak{z}}_{s}({\mathfrak{h}})\cap{\mathfrak{k}})_{\bar{0}}\not=0$, then
the Manin triple $({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{k}})$ is a double
extension of a Manin triple $({\mathfrak{c}},{\mathfrak{a}},{\mathfrak{b}})$
as in Theorem 5.4.1. Theorem.

###### Proof.

Let us assume that
$({\mathfrak{z}}_{s}({\mathfrak{h}})\cap{\mathfrak{g}})_{\bar{0}}\not=0$. The
case $({\mathfrak{z}}_{s}({\mathfrak{h}})\cap{\mathfrak{k}})_{\bar{1}}\not=0$
is absolutely the same. Let us choose a non-zero
$x\in({\mathfrak{z}}_{s}({\mathfrak{h}})\cap{\mathfrak{g}})_{\bar{0}}$ and
write ${\mathscr{K}}=\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$. By
construction, ${\mathscr{K}}$ and ${\mathscr{K}}^{\perp}$ are both ideals.
Since ${\mathscr{B}}_{\mathfrak{h}}({\mathfrak{g}},{\mathfrak{g}})=\\{0\\}$,
we obtain ${\mathfrak{g}}\subseteq{\mathscr{K}}^{\perp}$. Since
${\mathscr{B}}_{\mathfrak{h}}$ is non-degenerate and
${\mathscr{B}}_{\mathfrak{h}}({\mathfrak{g}},{\mathfrak{g}})=\\{0\\}$, then
there exists $x^{*}\in{\mathfrak{k}}_{\bar{0}}$ such that
${\mathscr{B}}_{\mathfrak{h}}(x^{*},x)\not=0$. We may assume that
${\mathscr{B}}_{\mathfrak{h}}(x^{*},x)=1$. Let us write
${\mathscr{K}}^{*}=\mathop{\mathrm{{Span}}}\nolimits\\{x^{*}\\}$. Since
$\mathscr{K}\subset\mathscr{K}^{\perp}$, it follows that
$\dim(\mathscr{K}^{\perp})=\dim({\mathfrak{h}})-1$. We can then write
${\mathfrak{h}}={\mathscr{K}}^{\perp}\oplus{\mathscr{K}}^{*}$ since
$x^{*}\not\in{\mathscr{K}}^{\perp}$. There exists a vector subsuperspace
${\mathfrak{c}}:=({\mathscr{K}}\oplus{\mathscr{K}}^{*})^{\perp}\subseteq{\mathfrak{h}}$
such that ${\mathscr{K}}^{\perp}={\mathscr{K}}\oplus{\mathfrak{c}}$ and
${\mathscr{B}}_{\mathfrak{h}}({\mathfrak{c}},x^{*})=\\{0\\}$. We easily see
that
${\mathscr{B}}_{{\mathfrak{c}}}:={\mathscr{B}}_{{\mathfrak{h}}}|_{{\mathfrak{c}}\times{\mathfrak{c}}}$
is non-degenerate. Since ${\mathfrak{g}}\subseteq{\mathscr{K}}^{\perp}$ we
have ${\mathfrak{g}}={\mathscr{K}}\oplus{\mathfrak{a}}$, where
${\mathfrak{a}}={\mathfrak{g}}\cap{\mathfrak{c}}$.

On the other hand, as ${\mathfrak{h}}={\mathfrak{g}}\oplus{\mathfrak{k}}$ and
${\mathfrak{g}}\subseteq{\mathscr{K}}^{\perp}$ then
${\mathscr{K}}^{\perp}={\mathfrak{g}}\oplus{\mathfrak{b}}$, where
${\mathfrak{b}}={\mathfrak{k}}\cap{\mathscr{K}}^{\perp}$. Since
${\mathscr{B}}_{\mathfrak{h}}({\mathfrak{k}},{\mathfrak{k}})=\\{0\\}$,
${\mathfrak{c}}=({\mathscr{K}}\oplus{\mathscr{K}}^{*})^{\perp}$ and
${\mathfrak{b}}={\mathfrak{k}}\cap{\mathscr{K}}^{\perp}$, we get
${\mathfrak{b}}\subseteq{\mathfrak{c}}$. Then,
${\mathfrak{h}}={\mathscr{K}}\oplus{\mathfrak{c}}\oplus{\mathscr{K}}^{*}$ and
${\mathfrak{c}}={\mathfrak{a}}\oplus{\mathfrak{b}}$ by the dimension
considerations. Recall that ${\mathfrak{c}}\subseteq{\mathscr{K}}^{\perp}$. We
get

$[f,g]_{\mathfrak{h}}=\beta(f,g)+\phi(f,g)x\text{ for all
$f,g\in{\mathfrak{c}}$},$

where $\beta(f,g)\in{\mathfrak{c}}$ and $\phi(f,g)\in{\mathbb{K}}$. Moreover,

$[x^{*},f]_{\mathfrak{h}}={\mathscr{D}}(f)+\psi(f)x\text{ for all
$f\in{\mathfrak{c}}$},$

with ${\mathscr{D}}(f)\in{\mathfrak{c}}$ and $\psi(f)\in{\mathbb{K}}$.

Moreover, the squaring reads as follows

$s_{{\mathfrak{h}}}(f)=s_{{\mathfrak{c}}}(f)+\alpha(f)x\text{ for all
$f\in{\mathfrak{c}}_{\bar{1}}$,}$

where $s_{\mathfrak{c}}(f)\in{\mathfrak{c}}_{\bar{0}}$.

Now, let $f\in{\mathfrak{b}}_{\bar{1}}\subset{\mathfrak{k}}_{\bar{1}}$. Since
$s_{\mathfrak{h}}(f)\in{\mathfrak{k}}_{\bar{0}}$, then
${\mathscr{B}}_{\mathfrak{h}}(s_{\mathfrak{h}}(f),x^{*})=0$ as
${\mathscr{B}}_{\mathfrak{h}}({\mathfrak{k}},{\mathfrak{k}})=0$. It follows
that

$0={\mathscr{B}}_{\mathfrak{h}}(s_{\mathfrak{h}}(f),x^{*})={\mathscr{B}}_{\mathfrak{h}}(s_{\mathfrak{c}}(f)+\alpha(f)x,x^{*})=\alpha(f).$

Therefore, $\alpha({\mathfrak{b}}_{\bar{1}})=\\{0\\}$.

Following the same proof as in [BeBou, Prop. 3.1.3], we can show that
$({\mathfrak{c}},\beta,s_{\mathfrak{c}},{\mathscr{B}}_{\mathfrak{c}})$ is a
double extension of
$({\mathfrak{h}},[\cdot,\cdot]_{\mathfrak{h}},s_{\mathfrak{h}},{\mathscr{B}}_{\mathfrak{h}})$
by means of the derivation ${\mathscr{D}}$ and the quadratic form $\alpha$.
The proof also implies that $\psi=0$ and that

$\phi(f,g)={\mathscr{B}}_{\mathfrak{c}}({\mathscr{D}}(f),g)\text{ for all
$f,g\in{\mathfrak{c}}$}.$

Let us show that
${\mathscr{D}}({\mathfrak{b}}_{\bar{0}})\subseteq{\mathfrak{b}}_{\bar{0}}$.
Since ${\mathfrak{b}}={\mathfrak{k}}\cap{\mathscr{K}}^{\perp}$, but
${\mathscr{K}}^{\perp}$ is an ideal in ${\mathfrak{h}}$ and ${\mathfrak{k}}$
is a sub-superalgebra of ${\mathfrak{h}}$, it follows that for all
$f\in{\mathfrak{b}}_{\bar{0}}$, we have

${\mathscr{D}}(f)=[x^{*},f]\in({\mathfrak{k}}\cap{\mathscr{K}}^{\perp})_{\bar{0}}={\mathfrak{b}}_{\bar{0}}.$

It remains to show that
${\mathscr{B}}_{\mathfrak{c}}({\mathfrak{a}},{\mathfrak{a}})={\mathscr{B}}_{\mathfrak{c}}({\mathfrak{b}},{\mathfrak{b}})=0$,
but this is obvious since ${\mathfrak{a}}\subset{\mathfrak{g}}$ and
${\mathfrak{b}}\subset{\mathfrak{k}}$.∎

