I was trying to prove the following statement(#9(a) in Guillemin & Pollack 1.2) but I couldn't make much progress.

"Show that for any manifolds $X$ and $Y$, $$T_{(x,y)}(X\times Y)=T_x(X)\times T_y(Y).$$"

My attempt so far: Parametrise X and Y locally with $U\overset{\phi}{\longrightarrow}X$ and $V\overset{\psi}{\longrightarrow}Y$ where $U\subset \mathbf R^m$ and $V\subset \mathbf R^n$.

Now we can parametrise $U\times V\overset{\phi\times \psi}{\longrightarrow}X\times Y$. By taking the derivative map, we have the tangent plane.

$\mathbf R^{m+n}\overset{d(\phi\times \psi)}\longrightarrow T_{(x,y)}(X\times Y)$. I don't know what to do after this... Apparently we are supposed to set up some relation between $T_{(x,y)}(X\times Y)$ and $T_x(X)\times T_y(Y)$...

Anyone would like to help me out? Thanks!


I don't understand what you are trying to do and here's what I would do. Consider the canonical projections $\pi_X,\pi_Y$ from $X \times Y$ to $X$ and $Y$ respectively. Let $(p,q) \in X \times Y$ and now consider the map

$$f : T_{(p,q)}(X \times Y) \to T_p X \times T_q Y$$

that sends a vector $v$ to elements $\Big(d(\pi_X)_{(p,q)}(v), d(\pi_Y)_{(p,q)}(v) \Big)$. You can easily check that this map is a linear map that is an isomorphism with the inverse given by $g : T_p X \times T_q Y \to T_{(p,q)} (X \times Y)$ that sends a pair of vectors $(v,w)$ to $d(\iota_X)_p(v) +d(\iota_Y)_q(w)$, where $\iota_X : X \to X\times Y$ sends $X$ to the slice $X \times \{q\}$ and similarly for $\iota_Y$.

  • $\begingroup$ Thanks Ben! I wasn't sure what I was trying to do either. $\endgroup$ – Evariste Jun 7 '13 at 23:56
  • 1
    $\begingroup$ $d(\imath_X)_p(v)+d(\imath_Y)_q(w)$ $\endgroup$ – Arimakat Oct 23 '17 at 22:48

I'm assuming you want to show that there is an isomorphism between the two spaces, if so, I would refer you Loring Tu's excellent book introduction to manifolds. Problem 8.7 is the question you are looking for and the solution is at the back. In a nutshell, he showed an isomorphism by sending basis elements to basis elements using the differential of the projection maps.

(Tu's book can be found using Google, just google "tu introduction to manifolds")

Hope this helps.

  • $\begingroup$ How is your answer different from mine? $\endgroup$ – user38268 Jun 7 '13 at 14:23
  • $\begingroup$ Thanks yanbo. I do have Tu's book in my hand. I'll give it a read. $\endgroup$ – Evariste Jun 8 '13 at 0:31
  • $\begingroup$ In Tu's, the proof using coordinates chart. $\endgroup$ – kelvinn aja Nov 19 '17 at 19:03

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