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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!

Evariste
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3 Answers3

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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 $\Big(d(\pi_X)_{(p,q)}(v), d(\pi_Y)_{(p,q)}(v) \Big)$. This is linear since $d(\pi_X)_{(p,q)}$ and $d(\pi_Y)_{(p,q)}$ are both linear. On the other hand, define $$g : T_p X \times T_q Y \to T_{(p,q)} (X \times Y)$$ 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 $\iota_Y$.

Using that \begin{align} \pi_X \circ \iota_X =\operatorname{id}_X, \ \ \pi_Y\circ \iota_Y=\operatorname{id}_Y, \end{align}

$\pi_Y \circ \iota_X, \ \ \pi_X\circ \iota_Y$ are constant maps and chain rule, we have

\begin{align}(f \circ g) (v, w) &= f ( d(\iota_X)_p(v) +d(\iota_Y)_q(w)) \\ &= \Big(d(\pi_X)_{(p,q)}( d(\iota_X)_p(v) +d(\iota_Y)_q(w)), d(\pi_Y)_{(p,q)}( d(\iota_X)_p(v) +d(\iota_Y)_q(w)) \Big) \\ &=\Big(d(\pi_X\circ \iota_X)_p(v) +d(\pi_X\circ\iota_Y)_q(w)), d(\pi_Y\circ\iota_X)_p(v) +d(\pi_Y\circ\iota_Y)_q(w)) \Big) \\ &= (v, w) \end{align}

Thus $f$ is surjective. Since $T_{(p,q)}(X \times Y)$ and $ T_p X \times T_q Y$ have the same dimension, $f$ is a linear isomorphism.

Arctic Char
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    $d(\imath_X)_p(v)+d(\imath_Y)_q(w)$ – Arimakat Oct 23 '17 at 22:48
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    To add to this, it is only obvious that the map provided is a right inverse of $f$. One needs to then argue that the dimensions of the two vector spaces under consideration are equal in order to show it is an isomorphism. – amc May 18 '20 at 08:24
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    Is there a way to show that g is also a left inverse without using dimensional arguments? I can't find out why $g\circ f=\text{id}$ algebraically. – user615345 Jul 28 '22 at 07:49
  • Just for anyone else who wondered it, as myself, let me note that the pushforward of a composition is equal to the composition of the pushforwards. That is $$ (\phi\circ\psi)_{\ast}V = V(f\circ\psi\circ\phi) = (\phi_{\ast}V)(\psi^{\ast}f) = \psi_{\ast}(\phi_{\ast} V). $$ – Albert Jun 11 '24 at 09:46
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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.

BlackAdder
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Abusing a little. Nowhere in the problem does it say exactly what to use and what not to so, to prove that $$T_{(x,y)}(X\times Y)=T_{x}X\times T_{y}Y$$ you can use a dimensional argument. Given that dim$(T_{(x,y)}(X\times Y))=$dim$(T_{x}X\times T_{y}Y)$ and that both vectorial spaces are of finite dimension, then both are isomorphic.

I know, I know, It's ugly... But it is correct.

  • This doesn't show they're isomorphic – General Grievous Jun 12 '20 at 15:35
  • Actually does it. Two vector spaces with the same dimension are isomorphic. – Carlos Beltran Jun 13 '20 at 01:16
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    That's not what the OP is asking. The OP wants a canonical isomorphism (basis indipendent isomorphism) – General Grievous Jun 13 '20 at 04:58
  • No, that's not true. The problem says: "Show that for any manifolds X and Y, T(x,y)(X×Y)=Tx(X)×Ty(Y)". Nowhere says absolutely nothing about a "canonical isomorphism" – Carlos Beltran Jun 14 '20 at 01:10
  • Well, yes, but then the problem doesn’t make much sense. Keep in mind that it is possible to have equal fibers of line bundles, whereas they are not isomorphic (AG contr-example: there is only trivial morphism $\mathcal{O}(1) \to \mathcal{O}$, stalks are equal, but the induced maps on them are simply zero maps. Remember that the global maps will help thoroughly understand the geometry of your space/bundle/etc. – Nicholas S Mar 21 '21 at 12:06