######  Theorem (${\mathscr{D}}_{\bar{1}}$-extension for Manin triples).

Let $({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{k}})$ be a Manin triple in
charac-teristic $2$ with a NIS ${\mathscr{B}}_{\mathfrak{h}}$. Let
${\mathscr{D}}\in{\mathfrak{der}}({\mathfrak{h}})_{\bar{1}}$ satisfy Eqns.
(42), (43), (44) and ${\mathscr{D}}({\mathfrak{k}})\subseteq{\mathfrak{k}}$.
Let
$(\tilde{\mathfrak{h}}={\mathscr{K}}\oplus{\mathfrak{h}}\oplus{\mathscr{K}}^{*},\tilde{\mathscr{B}}_{\tilde{\mathfrak{h}}})$
be the double extension of $({\mathfrak{h}},{\mathscr{B}})$ by the one-
dimensional Lie algebra ${\mathscr{K}}$ (by means of ${\mathscr{D}}$). If
$a_{\bar{0}}\in{\mathfrak{k}}_{\bar{0}}$, then
$(\tilde{\mathfrak{h}},\tilde{\mathfrak{g}},\tilde{\mathfrak{k}})$ is a Manin
triple with a NIS $\tilde{\mathscr{B}}_{\tilde{\mathfrak{h}}}$, where
$\tilde{\mathfrak{g}}={\mathscr{K}}\oplus{\mathfrak{g}}$ and
$\tilde{\mathfrak{k}}={\mathscr{K}}^{*}\ltimes{\mathfrak{k}}$.

###### Proof.

Similar to that of Theorem 5.4.1. Theorem. ∎

######  Theorem (Converse of Theorem 5.4.3. Theorem).

Let $({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{k}})$ be a Manin triple such
that $\dim({\mathfrak{h}})>1$. If
$({\mathfrak{z}}({\mathfrak{h}})\cap{\mathscr{C}}({\mathfrak{h}},{\mathscr{B}}_{\mathfrak{h}})\cap{\mathfrak{g}})_{\bar{1}}\not=0$
or
$({\mathfrak{z}}({\mathfrak{h}})\cap{\mathscr{C}}({\mathfrak{h}},{\mathscr{B}}_{\mathfrak{h}})\cap{\mathfrak{k}})_{\bar{1}}\not=0$
then the Manin triple $({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{k}})$ is a
double extension of a Manin triple
$({\mathfrak{c}},{\mathfrak{a}},{\mathfrak{b}})$ as in Theorem 5.4.3. Theorem.

###### Proof.

Similar to that of Theorem 5.4.2. Theorem. ∎

######  Remark.

It has been proved in [BaB] that every NIS-Lie superalgebra equiped with a
closed antisymmetric ortho-symplectic form over an algebraically closed field
of characteristic zero is actually a Manin triple. The proof makes use of the
weight space decomposition of a Lie superalgebra that is only valid over a
field of characteristic zero. We do not know whether the result of [BaB] is
true or not in our context, perhaps ideas from [BLLS] could help.

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# Minimum Weight Random Graphs with Edge Constraints

Ghurumuruhan Ganesan1
1IISER, Bhopal E-mail<EMAIL_ADDRESS>
Abstract: In this paper, we study two examples of minimum weight random graphs
with edge constraints. First we consider the complete graph on $n$ vertices
equipped with uniformly heavy edge weights and use iteration methods to obtain
deviation estimates for the minimum weight of subtrees with a given number of
edges. Next we analyze edge constrained minimum weight paths in the integer
lattice $\mathbb{Z}^{d}$ and employ martingale difference techniques to
describe the behaviour of the scaled minimum weight in terms of the edge
constraint.
Keywords: Minimum spanning tree, heavy weights, minimum passage time paths,
edge constraints.
2010 Mathematics Subject Classification: Primary: 60J10, 60K35; Secondary:
60C05, 62E10, 90B15, 91D30.

##  1 Introduction

Trees of complete graphs with random edge weights are important from both
theoretical and practical perspectives. For independent and identically
distributed (i.i.d.) edge weights with a common cumulative distribution
function (cdf) $F(.)$ that varies linearly close to zero, [5] studied
convergence of the weight of the minimum spanning tree (MST) of the complete
graph $K_{n}$ on $n$ vertices. Later [2] studied asymptotics for the
_expected_ value of the MST weight, when the edge weight distributions follow
a power law distribution. The paper [6] studied central limit theorems for a
scaled and centred version of the MST weight and more recently [1] studied
bounds on the diameter of the MST. The methods involve a combination of graph
evolution via Kruskal’s agorithm along with a component analysis of random
graphs. For MSTs with nonidentical edge weight distributions, [10] use the
Tutte polynomial approach [11] to compute expressions for the expected value
of $MST_{n}.$

In Section 2 of our paper, we study “approximate” MSTs containing $O(n)$
edges, obtained by placing random heavy weights in each edge of $K_{n}$ that
are not necessarily identically distributed but are uniformly heavy. We use
stochastic domination to obtain deviation type estimates for the minimum
weight and use the martingale method to bound the variance (see Theorem 2.1).

Next we study constrained paths in the integer lattice. Consider the following
scenario where each edge in the square lattice $\mathbb{Z}^{d}$ is associated
with a random passage time and it is of interest to determine the minimum
passage time $T_{n}$ between the origin and $(n,0,\ldots,0).$ The case of
independent and identically distributed (i.i.d.) passage times has been well-
studied and detailed results are known regarding the almost sure convergence
and convergence in mean of the scaled passage time $\frac{T_{n}}{n}$ (see
[8]). Later [4] studied central limit theorems for first passage across thin
cylinders and recently [7] have studied critical first passage percolation in
the triangular lattice.

In many applications, the passage times may not be i.i.d. For example, if we
model vertices of $\mathbb{Z}^{d}$ as mobiles stations and the edge passage
times as the delay in sending a packet between two adjacent stations, then
depending on external conditions, the edges may have different passage time
distributions. In such cases, it is of interest to study convergence
properties of the minimum passage time $T_{n},$ with appropriate centering and
scaling. In Section 3 our paper, we state and prove our result (Theorem 3.1)
regarding the behaviour of the constrained minimum passage times as a function
of the edge constraint.

The paper is organized as follows. In Section 2, we state and prove our result
regarding the asymptotic behaviour of weighted trees of the complete graph,
with edge constraints. Finally, in Section 3, we describe the behaviour of
constrained minimum passage time paths in the integer lattice
$\mathbb{Z}^{d}.$

##  2 Edge Constrained Minimum Weight Trees

For $n\geq 1,$ let $K_{n}$ be the complete graph with vertex set
$\\{1,2,\ldots,n\\}.$ Let $\\{w(i,j)\\}_{1\leq i<j\leq n}$ be independent
random variables with corresponding cumulative distribution functions (cdfs)
$\\{F_{i,j}\\}_{1\leq i<j\leq n}$ and for $1\leq j<i\leq n,$ set
$w(i,j):=w(j,i).$ We define $w(e):=w(i,j)$ to be the _weight_ of the edge
$e=(i,j)\in K_{n}$ and assume throughout that $w(e)\leq 1,$ for simplicity.

A tree in $K_{n}$ is a connected acyclic subgraph. For a tree ${\cal T}$ with
vertex set $\\{v_{1},\ldots,v_{t}\\},$ the weight of ${\cal T}$ is the sum of
the weights of the edges in ${\cal T};$ i.e., $W({\cal T}):=\sum_{e\in{\cal
T}}w(e).$ For $1\leq\tau\leq n-1$ we define

$M_{n}=M_{n}(\tau):=\min_{{\cal T}}W({\cal T}),$ (2.1)

where the minimum is taken over all trees ${\cal T}\subset K_{n}$ having at
least $\tau$ edges.

