# The Unapologetic Mathematician

## A Trace Criterion for Nilpotence

We’re going to need another way of identifying nilpotent endomorphisms. Let $A\subseteq B\subseteq\mathfrak{gl}(V)$ be two subspaces of endomorphisms on a finite-dimensional space $V$, and let $M$ be the collection of $x\in\mathfrak{gl}(V)$ such that $\mathrm{ad}(x)$ sends $B$ into $A$. If $x\in M$ satisfies $\mathrm{Tr}(xy)=0$ for all $y\in M$ then $x$ is nilpotent.

The first thing we do is take the Jordan-Chevalley decomposition of $x$ $x=s+n$ — and fix a basis that diagonalizes $x$ with eigenvalues $a_i$. We define $E$ to be the $\mathbb{Q}$-subspace of $\mathbb{F}$ spanned by the eigenvalues. If we can prove that this space is trivial, then all the eigenvalues of $s$ must be zero, and thus $s$ itself must be zero.

We proceed by showing that any linear functional $f:E\to\mathbb{Q}$ must be zero. Taking one, we define $y\in\mathfrak{gl}(V)$ to be the endomorphism whose matrix with respect to our fixed basis is diagonal: $f(a_i)\delta_{ij}$. If $\{e_{ij}\}$ is the corresponding basis of $\mathfrak{gl}(V)$ we can calculate that \displaystyle\begin{aligned}\left[\mathrm{ad}(s)\right](e_{ij})&=(a_i-a_j)e_{ij}\\\left[\mathrm{ad}(y)\right](e_{ij})&=(f(a_i)-f(a_j))e_{ij}\end{aligned}

Now we can find some polynomial $r(T)$ such that $r(a_i-a_j)=f(a_i)-f(a_j)$; there is no ambiguity here since if $a_i-a_j=a_k-a_l$ then the linearity of $f$ implies that \displaystyle\begin{aligned}f(a_i)-f(a_j)&=f(a_i-a_j)\\&=f(a_k-a_l)\\&=f(a_k)-f(a_l)\end{aligned}

Further, picking $i=j$ we can see that $r(0)=0$, so $r$ has no constant term. It should be apparent that $\mathrm{ad}(y)=r\left(\mathrm{ad}(s)\right)$.

Now, we know that $\mathrm{ad}(s)$ is the semisimple part of $\mathrm{ad}(x)$, so the Jordan-Chevalley decomposition lets us write it as a polynomial in $\mathrm{ad}(x)$ with no constant term. But then we can write $\mathrm{ad}(y)=r\left(p\left(\mathrm{ad}(x)\right)\right)$. Since $\mathrm{ad}(x)$ maps $B$ into $A$, so does $\mathrm{ad}(y)$, and our hypothesis tells us that $\displaystyle\mathrm{Tr}(xy)=\sum\limits_{i=1}^{\dim V}a_if(a_i)=0$

Hitting this with $f$ we find that the sum of the squares of the $f(a_i)$ is also zero, but since these are rational numbers they must all be zero.

Thus, as we asserted, the only possible $\mathbb{Q}$-linear functional on $E$ is zero, meaning that $E$ is trivial, all the eigenvalues of $s$ are zero, and $x$ is nipotent, as asserted.

August 31, 2012 - Posted by | Algebra, Lie Algebras, Linear Algebra

## 1 Comment »

1. […] tells us that if each is ad-nilpotent, then is itself nilpotent. We can now combine this with our trace criterion to get a convenient way of identifying solvable Lie […]

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