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If $\triangle$ is the diagonal of $X \times X$, show that its tangent space $T_{(x,x)}(\triangle)$ is the diagonal of $T_x(X) \times T_x(X).$

I don't have the slightest idea on how to do this.

By definition, the tangent space of $X$ at $x$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow X$ at 0. $\phi: V \rightarrow X$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$.

Thanks.

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  • $\begingroup$ The diagonal $\bigtriangleup \,= \{(x,y) \in X\times X \colon x = y\} = \{(x,x) \in X \times X\}$. In other words, it's (A). $\endgroup$ – Jesse Madnick Jun 26 '13 at 2:56
  • $\begingroup$ Thanks @JesseMadnick, that is what I've guess. But then how should I continue? Could you give me some hint? Thanks! $\endgroup$ – WishingFish Jun 26 '13 at 2:57
  • $\begingroup$ This depends on your definition of tangent space; there are several equivalent definitions. Could you edit your post to include your definition? $\endgroup$ – Jesse Madnick Jun 26 '13 at 2:57
  • $\begingroup$ 筱簵之箭 xiǎo lù zhī jiàn 8-) @JesseMadnick $\endgroup$ – WishingFish Jun 26 '13 at 3:03
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All tangent vectors at the point $(x,x)$ in $\Delta$ are described by velocity vectors of curves passing through $(x,x)$. Suppose $c(t)$ is a curve through $(x,x)$ such that $c(0) = (x,x)$, then the velocity vector is given by $c'(0)$. Any $c(t)$ that lies in the diagonal looks like $c(t) = (\gamma(t), \gamma(t))$, where $\gamma(t)$ is a curve in $X$ through the point $x$ i.e., $\gamma(0) = x$ and $\gamma'(0) = v \in T_x X$. Then $c'(t) = (\gamma'(t), \gamma'(t))$ and $c'(0) = (v,v) \in \Delta \subseteq T_x X \times T_x X$. Strictly speaking this just shows that $T_{(x,x)} \Delta \subseteq \Delta \subseteq T_x X \times T_x X$, but showing the reverse containment amounts to running the above argument backwards. Hope that helps.

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