# Product Manifold: Tangent Spaces

Problem

Given a product manifold.

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

Attempts

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??

• – Watson Apr 29 '17 at 15:44

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 $$s_N$$...
• I know why $F\circ g=$id. But why $g\circ F=$id? – No One May 1 '16 at 4:54
• It is really easy to see that this function $g$ is a isomorphism without use of the function $\Phi$ :-) – Gustavo Jun 29 '16 at 17:53
• @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. – Jānis Lazovskis Nov 20 '16 at 22:36
• @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. – Shiyu Liang Jul 18 '19 at 19:55