14

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.

t.b.
  • 78,116
Topologieeeee
  • 695
  • 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 Answers1

22

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.

Georges Elencwajg
  • 150,790
  • dear Georges, we are ok that more generally $TT^M$ is naturally isomorphic to $\pi^{-1}TM\oplus \pi^{-1}T^M$ where $\pi:T^*M\to M$ ? – epsilones Mar 26 '19 at 22:13
  • 1
    Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context. – Georges Elencwajg Mar 27 '19 at 07:56
  • 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. – Tobias Diez Aug 16 '20 at 13:55
  • Just to elaborate on Tobias' point: the sequence $(***)$ actually does split canonically** (i.e. restricted to the zero section). It's enough to provide a section of the projection $T(E)\rvert M\to T(M)$; the derivative of the zero section itself suffices for this purpose. If $[\gamma(t)]$ is the equivalence class of a curve $\gamma(t):\mathbb{R}\to E$ representing a tangent vector in $T(E)\rvert M$, then the corresponding map $T(E)\rvert M\to E$ is defined by$$[\gamma(t)]\mapsto\lim_{t \to 0}\frac{\gamma(t)}{t}.$$ Since $\gamma(0)=0$ (i.e. lies in the zero section), this is well-defined. – Keeley Hoek Mar 03 '24 at 04:42