# Characterising Simple Lie Algebras by Their Borel Nilpotent Radical

Guy Kapon, Lior Hadassy

###### Abstract

We show that simple lie algebras can be characterised by their maximal
nilpotent subalgebra, which is the same as the nilpotent radical of a borel
subalgebra.

##  1 Introduction

The classification of simple lie algebras is well known, and is made in terms
of the combinatorics of roots system. While this is satisfactory for many
purposes, one might wonder if they can also be classified by more algebraic
means, or at least characterised that way. Indeed such a theorem can perhaps
be extended to cases where the theory of root system is less applicable, like
the classification of finite groups. A first option could be to characterise
the algebra from a borel subalgebra, which is quite easy. A less obvious
question is if one can do this from the nilpotent radical of the borel, which
is also a maximal nilpotent subalgebra, see [DKK06], and we prove in this
note:

######  Main Theorem.

Let $\mathfrak{g}_{1},\mathfrak{g}_{2}$ be two simple lie algebras, let
$\mathfrak{b}(\mathfrak{g}_{1}),\mathfrak{b}(\mathfrak{g}_{2})$ be borel
subalgebras, and let
$\mathfrak{n}(\mathfrak{g}_{1}),\mathfrak{n}(\mathfrak{g}_{2})$ be the
nilpotent radical of those borels, assume that
$\mathfrak{n}(\mathfrak{g}_{1})\cong\mathfrak{n}(\mathfrak{g}_{2})$ then
$\mathfrak{g}_{1}\cong\mathfrak{g}_{2}$

The nilpotent radical is a nilpotent algebra, and while classifying such
algebras is a very hard problem, distinguishing between them is easier, as by
definition they have a lot of ideals and invariants.

###  1.1 Idea of The Proof

The idea of The proof is quite simple, our main tool is the lower central
sequence, defined by
$\mathfrak{n}_{1}=\mathfrak{n},\mathfrak{n}_{i+1}=[\mathfrak{n}_{i},\mathfrak{n}]$,
which are obviously invariants of the algebra, already
$\rm{dim}(\mathfrak{n}_{1}/\mathfrak{n}_{2})$ recovers the rank of the
original simple algebra, and together with the dimension of the algebra this
almost completes the proof. to deal with the remaining cases we consider
product
$\mathfrak{n}_{i}/\mathfrak{n}_{i+1}\times\mathfrak{n}_{j}/\mathfrak{n}_{j+1}\rightarrow\mathfrak{n}_{i+j}/\mathfrak{n}_{i+j+1}$
induced by the lie bracket, and show that in the remaining cases these are not
always the same.

##  2 Proof of Main Theorem

We follow the notations of [Kir08], everything is done over $\mathbb{C}$.

We have a simple algebra $\mathfrak{g}$, it’s cartan subalgebra is
$\mathfrak{h}(\mathfrak{g})$, it’s root system is denoted $R(\mathfrak{g})$,
and it’s root spaces are $\mathfrak{g}_{\alpha}$, we will assume that some
borel $\mathfrak{b}(\mathfrak{g})$ containing $\mathfrak{h}(\mathfrak{g})$ was
chosen, and we denote by
$\mathfrak{n}(\mathfrak{g})=[\mathfrak{b}(\mathfrak{g}),\mathfrak{b}(\mathfrak{g})]$
it’s nilpotent radical. We have:

$\mathfrak{n}(\mathfrak{g})=\bigoplus_{\alpha\in R_{+}}\mathfrak{g}_{\alpha}$

Where $R_{+}$ is the set of positive roots with respect to our chosen borel.
When there is no confusion about the simple lie algebra, we will drop it from
the notation.

Our Goal is to prove that we can characterise a simple lie algebra from the
structure of $\mathfrak{n}$. Two invariants that we will explain now will
suffice.

###  2.1 Lower Central Series

The Lower Central Series of $\mathfrak{n}$ is defined by
$\mathfrak{n}_{1}=\mathfrak{n}$ and
$\mathfrak{n}_{i+1}=[\mathfrak{n}_{i},\mathfrak{n}]$. Let us calculate the
lower central series in our case. For a positive root $\alpha\in R_{+}$ we
define it’s degree $\rm{deg}(\alpha)$ as the sum of coefficients when we write
$\alpha$ as a sum of simple roots.

