---
title: "Problem Sets: Lie Algebras"
subtitle: "Problem Set 2"
author:
- name: D. Zack Garza
  affiliation: University of Georgia 
  email: dzackgarza@gmail.com 
date: Fall 2022
order: 2
---
 
# Problem Set 2

## Section 3

:::{.problem title="Humphreys 3.1"}
Let $I$ be an ideal of $L$. Then each member of the derived series or descending central series of $I$ is also an ideal of $L$.
:::

:::{.solution}
To recall definitions:

- The derived series of $L$ is $L \contains L^{(0)} \da [LL] \contains L^{(1)} \da [[LL] [LL]] \contains \cdots$ and termination implies solvability.
- The descending central series of $L$ is $L \contains L^1 \da [LL] \contains L^2 \da [L [LL]] \contains \cdots$, and termination implies nilpotency (and hence solvability since $[LL] \subseteq L \implies L^{(i)} \subseteq L^i$).
- $I \normal L \iff [L,I] \subseteq I$.

For the derived series,
inductively suppose $I \da I^{(i)}$ is an ideal, so $[L I] \subseteq I$.
We then want to show $I^{(i+1)} \da [I, I]$ is an ideal, so $[L, [I, I] ] \subseteq [I, I]$.
Letting $l\in L$, and $i,j\in I$, one can use the Jacobi identity, antisymmetry of the bracket, and the fact that $[I, I] \da L^{(i+1)} \subseteq I$ to write
\[
[L, [I, I]] &\ni [l[ij]] \\
&= [[li]j] - [ [lj] i] \\
&\in [[L,I], I] - [[L,I], I] \\
&\subseteq [[L,I], I] \subseteq [I,I]
.\]

Similarly, for the lower central series, inductively suppose $I\da I^i$ is an ideal, so $[L, I] \subseteq I$; we want to show $[L, [L, I]] \subseteq [L, I]$.
Again using the Jacobi identity and antisymmetry, we have
\[
[L, [L, I]] 
&\ni [l_1, [l_2, i]] \\
&= [[i, l_1], l_2] + [[l_2, l_1], i] \\
&\subseteq [[I,L], L] + [ [L, L], I] \\
&\subseteq [I, L] + [L, I] \subseteq [L, I]
.\]
:::

:::{.problem title="Humphreys 3.4"}
Prove that $L$ is solvable (resp. nilpotent) if and only $\ad(L)$ is solvable (resp. nilpotent).
:::

:::{.solution}

$\implies$:
By the propositions in Section 3.1 (resp. 3.2), the homomorphic image of any solvable (resp. nilpotent) Lie algebra is again solvable (resp. nilpotent).

$\impliedby$:
There is an exact sequence
\[
0 \to Z(L) \to L \mapsvia{\ad}  \ad(L) \to 0
,\]
exhibiting $\ad(L)\cong L/Z(L)$.
Thus if $\ad(L)$ is solvable, noting that centers are always solvable, we can use the fact that the 2-out-of-3 property for short exact sequences holds for solvability.
Moreover, by the proposition in Section 3.2, if $L/Z(L)$ is nilpotent then $L$ is nilpotent.
:::

:::{.problem title="Humphreys 3.6"}
Prove that the sum of two nilpotent ideals of a Lie algebra $L$ is again a nilpotent ideal. Therefore, $L$ possesses a unique maximal nilpotent ideal. Determine this ideal for the nonabelian 2-dimensional algebra $\FF x + \FF y$ where $[xy]=x$, and the 3-dimensional algebra $\FF x + \FF y + \FF z$ where

- $[xy] = z$
- $[xz]=y$
- $[yz] = 0$

:::

:::{.solution}
To see that sums of nilpotent ideals are nilpotent, suppose $I^N = J^M = 0$ are nilpotent ideals.
Then $(I+J)^{M+N} \subseteq I^M + J^N$ by collecting terms and using the absorbing property of ideals.
One can now construct a maximal nilpotent ideal in $L$ by defining $M$ as the sum of all nilpotent ideals in $L$.
That this is unique is clear, since $M$ is nilpotent, so if $M'$ is another maximal nilpotent ideal then $M \subseteq M'$ and $M' \subseteq M$.

Consider the 2-dimensional algebra $L \da \FF x + \FF y$ where $[xy] = x$ and let $I$ be the maximal nilpotent ideal.
Note that $L$ is not nilpotent since $L^k = \FF x$ for all $k\geq 0$, since $L^1 = \FF x$ and $[L, \FF x] = \FF x$ (since all brackets are either zero or $\pm x$).
However, this also shows that the subalgebra $\FF x$ is an ideal, and is in fact a nilpotent ideal since $[\FF x, \FF x] = 0$.
Although $\FF y$ is a nilpotent subalgebra, it is not an ideal since $[L, \FF y] = \FF x$.
So $I$ is at least 1-dimensional, since it contains $\FF x$, and at most 1-dimensional, since it is not all of $L$, forcing $I = \FF x$.

