The convexity package for Hamiltonian actions on conformal symplectic manifolds

Abstract

Consider a Hamiltonian action of a compact connected Lie group on a conformal symplectic manifold. We prove a convexity theorem for the moment map under the assumption that the action is of Lee type, which establishes an analog of Kirwan’s convexity theorem in conformal symplectic geometry.

Appendices

Appendix A. An instance of real symplectic convexity

The next theorem, which is a special case of a result from [34], is used in Sect. 7 (contact hyperboloids). Let K be a compact connected Lie group with a choice of maximal torus T and a closed chamber C in $\mathfrak {t}^*$. The result concerns certain real algebraic subvarieties of a finite-dimensional unitary K-module V. The imaginary part of the Hermitian inner product $\langle {\cdot },{\cdot }\rangle _V$ on V is a symplectic form $\omega _V$, and a moment map for the K-action on V is given by

$$\begin{aligned} \Phi _V^\xi (v)=\frac{1}{2}\omega _V(\xi (v),v)=\frac{i}{2}\langle v,\xi (v)\rangle _V \end{aligned}$$

for $\xi \in \mathfrak {k}$ and $v\in V$. Let $W=V^\mathbb {C}=\mathbb {C}\otimes V$ be the complexified K-module, let J be the complex structure of V, and let $W=V^{1,0}\oplus V^{0,1}$ be the usual eigenspace decomposition of W respect to J. We have $V^{0,1}=\overline{V^{1,0}}$, where the conjugation is defined with respect to the totally real subspace V of W. The map $f(v)=\frac{1}{2}(v-iJv)$ is a complex linear isomorphism $V\cong V^{1,0}$. There is a unique K-module structure on $V^{1,0}$ that makes f a K-module isomorphism. Likewise, the map ${\bar{f}}(v)=\frac{1}{2}(v+iJv)$ is a complex antilinear isomorphism $V\cong V^{0,1}$, and there is a unique K-module structure on $V^{0,1}$ that makes ${\bar{f}}$ an anti-isomorphism of K-modules. Thus $W=V^{1,0}\oplus V^{0,1}$ becomes a complex $K\times K$-module isomorphic to $V\oplus V^*$. We will write elements of W as pairs $\bigl (f(v_1),{\bar{f}}(v_2)\bigr )$ with $v_1$, $v_2\in V$. In this notation complex conjugation is written as

$$\begin{aligned} \bigl (f(v_1),{\bar{f}}(v_2)\bigr )\longmapsto \bigl (f(v_2),{\bar{f}}(v_1)\bigr ). \end{aligned}$$

The real inner product ${\text {Re}}\langle {\cdot },{\cdot }\rangle _V$ on V extends to a Hermitian inner product $\langle {\cdot },{\cdot }\rangle _W$ on W, and the symplectic form $\omega _W={\text {Im}}\langle {\cdot },{\cdot }\rangle _W$ satisfies $f^*\omega _W=\frac{1}{2}\omega _V$ and ${\bar{f}}^*\omega _W=-\frac{1}{2}\omega _V$. Therefore the moment map for the $K\times K$-action on W is given by

$$\begin{aligned} \Phi _W^{(\xi _1,\xi _2)}\bigl (f(v_1),{\bar{f}}(v_2)\bigr )= \frac{1}{2}\bigl (\Phi _V^{\xi _1}(v_1)-\Phi _V^{\xi _2}(v_2)\bigr ). \end{aligned}$$

(7.2)

Theorem A.1

Let V be a unitary K-module. Let X be a K-invariant irreducible real algebraic subvariety of V. Let $X^\mathbb {C}\subseteq V^\mathbb {C}$ be the complexification of X. Suppose that $X^\mathbb {C}$ is $K\times K$-invariant and that X contains at least one smooth point of $X^\mathbb {C}$. Then the moment body $\Delta (X)=\Phi _V(X)\cap C$ of X is the convex hull of all dominant weights $\lambda \in C$ such that the $K\times K$-module with highest weight $(\lambda ,-\lambda )$ occurs in the coordinate ring of $X^\mathbb {C}$. In particular $\Delta (X)$ is a rational convex polyhedral cone in $\mathfrak {k}^*$.

