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Let $M$ be a differentiable manifold with cotangent bundle $T^*M$.

How can I prove that $T_{(p,0)}T^*M$ is naturally isomorphic to $T_pM\oplus T_pM^*$?

If this true, then I think I could prove that the Hessian of $f\colon M\to \mathbb{R}$ is well-defined (I mean, without choice of Connection or Riemannian metric) at critical point of $f$.

This is not any homework.

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Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition. –  t.b. Jul 25 '11 at 10:49
    
Your comment about the Hessian is also correct. This allows you to say, for instance, that a function is Morse independently of the choice of connection. –  Sam Lisi Dec 22 '11 at 13:00

1 Answer 1

up vote 9 down vote accepted

Things are much nicer i.e. more general and canonical than that!

Consider any vector bundle $\pi:V\to M$ on $M$. You then have the canonical exact sequence of vector bundles on $V\;$ (yes, vector bundles on a vector bundle!) :

$$ 0\to T_{vert}(V) \to T(V) \stackrel {d\pi}{\to} \pi^*T(M) \to 0 \quad (*)$$

Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $\pi: V\to M$. In order that both tangent bundles of $V$ and $M$ live on $V$, you have to pull back $T(M)$ to $V$: that is why we have $\pi^* T(M)$ on the right. The kernel is the set of vertical tangent vectors to $V$, those that lie along the fibers of $V$.
Now if you restrict (*) to the zero section of $V$, identified to $M$, you get

$$ 0\to V \to T(V)|M \stackrel {d\pi}{\to} T(M) \to 0 \quad (**) $$

The only point worth mentioning is the identification of the vector bundles $T_{vert}(V)|M$ and $V$. It boils down to the fact that for a vector space $E$ the tangent bundle $T_a(E)$ at any point $a\in E$ is canonically isomorphic to $E$ itself.
Needless to say, taking the fibers of the bundles in (**) at a point of $p\in M$ will give you what you wanted if you take $V=T^*(M)$.

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