The following is the main result of this section. Constants throughout do not
depend on $n.$

######  Theorem 2.1.

Suppose $\tau\geq\rho n$ for some $0<\rho\leq 1$ and also suppose there are
positive constants $D_{1}\leq D_{2}$ and $0<\alpha<1$ such that

$D_{1}x^{\frac{1}{\alpha}}\leq F_{i,j}(x)\leq D_{2}x^{\frac{1}{\alpha}}$ (2.2)

for $0\leq x\leq 1.$ There are positive constants $C_{i},1\leq i\leq 3$ such
that

$\mathbb{P}\left(C_{1}n^{1-\alpha}\leq M_{n}\leq C_{2}n^{1-\alpha}\right)\geq
1-e^{-C_{3}n^{1-\alpha}}$ (2.3)

and $C_{1}n^{1-\alpha}\leq\mathbb{E}M_{n}\leq C_{2}n^{1-\alpha}.$ Moreover,
$var(M_{n})\leq 2n.$

To prove Theorem 2.1, we use the following preliminary Lemma regarding the
behaviour of the exponential moments of small edge weights. For $1\leq j\leq
n$ and distinct deterministic integers $1\leq a_{1},\ldots,a_{j}\leq n$ let

$Y_{j}=Y_{j}(a_{1},\ldots,a_{j}):=\min_{a\notin\\{a_{1},\ldots,a_{j-1}\\}}w(a_{j},a).$
(2.4)

######  Lemma 2.2.

There are positive constants $C_{1}$ and $C_{2}$ not depending on the choice
of $\\{a_{i}\\}$ or $j$ such that for any $1\leq j\leq n-1,$

$\frac{C_{1}}{(n-j)^{\alpha}}\leq\mathbb{E}Y_{j}\leq\frac{C_{2}}{(n-j)^{\alpha}}.$
(2.5)

Moreover for every $s>1$ there are constants $K,C\geq 1$ not depending on the
choice of $\\{a_{i}\\}$ or $j$ such that for all $1\leq j\leq n-K$

$\mathbb{E}e^{sY_{j}}\leq\exp\left(\frac{C}{(n-j)^{\alpha}}\right).$ (2.6)

_Proof of Lemma 2.2_: In what follows, we use the following standard deviation
estimate. Suppose $W_{i},1\leq i\leq m$ are independent Bernoulli random
variables satisfying
$\mathbb{P}(W_{1}=1)=1-\mathbb{P}(W_{1}\leavevmode\nobreak\
=\leavevmode\nobreak\ 0)\leq\mu_{2}.$ For any $0<\epsilon<\frac{1}{2},$

$\mathbb{P}\left(\sum_{i=1}^{m}W_{i}>m\mu_{2}(1+\epsilon)\right)\leq\exp\left(-\frac{\epsilon^{2}}{4}m\mu_{2}\right).$
(2.7)

For a proof of (2.7, we refer to Corollary A.1.14, pp. 312 of Alon and Spencer
(2008).

We first find the lower bound for $\mathbb{E}Y_{j}$ and then upper bound
$\mathbb{E}Y_{j}$ and $\mathbb{E}e^{sY_{j}}$ for constant $s>1$ in that order.
The term $Y_{j}$ is the minimum of $n-j$ edge weights and so for
$0<x<(n-j)^{\alpha}$ we use the upper bound for the cdfs in (2.2) to get that
$\mathbb{P}\left(Y_{j}>\frac{x}{(n-j)^{\alpha}}\right)\geq\left(1-D_{2}\frac{x^{\frac{1}{\alpha}}}{n-j}\right)^{n-j},$
where $D_{2}\geq 1$ is as in (2.2). Thus

$(n-j)^{\alpha}\mathbb{E}Y_{j}=\int_{0}^{(n-j)^{\alpha}}\mathbb{P}\left(Y_{j}>\frac{x}{(n-j)^{\alpha}}\right)dx\geq\int_{0}^{(n-j)^{\alpha}}\left(1-D_{2}\frac{x^{\frac{1}{\alpha}}}{n-j}\right)^{n-j}dx.$
(2.8)

To evaluate the integral in (2.8), we use $1-y\geq e^{-2y}$ for all
$0<y<\frac{1}{2}.$ Letting $y=\frac{D_{2}x^{\frac{1}{\alpha}}}{n-j}$ for
$0<x<\left(\frac{n-j}{2D_{2}}\right)^{\alpha}$ we then have $(1-y)^{n-j}\geq
e^{-2D_{2}x^{\frac{1}{\alpha}}}$ and substituting this in (2.8) and using
$D_{2}\geq 1,$ we get

$(n-j)^{\alpha}\mathbb{E}Y_{j}\geq\int_{0}^{\left(\frac{n-j}{2D_{2}}\right)^{\alpha}}e^{-2D_{2}x^{\frac{1}{\alpha}}}dx\geq\int_{0}^{\left(\frac{1}{2D_{2}}\right)^{\frac{1}{\alpha}}}e^{-2D_{2}x^{\frac{1}{\alpha}}}dx=:C_{1}$

for all $1\leq j\leq n-1.$

For upper bounding $\mathbb{E}Y_{j},$ we again use the fact that the term
$Y_{j}$ is the minimum of $n-j$ edge weights and so for $0<x<(n-j)^{\alpha}$
we use the lower bound for the cdfs in (2.2) to get
$\mathbb{P}\left(Y_{j}>\frac{x}{(n-j)^{\alpha}}\right)\leq\left(1-D_{1}\frac{x^{\frac{1}{\alpha}}}{n-j}\right)^{n-j}\leq
e^{-D_{1}x^{\frac{1}{\alpha}}}.$ Thus

$(n-j)^{\alpha}\mathbb{E}Y_{j}=\int_{0}^{(n-j)^{\alpha}}\mathbb{P}\left(Y_{j}>\frac{x}{(n-j)^{\alpha}}\right)dx\leq\int_{0}^{\infty}e^{-D_{1}x^{\frac{1}{\alpha}}}dx=:C_{2},$
(2.9)

a finite positive constant not depending on the choice of $\\{a_{i}\\}.$

To compute $\mathbb{E}e^{sY_{j}}$ we split $\mathbb{E}e^{sY_{j}}=I_{1}+I_{2},$
where $I_{1}=\mathbb{E}e^{sY_{j}}1{1}\left(Y_{j}\leq\frac{1}{2s}\right)$ and
$I_{2}=\mathbb{E}e^{sY_{j}}1{1}\left(Y_{j}>\frac{1}{2s}\right)$ and estimate
each term separately. To evaluate $I_{1},$ we bound $e^{x}\leq 1+2x$ for
$x\leq\frac{1}{2}$ and set $x=sY_{j}\leq\frac{1}{2}$ to get that
$e^{sY_{j}}\leq 1+2sY_{j}.$ Thus

$I_{1}\leq 1+2s\mathbb{E}Y_{j}\leq 1+\frac{2sC_{2}}{(n-j)^{\alpha}},$ (2.10)

using (2.9).

To evaluate $I_{2},$ we recall that
$Y_{j}=\min_{a\notin\\{a_{1},\ldots,a_{j-1}\\}}w(a_{j},a)\leq 1$ is the
minimum of $n-j$ independent edge weights. Using the upper bound for the cdfs
in (2.2) we have $\mathbb{P}\left(w(a_{j},a)>\frac{1}{2s}\right)\leq
1-\frac{D_{1}}{(2s)^{\frac{1}{\alpha}}}<1,$ since $D_{1}\leq 1$ and
$s\leavevmode\nobreak\ >\leavevmode\nobreak\ 1.$ Setting
$e^{-\theta}=1-\frac{D_{1}}{(2s)^{\frac{1}{\alpha}}},$ we therefore have
$\theta>0$ and that $\mathbb{P}\left(Y_{j}>\frac{1}{2s}\right)\leq
e^{-\theta(n-j)}.$ Finally using $Y_{j}\leq 1$ (since all edge weights are at
most one) we get

$I_{2}=\mathbb{E}e^{sY_{j}}1{1}\left(Y_{j}>\frac{1}{2s}\right)\leq
e^{s}\mathbb{P}\left(Y_{j}>\frac{1}{2s}\right)\leq
e^{s}e^{-\theta(n-j)}\leq\frac{e^{s}C_{2}}{(n-j)^{\alpha}},$ (2.11)

where $C_{2}>0$ is as in (2.10), provided $n-j\geq K+1$ and $K=K(s,C_{2})$ is
large. The final estimate in (2.11) is obtained using $x^{\alpha}e^{-\theta
x}\longrightarrow 0$ as $x\rightarrow\infty.$

From (2.10) and (2.11), we therefore get for $1\leq j\leq n-K$ that

$\mathbb{E}e^{sY_{j}}\leq
1+\frac{2sC_{2}}{(n-j)^{\alpha}}+\frac{e^{s}C_{2}}{(n-j)^{\alpha}}\leq\exp\left(\frac{2sC_{2}+e^{s}C_{2}}{(n-j)^{\alpha}}\right),$

proving (2.6).

