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Edit: According to the comments to the previous version of this question, I remove my essential errors in the question. I thank the commenters very much.

Let $M$ be a n dimensional manifold. For $k<n$, $G_{k}TM$ is the space of all $k$ dimensional subspace of $T_{x}M, x\in M$. It is a $n+k(n-k)$ dimensional manifold and there is a natural fibre bundle structure $p:G_{k}TM \to M$. For what values of $k,n$, there is a natural symplectic structure on $G_{k}TM$?

In particular is $G_{2}TM$ a symplectic manifold when $M$ is an even dimensional manifold? What about if $M$ has already a symplectic structure?

If the answer is yes, assume that $(M,g)$ is an even dimensional riemannian manifold with the Levi-civita connection $\nabla$.

Define $H:G_{2}TM\to \mathbb{R}$ as follows: $$H(Y_{x})=\text{The sectional curvature of $Y_{x}$}$$ where $Y_{x}$ is a 2 dimensional subspace of $T_{x}M$.

Are there some research or results for a relation between the geometric properties of $(M,g)$ and dynamical properties of the hamiltonian vector field associated with the above $H$?

In particular, for what type of riemannian manifold this Hamiltonian vector field has periodic solution? For what type of Riemannian manifold the flow of this Hamiltonian is ergodic?

Assume that $\gamma\subset G_{2}TM$ is a solution curve to this hamiltonian vector field. What is the geometric interpretation for $P(\gamma)\subset M$.

The last question is motivated by the following: The geodesic flow is a vector field on $TM$ and the geometric interpretation for the projection of the solution on $M$ is " length minimizing". Now for this special hamiltonian "sectional curvature", we search for an appropriate geometric concept in $M$

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    $\begingroup$ I'm afraid that you have a wrong formula for the dimension of $G_2(TM)$. Its dimension is $n + 2(n-2) = 3n-2$, not $2n(n-2)$, so, in particular, it is odd when $n$ is odd and thus is not symplectic. Even when $n$ is not odd, there is no natural symplectic structure on this space, so I don't know what you mean by 'Hamiltonian'. Unless you can construct a symplectic structure on this space, I'm afraid that the rest of your question doesn't really make sense. $\endgroup$ – Robert Bryant Dec 4 '14 at 15:04
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    $\begingroup$ Its dimension is $3n-4$. $\endgroup$ – Ben McKay Dec 4 '14 at 16:51
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    $\begingroup$ @BenMcKay: Thanks for the correction; $3n-4$ it is! I really must get around to learning arithmetic some day; so many seem to find it useful. $\endgroup$ – Robert Bryant Dec 4 '14 at 20:39
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    $\begingroup$ @AliTaghavi: Unfortunately, I don't know what 'linked paper' you mean. I don't see a link. Generally, $G(k,n)$ has no symplectic structure that is invariant under $\mathrm{GL}(n,\mathbb{R})$. $\endgroup$ – Robert Bryant Dec 4 '14 at 20:42
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    $\begingroup$ My institution doesn't subscribe to the journal, but apparently the "symplectic form" is a section of $\Lambda^2 U^* \otimes (TM/U)$ where $U$ is the universal subbundle and $TM/U$ the universal quotient bundle. It isn't really a 2-form, not even a vector-bundle valued 2-form, because it only eats pairs of vectors from $U$. So you might not want to call it symplectic. $\endgroup$ – Ben McKay Dec 5 '14 at 6:56
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This isn't an answer, other than a general set of comments to explain why there is no answer in the form that the OP wants.

First of all, when $V$ is a vector space of dimension $n$ and $k$ is an integer satisfying $0<k<n$, there is no symplectic structure on $\mathrm{Gr}_k(V)$, the space of $k$-dimensional subspaces of $V$, that is invariant under the natural action of $\mathrm{GL}(V)$. The reason is simple: If $E\in \mathrm{Gr}_k(V)$ is a $k$-plane, then there a natural isomorphism $$ T^\ast_E\mathrm{Gr}_k(V) = (V/E)^\ast\otimes E = \mathrm{Hom}(V/E,E) $$ and the action of the $E$-stabilizing subgroup $P_E = \{ A\in \mathrm{GL}(V)\ \mid\ A(E) = E\ \}$ on this tangent space is the tensor product representation of $\mathrm{GL}(V/E)\times\mathrm{GL}(E)$. Under this action, we have the irreducible decomposition $$ \mathsf{\Lambda}^2\bigl((V/E)^\ast\otimes E\bigr) = \bigl(\mathsf{\Lambda}^2((V/E)^\ast)\otimes \mathsf{S}^2(E)\bigr) \oplus \bigl(\mathsf{S}^2((V/E)^\ast)\otimes \mathsf{\Lambda}^2(E)\bigr), $$ and neither of these two irreducible summands is the trivial representation on a $1$-dimensional vector space. Thus, there cannot be any $2$-form (whether degenerate or not) on $\mathrm{Gr}_k(V)$ that is invariant under $\mathrm{GL}(V)$.