Consider now the 3-dimensional algebra $L \da \FF x + \FF y + \FF z$ with the multiplication table given in the problem statement above.
Note that $L$ is not nilpotent, since $L^1 = \FF y + \FF z = L^k$ for all $k\geq 2$. This follows from consider $[L, \FF y + \FF z]$, where choosing $x\in L$ is always a valid choice and choosing $y$ or $z$ in the second slot hits all generators; however, no element brackets to $x$.
So similar to the previous algebra, the ideal $J \da \FF x + \FF y$ is an ideal, and it is nilpotent since all brackets between $y$ and $z$ vanish.
By similar dimensional considerations, $J$ must equal the maximal nilpotent ideal.
:::

:::{.problem title="Humphreys 3.10"}
Let $L$ be a Lie algebra, $K$ an ideal of $L$ such that $L / K$ is nilpotent and such that $\ro{\ad_x}{K}$ is nilpotent for all $x \in L$. Prove that $L$ is nilpotent.
:::

:::{.solution}
Suppose that $M \da L/K$ is nilpotent, so the lower central series terminates and $M^n = 0$ for some $n$.
Then $L^n \subseteq K$ for the same $n$, and the claim is that $L^n$ is nilpotent.
This follows from applying Engel's theorem: let $x\in L^n \subseteq K$, then $\ro{\ad_x}{L^n} = 0$ by assumption.
So every element of $L^n$ is ad-nilpotent, making it nilpotent.
Since $0 = (L^n)^k = L^{n+k}$ for some $k$, this forces $L$ to be nilpotent as well.
:::

## Section 4

:::{.problem title="Humphreys 4.1"}
Let $L= \liesl(V)$. Use Lie's Theorem to prove that $\operatorname{Rad} L=Z(L)$; conclude that $L$ is semisimple.

> Hint: observe that $\operatorname{Rad} L$ lies in each maximal solvable subalgebra $B$ of $L$. Select a basis of $V$ so that $B=L \cap \mathfrak{t}(n, \mathrm{~F})$, and notice that $B^t$ is also a maximal solvable subalgebra of $L$. Conclude that $\operatorname{Rad} L \subset L \cap \mathfrak{d}(n, \mathrm{~F})$ (diagonal matrices), then that $\operatorname{Rad} L=Z(L) .]$

:::

:::{.solution}
Let $R = \Rad(L)$ be the radical (maximal solvable ideal) of $L$.
Using the hint, if $S \leq L$ is a maximal solvable subalgebra then it must contain $R$.
By (a corollary of) Lie's theorem, $S$ stabilizes a flag and thus there is a basis with respect to which all elements of $S$ (and thus $R$) are upper triangular.
Thus $S \subseteq \lieb$; however, taking the transpose of every element in $S$ again yields a maximal solvable ideal which is lower triangular and thus contained in $\lieb^-$.
Thus $R \subseteq S \subseteq \lieb \intersect \lieb^- = \lieh$, which consists of just diagonal matrices.

We have $Z(L) \subseteq R$ since centers are solvable, and the claim is that $R \subseteq \lieh \implies R \subseteq Z(L)$.
It suffices to show that $R$ consists of scalar matrices, since it is well-known that $Z(\liegl_n(\FF))$ consists of precisely scalar matrices, and this contains $Z(L)$ since $L \leq \liegl_n(\FF)$ is a subalgebra.
This follows by letting $\ell = \sum a_i e_{i,i}$ be an element of $\Rad(L)$ and considering bracketing elements of $\liesl_n(\FF)$ against it.
Bracketing elementary matrices $e_{i, j}$ with $i\neq j$ yields
\[
[e_{i,j}, \ell] = a_j e_{i, j} - a_i e_{i, j}
,\]
which must be an element of $\Rad(L)$ and thus diagonal, which forces $a_j = a_i$ for all $i, j$.

To conclude that $L$ is semisimple, note that a scalar traceless matrix is necessarily zero, and so $Z(\liesl(V)) = 0$.
This suffices since $\Rad(L) = 0 \iff L$ is semisimple.
:::

:::{.problem title="Humphreys 4.3, Failure of Lie's theorem in positive characteristic"}
Consider the $p \times p$ matrices:
\[
x=\left[\begin{array}{cccccc}
0 & 1 & 0 & . & . & 0 \\
0 & 0 & 1 & 0 & . & 0 \\
. & . & \cdot & \cdot & \cdot & \cdot \\
0 & \cdot & \cdot & \cdot & \cdot & 1 \\
1 & . & \cdot & \cdot & \cdots & 0
\end{array}\right]
,\qquad y = \diag(0,1,2,3,\cdots,p-1)
.\]

