# Cotangent bundle of coadjoint orbit is stein manifold?

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 $O_\lambda$.

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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.

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 $$SL(2)$$.

• 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. Apr 8, 2014 at 18:23