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Let $X$ be a compact, oriented Riemann manifold. Let $\pi_{P}: P \rightarrow X$ be a principal $G$-bundle over $X$, for a compact Lie group $G$. Let $(M, \omega)$ be a symplectic manifold endowed with a symplectic action of $G$. Denote by $\mathcal{N}:=C^{\infty}(P,M)^{U(1)}$ the space of smooth $G$-equivariant maps $u:P \rightarrow M$. Then $C^{\infty}(P,M)^{G}$ is a smooth Frechet manifold. The total space of the tangent bundle $T\mathcal{N} = C^{\infty}(P,TM)^{G}$. At a point $u \in \mathcal{N}$, $T_{u}\mathcal{N} = \Gamma(P, u^{\ast}TM)^{G}$.

For $\xi_{1}, \xi_{2} \in T_{u}\mathcal{N}$, define $\Omega(\xi_{1}, \xi_{2}) = \displaystyle \int_{X} \omega_{u}(\xi_{1}, \xi_{2}) ~ dvol_{\scriptscriptstyle X}$, where $\omega_{u}(\cdot, \cdot)$ denotes the restric tion of $\omega$ along $u$. Will $\Omega(\cdot, \cdot)$ be a symplectic form on $\mathcal{N}$? More precisely, is $\Omega(\cdot, \cdot)$ closed?

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Yes, I think $\Omega$ is closed. I will add a proof later. Yes, as a mapping $T\mathcal N \to T^*\mathcal N$ it is injective, but it can never be surjective, since $T_n\mathcal N$ is a Frechet space, whereas its dual $T^*_u\mathcal N$ is a DF-space (generalized functions of distributions) which can never be isomorphic to a Frechet space. So $\Omega$ is a weak symplectic structure. See section 48 (called: Weak Symplectic Manifolds) of here, where 48.2 and 48.8 have to be corrected as described in the Errata.

Edit:

Answering your comment: It is not necessary to work with Sobolev completions. I your case you can do it. If the structure group is a diffeomorphism group, you loose smoothness of the action.

You can work with the image under $\Omega$ of $T\mathcal N$ as "symplectic dual". See 2.5 of this paper for an example of symplectic reduction, which in this case is equivalent to constructing a Riemannian submersion. Also this paper might be of interest.

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  • $\begingroup$ Thanks for the comment and the reference! You mention in section 48.2 that "..in order to have a strongly symplectic structure, the fibres of $T\mathcal{N}$ should be reflexive". So I presume that in order to have a strongly symplectic structure one may have to resort to Sobolev setting, especially if one wishes to carry out the Marsden-Weinstein symplectic reduction in infinite dimensions? $\endgroup$ – Varun Feb 5 '16 at 6:08
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Yes, the form $\Omega$ is closed and defines indeed a weak symplectic structure. This can be verified by a direct but a bit messy calculation. A cleaner way would be to generalize the ideas of Vizman: Induced differential forms on manifolds of functions to the case of sections of non-trivial bundles.

In case you wonder what the momentum maps are:

  • For the gauge group of $P$ the momentum map $\Gamma^\infty(P \times_G M) \to \Gamma(P \times_G \mathfrak{g}^*)$ is the composition with the momentum map $J_M: M \to \mathfrak{g}^*$ of the $G$-action on $M$.

  • Assume that the symplectic form on $M$ is exact, $\omega = d \theta$. Then the action of the group of volume-preserving diffeomorphisms on $X$ has a momentum map $\Gamma^\infty(P \times_G M) \to \Omega^1(X) / d \Omega^0(X)$ which is essentially the pullback of $\alpha$ by a section, twisted by the above momentum map for the gauge group. For more details see Donaldson: Moment maps in differential geometry.

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  • $\begingroup$ Thanks for the reply Tobias! I am in the situation, similar to the one described by Donaldson in Section 1.2 of the second reference you mention: "Donalsdon: Moment maps in different". He mentions that "it is straightforward to see that this yields a closed 2-form..". Somehow, I am unable to see this. $\endgroup$ – Varun Feb 5 '16 at 16:42

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