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.

  • $\begingroup$ Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition. $\endgroup$
    – t.b.
    Jul 25, 2011 at 10:49
  • $\begingroup$ 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. $\endgroup$
    – Sam Lisi
    Dec 22, 2011 at 13:00

1 Answer 1


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.

  • $\begingroup$ dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $\pi^{-1}TM\oplus \pi^{-1}T^*M$ where $\pi:T^*M\to M$ ? $\endgroup$
    – epsilones
    Mar 26, 2019 at 22:13
  • 1
    $\begingroup$ Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context. $\endgroup$ Mar 27, 2019 at 7:56
  • $\begingroup$ 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. $\endgroup$ Aug 16, 2020 at 13:55

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