######  Claim 2.1.1.

for $i\geq 1$, $\mathfrak{n}_{i}$ is the subspace generated by
$\mathfrak{g}_{\alpha}$ for $\alpha$ with $\rm{deg}(\alpha)\geq i$

###### Proof.

For $i=1$ this is obvious, we proceed by induction.

Let us denote the subspace generated by $\mathfrak{g}_{\alpha}$ for
$\rm{deg}(\alpha)\geq i$ by $\mathfrak{n}_{i}^{\prime}$, and assume we have
proved $\mathfrak{n}_{i}=\mathfrak{n}_{i}^{\prime}$. Since
$[\mathfrak{g}_{\alpha},\mathfrak{g}_{\beta}]\subseteq\mathfrak{g}_{\alpha+\beta}$,
and the representation of a root as a sum of simple roots is unique, it is
obvious that
$\mathfrak{n}_{i+1}=[\mathfrak{n}_{i},\mathfrak{n}]\subseteq\mathfrak{n}_{i+1}^{\prime}$.
On the other hand, let $\alpha$ be a root with $\rm{deg}(\alpha)\geq i+1$,
then we know there is a simple root $\beta$ s.t $\alpha-\beta$ is also a root,
in that case
$[\mathfrak{g}_{\alpha-\beta},\mathfrak{g}_{\beta}]=\mathfrak{g}_{\alpha}$,
obviously $\rm{deg}(\alpha-\beta)=\rm{deg}(\alpha)-1\geq i$ Since $\rm{deg}$
is linear, and so
$[\mathfrak{g}_{\alpha-\beta},\mathfrak{g}_{\beta}]\subseteq[\mathfrak{n}_{i},\mathfrak{n}]=\mathfrak{n}_{i+1}$
and so we are done. ∎

The lower central series has an additional structure in our case:

######  Notation 2.1.2.

For $i\geq 1$ denote $\mathfrak{n}^{i}=\mathfrak{n}_{i}/\mathfrak{n}_{i+1}$.

######  Claim 2.1.3.

For any $i,j\geq 0$ We have a bilinear map
$\mathfrak{n}^{i}\times\mathfrak{n}^{j}\rightarrow\mathfrak{n}^{i+j}$

###### Proof.

By our description of $\mathfrak{n}_{i}$ it’s obvious that
$[\mathfrak{n}_{i},\mathfrak{n}_{j}]\subseteq\mathfrak{n}_{i+j}$, which
defines a product
$\mathfrak{n}_{i}\times\mathfrak{n}_{j}\rightarrow\mathfrak{n}_{i+j}$, of
course we can then quotient out $\mathfrak{n}_{i+j+1}$, it remains only to see
that $\mathfrak{n}_{i+1}\times\mathfrak{n}_{j}$ and
$\mathfrak{n}_{i}\times\mathfrak{n}_{j+1}$ go to zero, but we already
explained that their commutator is in $\mathfrak{n}_{i+j+1}$ as required. ∎

######  Remark 2.1.4.

Both of the claims can be described by saying that
$gr(\mathfrak{n}):=\bigoplus_{i}\mathfrak{n}^{i}$ is a lie algebra by itself,
which in our case is in fact (non-canonically) isomorphic to $\mathfrak{n}$.

###  2.2 Rank

From the previous section we see that $\mathfrak{n}^{i}$ is isomorphic to the
space generated by $\mathfrak{g}_{\alpha}$ with $\rm{deg}(\alpha)=i$, so the
number of such roots is an invariant of $\mathfrak{n}$. in particular the
number of roots of degree $1$, i.e simple roots, is an invariant, this is of
course equal to the rank of the original lie algebra. This means we only have
to distinguish between $A_{n},B_{n},C_{n},D_{n}$ and the exceptionals for a
constant $n$. now we can compare dimensions, since
$\rm{dim}(\mathfrak{g})=2\cdot\rm{dim}(\mathfrak{n})+n$ we will compare the
dimension of $\mathfrak{g}$ instead, which is well known:

Lie Algebra | Rank | Dimension
---|---|---
$A_{n}$ | $n$ | $n(n+2)$
$B_{n}$ | $n$ | $n(2n+1)$
$C_{n}$ | $n$ | $n(2n+1)$
$D_{n}$ | $n$ | $n(2n-1)$
$E_{6}$ | $6$ | $78$
$E_{7}$ | $7$ | $133$
$E_{8}$ | $8$ | $248$
$F_{4}$ | $4$ | $52$
$G_{2}$ | $2$ | $14$

This table can be found for example at [MFM08, Table 2].

