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This question actually comes from the question I asked before:

Derivative map of the diagonal inclusion map on manifolds

And I repeat it 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.

I thought of a new way of doing it without going through the painful commutative squares. But it was so simple that I suspect it might be wrong.

I was thinking of this: Pick a vector $v$ in the tangent plane of $X$ at point $x$($v\in T_x(X)$). Then according to the definition of derivative map, we can compute

$$\lim_{t\to 0}\frac{f(x+tv)}{t}=\lim_{t \to 0} \frac{(x+tv,x+tv)-(x,x)}{t}=(v,v)$$

Therefore the derivative map $df_x(v)=(v,v)$.

I was wondering if there a loophole in my argument? It just looks so much simpler...

Also just for everyone's convenience, so that you don't have to cross reference, I post the accepted answer for my previous question as follows(it's given by Giuseppe):

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.: $$\begin{array}{ccc} v\in T_xM&\overset{d_xf}{\longrightarrow}&(v,v)\in T_{(x,x)}X\times X\\ \downarrow d_x\phi&&\downarrow d_{(x,x)}(\phi\times\phi)\\ \tilde v\in\mathbb R^m&\overset{d_0\widetilde f}{\longrightarrow}&(\tilde v,\tilde v)\in\mathbb R^m\times\mathbb R^m\end{array}$$

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  • $\begingroup$ Your argument is correct! The point of that complicated looking diagram is you may suppose $M=\mathbb{R}^n$ locally, then life becomes easier. $\endgroup$ – Ma Ming Jun 1 '13 at 9:06
  • $\begingroup$ Hi Ma, at this stage I don't think it was correct. The main reason for me to believe so is that $f(x+tv)$ may not be on the manifold $X$. Thanks for your comment! $\endgroup$ – Evariste Jun 1 '13 at 9:34
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Unfortunately, your attempt only makes sense once you're working in local coordinates. When you pick $v\in T_xX$, the point $x+tv$ rarely lives in $X$, so you cannot make sense of $f(x+tv)$.

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  • $\begingroup$ Hi Prof. Shifrin! Yes, you're absolutely right! I realised this 4 hours ago(see my comment on Ma's post below my question.). Before that I was stuck on this for pretty much the entire day. And I even tried to show that the two methods are the same :p $\endgroup$ – Evariste Jun 1 '13 at 14:05
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    $\begingroup$ Sorry ... I didn't read the comments. :( $\endgroup$ – Ted Shifrin Jun 1 '13 at 14:38
  • $\begingroup$ Don't worry about it. I'm glad that you confirmed my suspicion(without your answer, I would probably still be tossing between those two ways). $\endgroup$ – Evariste Jun 1 '13 at 17:29

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