Check that $[x, y]=x$, hence that $x$ and $y$ span a two dimensional solvable subalgebra $L$ of $\mathfrak{g l}(p, F)$. Verify that $x, y$ have no common eigenvector.
:::

:::{.solution}
Note that $x$ acts on the left on matrices $y$ by cycling all rows of $y$ up by one position, and similar yacts on the right by cycling the columns to the right.
Thus 
\[
xy - yx &= 
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & \vdots & \vdots & \ddots & 0 \\
0 & 0 & \cdots & 0 & p-1 \\
0 & 0 & \cdots & 0 & 0 
\end{bmatrix}
-
\begin{bmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
\vdots & \cdots & 0 & 2 & 0 \\
0 & 0 & \cdots & 0 & 3 \\
p-1 & 0 & \cdots & 0 & 0
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
-(p-1) & 0 & 0 & 0 & 0
\end{bmatrix}\\
&\equiv
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix} \qquad \in \GL_n(\FF_p) \\
&= x
.\]
Thus $L \da \FF x + \FF y$ span a solvable subalgebra, since $L^{(1)} = \FF x$ and so $L^{(2)} = 0$.

Moreover, note that every basis vector $e_i$ is an eigenvector for $y$ since $y(e_i) = i e_i$, while no basis vector is an eigenvector for $x$ since $x(e_i) = e_{i+1}$ for $1\leq i \leq p-1$ and $x(e_p) = e_1$, so $x$ cycles the various basis vectors.
:::

:::{.problem title="Humphreys 4.4"}
For arbitrary $p$, construct a counterexample to Corollary $\mathrm{C}$[^cor_c]
as follows: Start with $L \subset \liegl_p(\FF)$ as in Exercise 3. Form the vector space direct sum $M=L \oplus \FF^p$, and make $M$ a Lie algebra by decreeing that $\FF^p$ is abelian, while $L$ has its usual product and acts on $\FF^p$ in the given way. 

Verify that $M$ is solvable, but that its derived algebra $\left(=\FF x+ \FF^p\right)$ fails to be nilpotent.

[^cor_c]: 
Corollary C states that if $L$ is solvable then every $x\in L^{(1)}$ is ad-nilpotent, and thus $L^{(1)}$ is nilpotent.

:::

:::{.solution}
For pairs $A_1 \oplus v_1$ and $A_2 \oplus v_2$ in $M$, we'll interpret the given definition of the bracket as
\[
[A_1 \oplus v_1, A_2 \oplus v_2] \da [A_1, A_2] \oplus (A_1(v_2) - A_2(v_1))
,\]
where $A_i(v_j)$ denotes evaluating an endomorphism $A\in \liegl_p(\FF)$ on a vector $v\in \FF^p$.
We also define $L = \FF x + \FF y$ with $x$ and $y$ the given matrices in the previous problem, and note that $L$ is solvable with derived series 
\[
L = \FF x \oplus \FF y \contains L^{(1)} = \FF x \contains L^{(2)} = 0
.\]

Consider the derived series of $M$  -- by inspecting the above definition, we have 
\[
M^{(1)} \subseteq L^{(1)} \oplus \FF^p = \FF x \oplus \FF^p
.\]
Moreover, we have
\[
M^{(2)} \subseteq L^{(2)} \oplus \FF^p = 0 \oplus \FF^p
,\]
which follows from considering considering bracketing two elements in $M^{(1)}$: set $w_{ij} \da A_i(v_j) - A_j(v_i)$, then
\[
[ [A_1, A_2] \oplus w_{1,2}, \,\, [A_3, A_4] \oplus w_{3, 4} ] \hspace{16em}
\\
= [ [A_1, A_2], [A_3, A_4] ] \oplus [A_1, A_2](w_{3, 4}) - [A_3, A_4](w_{1, 2})
.\]
We can then see that $M^{(3)} = 0$, since for any $w_i \in \FF^p$,
\[
[0 \oplus w_1, \, 0 \oplus w_2] = 0 \oplus 0(w_2)-0(w_1) = 0 \oplus 0
,\]
and so $M$ is solvable.

Now consider its derived subalgebra $M^{(1)} = \FF x \oplus \FF^p$.
If this were nilpotent, every element would be ad-nilpotent, but let $v = \tv{1,1,\cdots, 1}$ and consider $\ad_{x \oplus 0}$. We have
\[
\ad_{x \oplus 0}(0 \oplus v) = [x \oplus 0, 0 \oplus v] = 0 \oplus xv = 0 \oplus v
,\]
where we've used that $x$ acts on the left on vectors by cycling the entries.
Thus $\ad_{x \oplus 0}^n (0 \oplus v) = 0 \oplus v$ for all $n\geq 1$ and $x \oplus 0 \in M^{(1)}$ is not ad-nilpotent.
:::