Proof

The strategy is to interpret $\mathfrak {k}$ as the tangent space to the symmetric space $(K\times K)/K$ and X as a totally real Lagrangian in $X^\mathbb {C}$. Let $U=K\times K$ and $W=V^\mathbb {C}$. Define involutions $\sigma =\sigma _U$ of U and $\sigma =\sigma _W$ of W by $\sigma _U(k_1,k_2)=(k_2,k_1)$ and $\sigma _W(w)={\bar{w}}$. Then $\sigma _W$ is antilinear and antisymplectic, and $X=(X^\mathbb {C})^{\sigma _W}$, the set of $\sigma _W$-fixed points of X. Moreover,

$$\begin{aligned} \sigma _W\bigl ((k_1,k_2)\cdot (f(v_1),{\bar{f}}(v_2))\bigr )&= \sigma _W\bigl (f(k_1v_1),{\bar{f}}(k_2v_2)\bigr )\\&=\bigl (f(k_2v_2),{\bar{f}}(k_1v_1)\bigr )\\&=\bigl (k_2,k_1)\cdot (f(v_2),{\bar{f}}(v_1)\bigr ), \end{aligned}$$

which shows that $\sigma _W(u\cdot w)=\sigma _U(u)\cdot \sigma _W(w)$ for all $u\in U$ and $w\in W$. It follows that $\Phi _W(\sigma _W(w))=-\sigma _U^*\bigl (\Phi _W(w)\bigr )$ for $w\in W$, where $\sigma _U^*$ is the transpose of the involution $\sigma _{U,*}$ of $\mathfrak {u}=\mathrm{Lie}(U)$ induced by $\sigma _U$. Let $\mathfrak {p}^*=\{\,\mu \in \mathfrak {u}^*\mid \sigma _U^*(\mu )=-\mu \,\}$. Then

$$\begin{aligned} \mathfrak {p}^*= \{\,(\lambda ,-\lambda )\mid \lambda \in \mathfrak {k}^*\,\}\cong \mathfrak {k}^*, \end{aligned}$$

and the image $\Phi _W(X)$ is a subset of $\mathfrak {p}^*$. It follows from (7.2) that

$$\begin{aligned} \Phi _W^{(\xi ,-\xi )}\bigl (f(v),{\bar{f}}(v)\bigr )= \frac{1}{2}\bigl (\Phi _V^{\xi }(v)-\Phi _V^{-\xi }(v)\bigr )=\Phi _V^{\xi }(v), \end{aligned}$$

so $\Phi _V=\Phi _W|_V$ modulo the identification $\mathfrak {p}^*\cong \mathfrak {k}^*$. Hence $\Phi _V(X)=\Phi _W(X)$. The theorem now follows from [34, Theorem 6.3]. $\square$

Example A.2

Let $V=T^*\mathbb {R}^n=\mathbb {R}^{2n}$ equipped with the standard symplectic structure $\omega$ and complex structure $J=\bigl ({\begin{matrix}0&{}-I\\ I&{}0\end{matrix}}\bigr )$. Then V is a unitary K-module for any closed subgroup K of ${{\mathbf{O}}}(n,\mathbb {R})$. Let $\zeta ={{\,\mathrm{diag}\,}}(\zeta _1,\zeta _2,\dots ,\zeta _n)$, where $\zeta _1\ge \zeta _2\ge \cdots \ge \zeta _n>0$ and suppose that K centralizes $\zeta$. Then the action of K preserves the hyperboloid

$$\begin{aligned} X=\{\,(x,y)\in V\mid x^\top \zeta y=1\,\}. \end{aligned}$$

To complexify the hyperboloid we substitute new variables $x=\frac{1}{2}(z+{\bar{z}})$ and $y=\frac{1}{2i}(z-{\bar{z}})$ and obtain

$$\begin{aligned} X^\mathbb {C}=\{\,(z,{\bar{z}})\in V^{1,0}\oplus V^{0,1}\mid z^\top \zeta z-{\bar{z}}^\top \zeta {\bar{z}}=4i\,\}. \end{aligned}$$

Evidently $X^\mathbb {C}$ is unchanged by substitutions of the form $z\mapsto k_1z$, ${\bar{z}}\mapsto k_2{\bar{z}}$ with $k_1$, $k_2\in K$. The variables z are coordinates on $V^{1,0}$ and the variables ${\bar{z}}$ are coordinates on $V^{0,1}$, so we see that $X^\mathbb {C}$ is preserved by the action of $K\times K$. It now follows from Theorem A.1 that the moment body $\Delta (X)$ of X with respect to the K-action is a rational convex polyhedral cone.

