Given a product manifold.

How to prove that its tangent spaces split into direct sums: $$T_{(p,q)}(M\times N)\cong T_pM\oplus T_qN$$


One could try the geometric perspective: $$\Phi:T_{(p,q)}(M\times N)\to T_pM\oplus T_qN:[(\alpha,\beta)]\mapsto([\alpha],[\beta])$$ $$\Psi:T_pM\oplus T_qN\to T_{(p,q)}(M\times N):([\alpha],[\beta])\mapsto[(\alpha,\beta)]$$ Then bijectivity becomes pretty easy but linearity quite nasty.

(Besides it is well-defined.)

One could also try the algebraic perspective: $$\Phi:T_{(p,q)}(M\times N)\to T_pM\oplus T_qN:\delta\mapsto(d_{(p,q)}\pi_M\delta,d_{(p,q)}\pi_N\delta)$$ Then linearity becomes evident but bijectivity becomes pain.

(Besides there's an explicit but ugly inverse.)

Is there maybe some nice trick??


1 Answer 1


Take "coordinate" projections $\pi_X,\pi_Y$ from $M \times N$ to $M$ and $N$. Let $(p,q) \in M \times N$, so that you have a map $F : T_{(p,q)}(M \times N) \to T_p M \times T_q N$ sending $v$ to $\left(d(\pi_X)_{(p,q)}(v), d(\pi_Y)_{(p,q)}(v) \right)$. This map is a linear map that is an isomorphism with inverse given by the linear map $g : T_p M \times T_q N \to T_{(p,q)} (M \times N)$ that sends $(v,w)$ to $d(s_M)_p(v) +d(s_N)_q(w)$ where $s_M : M\to M\times N$ sends $M$ to $M \times \{q\}$ and where $s_N$...

  • $\begingroup$ Great, thanks. :) $\endgroup$ Jan 30, 2015 at 2:00
  • $\begingroup$ I know why $F\circ g=$id. But why $g\circ F=$id? $\endgroup$
    – No One
    May 1, 2016 at 4:54
  • $\begingroup$ It is really easy to see that this function $g$ is a isomorphism without use of the function $\Phi$ :-) $\endgroup$
    – Gustavo
    Jun 29, 2016 at 17:53
  • 8
    $\begingroup$ @TiWen the composition $g\circ F$ is not the identity (as far as I can see). However, since $F\circ g$ is the identity, $g$ is injective and $F$ is surjective. Using either of these facts, since domain and range have the same dimension and both $F$ and $g$ are linear (as they are defined in terms of derivatives), they both are isomorphisms. $\endgroup$ Nov 20, 2016 at 22:36
  • $\begingroup$ @Jānis Lazovskis I appreciate your hint about dimension, since I could not work out $g\circ F = id$ either. However, as far as I am concerned, $g\circ F$ should be identity, because $g$ is a right inverse of $F$, and $F$ is invertible by the dimension argument. $\endgroup$ Jul 18, 2019 at 19:55

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