Posted by: Steven Sam | February 15, 2010

Nilpotent orbits in classical Lie algebras

One of the most important things taught in linear algebra is Jordan normal form. When we work over an algebraically closed field, every ${n \times n}$ matrix can be expressed as a diagonal matrix, possibly with some of the superdiagonal entries containing a 1, in some basis. So, for instance, if one wants to classify nilpotent matrices up to conjugation, the diagonal entries will be 0, and we only need to specify where the 1s go. Thus, the conjugacy classes are parametrized by partitions of ${n}$ in a natural way.

This can be seen as part of a more general situation. Start with a semisimple complex Lie group ${G}$ and let ${\mathfrak{g}}$ be its Lie algebra. The adjoint representation ${\mathrm{ad} \colon \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g})}$ given by ${x \mapsto (y \mapsto [x,y])}$ is a faithful representation and hence we can define an element ${x}$ to be nilpotent if ${\mathrm{ad}(x)}$ is a nilpotent matrix. This is equivalent to asking that ${\rho(x)}$ is nilpotent for all finite dimensional representations ${\rho \colon \mathfrak{g} \rightarrow \mathfrak{gl}(V)}$. The group ${G}$ acts on ${\mathfrak{g}}$ by conjugation (more precisely, ${G}$ acts on itself by conjugation, and hence acts on ${\mathfrak{g}}$ by taking the derivative of this action), and this action preserves the notion of nilpotency. So we can ask what the orbits look like.

The situation we first mentioned is when ${G}$ is the general linear group (which isn’t semisimple, but the story with the special linear group is identical). So we know what the nilpotent orbits look like in this case. What about for other Lie algebras? For this post, we’ll look at the case of the Lie algebras ${\mathfrak{so}_n}$ and ${\mathfrak{sp}_{2m}}$, where the conjugacy classes are parametrized by special kinds of partitions.

So let’s start with generalities. For us, ${V}$ will be a complex vector space with a nondegenerate form ${\omega}$ which is either skew-symmetric or symmetric. We’ll let ${\epsilon}$ stand for ${-1}$ in the first case, and for 1 in the second case. The dimension of ${V}$ is either ${2m}$ or ${2m+1}$. We’ll take ${G}$ to be the subgroup of ${{\bf GL}(V)}$ that preserves this form, and ${\mathfrak{g}}$ is the Lie algebra of ${G}$. So we’re dealing with either an orthogonal group or a symplectic group. Later we will replace the orthogonal group with the special orthogonal group. For this post, I am mostly following Section 5.1 in Collingwood and McGovern, Nilpotent Orbits in Semisimple Lie Algebras, but I am trying to add in more details.

For our work, we’ll only need to know a few facts about ${\mathfrak{sl}_2}$-representations. The main result we need is the following.

Theorem (Jacobson–Morozov). Let ${\mathfrak{g}}$ be a semisimple Lie algebra, and let ${X \in \mathfrak{g}}$ be a nilpotent element. Then there exists ${Y,H \in \mathfrak{g}}$ such that the subalgebra generated by ${\{Y,H,X\}}$ is isomorphic to ${\mathfrak{sl}_2}$.

With this we denote this copy of ${\mathfrak{sl}_2}$ by ${\mathfrak{a}(X)}$, and we can view ${V}$ as a representation of ${\mathfrak{a}(X)}$. Finite-dimensional representations of ${\mathfrak{sl}_2}$ may be written as a direct sum of irreducible representations, and the irreducible representations are classified by their dimension. Any two conjugate nilpotent matrices give rise to isomorphic representations, so let’s examine what kind of dimensions show up. Then we will show that the irreducible decomposition determines the orbit.

