I've been working out a nice example of symplectic reduction, and have come to a solution only after quite a lot of effort. So I was wondering if anyone knew a more straightforward route to the answer.

This example is about the space of oriented lines in $\mathbb{R}^{n+1}$.

Start by considering the symplectic manifold $(\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}, dp_{i} \wedge dq_{i})$, where $(q_{i},p_{i})$ are the coordinates on $\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$. We let $\mathbb{R}$ act on $\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$ as follows: \begin{equation} t \ast (q,p) = (q+tp, p). \end{equation} This action is evidently symplectic. Then define the following moment map \begin{equation} \mu : \mathbb{R}^{n+1} \times \mathbb{R}^{n+1} \to \mathbb{R}, \ \ \ \ (q,p) \mapsto \frac{1}{2}(|p|^2 - 1). \end{equation} The symplectic quotient is the space $(\mu^{-1}(0)/\mathbb{R}, \omega_{r})$, where $\omega_{r}$ is the reduced symplectic form, which is the unique symplectic form on the quotient $\mu^{-1}(0)/\mathbb{R}$, satisfying \begin{equation} \pi^{*}(\omega_{r}) = \iota^{*}(dp_{i} \wedge dq_{i}), \end{equation} where $\pi : \mu^{-1}(0) \to \mu^{-1}(0)/\mathbb{R}$ is the orbit projection map, and $\iota : \mu^{-1}(0) \to \mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$ is the inclusion.

We have the following interpretation of the resulting space: If we restrict to $|p| = 1$, then $p$ specifies a direction in $\mathbb{R}^{n+1}$. The coordinate $q$ specifies a point in $\mathbb{R}^{n+1}$. Therefore, the pair $(q,p)$ singles out the line passing through $q$ in the direction of $p$. Hence $\mu^{-1}(0)$ consists of all the directed lines in $n+1$ dimensional space. However, there is a degeneracy: pairs $(q,p)$, where the $q$ lie along the same line, all pick out the same oriented line. In fact, our action of $\mathbb{R}$ is a symmetry of this interpretation because all points $t \ast (q,p)$ correspond to the same line; each orbit corresponds to a single line. Therefore $\mu^{-1}(0)/\mathbb{R}$ is the space of oriented lines.

Now we want to describe the geometry of this space, and to do so we should pick a representative from each orbit. The natural choice for each line is to pick the closest point to the origin. Given a line specified by $(q,p)$, this point is $(q - (q \cdot p)p, p)$. Note that $(q - (q \cdot p)p) \cdot p = 0$. The two coordinates of this representative are orthogonal. So if we define $N$ to be the following embedded submanifold of $ \mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$: \begin{equation} N = \{ (q,p) \in \mathbb{R}^{n+1} \times \mathbb{R}^{n+1} \ | \ |p| = 1, \ q \cdot p = 0\}, \end{equation} then we can describe the orbit projection map $\pi$ as the smooth submersion \begin{equation} \pi : \mu^{-1}(0) \to N, \ \ (q,p) \mapsto (q - (q \cdot p)p, p). \end{equation}

Now note that the image $N$ is nothing but the tangent bundle of the n-sphere $TS^n$, and that this is isomorphic to the cotangent bundle of the n-sphere $T^{*}S^{n}$. I want to show that in fact, $N$ is symplectomorphic to $T^{*}S^{n}$ with the canonical symplectic structure (http://en.wikipedia.org/wiki/Cotangent_bundle#Symplectic_form).

This is where I'm a little less sure about how to proceed.

My idea is as follows: We can use the standard Euclidean metric $g$ on the ambient space $\mathbb{R}^{n+1}$ to induce a metric on the sphere $S^n$. This induces a vector bundle isomorphism between the tangent and cotangent bundles of $S^n$ which covers the identity on $S^n$ \begin{equation} \hat{g} : TS^n \to T^{*}S^n, \ \ v_{s} \mapsto g_{s}(v_{s}, -). \end{equation} Note that this means that if $\Pi : TS^n \to S^n$ and $\tilde{\Pi} : T^{*}S^n \to S^n$ are the projection maps, then $\tilde{\Pi} \circ \hat{g} = \Pi$. I want to now pull back the canonical symplectic structure on $T^{*}S^n$ with $\hat{g}$ and compare it to the reduced symplectic form $\omega_{r}$. If they turn out to agree, then this proves that $\hat{g}$ is a symplectomorphism between $\mu^{-1}(0)/\mathbb{R}$ and $T^{*}S^n$.

