# Tangent space of Cotangent bundle at zero section?

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.

• Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition.
– t.b.
Jul 25, 2011 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. Dec 22, 2011 at 13:00

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

a) Let $$\pi:N\to M$$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $$N$$: $$0\to T^{vert}(N) \to T(N) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (*)$$ Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $$\pi: N\to M.$$
In order that both the tangent bundles to $$N$$ and to $$M$$ live on $$N$$, you have to pull back $$T(M)$$ to $$N$$: that is why we have $$\pi^{-1} T(M)$$ on the right.
The kernel is the set of vertical tangent vectors to $$N$$, those that lie along the level lines $$N[n]\stackrel {def}{=}\pi^{-1}(\pi (n))$$ of $$\pi$$.
In other words at a point $$n\in N$$ the fibre of $$T^{vert}(N)$$ is $$T^{vert}_n(N)=T_n(N[n]) \quad (**)$$

b) Consider now a 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!) :

$$0\to T^{vert}(V) \to T(V) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (***)$$

In this new set-up you have the interesting identification $$T^{vert}(V)=\pi^{-1}(V)$$.
This boils down to the fact that the tangent space at any point $$e\in E$$ of a vector space $$E$$ is that vector space itself: $$T_e(E)=E.$$
Hence $$(***)$$ becomes $$0\to \pi^{-1}(V) \to T(V) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (****)$$

c) Finally, if you restrict this last exact sequence (****) to the zero section of $$V$$, identified to $$M$$, you get

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

d) Although this is not very helpful, you can if the manifold $$M$$ is paracompact (see here) non-canonically split the exact sequence (*****) and obtain the isomorphism $$T(V)|M \cong V \oplus T(M)$$ At a point $$(m,0)\in M\subset V$$ this gives the decomposition $$T_{(m,0)}(V) \cong V_m \oplus T_m(M)$$ and for $$V=T^*(M)$$ this is what you were looking for.

• dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $\pi^{-1}TM\oplus \pi^{-1}T^*M$ where $\pi:T^*M\to M$ ? Mar 26, 2019 at 22:13
• Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context. Mar 27, 2019 at 7:56
• Concerning point d), the derivative of the zero section $M \to V$ does provide a natural splitting of the exact sequence (*****). It's a bit of the crux, that the sequence (****) does nott natually split, but if you restrict it to the zero section it does. For this reason, there is no intrinsic definition of a second derivative of a function on a manifold, except at a critical point. Aug 16, 2020 at 13:55