Why are the orbits of elements of Lie algebra under adjoint action of the Lie group even dimension manifolds?

Question

I can't find a proof of this statement anywhere - in fact I can't even find a statement of this statement anywhere. My advisor told me this in our last meeting, but I am having trouble verifying it. I know that they are manifolds, but I cannot figure out the even dimension part.

Answer 1

The statement is only true for semisimple Lie groups. I will assume the orbits are manifolds, since you know this already.

Fix some $x$ and consider the orbit map $G \to \mathfrak{g}$ given by $g \mapsto \mathrm{Ad}(g) x$. Its derivative is the map $\mathfrak{g} \to \mathfrak{g}$ given by $y \mapsto \mathrm{ad}(y) x$; this is simply the map $-\mathrm{ad}(x)$ and we only need to know the dimension of its image which is $\mathrm{dim}\, \mathfrak{g} - \mathrm{dim}\, \mathrm{ker}\, \mathrm{ad}(x)$ and $\mathrm{ker} \, \mathrm{ad}(x)$ is just the centralizer of $x$, denoted $Z(x)$.

Consider the following counterexample: Let $\mathfrak{n}$ be the strictly upper triangular $3 \times 3$ matrices and let $x$ be a generic element, so that $x^3=0$ but $x^2$ is not zero. It follows that the vector space of all matrices commuting with $x$ is just given by polynomials in $x$; of these the matrices in $\mathfrak{n}$ are given by the span of $\{x,x^2\}$. Hence $\mathrm{dim} \, Z(x) = 2$, while $\mathrm{dim}\,\mathfrak{n}=3$, so the orbit is $1$-dimensional.

The result does hold for semi-simple Lie algebras though: assume now $\mathfrak{g}$ is semisimple and let $\mathfrak{a}$ be a Cartan subalgebra containing $x$.

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EDIT: as Claudius mentions, this assumption is erroneous because $x$ may not be semisimple. If we do assume $x$ is semisimple we can see why its orbit is even-dimensional, as I explain at the end of the post. If we can do a similar analysis for nilpotent elements, we should be able to prove the result, because every element $x$ admits a Jordan-Chevalley decomposition into unique semisimple and nilpotent parts $x=s+n$ and $Z(x)=Z(s) \cap Z(n)$ since $s$ and $n$ are polynomials in $x$. However computing the dimension of $Z(n)$ seems to be a little more difficult.

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Now, assuming $x$ is semisimple we get the root space decomposition $$ \mathfrak{g} =\mathfrak{g}_0 \oplus \bigoplus_{\alpha} \mathfrak{g}_\alpha $$ and $$ Z(x) = \mathfrak{g}_0 \oplus \bigoplus_{\alpha \, | \, x \in \ker \alpha} \mathfrak{g_\alpha}$$ where the sum is over all roots $\alpha$ such that $\alpha(x)=0$. The point is that if $\alpha(x)=0$, we also have $(-\alpha)(x)=0$. Since every root space is one-dimensional, its clear that the dimension of every centralizer $Z(x)$ has the same parity, including the zero vector whose centralizer is all of $\mathfrak{g}$. Taking a look at the formula above, we see that the dimension of the image is always even.