_Proof of Theorem 2.1_: We obtain the lower deviation bound by counting the
number of edges with large enough weight and the upper deviation bound by
constructing a spanning path with low weight analogous to Aldous (1990). The
expectation bounds then follow from (2.3). To compute the variance, we use the
martingale difference method. In fact, from the variance bound, we get that
$var\left(\frac{M_{n}}{n^{1-\alpha}}\right)\leq\frac{C}{n^{1-2\alpha}}$ and so
if $\alpha<\frac{1}{2},$ then $\frac{M_{n}-\mathbb{E}M_{n}}{n^{1-\alpha}}$
converges to zero in probability. Details follow.

We begin with the proof of the lower deviation bound. For $\gamma>0$ a small
constant, let $R_{tot}:=\sum_{e\in
K_{n}}1{1}\left(w(e)<\left(\frac{\gamma}{n}\right)^{\alpha}\right)$ be the
number of edges of weight at most $\left(\frac{\gamma}{n}\right)^{\alpha}.$ We
estimate $R_{tot}$ using the standard deviation bound (2.7). First, we have
from the bounds for the cdfs in (2.2) that
$\mathbb{P}\left(w(e)<\left(\frac{\gamma}{n}\right)^{\alpha}\right)<\frac{D_{2}\gamma}{n}$
and since the edge weights are independent, we use (2.7) with $m={n\choose
2},\mu_{2}=\frac{D_{2}\gamma}{n}$ and $\epsilon=\frac{1}{4}$ to get that

$\mathbb{P}\left(R_{tot}>\frac{5mD_{2}\gamma}{4n}\right)\leq\exp\left(-m\frac{D_{2}\gamma}{64n}\right)\leq\exp\left(-\frac{nD_{2}\gamma}{256}\right),$

since $m=\frac{n(n-1)}{2}>\frac{n^{2}}{4}.$ Let ${\cal T}_{n}$ be any tree
with weight $M_{n}$ and containing at least $\tau\geq\rho n$ edges. Using
$m<\frac{n^{2}}{2},$ we get that with probability at least
$1-e^{-\frac{nD_{2}\gamma}{256}},$ the weight of ${\cal T}_{n}$ is at least

$\left(\rho
n-\frac{5mD_{2}\gamma}{4n}\right)\cdot\left(\frac{\gamma}{n}\right)^{\alpha}\geq\left(\rho
n-\frac{5nD_{2}\gamma}{8}\right)\cdot\left(\frac{\gamma}{n}\right)^{\alpha}\geq
Cn^{1-\alpha}$

for some constant $C>0,$ provided $\gamma>0$ is small. This completes the
proof of the lower deviation bound in (2.3).

For the upper deviation bound, we consider the spanning path obtained by an
incremental approach similar to Aldous (1990). Let $i_{1}=1$ and among all
edges with endvertex $i_{1},$ let $i_{2}$ be the index such that
$w(i_{1},i_{2})$ has the least weight. Similarly, among all edges with
endvertex in $i_{2}\setminus\\{i_{1}\\},$ let $i_{3}$ be such that
$w(i_{2},i_{3})$ has the least weight. Continuing this way, the path ${\cal
P}_{iter}:=(i_{1},\ldots,i_{n})$ is a spanning path containing all the nodes
and so letting $Z_{j}=w(X_{i_{j-1}},X_{i_{j}})$ be the weight of the $j^{th}$
edge in ${\cal P}_{iter},$ we have $M_{n}\leq W({\cal
P}_{iter})=\sum_{j=1}^{n}Z_{j}.$ For $s>0$ we therefore have

$\mathbb{E}e^{sM_{n}}\leq\mathbb{E}e^{\sum_{j=1}^{n-1}sZ_{j}}$ (2.12)

and in what follows, we find an upper bound for the right hand side of (2.12).

Let $a_{1}:=1$ and $a_{l},2\leq l\leq j-1$ be deterministic numbers and
suppose the event $\\{i_{1}=a_{1},\ldots,i_{j-1}=a_{j-1}\\}$ occurs so that
$Z_{l}=w(a_{l},a_{l+1})$ for $1\leq l\leq j-1$ and
$Z_{j}=Y_{j}=Y_{j}(a_{1},\ldots,a_{j})=\min_{a\notin\\{a_{1},\ldots,a_{j-1}\\}}w(a_{j},a)$
is as in (2.4). The event $\\{i_{1}=a_{1},\ldots,i_{j-1}=a_{j-1}\\}$ and the
random variables $w(a_{l},a_{l+1}),1\leq l\leq j-1$ depend only the state of
edges having at least one endvertex in $\\{a_{1},\ldots,a_{j-1}\\}.$ On the
other hand, the random variable $Y_{j}$ depends only on the state of edges
having both endvertices in
$\\{1,\ldots,n\\}\setminus\\{a_{1},\ldots,a_{j-1}\\}.$ Thus

$\displaystyle\mathbb{E}e^{s\sum_{l=1}^{j}Z_{l}}1{1}(i_{1}=a_{1},\ldots,i_{j-1}=a_{j-1})$
$\displaystyle\;\;\;\;=\mathbb{E}e^{sY_{j}}e^{\sum_{l=1}^{j-1}sZ_{l}}1{1}(i_{1}=a_{1},\ldots,i_{j-1}=a_{j-1})$
$\displaystyle\;\;\;\;=\mathbb{E}e^{sY_{j}}\mathbb{E}e^{\sum_{l=1}^{j-1}sZ_{l}}1{1}(i_{1}=a_{1},\ldots,i_{j-1}=a_{j-1}).$
(2.13)

Using (2.5) we have
$\mathbb{E}e^{sY_{j}}\leq\exp\left(\frac{C}{(n-j)^{\alpha}}\right)$ for all
$1\leq j\leq n-K,$ where $K$ and $C$ do not depend on the choice of
$\\{a_{i}\\}.$ Thus summing (2.13) over all possible $a_{1},\ldots,a_{j-1},$
we get
$\mathbb{E}e^{s\sum_{l=1}^{j}Z_{l}}\leq\exp\left(\frac{C}{(n-j)^{\alpha}}\right)\mathbb{E}e^{s\sum_{l=1}^{j-1}Z_{l}}$
and continuing iteratively, we get
$\mathbb{E}e^{s\sum_{l=1}^{j}Z_{l}}\leq\exp\left(C\sum_{l=1}^{j}\frac{1}{(n-l)^{\alpha}}\right)$
for $n-j\geq K.$ For $n-j<K,$ we use the bound $\mathbb{E}e^{sY_{j}}\leq
e^{s}$ since $Y_{j}\leq 1$ (all the edge weights are at most one) and argue as
before to get that

$\mathbb{E}e^{s\sum_{l=1}^{n-1}Z_{l}}\leq\exp\left(C\sum_{l=1}^{n-K}\frac{1}{(n-l)^{\alpha}}\right)e^{sK}.$
(2.14)

Comparing with integrals, the term

$\sum_{l=1}^{n-K}\frac{1}{(n-l)^{\alpha}}=\sum_{j=K+1}^{n-1}\frac{1}{j^{\alpha}}\leq
C_{3}\int_{K}^{n-1}\frac{1}{x^{\alpha}}dx\leq C_{4}n^{1-\alpha}$

for some positive constants $C_{3},C_{4}.$ We therefore get from (2.14) and
(2.12) that
$\mathbb{E}e^{sM_{n}}\leq e^{C_{5}n^{1-\alpha}}$ for some constant
$C_{5}=C_{5}(s).$ Therefore by Chernoff estimate, we have
$\mathbb{P}(M_{n}\geq C_{6}n^{1-\alpha})\leq e^{-C_{7}n^{1-\alpha}}$ for some
positive constants $C_{6},C_{7}.$ This completes the proof of the lower
deviation bound in (2.3).

