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Hi, I am reading "Introduction to symplectic topology" by McDuff and salamon. At some point I cant go further. My question is: Let $(M,g)$ be a Riemannian manifold and consider the cotangent bundle $T^{\star}M$. My question is now, how one can show that there is an bundle isomorphism $\varphi : TM \oplus T^{\star}M \rightarrow T(T^{\star}M)$, s.t. $d\pi \circ \varphi$ is the identily on $TM$ and it preserves the symplectic forms. In the book there is a hint giving, by saying that one can choose $\varphi$ via a connection on the cotangen bundle.

edited May 12 2011 at 7:24

asked May 12 2011 at 6:15

marco
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What is $T^M$?? – Olivier Bégassat May 12 2011 at 6:18

Sorry I did some mistake. Its the cotangent bundle. Now I have changed it. – marco May 12 2011 at 7:27

The proposed isomorphism can't be right. Let $M$ be $n$-dimensional; then $\mathrm T M$ is rank-$n$, as is $\mathrm T^\ast M$, so $\mathrm T \oplus \mathrm T^\ast$ is rank $n+n$, so the total space of $\mathrm TM \oplus_M \mathrm T^\ast M$ has dimension $3n$. On the other hand, the total space of $\mathrm T(\mathrm T^\ast M)$ has dimension $2(2n) = 4n > 3n$ if $n\neq 0$. I agree that there is a map, however. (Side remark, at the risk of being a little rude: please correct spelling, capitalization, and punctuation; MO should be more professional than emails.) – Theo Johnson-Freyd May 12 2011 at 16:28

Bugs Bunny · Answer 1 · 2011-05-12 10:29:50Z UTC

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Your Maths does not add up. Let $n$ be the dimensional of $M$. The LHS is a vector bundle of dimension $2n$ on $M$. The RHS is a vector bundle of dimension of $2n$ on $T^*M$. If you just consider it as a bundle on $M$ , its dimension is $3m$, so no iso possible.

answered May 12 2011 at 10:29

Bugs Bunny
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But is it possible to decompose $T(T^{\star}M)$ in such a way ? – marco May 12 2011 at 10:47

Stefan Waldmann · Answer 2 · 2011-05-12 12:02:56Z UTC

I guess what you want to prove is the following: for a given vector bundle $E$ over $M$ there is a canonical subbundle, the vertical bundle $\mathrm{Ver}(E) = \ker T\pi \subseteq TE$ of the tangent bundle of the total space., where $\pi\colon E \longrightarrow M$ is the bundle projection. Then the choice of a connection gives a complementary subbundle $\mathrm{Hor}(E)$, the horizontal bundle. Depending on your favorite definition of a connection, this is just the definition ;) Now what is true is that the horizontal bundle is isomorphic to the pull-back bundle $\pi^\sharp TM$ of the tangent bundle of the base $M$. This is intrinsically defined as vector bundle over $E$ but it is not a subbundle of $TE$ in an intrinsic way. Here you need the connection. But then you have $\pi^\sharp TM \oplus \ker T\pi \cong TE$.

Now you can unwind this for $E = T^*M$ and get your result (stated correctly). I hope that helps.

EDIT: for $T^*M$ one then has a canonical pairing of the horizontal and vertical spaces at every point. This you can either symmetrize to get a pseudo-riemannian metric or antisymmetrize to get a symplectic form (guess which). Interestingly enough, the metric depends on the connection while the antisymmetric form does not (you may need torsion-free, I forgot). Also as a side remark: the Laplacian (better d'Alembertian) of the metric plays an important role in quantization theory when passing from standard to Weyl ordering on a cotangent bundle.

You can find some background info in the book of Yano&Ishihara or, if you prefer german, in Sect 5.4 of my book ;)

tangent and cotangent bundle

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