Example A.3

Let $V=\mathbb {C}^n$ equipped with the standard unitary structure. Then V is a unitary K-module for any closed subgroup K of ${\mathbf{U}}(n)$. Let $\zeta$ be as in Example A.2 and suppose that K centralizes $\zeta$. Then K acts on the ellipsoid

$$\begin{aligned} X=\{\,z\in V\mid z^*\zeta z=1\,\}. \end{aligned}$$

The complexified ellipsoid is

$$\begin{aligned} X^\mathbb {C}=\{\,(z,{\bar{z}})\in V^{1,0}\oplus V^{0,1}\mid {\bar{z}}^\top \zeta z=1\,\}, \end{aligned}$$

which is not preserved by the action of $K\times K$, so Theorem A.1 does not apply. Obviously the moment body $\Delta (X)$ cannot be a cone because X is compact. However, the contact manifold X is the level set of the $\zeta$-component of the ${\mathbf{U}}(n)$-action on $\mathbb {C}^n$, so it follows from the contact convexity theorem [30, Theorem 4.5.1] that $\Delta (X)$ is a (not necessarily rational) convex polytope.

Appendix B. Conformal presymplectic convexity

The conformal symplectic convexity theorem, Theorem 1.1, extends without difficulty to conformal presymplectic manifolds, where the 2-form is of constant rank less than the dimension. The purpose of this appendix is to briefly explain this generalization, Theorem B.1 below, which uses the techniques of [30].

The kernel of a conformal presymplectic form determines a foliation. If the leaf space of this foliation is a manifold, the conformal presymplectic structure descends to a conformal symplectic structure on the leaf space, and Theorem B.1 is just a restatement of Theorem 1.1. If the leaf space is not a manifold, Theorem B.1 can be viewed as a generalization of Theorem 1.1 to conformal symplectic stacks. We will not develop this point of view here, but refer to [18, Theorem 7.6] for a discussion of the symplectic case.

1.1 Transverse vector fields and basic differential forms

Let $\mathcal {F}$ be a foliation of a manifold P and $\mathfrak {X}(\mathcal {F})$ the Lie algebra of vector fields tangent to $\mathcal {F}$. Recall from Sect. 5 that the Lie algebra of foliate vector fields is the normalizer $N_{\mathfrak {X}(P)}(\mathfrak {X}(\mathcal {F}))$ of $\mathfrak {X}(\mathcal {F})$ inside the Lie algebra $\mathfrak {X}(P)$ of all vector fields. The quotient Lie algebra

$$\begin{aligned} \mathfrak {X}(P,\mathcal {F})=N_{\mathfrak {X}(P)}(\mathfrak {X}(\mathcal {F}))/\mathfrak {X}(\mathcal {F}) \end{aligned}$$

is the Lie algebra of transverse vector fields of the foliation. A transverse vector field X is not a vector field, but an equivalence class of vector fields. The flow of X maps leaves of $\mathcal {F}$ to leaves of $\mathcal {F}$, but is well-defined only up to a flow along the leaves of $\mathcal {F}$. On every transversal of the foliation X induces a genuine vector field. (Hence $\mathfrak {X}(M,\mathcal {F})=\mathfrak {X}(M/\mathcal {F})$ if the leaf space $M/\mathcal {F}$ is a manifold.)

A differential form $\alpha \in \Omega ^*(M)$ is $\mathcal {F}$-basic if $\iota (X)\alpha =\iota (X)d\alpha =0$ for all $X\in \mathfrak {X}(\mathcal {F})$. We denote the complex of $\mathcal {F}$-basic forms by $\Omega ^*(M,\mathcal {F})$ and the algebra of $\mathcal {F}$-basic functions by $\mathcal {C}^\infty (M,\mathcal {F})=\Omega ^0(M,\mathcal {F})$.

1.2 Conformal presymplectic forms

A conformal presymplectic structure on a manifold M is a pair $(\omega ,\theta )$, where $\omega$ is a 2-form of constant rank and $\theta$ a closed 1-form satisfying $d_\theta \omega =0$ and $\iota (X)\theta =0$ for all vector fields X with $\iota (X)\omega =0$. As in the nondegenerate case we call $\theta$ the Lee form and its cohomology class the Lee class, and we say $(\omega ,\theta )$ is strict if $\theta$ is not exact, and global if $\theta$ is exact.

Let $(M,\omega ,\theta )$ be a conformal presymplectic manifold. The kernel of $\omega$ is an involutive subbundle of TM, which generates a (regular) foliation $\mathcal {F}=\mathcal {F}_\omega$ called the null foliation of $\omega$. The forms $\omega$ and $\theta$ are basic with respect to this foliation, and $(\omega ,\theta )$ induces a conformal symplectic structure on every transversal of the foliation. In this sense a conformal presymplectic structure is a transverse analogue of a conformal symplectic structure.