For each nonnegative integer ${r}$, we’ll let ${V(r)}$ denote the subspace of ${V}$ consisting of highest weight vectors with weight ${r}$, i.e., ${V(r) = \{v \in V \mid Hv = rv,\ Xv = 0\}}$, and ${V^-(r)}$ denote the subspace of ${V}$ consisting of lowest weight vectors with weight ${-r}$, i.e., ${V^-(r) = \{v \in V \mid Hv = -rv,\ Yv = 0\}}$. We define a form on ${V(r)}$ via ${(v,w)_r = \omega(v, Y^rw)}$. Note that the action of ${\mathfrak{g}}$ satisfies ${\omega(Av, w) = -\omega(v,Aw)}$ for all ${v,w \in V}$ and ${A \in \mathfrak{g}}$, so

$\displaystyle (v,w)_r = \omega(v, Y^rw) = (-1)^r \omega(Y^rv, w) = (-1)^{r + \epsilon} (w,v)_r,$

and we see that ${(-,-)_r}$ is skew-symmetric if ${r+\epsilon}$ is odd and symmetric if ${r+\epsilon}$ is even.

We claim that ${(-,-)_r}$ is nondegenerate. Pick ${v \in V(r)}$, and pick ${w}$ such that ${\omega(v,w) \ne 0}$. We may assume without loss of generality that ${w}$ is a weight vector of weight ${s}$. We claim that ${w \in V^-(r)}$. First, ${r\omega(v,w) = \omega(Hv,w) = -\omega(v,Hw) = -s\omega(v,w)}$, so ${s = -r}$. Now ${0 = \omega(Xv, Yw) = \omega(v, XYw)}$. However, ${XYw}$ is a scalar multiple of ${w}$, so it must be 0, which gives the claim.

In particular, we have just concluded that ${\dim V(r)}$ has to be even if ${r+\epsilon}$ is odd since ${V(r)}$ has a symplectic structure. Since an irreducible representation weight highest weight ${r}$ has dimension ${r+1}$, we know that

$\displaystyle n = \sum_{r \ge 0} (r+1) \dim V(r).$

This gives a partition of ${n}$, where ${r+1}$ appears with multiplicity ${\dim V(r)}$. In the orthogonal case, ${r+1}$ always appears with even multiplicity when ${r+1}$ is even, and in the symplectic case, ${r+1}$ always appears with even multiplicity when ${r+1}$ is odd. This leads us to the main point.

Theorem (Gerstenhaber). The nilpotent conjugacy classes of ${{\bf O}(n)}$ on ${\mathfrak{so}_n}$ are indexed by partitions of ${n}$ for which each even integer appears with even multiplicity. Similarly, the nilpotent conjugacy classes of ${{\bf Sp}(2m)}$ are indexed by partitions of ${2m}$ for which each odd integer appears with even multiplicity.

The thing left to do is first prove that all such partitions arise from ${\mathfrak{a}(X)}$-representations, and that it uniquely determines the conjugacy class.

First we handle existence. Let ${\lambda}$ be a partition such that all even parts appear with even multiplicity. We’ll just work backwards through the first part of the theorem. Namely, for each integer ${r}$, let ${m_r}$ be its multiplicity in ${\lambda}$. Let ${W(r)}$ be the direct sum of ${m_{r+1}}$ copies of the irreducible representation of ${\mathfrak{sl}_2}$ of highest weight ${r}$, so ${\dim W(r) = m_{r+1}(r+1)}$, and let ${V(r)}$ be the space of highest weight vectors. Equip ${V(r)}$ with a nondegenerate symmetric (resp. skew-symmetric) form ${(-,-)_r}$ if ${r}$ is odd (resp. even). We are going to use ${(-,-)_r}$ to define a nondegenerate symmetric bilinear form ${\omega}$ on ${W(r)}$. By prior calculations, we know that for weight vectors ${v}$ and ${w}$, we must have ${\omega(v,w) = 0}$ unless their weights add to 0. So first we set ${\omega(v,Y^rw) = (v,w)_r}$. Then the rest is uniquely determined since one has

$\displaystyle \omega(Y^iv, Y^{r-i}w) = -\omega(Y^{i-1}v, Y^{r-i+1}w) = (-1)^i(v,w)_r.$

The form is symmetric since

$\displaystyle \omega(Y^iv, Y^{r-i}w) = (-1)^i(v,w)_r = (-1)^{r-i}(w,v)_r = \omega(Y^{r-i}w, Y^iv)$

Since every vector in ${W(r)}$ is of the form ${Y^iv}$ for some ${v \in V(r)}$, it follows that ${\omega}$ is also nondegenerate.

