Let me first define stein manifolds and coadjoint orbits.

A complex manifold $X$ of complex dimension $n$ is called a Stein manifold if the following conditions hold:

$X$ is holomorphically convex, i.e. for every Compact space|compact subset $K \subset X$, the so-called ''holomorphic convex hull'',

$\bar K = \{z \in X: |f(z)| \leq \sup_K |f| \ \forall f \in \mathcal O(X) \},$

is again a ''compact'' subset of $X$. Here $\mathcal O(X)$ denotes the ring of holomorphic functions on $X$.

$X$ is holomorphically separable, i.e. if $x \neq y$ are two points in $X$, then there is a holomorphic function $f \in \mathcal O(X)$.

Now, Let $G$ be a Lie group, and $\mathfrak g$ its Lie algebra. Then $G$ has a natural action on $\mathfrak g^*$ called the coadjoint action, since it is dual to the adjoint action of $G$ on $\mathfrak g$. The orbits of this action are submanifolds of $\mathfrak g^*$ which carry a natural symplectic structure, and are in a certain sense, the minimal symplectic manifolds on which $G$ acts. The orbit through a point $\lambda\in\mathfrak g^*$ is typically denoted $O_\lambda$.


Now my question is If G be a compact Lie group then $O_\lambda$ is complex manifold and when $T^*(O_\lambda)$ (cotangent bundle of coadjoint orbit)is Stein manifold? Note that $T^*(O_\lambda)\cong \frac{G^\mathbb C}{G_\lambda^\mathbb C} $ which $G^\mathbb C$ is complexification of compact lie group $G$ and it is not compact.


1 Answer 1


Not with the obvious complex structure. Notice that $O_{\lambda}$ is a closed subvariety of $T^{\ast}(O_{\lambda})$ (namely the zero section). Closed subvarieties of Stein varieties are Stein. However, positive dimensional Stein varieties are never compact, and $O_{\lambda}$ is compact.

However, there is a sense in which $T^{\ast} O_{\lambda}$ is almost an affine variety, which I will now sketch. Before I get into the details, let me the first example.

When $G=SU(2)$, then $O_{\lambda}$ is $\mathbb{CP}^1$ and $T^{\ast} O_{\lambda}$ is the total space of the line bundle $\mathcal{O}(-2)$. This can be viewed as the blow up of the singularity $xz+y^2=0$, and the global holomorphic functions on $T^{\ast} O_{\lambda}$ are all pulled back from this singular cone. Notice that $xz+y^2=0$ is the same equation as $\left( \begin{smallmatrix} y & x \\ z & -y \end{smallmatrix} \right)^2=0$. This will be relevant later.

Consider the variety $xz+y^2 = 1$, which is also $\left( \begin{smallmatrix} y & x \\ z & -y \end{smallmatrix} \right)^2=\mathrm{Id}$. This is a smooth Stein variety and has a map to $\mathbb{CP}^1$ sending $\left( \begin{smallmatrix} y & x \\ z & -y \end{smallmatrix} \right)$ to its first column. The fibers of this map are affine spaces and, in fact, this is an affine bundle. The corresponding vector bundle is, indeed, $\mathcal{O}(-2)$. Since an affine bundle is diffeomorphic to its corresponding vector bundle, choosing such a diffeomorphism gives a new complex structure on $T^{\ast} O_{\lambda}$ and that structure is Stein.

Okay, now the general case. I'll write $K$ for the compact group, $S$ for a maximal torus. I'll write $G$ for the complexification of $K$, $T$ for a complexification of $S$ within $G$, $B$ for a Borel containing $T$ and $N$ for the unipotent radical of $B$. I'll use corresponding Fraktur letters for the Lie algebras. To make life simple, I'll assume that your orbit goes through a regular element of $\mathfrak{k}$ (one whose stabilizer for the adjoint action is just a torus). Otherwise, I'd also need to introduce a parabolic $P$.