Finally, the lower bound on the expectation $\mathbb{E}M_{n}$ follows directly
from the lower deviation bound (2.3). For the expectation upper bound, we use
the fact that the edge weights are at most one and so total weight of any tree
containing at least $\rho n$ edges is at most $n.$ Consequently from the upper
deviation bound in (2.3), we get that $\mathbb{E}M_{n}\leq
C_{2}n^{1-\alpha}+n\cdot e^{-Cn^{1-\alpha}}\leq 2C_{2}n^{1-\alpha}.$ The proof
of the variance bound is analogous to the pivotal edge argument in Kesten
(1993) together with the fact that the number of edges in a spanning tree is
at most $n.$ This completes the proof of the Theorem.

##  3 Edge Constrained Minimum Passage Time Paths

Consider the square lattice $\mathbb{Z}^{d},$ where two vertices
$w_{1}=(w_{1,1},\ldots,w_{1,d})$ and $w_{2}=(w_{2,1},\ldots,w_{2,d})$ are
_adjacent_ if $\sum_{i=1}^{d}|w_{1,i}-w_{2,i}|=1$ and adjacent vertices are
joined together by an edge. Let $\\{q_{i}\\}_{i\geq 1}$ denote the set of
edges. Each edge $q_{i}$ is equipped with a random passage time $t(q_{i})$ and
we define the random sequence $(t(q_{1}),t(q_{2}),\ldots)$ on the probability
space $(\Omega,{\cal F},\mathbb{P}).$

A _path_ $\pi$ is a sequence of distinct adjacent vertices
$(w_{1},\ldots,w_{r+1}).$ If $e_{i},1\leq i\leq r$ is the edge with
endvertices $w_{i}$ and $w_{i+1},$ then we denote $\pi=(e_{1},...,e_{r}).$ By
definition, $\pi$ is self-avoiding and $w_{1}$ and $w_{r+1}$ are said to be
the _endvertices_ of $\pi.$ The length of $\pi$ is the number of edges in
$\pi$ and the passage time of $\pi$ is defined as
$T(\pi):=\sum_{i=1}^{r}t(e_{i}).$

######  Definition 3.1.

For $k\geq 1$ we define the $k-$_constrained_ minimum passage time between the
origin and the vertex $(n,\mathbf{0})$ as $T_{n}(k):=\min_{\pi}T(\pi),$ where
the minimum is over all paths $\pi$ of length at most $k$ and with endvertices
$(0,\mathbf{0})$ and $(n,\mathbf{0}).$ We define the _unconstrained_ minimum
passage time as $T_{n}:=\inf_{k\geq 1}T_{n}(k).$

By definition $T_{n}(k)\downarrow T_{n}$ a.s. as $k\rightarrow\infty.$ In this
section, we are primarily interested in studying how $T_{n}(k)$ varies as the
constraint parameter $k$ increases and also how fast $T_{n}(k)$ converges to
$T_{n}.$

The following are the main results of this section. Throughout constants do
not depend on $n.$

######  Theorem 3.1.

Suppose

$\sup_{i\geq 1}\mathbb{P}(t(q_{i})\leq\epsilon)\longrightarrow 0\text{ as
}\epsilon\downarrow 0\text{ and }\mu_{2}:=\sup_{i\geq
1}\mathbb{E}t^{2}(q_{i})<\infty.$ (3.1)

$(a)$ There are constants $C_{1},C_{2}>0$ such that for every $k\geq n:$

$\mathbb{P}\left(C_{1}n\leq T_{n}\leq T_{n}(k)\leq C_{2}n\right)\geq
1-\frac{C_{2}}{n},\;\;\;\;\;\;\;var(T_{n}(k))\leq C_{2}n$ (3.2)

and $C_{1}n\leq\mathbb{E}T_{n}\leq\mathbb{E}T_{n}(k)\leq C_{2}n.$
$(b)$ There exists a constant $C_{3}>0$ such that if $k\geq C_{3}n$ then
$\mathbb{P}(T_{n}\neq T_{n}(k))\leq\frac{C_{3}}{k}.$ If $k\geq n^{1+\epsilon}$
for some $\epsilon>0,$ then both $\frac{T_{n}(k)-\mathbb{E}T_{n}(k)}{n}$ and
$\frac{T_{n}-\mathbb{E}T_{n}}{n}$ converge to zero a.s. as
$n\rightarrow\infty.$
$(c)$ If the edge weights are uniformly square integrable in the sense that
$\sup_{i\geq 1}\mathbb{E}t^{2}(q_{i})1{1}(t(q_{i})\geq M)\longrightarrow 0$ as
$M\rightarrow\infty,$ then $var\left(\frac{T_{n}}{n}\right)\longrightarrow 0$
as $n\rightarrow\infty.$
If $\sup_{i\geq 1}\mathbb{E}t^{p}(q_{i})<\infty$ for some $p>2,$ then
$var(T_{n})\leq Cn$ for some constant $C>0.$

_Proof of Theorem 3.1$(a)$_: Let $\mu:=\sup_{f}\mathbb{E}t(f)$ be the maximum
expected passage time of an edge. We begin by showing that there exists a
constant $0<\beta\leq\mu$ such for any integer $m\geq 1$ and any path $\pi$
containing $m$ edges,

$\mathbb{P}\left(T(\pi)\leq\beta m\right)\leq e^{-dm}$ (3.3)

for some positive constant $\beta=\beta(d)\leq\mu,$ not depending on $m$ or
$\pi.$ Here $d$ is the dimension of the integer lattice under consideration.

Indeed, let $\pi=(e_{1},\ldots,e_{m})$ so that
$T(\pi)=\sum_{i=1}^{m}t(e_{i}).$ Using the Chernoff bound we obtain for
$\delta,s>0$ that

$\mathbb{P}(T(\pi)\leq\delta
m)=\mathbb{P}\left(\sum_{i=1}^{m}t(e_{i})\leq\delta m\right)\leq e^{s\delta
m}\prod_{i=1}^{m}\mathbb{E}\left(e^{-st(e_{i})}\right).$ (3.4)

For a fixed $\eta>0,$ we write
$\mathbb{E}e^{-st(e_{i})}=\int_{t(e_{i})<\eta}e^{-st(e_{i})}d\mathbb{P}+\int_{t(e_{i})\geq\eta}e^{-st(e_{i})}d\mathbb{P}$
and use

$\int_{t(e_{i})<\eta}e^{-st(e_{i})}d\mathbb{P}\leq\mathbb{P}(t(e_{i})<\eta)\text{
and }\int_{t(e_{i})\geq\eta}e^{-st(e_{i})}d\mathbb{P}\leq e^{-s\eta}$

to get that

$\mathbb{E}e^{-st(e_{i})}\leq\mathbb{P}(t(e_{i})<\eta)+e^{-s\eta}.$

Since $F(0)=0,$ we choose $\eta>0$ small so that
$\mathbb{P}(t(e_{i})<\eta)\leq\frac{e^{-6\epsilon}}{2}.$ Fixing such an $\eta$
we choose $s=s(\eta,\epsilon)>0$ large so that the second term
$e^{-s\eta}<\frac{e^{-6\epsilon}}{2}.$ This implies that
$\mathbb{E}e^{-st(e_{i})}\leq e^{-6\epsilon}$ and so from (3.4) we then get
that $\mathbb{P}(W({\cal P})\leq\delta m)\leq e^{s\delta m}e^{-6\epsilon
m}\leq e^{-2\epsilon m}$ for all $m\geq 1,$ provided
$\delta=\delta(s,\epsilon)>0$ is small. This completes the proof of (3.3).

