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It is well known that $T\mathbb{S}^{n-1}$ is diffeomorphic to $M= f^{-1}(1)$ where $f:\mathbb{C}^n\rightarrow \mathbb{C}$ is $f(z):=\sum_{i=1}^{n} z_{i}^{2}$. Two questions:

1) Is $M$ a symplectic submanifold of $\mathbb{C}^n\sim \mathbb{R}^{2n}$ (with the standard symplectic structure)?

If the answer is affirmative, we can consider two symplectic structures for $T\mathbb{S}^{n-1}$. The first is the original structure of the tangent or cotangent bundle, the second one is the pull back structure of $M$.

2) Are these structures equivalent?

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Here is a formula for an explicit symplectomorphism $F$ from $T^*S^{n-1}$ to the affine quadric $\{\sum z_{j}^{2}=1\}$ in $\mathbb{C}^n$:

$$ F(p,q) = \left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{1/2} q - i\left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{-1/2}p $$

Here I view $T^*S^{n-1}$ as consisting of pairs $(p,q)\in \mathbb{R}^n\times\mathbb{R}^n$ with $p\cdot q=0$ and $|q|=1$. To check that $F$ is a symplectomorphism one can just check that $F$ pulls back the one-form $\alpha=-\sum y_jdx_j$ on $\mathbb{C}^n$ (which is a primitive for the standard symplectic form) to the canonical one-form on $T^*S^{n-1}$. The coefficients involving $\sqrt{1+4|p|^2}$ are designed to cause the map to have image in the quadric.

To indicate how I came up with this formula, the reasoning was that a suitably natural symplectomorphism ought to be equivariant with respect to the natural Hamiltonian $O(n)$-actions on both spaces. The easiest way to do this seems to be to take $F(p,q)=f(|p|)q-ig(|p|)p$ for some real functions $f$ and $g$. For the image to be in the quadric one needs $$f(|p|)^2- |p|^2g(|p|)^2=1.$$ Meanwhile if this is to be a symplectomorphism it should pull back the moment map for the $O(n)$-action on $\mathbb{C}^n$ to the moment map for the $O(n)$-action on $T^*S^{n-1}$. The norms of these moment maps are, respectively $(x+iy)\mapsto |x||y|$ and $(p,q)\mapsto |p|$, giving an equation $|f||g||p|=|p|$, i.e., $$|g|=\frac{1}{|f|}.$$ Solving these for $f$ and $g$ yields the formula at the top, which can then be directly confirmed to have the required properties.

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  • $\begingroup$ I didn't check your computation, but that's not the point. I just want to congratulate you for using the wording "Moment Map" :-}) $\endgroup$ – Patrick I-Z Nov 30 '13 at 17:58
  • $\begingroup$ Mike thank you for your beutiful answer. But do you mean that an arbitrary symplectomorphism is necessarily equivariant with respect to O(n) action? $\endgroup$ – Ali Taghavi Dec 1 '13 at 18:21
  • $\begingroup$ No, that certainly wouldn't be the case (for instance if you composed the symplectomorphism in the answer with a non-equivariant symplectomorphism of the domain--of which there are many--the result wouldn't be equivariant). The idea of asking for O(n)-equivariance was just to narrow down the search, and also to take advantage of the fact that an equivariant symplectomorphism will respect the moment maps (thus giving a function that the map should preserve, which is usually easier to arrange than preserving a form). $\endgroup$ – Mike Usher Dec 1 '13 at 21:13
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$ \def\C{{\mathbf C}} \def\d{\delta} \def\e{{\mathbf e}} \def\r{{\mathbf r}} \def\u{{\mathbf u}} \def\x{{\mathbf x}} \def\y{{\mathbf y}} \def\z{{\mathbf z}} \def\<{\langle} \def\>{\rangle} $ If I understand Now that I understand your questions correctly, the answers are 1) Yes 2) No Yes.

1) Writing $\z=\x+i\y$, we have $f(\z) = \|\x\|^2 - \|\y\|^2+2i\<\x,\y\>$. So $M=f^{-1}(1)$ (resp. $\mathrm T_\z M$) consists of all $\z$ (resp. $\d\z=\d\x+i\d\y$) such that $$ \|\x\|^2 - \|\y\|^2= 1, \qquad \<\x,\y\>=0, \tag1 $$ $$ \llap{(\text{resp.}\qquad}\<\x,\d\x\>-\<\y,\d\y\>=0,\qquad\<\d\x,\y\>+\<\x,\d\y\>=0.) \tag2 $$ Given $\z$ in (1), we can choose an orthonormal basis $(\e_1,\dots,\e_n)$ such that $\x = \|\x\|\e_1$ and $\mathbf y = \|\y\|\e_2$. Then (2) says that $(\d x_1,\d y_1)=\frac{\|\y\|}{\|\x\|}(\d y_2,-\d x_2)$ and hence \begin{align} \omega(\d\z,\d'\z) &= \sum_{i=1}^n(\d x_i\d'y_i-\d'x_i\d y_i)\\ &=\Bigl(1+\tfrac{\|\y\|^2}{\|\x\|^2}\Bigr)(\d x_2\d'y_2-\d'x_2\d y_2) + \sum_{i=3}^n(\d x_i\d'y_i-\d'x_i\d y_i) \end{align} which is obviously nondegenerate. (Or as Tim Perutz says, just note that $M$ is a complex submanifold.)