Now note that the canonical symplectic structure is the (negative of the) differential of the tautological one form $\alpha \in \Omega^{1}(T^{*}S^n)$, which is defined as follows: \begin{equation} \xi_{s} \in T^{*}_{s}S^n, \ \ \alpha(\xi_{s}) = d\tilde{\Pi}^{*}_{\xi_{s}}(\xi_{s}) \in T^{*}_{\xi_{s}}(T^{*}S^n). \end{equation} So we pull this back instead, since we can always take the differential later on. Now $\hat{g}^{*}(\alpha) \in \Omega^{1}(TS^n)$ and at a point $v_{s} \in T_{s}S^n$ we have \begin{equation} \hat{g}^{*}(\alpha)_{v_{s}} = \alpha_{\hat{g}(v_{s})} \circ d\hat{g}_{v_{s}} = \hat{g}(v_{s}) \circ d(\tilde{\Pi} \circ \hat{g})_{v_{s}} = \hat{g}(v_{s}) \circ d\Pi_{v_{s}} = v_{s} \cdot (d\Pi_{v_{s}}(-)), \end{equation} where the last term means taking a dot product in euclidean space. Now how does this act on specific vectors: Take coordinates $(u_{i})$ for $S^n$ around a point $s$. This gives local frame $\frac{\partial}{\partial u_{i}}$ for the tangent bundle. And this in turn gives coordinates $(u_{i}, v_{i})$ for the tangent bundle. That is to say $(u_{i}, v_{i}) \mapsto v_{i}\frac{\partial}{\partial u_{i}}|_{u}$. Now take a vector $(t_{i} \frac{\partial}{\partial u_{i}} + l_{i} \frac{\partial}{\partial v_{i}})$ at a point $(u,v)$. Then \begin{equation} \hat{g}^{*}(\alpha)_{(u,v)}(t_{i} \frac{\partial}{\partial u_{i}} + l_{i} \frac{\partial}{\partial v_{i}}) = v_{u} \cdot d\Pi_{(u,v)}(t_{i} \frac{\partial}{\partial u_{i}} + l_{i} \frac{\partial}{\partial v_{i}}) = v_{u} \cdot t_{u}. \end{equation} In other words, if we note that $T_{(u,v)}(TS^n) \cong T_{u}S^n \times \mathbb{R}^{n}$, then $\hat{g}^{*}(\alpha)_{(u,v)}(t,l) = v \cdot t$.

Now we also have the standard 1-form on $\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$ which is given by $q_{i} dp_{i}$. We can pull this back to $TS^n = N$ using the inclusion $i : N \to \mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$. Since $T_{(q,p)}N \cong \mathbb{R}^{n} \times T_{p}S^{n} \subseteq \mathbb{R}^{n+1}_{q} \times \mathbb{R}^{n+1}_{p}$, for $(l,t) \in T_{(q,p)}N$ we have $i^{*}(q \cdot dp)_{(q,p)}(l,t) = q \cdot dp_{(q,p)} (l + t) = q \cdot dp_{(q,p)}(t) = q \cdot t$. Therefore, $\hat{g}^{*}(\alpha) = i^{*}(q \cdot dp)$ since they have the same effect on vectors.