Next, for integer $m\geq 1$ define the event $E_{m}$ as

$E_{m}:=\bigcap_{r\geq\frac{3\mu}{\beta}m}\;\;\bigcap_{\pi}\left\\{T(\pi)\geq\beta
r\right\\}$ (3.5)

where the second intersection is over all paths with origin as an endvertex
and consisting of $r$ edges. Thus $E_{m}$ is the event that every path $\pi$
with origin as an endvertex and consisting of $r\geq\frac{3\mu}{\beta}m$ edges
has passage time $T(\pi)\geq\beta r.$ Since there are at most $(2d)^{r}$ paths
of length $r$ starting from the origin, the estimate (3.3) gives

$\mathbb{P}(E_{m}^{c})\leq\sum_{r\geq
3\mu\beta^{-1}m}(2d)^{r}e^{-dr}\leq\sum_{r\geq
3\mu\beta^{-1}m}(2e^{-1})^{r}\leq\frac{e^{-\delta m}}{1-2e^{-1}}$ (3.6)

for all $m\geq 1$ and some positive constant $\delta=\delta(d,\mu),$ not
depending on $m.$ Here, the second inequality in (3.6) is obtained using the
fact that the function $xe^{-x}$ attains its maximum at $x=1$ and so
$2de^{-d}\leq 2e^{-1}<1$ for all $d\geq 2.$

Let $F_{m}$ be the event that $\sum_{i=1}^{m}t(f_{i})\leq 2\mu m$ where
$f_{i}$ is the horizontal edge with endvertices $(i-1,\mathbf{0})$ and
$(i,\mathbf{0}).$ Letting $X_{i}:=t(f_{i})-\mathbb{E}t(f_{i})$ and using the
fact that $\\{X_{i}\\}$ are independent, we then get from Chebychev’s
inequality that

$\mathbb{P}(F_{m}^{c})\leq\mathbb{P}\left(\sum_{i=1}^{m}X_{i}\geq\mu
m\right)\leq\frac{\sum_{i=1}^{m}var(X_{i})}{\mu^{2}m^{2}}\leq\frac{C}{m}$
(3.7)

for some constant $C>0.$ Now set $m=\frac{\beta n}{3\mu}<n$ and suppose
$E_{m}\cap F_{n}$ occurs. From (3.6), (3.7) and the union bound, we get that
$\mathbb{P}(E_{m}\cap F_{n})\geq 1-\frac{C}{n}$ for some constant $C>0.$ Since
$F_{n}$ occurs, we get that $T_{n}(k)\leq 2\mu n$ and since $E_{m}$ occurs, we
get that any path starting from the origin and containing
$r\geq\frac{3\mu}{\beta}m=n$ edges has weight at least $\beta r\geq 3\mu
m=\beta n.$ This obtains the first estimate (3.2).

Next, using the bounded second moment assumption in (3.1) and arguing as in
the variance estimate in Theorem $1,$ Kesten (1993) we get that
$var(T_{n}(k))\leq C\mathbb{E}N_{n}(k),$ where $N_{n}(k)$ is the number of
edges in the path with passage time $T_{n}(k).$ If $E_{m}\cap F_{n}$ occurs,
then from the discussion in the previous paragraph, we get that
$N_{n}(k)\leq\frac{3\mu}{\beta}n.$ For $x\geq\frac{3\mu}{\beta}n$ we assume
for simplicity that $y=\frac{\beta x}{3\mu}$ is an integer and write

$\displaystyle\mathbb{P}(N_{n}(k)\geq x)$ $\displaystyle\leq$
$\displaystyle\mathbb{P}\left(\\{N_{n}(k)\geq x\\}\cap
E_{y}\right)+\mathbb{P}\left(E_{y}^{c}\right)$ (3.8) $\displaystyle\leq$
$\displaystyle\mathbb{P}\left(\\{N_{n}(k)\geq x\\}\cap
E_{m}\right)+D_{1}e^{-D_{2}x}$

for some constants $D_{1},D_{2}>0$ by (3.6). If $N_{n}(k)\geq x$ and the event
$E_{y}$ occurs, then every path containing $r\geq\frac{3\mu}{\beta}y=x$ edges
has weight at least $\beta r\geq\beta x.$ Thus $T_{n}(k)\geq\beta x$ and so

$\mathbb{P}\left(\\{N_{n}(k)\geq x\\}\cap
E_{y}\right)\leq\mathbb{P}(T_{n}(k)\geq\beta
x)\leq\mathbb{P}\left(\sum_{i=1}^{n}t(f_{i})\geq\beta x\right).$

Since $\beta x\geq 3\mu n$ we have that $\beta
x-\sum_{i=1}^{n}\mathbb{E}t(f_{i})\geq\beta x-\mu n\geq\frac{2\beta x}{3}.$
Consequently, recalling that $X_{i}=t(f_{i})-\mathbb{E}t(f_{i})$ and using the
fact that $var(X_{i})\leq\mathbb{E}t^{2}(f_{i})\leq C$ for some constant
$C>0,$ we get from Chebychev’s inequality that

$\mathbb{P}\left(\\{N_{n}(k)\geq x\\}\cap
E_{y}\right)\leq\mathbb{P}\left(\sum_{i=1}^{n}X_{i}\geq\frac{2\beta
x}{3}\right)\leq\frac{D_{3}\sum_{i=1}^{n}var(X_{i})}{x^{2}}\leq\frac{D_{4}n}{x^{2}}$
(3.9)

for some constants $D_{3},D_{4}>0.$ Combining (3.8) and (3.9), we get that
$\mathbb{P}(N_{n}(k)\geq x)\leq\frac{D_{5}}{x^{2}}$ for
$x\geq\frac{3\mu}{\beta}n$ and so $\mathbb{E}N_{n}(k)\leq D_{6}n$ for some
constant $D_{6}>0.$ Plugging this into the variance estimate for $T_{n}(k)$
obtained in the previous paragraph, we get the second estimate in (3.2).

The lower expectation bound follows directly from the lower deviation bound in
(3.2). The upper expectation bound follows from the fact that
$\mathbb{E}T_{n}\leq\sum_{i=1}^{n}\mathbb{E}t(f_{i})\leq\mu n.$ This completes
the proof of part $(a).$

_Proof of Theorem 3.1$(b)$_: Let $\beta,\mu$ be as in part $(a)$ and for
$k\geq n$ suppose that the event $E_{k}\cap F_{k}$ occurs. From the discussion
following (3.7), we know that $\mathbb{P}(E_{k}\cap F_{k})\geq 1-\frac{C}{k}$
for some constant $C>0.$ The minimum passage time between the origin and
$(n,\mathbf{0})$ is at most $2\mu k$ and any path starting from the origin and
containing $r\geq\frac{3\mu}{\beta}k$ edges has passage time at least $\beta
r\geq 3\mu k.$ Thus $T_{n}\left(\frac{3\mu k}{\beta}\right)=T_{n}$ and this
obtains the probability estimate in $(b)$ with $C_{3}\leavevmode\nobreak\
=\leavevmode\nobreak\ \frac{3\mu}{\beta}.$

We prove the a.s. convergence in two steps. In the first step, we use a
subsequence argument to show that $\frac{T_{n}(k)-\mathbb{E}T_{n}(k)}{n}$
converges to zero a.s. In the second step we show that the _difference_
$\frac{T_{n}(k)-T_{n}}{n}$ converges to zero a.s. and in $L^{1},$ provided $k$
is sufficiently large. This then obtains the a.s. convergence for
$\frac{T_{n}-\mathbb{E}T_{n}}{n}.$