From the table we easily see that $B_{n},C_{n}$ always have the same
dimension, which is always different to $D_{n}$, $A_{n}$ has the same
dimension as $B_{n}$ only for $n=1$, and the same dimension as $D_{n}$ only
for $n=3$, both of which corresponds to the well known facts that
$A_{1}=B_{1}=C_{1}$ and $A_{3}=D_{3}$.

Finally the only exceptional that has the same dimension as a classic algebra
of the same rank is $E_{6}$ with $B_{6}$ and $C_{6}$.

In summary we proved:

######  Proposition 2.2.1.

If $\mathfrak{g}_{1},\mathfrak{g}_{2}$ are two simple lie algebras, and we
have $\mathfrak{n}(\mathfrak{g}_{1})\cong\mathfrak{n}(\mathfrak{g}_{2})$ then
either $\mathfrak{g}_{1}\cong\mathfrak{g}_{2}$,
$\\{\mathfrak{g}_{1},\mathfrak{g}_{2}\\}\cong\\{B_{n},C_{n}\\}$ or one of the
algebras is $E_{6}$ and the other is $B_{6}$ or $C_{6}$.

The next natural step is to look at the number of roots of height $i$ for
general $i$, as it turns out those numbers are the same for $B_{n}$ and
$C_{n}$ but do distinguish between $E_{6}$ and $B_{6},C_{6}$, specifically
$E_{6}$ has $5$ roots of degree $4$ while $B_{6},C_{6}$ have $4$ such roots.
Both those claims can be checked by writing down root systems. for
$B_{6},C_{6}$ we will do this later, for $E_{6}$ one can for example use the
representation in [Wik21], we also refer to [Rin18, Appendix] where the root
posets for those root system are drawn, in particular one can see this result.

###  2.3 $B_{n}$ and $C_{n}$

As we mentioned, the number roots of each degree for $B_{n}$ and $C_{n}$ is
the same. still the combinatorics of the roots is different, and we claim that
our bilinear product will distinguish between them.

First we write down the root systems, and our choice of simple roots. Both are
written inside the vector space with base $\\{e_{i}\\}_{1\leq i\leq n}$ like
in [Kir08, 205-206]:

$R(B_{n})=\\{\pm e_{i}\pm e_{j}\\}_{i\neq j}\cup\\{\pm e_{i}\\}$
$\Pi(B_{n})=\\{e_{i}-e_{i+1}\\}_{1\leq i\leq n-1}\cup\\{e_{n}\\}$

$R(C_{n})=\\{\pm e_{i}\pm e_{j}\\}_{i\neq j}\cup\\{\pm 2e_{i}\\}$
$\Pi(C_{n})=\\{e_{i}-e_{i+1}\\}_{1\leq i\leq n-1}\cup\\{2e_{n}\\}$

Now let us consider in both cases the bilinear product:

$\mathfrak{n}^{2}\times\mathfrak{n}^{2n-3}\rightarrow\mathfrak{n}^{2n-1}$

Notice that in both cases $\mathfrak{n}_{2n-1}$ is $1$ dimensional, while
$\mathfrak{n}_{2n}=0$. Indeed in $B_{n}$ the root of highest degree is
$e_{1}+e_{2}$ with degree $2n-1$, and in $C_{n}$ it is $2e_{1}$ also with
degree $2n-1$.

Next we see that $\mathfrak{n}^{2n-3}$ is two dimensional in both cases, in
the case of $B_{n}$ the roots of degree $2n-3$ are $e_{1}+e_{4},e_{2}+e_{3}$
and for $C_{n}$ there are $e_{1}+e_{3},2e_{2}$.

