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I was trying to work through a problem(#10 of $\S$1.2) in Guillemin and Pollack's book $\textit{differential topology. }$ The problem is given as follows.

Let $f: X\longrightarrow X\times X$ be the mapping $f(x)=(x,x)$. Check that $df_x(v)=(v,v)$. Here $X\subset \mathbf R^m$ is a manifold.

My attempt so far has been:

First we parametrise an open neighbourhood of $x\in X$ and $(x,x) \in X\times X$ locally by $\phi$ and $\phi \times \phi$ into open subsets $U\subset\mathbf R^m$ and $U\times U$ (we use $\phi(0)=x$ for simplicity). This gives the commuting diagram as follows:

$$ \begin{array}[c]{ccc} X\;\;&\stackrel{f}{\longrightarrow}&X\times X\\ \downarrow\scriptstyle{\phi}&&\downarrow\scriptstyle{\phi \times \phi}\\ U\;\;&\stackrel{h}{\longrightarrow}&U\times U \end{array} $$

$$ \begin{array}[c]{ccc} T_x(X)&\stackrel{df_x}{\longrightarrow}&T_{(x,x)}(X\times X)\\ \downarrow\scriptstyle{d\phi_0}&&\downarrow\scriptstyle{d\phi_0 \times d\phi_0}\\ \mathbf R^m\;\;&\stackrel{dh_0}{\longrightarrow}&\mathbf R^m \times \mathbf R^m \end{array} $$

According to the definition (or the commuting diagram above), $df_x=(d\phi_0 \times d\phi_0) \circ dh_0 \circ d\phi_0$.

However I have no idea how to proceed after that. If I want to calculate $df_x$, I have to know what $d\phi_0$ is first... But since we let $\phi(0)=x$, what would the derivative map of that be (since $\phi(0)=x$ just means we send a specific point $0$ to a specific point $x$, it doesn't tell us anything about the expression of this parameterisation)?

Thanks everyone for the help!

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The map $\phi$ is giving you local coordinates, so in those coordinates the differential of $f$ will be that of $h$. –  A. Bellmunt May 31 '13 at 11:24
    
Hi Mr. Bellmunt, would you mind being more specific? I agree that in those coordinates the $df$ will be $dh$. But I was wondering how that helps with finding $df$? Thanks a lot! –  Evariste May 31 '13 at 11:34
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@A.Bellmunt You edit changed the meaning of the question (replacing some instances of $X$ with a $V$). Could you perhaps ask Evariste if some of the $X$'s were meant to be $V$'s? If this was, indeed, a typo, then edit it again and it should be accepted. Otherwise, it is quite radical... –  user1729 May 31 '13 at 11:40
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@user1729: Thanks for your suggestion. My apologies, I didn't mean to be that 'radical'. I'll be more careful the next time. –  A. Bellmunt May 31 '13 at 13:16
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(@A.Bellmunt: I should say - "radical" was the system's word, not mine! I think I gave up using "radical when I was $16$...) –  user1729 May 31 '13 at 14:01

1 Answer 1

up vote 3 down vote accepted

Hoping it helps you, I am expanding A.Bellmunt's comment.
Let be $x$ a point of $X$, a submanifold of $\mathbb R^n$, and $\phi:X\to\mathbb R^m$ a local coordinate chart centered at $x$ (i.e. $\phi(x)=0$).

Therefore $\phi\times\phi:X\times X\to\mathbb R^m\times\mathbb R^m$ is a local coordinates chart centered at $(x,x)$ (i.e. $(\phi\times\phi)(x,x)=(0,0)$).

Now we get the local expression $f=(\phi\times\phi)^{-1}\circ \widetilde{f}\circ\phi$, where $\widetilde{f}$ is the linear map $$\widetilde{f}:u\in\mathbb R^m\to(u,u)\in\mathbb R^m\times\mathbb R^m.$$ Therefore:

  1. $\widetilde f$ is linear, so it coincides with $d_0\widetilde f$, and
  2. if $v\in T_xM \overset{d_x\phi}{\longrightarrow}\tilde v\in\mathbb R^m$, then $(v,v)\in T_{(x,x)}X\times X\overset{d_{(0,0)}(\phi\times\phi)}{\longrightarrow}(\tilde v,\tilde v)\in\mathbb R^m\times\mathbb R^m$,

and by 1. and 2. we get immediately the searched expression of $d_xf$, i.e.: $$\begin{array}{ccc} v\in T_xM&\overset{d_xf}{\longrightarrow}&(v,v)\in T_{(x,x)}X\times X\\ \downarrow d_x\phi&&\downarrow d_{(x,x)}(\phi\times\phi)\\ \tilde v\in\mathbb R^m&\overset{d_0\widetilde f}{\longrightarrow}&(\tilde v,\tilde v)\in\mathbb R^m\times\mathbb R^m\end{array}$$


Dear Evariste, answering the supplementary question in your comment:

given any curve $\gamma=(\gamma_1,\gamma_2)$ in $M\times M$, if we take the time-derivative of $(\phi\times\phi)\circ\gamma=(\phi\circ\gamma_1)\times(\phi\circ\gamma_2)$, then, by the chain rule, we get $$\begin{aligned}(d\phi\times d\phi)\circ\gamma'&=(d\phi\circ\gamma_1')\times(d\phi\times\gamma_2')=\dfrac{d}{dt}((\phi\circ\gamma_1)\times(\phi\circ\gamma_2))\\&=\dfrac{d}{dt}((\phi\times\phi)\circ\gamma)=d(\phi\times\phi)\circ\gamma'.\end{aligned}$$ By the arbitrariness of $\gamma$, we derive the desired identity.

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Thanks Giuseppe! However I was wondering how that helps with finding $df$? Working from your last line, I will get $df=d[(\phi \times \phi)^{-1}] \circ d\tilde{f}\circ d\phi$. However I still don't know what $d\tilde{f}$ and $d\phi$ are, do I? –  Evariste May 31 '13 at 13:27
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1) $\widetilde f$ is linear, so it coincides with $d_0\widetilde f$. 2) if $v\in T_xX$ is sent to $\widetilde v$ by $d_x\phi$, then $(v,v)\in T_{(x,x)}X$ is sent to $(\widetilde v,\widetilde v)$ by $d_{(x,x)}(\phi\times\phi)=d_x\phi\times d_x\phi$. Therefore by 1) and 2) you get $d_xf$. –  Giuseppe Tortorella May 31 '13 at 14:04
    
Great answer! Just one more question: How did you show that $d(\phi\times \phi)=d\phi \times d\phi$? I stared at this for like the entire evening but still haven't got a clue... –  Evariste May 31 '13 at 17:29
    
I think I can do it by writing down the Jacobian matrix... However is there a simpler way of doing it? Thanks :) –  Evariste May 31 '13 at 17:34
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If you like the answer, then you ought to select it, so the website shows that the question was answered. –  Sammy Black May 31 '13 at 20:36

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