_Step 1_ : We begin with a description of the sub-additivity property of the
unconstrained passage time $T_{n}.$ If $T_{n,m}$ is the minimum passage time
between $(n,\mathbf{0})$ and $(m,\mathbf{0}),$ then $T_{n}\leq T_{m}+T_{n,m}.$
This is because the concatenation of the minimum passage time path with
endvertices $(0,\mathbf{0})$ and $(m,\mathbf{0})$ and the minimum passage time
path with endvertices $(m,\mathbf{0})$ and $(n,\mathbf{0})$ contains a path
with endvertices $(0,\mathbf{0})$ and $(n,\mathbf{0}).$ Switching the roles of
$m$ and $n$ we therefore have that $|T_{n}-T_{m}|\leq T_{n,m}$ and we refer to
this estimate as the _sub-additive property_ of $T_{n}.$

Letting $k\geq n^{1+\epsilon}$ with $\epsilon>0,$ we now perform the
subsequence argument. Setting $U_{n}:=T_{n}(k)$ and
$S_{n}:=U_{n}-\mathbb{E}U_{n},$ we first show that
$\frac{S_{n}}{n}\longrightarrow 0$ a.s. as $n\leavevmode\nobreak\
\rightarrow\leavevmode\nobreak\ \infty.$ Indeed, from the variance estimate
for $U_{n}$ in (3.2), we know that $\mathbb{E}S^{2}_{n}\leq Cn$ for some
constant $C>0$ and so for a fixed $\delta>0,$ the sum

$\sum_{n\geq 1}\mathbb{P}(|S_{n^{2}}|>n^{2}\delta)\leq\sum_{n\geq
1}\frac{\mathbb{E}S^{2}_{n^{2}}}{\delta^{2}n^{4}}\leq\sum_{n\geq
1}\frac{C}{\delta^{2}n^{2}}<\infty.$

Since this is true for all $\delta>0,$ Borel-Cantelli Lemma implies that
$\frac{S_{n^{2}}}{n^{2}}$ converges to zero a.s. as $n\rightarrow\infty.$

To estimate the intermediate values of $S_{j},$ we let $n^{2}\leq j<(n+1)^{2}$
and set $R_{n}:=\max_{n^{2}\leq j<(n+1)^{2}}|S_{j}-S_{n^{2}}|$ and show below
that $\frac{R_{n}}{n^{2}}\longrightarrow 0$ a.s. as $n\leavevmode\nobreak\
\rightarrow\leavevmode\nobreak\ \infty.$ This would imply that for $n^{2}\leq
j<(n+1)^{2},$

$\frac{|S_{j}|}{j}\leq\frac{|S_{j}-S_{n^{2}}|}{j}+\frac{|S_{n^{2}}|}{j}\leq\frac{|S_{j}-S_{n^{2}}|}{n^{2}}+\frac{|S_{n^{2}}|}{n^{2}}\leq\frac{D_{n}}{n^{2}}+\frac{|S_{n^{2}}|}{n^{2}}$

and so $\frac{S_{j}}{j}$ converges to zero a.s. as $j\rightarrow\infty.$

To estimate $R_{n},$ use first the triangle inequality to get that

$|S_{j}-S_{n^{2}}|\leq|U_{j}-U_{n^{2}}|+\mathbb{E}|U_{j}-U_{n^{2}}|$ (3.10)

We know that $\mathbb{P}(T_{j}\neq
U_{j})\leq\frac{D_{1}}{k(j)}\leq\frac{D_{2}}{j^{1+\epsilon}}$ for some
constants $D_{1},D_{2}>0$ and so setting
$E_{tot}:=\bigcap_{j=n^{2}}^{(n+1)^{2}}\\{T_{j}=U_{j}\\},$ we get from the
union bound that

$\mathbb{P}(E_{tot})\geq
1-\sum_{j=n^{2}}^{(n+1)^{2}}\frac{D_{2}}{j^{1+\epsilon}}\geq
1-\frac{D_{3}n}{n^{2+2\epsilon}}=1-\frac{D_{3}}{n^{1+2\epsilon}}$ (3.11)

for some constant $D_{3}>0.$

We now write
$|U_{j}-U_{n^{2}}|=|T_{j}-T_{n^{2}}|1{1}(E_{tot})+|U_{j}-U_{n^{2}}|1{1}(E^{c}_{tot})$
and evaluate each term separately. For $n^{2}\leq j<(n+1)^{2}$ we know by the
subadditivity property that $|T_{j}-T_{n^{2}}|\leq
T_{j,n^{2}}\leq\sum_{i=n^{2}}^{(n+1)^{2}}t(f_{i})=:A_{n}$ and by definition,
we have that $U_{j}\leq\sum_{i=1}^{(n+1)^{2}}t(f_{i})=:J_{n}.$ Thus
$|U_{j}-U_{n^{2}}|\leq A_{n}+2J_{n}1{1}(E^{c}_{tot})$ for all $n^{2}\leq
j\leq(n+1)^{2}$ and so from (3.10), we see that

$R_{n}\leq
A_{n}+2J_{n}1{1}(E^{c}_{tot})+\mathbb{E}A_{n}+2\mathbb{E}J_{n}1{1}(E^{c}_{tot}).$
(3.12)

Based on (3.12), it suffices to show that both the terms $\frac{A_{n}}{n^{2}}$
and $\frac{J_{n}1{1}(E^{c}_{tot})}{n^{2}}$ converge to zero a.s. and in
$L^{1}$ as $n\rightarrow\infty.$ First, from the estimate
$\mathbb{E}A_{n}\leq\sum_{i=n^{2}}^{(n+1)^{2}}\mathbb{E}t(f_{i})\leq Cn$ for
some constant $C>0,$ we get that $\frac{\mathbb{E}A_{n}}{n^{2}}\longrightarrow
0$ as $n\leavevmode\nobreak\ \rightarrow\leavevmode\nobreak\ \infty.$ Next,
let $0<\theta<1$ be any constant. Using Chebychev’s inequality and arguing as
in (3.7), we get that $\mathbb{P}(A_{n}\geq 2\mu
n^{1+\theta})\leq\frac{C}{n^{1+2\theta}}$ for all $n$ large and so by the
Borel-Cantelli Lemma we get that $\mathbb{P}\left(A_{n}\leq 2\mu
n^{1+\theta}\text{ for all large }n\right)=1.$ Since $\theta<1,$ we get that
$\frac{A_{n}}{n^{2}}\longrightarrow 0$ a.s. as $n\rightarrow\infty.$

To evaluate $J_{n}1{1}(E^{c}_{tot}),$ we use (3.11) and the Borel-Cantelli
Lemma to get $1{1}(E^{c}_{tot})\longrightarrow 0$ a.s. as
$n\rightarrow\infty.$ Thus
$\frac{J_{n}1{1}(E^{c}_{tot})}{n^{2}}\longrightarrow 0$ a.s. as
$n\rightarrow\infty.$ Using the Cauchy-Schwartz inequality, we also get that
$\mathbb{E}J_{n}1{1}(E^{c}_{tot})\leq\left(\mathbb{E}J_{n}^{2}\right)^{1/2}\left(\mathbb{P}(E^{c}_{tot})\right)^{1/2}.$
By the AM-GM inequality and the bounded second moment assumption in (3.1), we
get that

$\mathbb{E}J_{n}^{2}\leq(n+1)^{2}\sum_{i=1}^{(n+1)^{2}}\mathbb{E}t^{2}(f_{i})\leq
C_{2}n^{4}$

for some constant $C_{2}>0$ and so using (3.11), we get that

$\mathbb{E}J_{n}1{1}(E^{c}_{tot})\leq\sqrt{C_{2}}n^{2}\cdot\left(\frac{D}{n^{1+2\epsilon}}\right)^{1/2}.$

Thus $\frac{\mathbb{E}J_{n}1{1}(E^{c}_{tot})}{n^{2}}\leavevmode\nobreak\
\longrightarrow\leavevmode\nobreak\ 0$ as $n\rightarrow\infty$ as well and
this completes the proof of $\frac{S_{n}}{n}\longrightarrow 0$ a.s. as
$n\rightarrow\infty.$

