The Unapologetic Mathematician

Mathematics for the interested outsider

Uses of the Jordan-Chevalley Decomposition

Now that we’ve given the proof, we want to mention a few uses of the Jordan-Chevalley decomposition.

First, we let A be any finite-dimensional \mathbb{F}-algebra — associative, Lie, whatever — and remember that \mathrm{End}_\mathbb{F}(A) contains the Lie algebra of derivations \mathrm{Der}(A). I say that if \delta\in\mathrm{Der}(A) then so are its semisimple part \sigma and its nilpotent part \nu; it’s enough to show that \sigma is.

Just like we decomposed V in the proof of the Jordan-Chevalley decomposition, we can break A down into the eigenspaces of \delta — or, equivalently, of \sigma. But this time we will index them by the eigenvalue: A_a consists of those x\in A such that \left[\delta-aI\right]^k(x)=0 for sufficiently large k.

Now we have the identity:

\displaystyle\left[\delta-(a+b)I\right]^n(xy)=\sum\limits_{i=0}^n\binom{n}{i}\left[\delta-aI\right]^{n-i}(x)\left[\delta-bI\right]^i(y)

which is easily verified. If a sufficiently large power of \delta-aI applied to x and a sufficiently large power of \delta-bI applied to y are both zero, then for sufficiently large n one or the other factor in each term will be zero, and so the entire sum is zero. Thus we verify that A_aA_b\subseteq A_{a+b}.

If we take x\in A_a and y\in A_b then xy\in A_{a+b}, and thus \sigma(xy)=(a+b)xy. On the other hand,

\displaystyle\begin{aligned}\sigma(x)y+x\sigma(y)&=axy+bxy\\&=(a+b)xy\end{aligned}

And thus \sigma satisfies the derivation property

\displaystyle\sigma(xy)=\sigma(x)y+x\sigma(y)

so \sigma and \nu are both in \mathrm{Der}(A).

For the other side we note that, just as the adjoint of a nilpotent endomorphism is nilpotent, the adjoint of a semisimple endomorphism is semisimple. Indeed, if \{v_i\}_{i=0}^n is a basis of V such that the matrix of x is diagonal with eigenvalues \{a_i\}, then we let e_{ij} be the standard basis element of \mathfrak{gl}(n,\mathbb{F}), which is isomorphic to \mathfrak{gl}(V) using the basis \{v_i\}. It’s a straightforward calculation to verify that

\displaystyle\left[\mathrm{ad}(x)\right](e_{ij})=(a_i-a_j)e_{ij}

and thus \mathrm{ad}(x) is diagonal with respect to this basis.

So now if x=x_s+x_n is the Jordan-Chevalley decomposition of x, then \mathrm{ad}(x_s) is semisimple and \mathrm{ad}(x_n) is nilpotent. They commute, since

\displaystyle\begin{aligned}\left[\mathrm{ad}(x_s),\mathrm{ad}(x_n)\right]&=\mathrm{ad}\left([x_s,x_n]\right)\\&=\mathrm{ad}(0)=0\end{aligned}

Since \mathrm{ad}(x)=\mathrm{ad}(x_s)+\mathrm{ad}(x_n) is the decomposition of \mathrm{ad}(x) into a semisimple and a nilpotent part which commute with each other, it is the Jordan-Chevalley decomposition of \mathrm{ad}(x).

August 30, 2012 Posted by | Algebra, Lie Algebras, Linear Algebra | 3 Comments

   

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