Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
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
@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
@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
(@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.

share|improve this answer
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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.