_Step 2_ : We now set $k=n^{1+\epsilon}$ and show that
$\frac{T_{n}(k)-T_{n}}{n}$ converges to zero a.s. and in $L^{1}.$ First using
$\mathbb{P}(T_{n}(k)\neq T_{n})\leq\frac{D_{1}}{n^{1+\epsilon}}$ for some
constant $D_{1}>0,$ we get from the Borel-Cantelli Lemma that
$\frac{T_{n}(k)-T_{n}}{n}\longrightarrow 0$ a.s. as $n\rightarrow\infty.$ Next
using the fact that $T_{n}(k)$ and $T_{n}$ are both bounded above by
$\sum_{i=1}^{n}t(f_{i})$ we have that
$\mathbb{E}|T_{n}(k)-T_{n}|=\mathbb{E}|T_{n}(k)-T_{n}|1{1}(T_{n}(k)\neq
T_{n})$ is bounded above by

$\displaystyle\mathbb{E}\sum_{i=1}^{n}t(f_{i})1{1}(T_{n}(k)\neq T_{n})$
$\displaystyle\leq$
$\displaystyle\mathbb{E}^{1/2}\left(\sum_{i=1}^{n}t(f_{i})\right)^{2}\mathbb{P}^{1/2}(T_{n}(k)\neq
T_{n})$ $\displaystyle\leq$
$\displaystyle\mathbb{E}^{1/2}\left(\sum_{i=1}^{n}t(f_{i})\right)^{2}\left(\frac{D_{1}}{n^{1+\epsilon}}\right)^{1/2}.$

Using $(\sum_{i=1}^{l}a_{i})^{2}\leq\l\sum_{i}a_{i}^{2}$ we have
$\mathbb{E}\left(\sum_{i=1}^{n}t(f_{i})\right)^{2}\leq
n\sum_{i=1}^{n}\mathbb{E}t^{2}(f_{i})\leq D_{2}n^{2}$ for some constant
$D_{2}>0.$ Thus $\mathbb{E}|T_{n}(k)-T_{n}|\leq
D_{3}(n^{1-\epsilon})^{1/2}=o(n)$ and so
$\frac{\mathbb{E}|T_{n}(k)-T_{n}|}{n}\longrightarrow 0$ as
$n\rightarrow\infty.$ This completes the proof of a.s. convergence in part
$(b).$

_Proof of Theorem 3.1$(c)$_: Using $(a+b)^{2}\leq 2(a^{2}+b^{2})$ for any two
real numbers $a$ and $b$ we have that the variance of the sum of any two
random variables $X$ and $Y$ satisfies

$var(X+Y)=\mathbb{E}\left((X-\mathbb{E}X)+(Y-\mathbb{E}Y)\right)^{2}\leq
2(var(X)+var(Y)).$ (3.13)

Setting $T=T_{n},U=T_{n}(k),X=T-U$ and $Y={U}$ we get that

$var(T)\leq 2var(T-U)+2var(U)\leq 2var(T-U)+D_{1}n$ (3.14)

for all $n$ large and some constant $D_{1}>0,$ by the variance estimate for
$U$ in part $(a)$ of this Theorem.

We estimate $var(T-U)$ as follows. Using $T\leq U$ we write

$\mathbb{E}(T-U)^{2}=\mathbb{E}(T-U)^{2}1{1}(T\neq U)\leq
2\mathbb{E}(T^{2}+U^{2})1{1}(T\neq U)\leq 4\mathbb{E}U^{2}1{1}(T\neq U).$

Since $U\leq\sum_{i=1}^{n}t(f_{i}),$ we have that $U^{2}\leq
n\sum_{i=1}^{n}t^{2}(f_{i})$ and so $var(T-U)$ is bounded above by

$\mathbb{E}(T-U)^{2}\leq 4n\sum_{i=1}^{n}\mathbb{E}t^{2}(f_{i})1{1}(T\neq
U)\leq 4n^{2}\sup_{i}\mathbb{E}t^{2}(f_{i})1{1}(T\neq U).$ (3.15)

Let $\theta>0$ be a constant and split

$\displaystyle\mathbb{E}t^{2}(f_{i})1{1}(T\neq U)$ $\displaystyle=$
$\displaystyle\mathbb{E}t^{2}(f_{i})1{1}(\\{T\neq
U\\}\cap\\{t(f_{i})<n^{\theta}\\})$ (3.16)
$\displaystyle\;\;\;\;\;\;\;+\;\;\;\mathbb{E}t^{2}(f_{i})1{1}(\\{T\neq
U\\}\cap\\{t(f_{i})\geq n^{\theta}\\}).$

From (3.2), we know that there are constants $D_{2},D_{3}>0$ such that if
$k\geq D_{2}n$ then
$\mathbb{P}(T\neq U)\leq\frac{D_{3}}{k}.$ With this choice of $k,$ the first
term in (3.16) is bounded above by

$\mathbb{E}t^{2}(f_{i})1{1}(\\{T\neq U\\}\cap\\{t(f_{i})<n^{\theta}\\})\leq
n^{2\theta}\mathbb{P}(T\neq
U)\leq\frac{D_{2}n^{2\theta}}{k}\leq\frac{1}{n^{3}}$ (3.17)

provided we choose $k$ larger if necessary so that $k\geq n^{2\theta}\log{n}.$

We now consider the case where the uniform square integrability condition
holds. For any $\eta>0$ and all $n$ large, the final term in (3.16) is then at
most $\mathbb{E}t^{2}(f_{i})1{1}(t(f_{i})\geq n^{\theta})\leq\eta.$ Combining
this estimate with (3.17), we get that $\mathbb{E}t^{2}(f_{i})1{1}(T\neq
U)\leq\frac{1}{n^{3}}+\eta\leq 2\eta$ for all $n$ large and so from (3.15) we
get that $\mathbb{E}(T-U)^{2}\leq 8n^{2}\eta.$ Plugging this into (3.14) we
get that $var(T)\leq D_{1}n+8n^{2}\eta$ and since $\eta>0$ is arbitrary, this
implies that $var\left(\frac{T}{n}\right)=o(1).$

Suppose now that bounded $p^{th}$ moment condition holds for some $p>2.$ Using
Hölder’s inequality and Markov inequality in succession, the final term in
(3.16) is at most

$\displaystyle\mathbb{E}t^{2}(f_{i})1{1}(t(f_{i})\geq n^{\theta})$
$\displaystyle\leq$
$\displaystyle\left(\mathbb{E}t^{p}(f_{i})\right)^{2/p}\mathbb{P}\left(t(f_{i})\geq
n^{\theta}\right)^{1-2/p}$ (3.18) $\displaystyle\leq$
$\displaystyle\left(\mathbb{E}t^{p}(f_{i})\right)^{2/p}\left(\frac{\mathbb{E}t^{p}(f_{i})}{n^{\theta
p}}\right)^{1-2/p}$ $\displaystyle\leq$
$\displaystyle\frac{D_{4}}{n^{\theta(p-2)}}$

for some constant $D_{4}>0,$ by the bounded $p^{th}$ moment assumption. We
choose $\theta>0$ large so that the final term in (3.18) is at most
$\frac{1}{n^{3}}.$ With this choice of $\theta$ we get from (3.17) that
$\mathbb{E}t^{2}(f_{i})1{1}(T\neq U)\leq\frac{2}{n^{3}}$ and plugging this
into (3.15), we get that $\mathbb{E}(T-U)^{2}\leq\frac{D_{5}}{n}$ for constant
$D_{5}>0.$ From (3.14), we then get that $var(T)\leq D_{6}n$ for some constant
$D_{6}>0.$ This completes the proof of part $(c).$

_Remark_ : For $p>2,$ the bounded $p^{th}$ moment condition in Theorem
3.1$(c)$ is stronger than the uniformly square integrable condition which in
turn is stronger than the bounded second moment condition in (3.1). For the
particular case of i.i.d. passage times, uniform square integrability is
implied by the bounded second moment condition and the first condition in
(3.1) above simply states that the passage times are a.s. positive.

### Acknowledgement

I thank Professors Rahul Roy, C. R. Subramanian and the referee for crucial
comments that led to an improvement of the paper. I also thank IMSc and IISER
Bhopal for my fellowships.

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