Note, however, that if one fixes a positive definite inner product on $V$ and considers the action of $\mathrm{SO}(V)$ on $\mathrm{Gr}^+_k(V)$, the Grassmannian of oriented $k$-planes in $V$, then this same calculation shows that there is a (nondegenerate and closed) $2$-form on $\mathrm{Gr}^+_k(V)$ that is $\mathrm{SO}(V)$-invariant exactly when $k=2$ or $n{-}2$.

This does give a way to define something like a Hamiltonian vector field associated to the sectional curvature of a Riemannian manifold: If $(M,g)$ is a Riemannian manifold, then the fibers of the bundle $\pi:\mathrm{Gr}^+_2(TM)\to M$ of oriented $2$-planes are symplectic manifolds in a natural way. Thus, if $\sigma:\mathrm{Gr}^+_2(TM)\to\mathbb{R}$ is the sectional curvature function, then there is a vector field $X$ on $\mathrm{Gr}^+_2(TM)$ that is tangent to the fibers of $\pi$ and that, in each fiber $\mathrm{Gr}^+_2(T_xM)$, is the Hamiltonian vector field associated to the function $\sigma_x:\mathrm{Gr}^+_2(T_xM)\to\mathbb{R}$. However, because the vector field $X$ is tangent to the $\pi$-fibers, its integral curves project to points in $M$, so they are not interesting. The condition that the flow of $X$ be periodic with some fixed period is very restrictive, of course. For example, in the first interesting dimension, which is $3$, this condition is that the eigenvalues of the Ricci tensor be constant and that one of them have multiplicity $2$. (If they are all equal, then $X=0$.) In higher dimensions, the condition is even more restrictive.

Second, the abstract of the article to which the OP linked explicitly says that the author is constructing a generalized symplectic form, but then goes on to muddy the waters by referring to it as a 'symplectic form' (without the word 'generalized') in the next sentence. Of course, this tensor is not anything like a symplectic form in the usual sense, and it cannot be used to define Hamiltonian vector fields or any of the other usual apparatus in symplectic geometry. In my opinion, it is a disservice to readers to introduce confusing terminology such as this.

In this particular case, it's not hard to figure out what this 'generalized symplectic form' is, since it is supposed to be invariant under the natural action of $\mathrm{Diff}(M)$ on the bundle $\mathrm{Gr}_k(TM)$. It was identified by É. Cartan (though not by that name) more than a century ago, as the canonical torsion associated to the contact system on $\mathrm{Gr}_k(TM)$. One can describe it as follows: If $\pi:\mathrm{Gr}_k(TM)\to M$ is the basepoint mapping (i.e., $\pi(E) = x$ when $E\in\mathrm{Gr}_k(T_xM)$), then there is a canonical subspace (known classically as the contact system, as it generalizes the contact plane field in the classical case $k=n{-}1$) $D_E\subset T_E\mathrm{Gr}_k(TM)$ of codimension $n{-}k$ such that $\pi'(E)(D_E) = E$, and there is also a canonical quotient bundle $$ Q = \bigl(T\mathrm{Gr}_k(TM)\bigr)/D, $$ whose fiber $Q_E$ is naturally isomorphic to $\bigl(T_{\pi(E)}M\bigr)/E$. For vector fields $X$ and $Y$ on $\mathrm{Gr}_k(TM)$ that are everywhere tangent to $D$, define a section $\delta(X,Y)$ of $Q$ by the formula $$ \delta(X,Y)(E) = \pi'(E)\bigl([X,Y]\bigr) + E \in \bigl(T_{\pi(E)}M\bigr)/E = Q_E. $$ Then $\delta$, which is clearly bilinear and antisymmetric, is also linear over the smooth functions on $\mathrm{Gr}_k(TM)$. Thus, $\delta$ is a section of the bundle $Q\otimes\mathsf{\Lambda}^2(D^\ast)$, and it is invariant under the natural induced (i.e., prolonged) action of $\mathrm{Diff}(M)$ on $\mathrm{Gr}_k(TM)$. This $\delta$ is what the author of the linked paper calls a 'generalized symplectic form'. It is not ever a genuine symplectic form, and essentially none of the usual constructions in symplectic geometry apply to it.

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    $\begingroup$ However, what if - as the op suggested - the manifold $(M, g)$ comes already with a symplectic structure $\omega$ (e.g. a Kähler manifold)? Then one could form $\pi^* \omega + \tau$ on $G_2(T M)$, where $\tau$ is the symplectic structure on the fibers. However, this is probably not closed unless $M$ is flat, I did not check this. $\endgroup$ – Matthias Ludewig Dec 7 '14 at 12:22
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    $\begingroup$ @MatthiasLudewig: Indeed, using the natural way (via the Levi-Civita connection) to extend the fiberwise $\tau$ to a $2$-form on $\mathrm{Gr}^+_2(TM)$, one does not get a closed $2$-form unless the metric is flat. $\endgroup$ – Robert Bryant Dec 7 '14 at 13:07

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