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A riemannian metric on $M$ is a smooth bilinear map $g:\mathfrak{X}(M)\times\mathfrak{X}(M)\to C^\infty(M)$ such that $g(X,Y) = g(Y,X)$ and $g(X,X)(x)>0$ if $X_x\neq0$

We can then define a scalar product $T_xM$ thanks to $g_x(X_x,Y_x) := g(X,Y)(x)$ where $X_x = X(x)$ and $Y_x = Y(x)$.

How do we show that this does not depend on the choice of $X$ and $Y$?

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  • $\begingroup$ It follows somehow by smoothness and bilinearity. $\endgroup$ – Berci Jun 8 '18 at 12:37
  • $\begingroup$ If the condition is $g(X,Y)(x)>0\iff X_x\neq 0$, then I can show it. However the condition is $X_x\neq 0\Rightarrow g(X,X)(x)$. Unless, I have miscopied form the blackboard... $\endgroup$ – tomak Jun 8 '18 at 12:47
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    $\begingroup$ Can you show that $g(X,Y)(x) = 0$ if $X_x = 0$? The statement $g(X,Y) > 0 \Leftrightarrow X_x \ne 0$ is definitely wrong because it would imply $X_x \ne 0 \Leftrightarrow Y_x \ne 0$ (use $g(X,Y) = g(Y,X)$). $\endgroup$ – Paul Frost Jun 8 '18 at 13:44
  • $\begingroup$ Yes I can. $X_x = 0\Rightarrow g_x(X_x,Y_x) = 0\Rightarrow g(X,Y)(x) = 0$ and this for all $Y$ and for all $X$ such that $X(x) = X_x$ $\endgroup$ – tomak Jun 12 '18 at 17:26
  • $\begingroup$ Ok. I get it. Now suppose that $X,X'$ are such that $X(x) = X'(x)$ $\endgroup$ – tomak Jun 12 '18 at 17:28
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First we show that if $X_x = 0$, we get that $g(X,Y)(x) = 0$ for all $X$ such that $X(x) = X_x$ and for all $Y$.

$X_x = 0\Rightarrow g_x(X_x,Y_x) = 0$ by linearity. So $g(X,Y)(x) = g_x(X_x,Y_x) = 0$

This implies our result since then, if we take $X,X'$ such that $X(x) = X'(x) = X_x$,

$$g(X,Y)(x) = g(X',Y) \iff g(X-X',Y)(x) = 0$$

Then, since $(X-X')_x = X(x)-X'(x) = X_x-X_x = 0$, we have our result using the first part.

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