Suppose a function $f\in \mathcal {C}^\infty (M)$ and a vector field $X\in \mathfrak {X}(M)$ satisfy $d_\theta f=\iota (X)\omega$. Then f must be $\mathcal {F}$-basic, and X is determined by f only up to an element of $\mathfrak {X}(\mathcal {F})$. Moreover, $L_\theta (X)\omega =0$, so

$$\begin{aligned} \omega ([X,Y],Z)=-\omega (Y,[X,Z])-\theta (X)\omega (Y,Z) \end{aligned}$$

for all Y, $Z\in \mathfrak {X}(M)$. This shows that $[X,Y]\in \mathfrak {X}(\mathcal {F})$ for all $Y\in \mathfrak {X}(\mathcal {F})$; in other words X is foliate and therefore represents a transverse vector field $X_f\in \mathfrak {X}(M,\mathcal {F})$, called the Hamiltonian vector field of f. Thus the Hamiltonian correspondence $f\mapsto X_f$ is a map

$$\begin{aligned} \mathcal {C}^\infty (M,\mathcal {F})\longrightarrow \mathfrak {X}(M,\mathcal {F}). \end{aligned}$$

In particular the Lee vector field is the transverse vector field $A\in \mathfrak {X}(M,\mathcal {F})$ defined by $A=X_1$.

Suppose K acts on M by foliate transformations. Then the infinitesimal action $\mathfrak {k}\rightarrow \mathfrak {X}(M)$ takes values in $N_{\mathfrak {X}(\mathcal {F})}(\mathfrak {X}(M))$. Composing this map with the quotient map gives a homomorphism $\mathfrak {k}\rightarrow \mathfrak {X}(M,\mathcal {F})$. As in the nondegenerate case, we say the action is weakly Hamiltonian if there is a linear lifting $\Phi ^\vee$ of this homomorphism,

and the map $\Phi :M\rightarrow \mathfrak {k^*}$ dual to $\Phi ^\vee$ is the moment map. The K-action is Hamiltonian, or M is a Hamiltonian conformal presymplectic K-manifold, if there exists a moment map $\Phi$ which is K-equivariant. The action is of Lee type if the transverse Lee vector field A has a foliate representative of the form $\zeta _M$ with $\zeta \in \mathfrak {k}$. We can now state the conformal presymplectic convexity theorem.

Theorem B.1

Let K be a compact connected Lie group which acts on a connected strict conformal presymplectic manifold $(M,\omega ,\theta )$ in a Hamiltonian fashion. Assume that the K-action is clean and of Lee type, and that the moment map $\Phi :M\rightarrow \mathfrak {k}^*$ is proper. Choose a maximal torus T of K and a closed Weyl chamber C in $\mathfrak {t}^*$, where $\mathfrak {t}=\mathrm{Lie}(T)$.

• (i) The fibres of $\Phi$ are connected and $\Phi :M\rightarrow \Phi (M)$ is an open map.

• (ii) $\Delta (M)=\Phi (M)\cap C$ is a closed convex polyhedral set.

Proof

(Outline of proof) The essential fact that $d_\theta :\mathcal {C}^\infty (M)\rightarrow \Omega ^1(M)$ is injective (Proposition 2.2) remains valid for conformal presymplectic manifolds, as is clear from the proof in [38]. None of the auxiliary results in Sect. 3 uses the nondegeneracy of $\omega$. The tubular neighbourhood theorem, Proposition 3.4, generalizes as follows: for every $m\in M$ and every equivariant tubular neighborhood V of $K\cdot m$ there exists a K-invariant smooth function f on V such that $\theta |_V=df$; the form $\Omega =e^f\cdot \omega |_V$ is presymplectic, the K-action on $(V,\Omega )$ is Hamiltonian with equivariant moment map $\Psi =e^f\cdot \Phi |_V$, and the null foliation of $\Omega$ is $\mathcal {F}_\Omega =\mathcal {F}_\omega |_V$. In particular, the K-action on V is clean with respect to the foliation $\mathcal {F}_\Omega$. Because of this fact, in the proof of the local convexity theorem, Theorem 4.2, instead of appealing to the symplectic version of the local convexity theorem, we can appeal to the presymplectic local convexity theorem [30, Theorem 2.12.1]. The remainder of the proof, and the proof of Theorem 1.1, then go through unchanged. $\square$