Here we have assumed $n\geq 4$, indeed we had $e_{1}+e_{4}$, this however also
works for $n=3$, by considering $e_{4}=0$ in this case. For $n=2$ or $n=1$ the
dimension calculation is wrong, but in those cases $B_{2}=C_{2},B_{1}=C_{1}$
so there is nothing to prove.

######  Claim 2.3.1.

In the case of $C_{n}$, $n\geq 3$, this bilinear product has non-trivial right
kernel, specifically it contains $\mathfrak{g}_{2e_{2}}$.

###### Proof.

For any $\alpha$ of degree $2$, $\alpha+2e_{3}\notin R$, indeed if it was it
would have to be $2e_{1}$ because it’s degree would be $2n-1$, but
$2e_{1}-2e_{2}$ is not a root.

This means that
$[\mathfrak{g}_{\alpha},\mathfrak{g}_{2e_{2}}]=\mathfrak{g}_{\alpha+2e_{2}}=0$
for any root of degree $2$, but those $\mathfrak{g}_{\alpha}$ exactly generate
$\mathfrak{n}^{2}$. ∎

So now to finish we only have to prove:

######  Claim 2.3.2.

In the case of $B_{n}$,$n\geq 3$, the bilinear product is non-degenerate on
the right.

###### Proof.

Similar to before, notice that
$(e_{1}+e_{4})+(e_{2}-e_{4})=(e_{2}+e_{3})+(e_{1}-e_{3})=e_{1}+e_{2}$, which
means that

$[\mathfrak{g}_{e_{2}-e_{4}}\mathfrak{g}_{e_{1}+e_{4}}]=\mathfrak{g}_{e_{1}+e_{2}}$ | $[\mathfrak{g}_{e_{1}-e_{3}}\mathfrak{g}_{e_{1}+e_{4}}]=0$
---|---
$[\mathfrak{g}_{e_{2}-e_{4}}\mathfrak{g}_{e_{2}+e_{3}}]=0$ | $[\mathfrak{g}_{e_{1}-e_{3}}\mathfrak{g}_{e_{2}+e_{3}}]=\mathfrak{g}_{e_{1}+e_{2}}$

Which obviously means that the bilinear product is non-degenerate on the
right.

(For $n=3$ the same calculation holds after erasing $e_{4}$) ∎

So now we are done.

###### proof of Main Theorem.

By proposition 2.2.1 with addition to the remarks after it, it only remains to
show that $\mathfrak{n}(B_{n})$ is not isomorphic to $\mathfrak{n}(C_{n})$ for
$n\geq 3$ as for $n\leq 2$, $B_{n}\cong C_{n}$. If they were isomorphic then
we also would have had an isomorphisim of the lower central series, and the
bilinear product we produced on it. but we just showed that for $B_{n}$ one of
those products is non-degenerate on the right, while for $C_{n}$ the same one
is, and therefore are not isomorphic. ∎

## References

  * [DKK06] Jan Draisma, Hanspeter Kraft and Jochen Kuttler  “Nilpotent subspaces of maximal dimension in semi-simple Lie algebras”  In _Compositio Mathematica_ 142.2 London Mathematical Society, 2006, pp. 464–476  DOI: 10.1112/S0010437X05001855
  * [Kir08] Alexander Kirillov  “An Introduction to Lie Groups and Lie Algebras”, Cambridge Studies in Advanced Mathematics  Cambridge University Press, 2008  DOI: 10.1017/CBO9780511755156
  * [MFM08] Paul de Medeiros, José Figueroa-O’Farrill and Elena Méndez-Escobar  “Lorentzian Lie 3-algebras and their Bagger-Lambert moduli space”  In _Journal of High Energy Physics_ 2008.07 Springer ScienceBusiness Media LLC, 2008, pp. 111–111  DOI: 10.1088/1126-6708/2008/07/111
  * [Rin18] Claus Michael Ringel  “The root posets and their rich antichains”, 2018  arXiv:1306.1593 [math.CO]
  * [Wik21] Wikipedia  “E6 (mathematics) — Wikipedia, The Free Encyclopedia” [Online; accessed 24-October-2021], http://en.wikipedia.org/w/index.php?title=E6%20(mathematics)&oldid=1040335753, 2021