Therefore $\pi^{*}(\hat{g}^{*}(\alpha)) = \pi^{*} i^* (q \cdot dp) = (i \circ \pi)^* (q \cdot dp)$. We wish to compare this with $\iota^{*}(q \cdot dp)$. We do this using coordinates on $\mu^{-1}(0) = \mathbb{R}^{n+1} \times S^n$. We choose coordinates $(q,u) \in \mathbb{R}^{n+1} \times \mathbb{B}^n$, with coordinate map $\phi^{\pm} : \mathbb{R}^{n+1} \times \mathbb{B}^n \to \mu^{-1}(0), \ \ (q,u) \mapsto (q,u,\pm\sqrt{1-|u|^2})$. We also write $q = (\overrightarrow{q}, q_{n+1})$, where $\overrightarrow{q}$ denotes the first $n$ coordinates. In these coordinates then we see that \begin{equation} \iota^{*}(q \cdot dp) = \overrightarrow{q} \cdot du \pm q_{n+1}d\sqrt{1-|u|^2} = \overrightarrow{q} \cdot du - (\pm 1)\frac{q_{n+1} u \cdot du}{\sqrt{1-|u|^2}}. \end{equation} And that (after some cancellation of terms) \begin{equation} \pi^{*}(\hat{g}^{*}(\alpha)) = (i \circ \pi)^* (q \cdot dp) = \overrightarrow{q} \cdot du - (\pm 1)\frac{q_{n+1} u \cdot du}{\sqrt{1-|u|^2}}. \end{equation} Therefore we have shown that $\iota^{*}(q \cdot dp) = \pi^{*} (\hat{g}^{*}(\alpha))$. Taking the differential of both sides and multiplying by $-1$ gives \begin{equation} \iota^{*}(dp_{i} \wedge dq_{i}) = \pi^{*} (\hat{g}^{*}(-d\alpha)), \end{equation} where $-d\alpha$ is the canonical symplectic form on $T^{*}S^n$. And so $\omega_{r} = \hat{g}^{*}(-d\alpha)$, showing that $\hat{g} : (N,\omega_{r}) \to (T^{*}S^n, -d\alpha)$ is a symplectomorphism.

The way I exhibit the symplectomorphism uses a lot of extra stuff like the tautological 1-form, and the ambient euclidean metric. I'm wondering if there is a more direct way to see the symplectomorphism.



1 Answer 1


First of all, you sure put a lot of effort in solving this problem, as well as in making your post as clear as possible. I thank you for that, it was interesting and fun to read.

It seems that your approach is the right one to begin with, but at some point you made things more complicated than they should be. Namely, the embedded submanifold $N$ can be thought of from the first place as naturally isomorphic to the cotangent bundle, rather than the tangent bundle. Consider some point $(q,p)\in N$. As you say, $p$ is just a point on the unit sphere $S^n$. The $q$ coordinate, however, is just a functional on $T_pS^n$, by the symplectic structure of $\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}$. In other words, given $v=\sum a_i\partial_{p_i}\in T_pS^n$, we define $q(v)=\omega(q,v)=\sum q_ia_i$. (This definition is natural, and the coordinates are not really necessary. I wrote them anyway only for clarity).

This gives us a natural way to identify $N$ with $T^*S^n$. Now we need to convince ourselves that the symplectic form induced via this identification coincides with the one inherited from the ambient space. The tautological $1$-form $\alpha$ is given in these coordinates by $\alpha=q_i\cdot dp_i$. Differentiating it yields the symplectic form induced by the inclusion into $\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}$, and the proof is complete.

  • $\begingroup$ Thanks for the response. I have comment and a question. First, I want to point out that you are still implicitly using the metric when you identify $N$ with $T^{*}S^n$ (you are, after all, using dot products). I think that for a general embedding of one manifold into another, it is only possible to induce a natural embedding between tangent bundles (not cotagent bundles). So I think that maybe using the metric is unavoidable. In any case, you give a very simple description of the tautological form, so that is an improvement on what I had. $\endgroup$ Sep 14, 2014 at 19:20
  • $\begingroup$ Second, my question has to do with your very last comment: I'm confused when you claim that the proof is complete once you show that the canonical symplectic form is induced by the inclusion into $\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$. You still need to compare this form pulled back to $\mu^{-1}(0)$ by $\pi$ with $dp \wedge dq$ pulled back to $\mu^{-1}(0)$ by $\iota$. How does your statement address that? $\endgroup$ Sep 14, 2014 at 19:23
  • $\begingroup$ I'll start with the question: Well, you're right, of course. But it is not hard to verify that both symplectic forms (the one induced by the inclusion, and the one obtained by the quotient) turn out to be the same. Yes, it needs to be said. $\endgroup$ Sep 14, 2014 at 19:29
  • $\begingroup$ And about the remark: I think it is the other way around! The natural identification is with the cotangent bundle, whereas the metric is needed for the identification with the tangent bundle. Think about it - the identification in my answer only uses the symplectic form of the ambient space. $\endgroup$ Sep 14, 2014 at 19:31
  • $\begingroup$ We have a pair $(q,p)$ such that $q$ is orthogonal to $p$. But $q$ is not a tangent vector, but a cotangent vector since to begin with the $q$s are functionals on the $p$s and vice versa. $\endgroup$ Sep 14, 2014 at 19:33

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