2) Consider (after Seidel's thesis pointed out by Daniel Pomerleano) the diffeo $\Phi:M\to\mathrm{TS}^{n-1}$ sending $\z=\x+i\y$ to the pair $(\u,\r)=(\frac{\x}{\|\x\|}, \|\x\|\y)$. Writing $\sigma$ for the canonical 2-form $d\<\u,d\r\>$ on $\mathrm{TS}^{n-1}$, we have \begin{align} \Phi^*\sigma &= d(\Phi^*\<\u,d\r\>)=d(\<\tfrac{\x}{\|\x\|},d(\|\x\|\y)\>)\\ &=d\<\x,d\y\> + \<\x,\y\>\tfrac{d\|\x\|}{\|\x\|} \end{align} and this equals $\omega$ since the second term vanishes by (1). Thus $\Phi$ is a symplectomorphism.

3) My initial (stupid and wrong) answer came from my thinking of another similar question. Namely $\mathrm{TS}^2$, thought of as the manifold of oriented affine lines in $\mathbf R^3$ (identify $(\u,\r)$ with the line $\r+\mathbf R\u$ with orientation $+\u$), carries a 2-parameter family of Euclid-invariant symplectic structures given by $$ \omega_{k,s} = d\<k\u,d\r\> - s\Omega,\qquad k>0, s\in \mathbf R, \tag3 $$ where the first term is the cotangent bundle 2-form, and the second ("magnetic") term is $-s$ times the area 2-form of the base $\mathrm S^2$.

On the other hand, identifying $\mathfrak{so}(3,\C)=\C^3$ with Lie bracket the vector product, the nonzero levels of your quadratic form $\<\z,\z\>=\sum_{i=1}^3 z_i^2$ are precisely the semisimple (co)adjoint orbits of $\mathrm{SO}(3,\C)$. As such they carry a (Kirillov-Kostant-Souriau) 2-form $\omega_{\mathrm{KKS}}(\d\z,\d'\z)=\mathrm{Re}\bigl(\frac1{s^2}\<\z,\d'\z\times\d\z\>\bigr)$ where $\<\z,\z\>=s^2$ is the equation of our orbit $\mathcal O$. So we can again ask if this is symplectomorphic with $\mathrm{TS}^2$ equipped with one of the forms (3). At least when $s^2$ is real positive, the answer is yes and the diffeomorphism $F:\mathrm{TS}^2\to\mathcal O$ I was considering (with $k=s=1$), $$ F(\u,\r) = k\r\times\u + s\u + ik\r $$ is symplectic: $F^*\omega_{\mathrm{KKS}}=\omega_{k,s}$. Being familiar with this diffeomorphism, I then made the mistake of thinking it must be "the one" of which you were asking if it was symplectic relative to your (different) 2-forms. And it isn't.

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  • $\begingroup$ Not many people use $\delta$ to denote tangent vectors... $\endgroup$ – Patrick I-Z Nov 29 '13 at 22:42
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    $\begingroup$ Only perverts like us! $\endgroup$ – Francois Ziegler Nov 29 '13 at 22:46
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    $\begingroup$ For 1), there's no need to compute: $M$ is a complex submanifold of $\mathbb{C}^n$, hence a symplectic submanifold. $\endgroup$ – Tim Perutz Nov 30 '13 at 15:14
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    $\begingroup$ I'm very confused: I think the answer to question 2 is "yes" if we mean "symplectomorphic". See e.g. Lemma 18.1 of Seidel's thesis or lemma 6.11 of arxiv.org/pdf/math/0006056v2.pdf. $\endgroup$ – Daniel Pomerleano Nov 30 '13 at 15:29
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    $\begingroup$ Part 2 of the answer successfully shows that a certain specific diffeomorphism between the spaces is not a symplectomorphism...of course this is not the same as showing that they are not symplectomorphic. $\endgroup$ – Mike Usher Nov 30 '13 at 17:39

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