Now set ${V = \bigoplus_r W(r)}$. Then ${\dim V = n}$ since ${|\lambda| = n}$, and we have a distinguished ${X \in \mathfrak{so}_n}$ for which the direct sum decomposition of ${V}$ into irreducibles consists of the dimensions given by ${\lambda}$. Furthermore, if ${X}$ and ${X'}$ are both nilpotent elements for which ${V}$ has the same dimensions in its direct sum decomposition, there is an element of ${{\bf O}(V)}$ which will take one direct sum decomposition to the other (since ${{\bf O}(V)}$ acts transitively on the hyperbolic bases of ${\omega}$). All of the above works for the symplectic case as well with a few sign changes, so we have proved the theorem.

There is one last thing to address. In the above, we looked at conjugacy classes of the orthogonal group ${{\bf O}(n)}$. What if we restrict to conjugacy classes of the special orthogonal group ${{\bf SO}(n)}$?. When ${n}$ is odd, one has ${{\bf O}(n) = {\bf SO}(n) \times \{\pm I_n\}}$, so the conjugacy classes don’t change. When ${n}$ is even, some of the classes do split up. Given ${X, X' \in \mathfrak{so}_n}$ such that ${gX = X'}$ for ${g \in {\bf SO}(n)}$ with ${\det g = -1}$, we have to ask if when we can replace ${g}$ by ${g'}$ where ${\det g' = 1}$. Let ${\lambda}$ be the corresponding partition. First suppose that ${\lambda}$ contains an odd part ${r}$. Then using the notation from before, ${V(r-1)}$ (with respect to ${X}$) has a nondegenerate symmetric bilinear form ${(-,-)_{r-1}}$. Find an orthonormal basis, and let ${v}$ be in this basis. The ${\mathfrak{a}(X)}$-representation generated by ${v}$ is odd dimensional, and the element ${h \in {\bf O}(n)}$ given by multiplying this subrepresentation by ${-1}$ and leaving its complement in ${V}$ fixed has determinant 1, and we can take ${g' = gh}$.

The last case to consider is when all ${n}$ is even and all parts of ${\lambda}$ are even (and occur with even multiplicity). Pick ${X}$ in this ${{\bf O}(n)}$-conjugacy class. Then each ${V(r)}$ has a symplectic form on it, and hence any ${g \in {\bf O}(n)}$ which preserves ${X}$ must preserve these symplectic forms. Thus the subgroup of all such ${g}$ is a product of symplectic groups, so ${\det g = 1}$. Given any ${h \in {\bf O}(n)}$ with ${\det h = -1}$, we conclude that ${X}$ and ${hX}$ cannot be conjugate under ${{\bf SO}(n)}$. Finally, we have the following theorem.

Theorem (Springer-Steinberg). The ${{\bf SO}(n)}$-conjugacy classes of ${\mathfrak{so}_{n}}$ are naturally indexed by partitions ${\lambda}$ of ${n}$ for which each even integer appears with even multiplicity, except when ${n}$ is divisible by 4 and ${\lambda}$ contains no odd parts: in this case, there are exactly two orbits indexed by this partition.

It is a general fact that in our original situation of a semisimple complex Lie group ${G}$ acting on the set of nilpotent elements of its Lie algebra ${\mathfrak{g}}$ that there will always be finitely many orbits. I don’t really want to go into the proof of this fact, but I’ll use it next time when I’ll discuss a technique due to Vinberg for studying orbits of ${G}$ inside of its various representations. For a preview, I’ll use it to show that the action of ${{\bf GL}(n)}$ on ${\bigwedge^3 {\bf C}^n}$ has finitely many orbits if and only if ${n \le 8}$.

-Steven