As I imagine you know, $O_{\lambda} \cong K/S \cong G/B$. The tangent space to the coset $B$ in $G/B$ is $\mathfrak{g}/\mathfrak{b}$. As explained above, $T^{\ast}(G/B)$ is not Stein.

However, $G/T$ is Stein. In general, quotients of linear algebraic groups by reductive subgroups (such as $T$) exist and are affine; I'll also give a direct embedding of $G/T$ into $\mathfrak{g}$ below. The map $G/T \longrightarrow G/B$ is an affine bundle, and the corresponding vector bundle is $T^{\ast}(G/B)$. So we can make $T^{\ast}(G/B)$ into a Stein space by using a diffeomorphism between an affine bundle and the corresponding vector bundle to give a new complex structure.

There is a beautiful concrete way to realize these spaces. The tangent space to $G/B$ at the coset $B$ is $\mathfrak{g}/\mathfrak{b}$. Use the Killing form to identify $\mathfrak{g}$ with its dual; then the cotangent space at the coset $B$ is $\mathfrak{b}^{\perp} = \mathfrak{n}$. I'll write $\phi$ for this isomorphism $T^{\ast}_{B} (G/B) \cong \mathfrak{n}$. Let $g \in G$ and let $v$ be a cotangent vector to $G/B$ at the coset $gB$. Define an element of $\mathfrak{g}$ by $Ad(g) \cdot \phi(g^{\ast} v)$. (We are using the action of $g$ on $G/B$ to pull back from $T^{\ast}_{gB}$ to $T^{\ast}_B$.) One can check that this construction is unaltered by replacing $g$ by $gb$ for $b \in B$, so this gives a map $T^{\ast} (G/B) \to \mathfrak{g}$.

The image of this map is $\mathcal{N} := \bigcup_{g \in G} Ad(g) \cdot \mathfrak{n}$. This space is known as the nilpotent cone and is a (singular) closed subvariety of $\mathfrak{g}$; explicitly, an element $x$ of $\mathfrak{g}$ is in $\mathcal{N}$ if the coefficients of the characteristic polynomial of $Ad(x)$ are $0$. The map from $T^{\ast}(G/B)$ to $\mathcal{N}$ is called the Springer resolution. There is a good discussion of this in chapters 3 and 4 of Chriss and Ginzburg's Representation Theory and Complex Geometry.

Take a regular element $t$ in $\mathfrak{t}$. Then $T$ is the stabilizer of $t$, so $G/T$ embeds in $\mathfrak{g}$ as the $G$ orbit through $t$. This is a closed embedding: an element $x$ of $\mathfrak{g}$ is in this orbit if and only if $Ad(x)$ and $Ad(t)$ have the same characteristic polynomial. So this gives an explicit embedding of $G/T$ into $\mathfrak{g}$ and thus gives a second proof that $G/T$ is Stein.

In summary: We can explicitly embed $G/T$ into $\mathfrak{g}$. The space $T^{\ast}(G/B)$ has an analogous map to $\mathfrak{g}$ which gives a resolution of the nilpotent cone.

The above example is this theory worked out for $SL(2)$.

  • $\begingroup$ Cher David, thanks a lot for your nice answer. $\endgroup$
    – user21574
    Apr 4, 2014 at 9:40
  • $\begingroup$ You said I'll write G for the complexification of G,. it must be edited as I'll write G for the complexification of K, $\endgroup$
    – user21574
    Apr 4, 2014 at 10:32
  • 2
    $\begingroup$ To spell out the connection between the spaces more tightly: consider the map $(G \times \mathfrak{b})/B \to \mathfrak g$, $[g,X] \mapsto g\cdot X$, where $B$ acts on the right of $G$ and by conjugation on $\mathfrak b$. Then map further to $\mathfrak g/G = \mathfrak t/W$. The fibers of this flat map are $T^* G/B$ over $0$ and $G/T$ over a general point. $\endgroup$ Apr 8, 2014 at 18:23

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