Definition of isometry in components form

[maxverywell]

From the book of Nakahara "Geometry, Topology and Physics":

A diffeomorfism $f:\mathcal{M}\to \mathcal{M}$ is an isometry if it preserves the metric:
$f^{*}g_{f(p)}=g_{p}$

In components this condition becomes:
$\frac{\partial y^{\alpha}}{\partial x^{\mu}}\frac{\partial y^{\beta}}{\partial a^{\nu}}g_{\alpha\beta}(f(p))=g_{\mu\nu}(p)$
where x and y are the coordinates of p and f (p), respectively.

I don't understand it. I know that a metric transforms under coordinate transformation $x^{\mu}\to x'^{\mu}$ (coordinates of the same point p - passive transformation). But here we have two different points. It doesn't make sense. I know that this has to do with passive vs active coordinate transformation but cannot understand it.

 

[WannabeNewton]

That definition is not exactly complete. An isometry is a diffeomorphism $\varphi: (M,g)\rightarrow (N,g')$, where $M,N$ are Riemannian manifolds, such that the pullback $\varphi^*$ satisfies $g = \varphi^{*}g'$. The meaning of this is quite simple and clear. Let $p\in M$ and $v,w\in T_pM$. If $\varphi$ is an isometry then by the above definition it immediately follows that $g_{p}(v,w) = g'_{\varphi(p)}(\varphi_{*}v,\varphi_{*}w)$ i.e. the isometry preserves the inner product between vectors when they are sent from one tangent space to another under the isometry.

 

[Bill_K]

$\frac{\partial y^{\alpha}}{\partial x^{\mu}}\frac{\partial y^{\beta}}{\partial a^{\nu}}g_{\alpha\beta}(f(p))=g_{\mu\nu}(p)$
where x and y are the coordinates of p and f (p), respectively.

The LHS is the metric in the new coordinate system at the new point. The equality just says the metric at the new point in the new coordinates is equal to the metric at the old point.

 

[maxverywell]

WannabeNewton, I have difficulty in proving the component form of the equation.

The LHS is the metric in the new coordinate system at the new point. The equality just says the metric at the new point in the new coordinates is equal to the metric at the old point.

Yeah. But this is what I want to prove.

Let me show you how I understand it and please let me know if I am correct.

Tensor transformation law:
$g'_{\mu\nu}(x')=\frac{\partial x^{\alpha}}{ \partial x'^{\mu}}\frac{\partial x^{\beta}}{\partial x'^{\nu}}g_{\alpha\beta}(x)$

where $\{x^{\mu}\}$ and $\{x'^{\mu}\}$ are the coordinate systems covering the neighborhood $U\subset M$ of the point $p$. We call the transformation of the coordinates $x\to x'$ "passive" because we are referring to the same point.

The diffeomorfism $f: M\to M$ is an "active" coordinate transformation because we are "moving" points on the manifold, $p\to f(p)$. Let's suppose that the coordinate system covering the neighborhood $V$ of the point $f(p)$, is $\{y^{\mu}\}$.

However, we can consider the diffeomorfism $f$ from an passive point of view (Wald) by defining a new coordinate system in the neighborhood $O=f^{-1}[V]$ of the point $p$, $\{x'^{\mu}\}$, by setting $x'(q)=y(f(q))$, $q\in O$.
For the metric tensor this means $g'_{\alpha\beta}(x')=g_{\alpha\beta}(y)$ (right?). So

$g_{\mu\nu}(x)=\frac{\partial x'^{\alpha}}{ \partial x^{\mu}}\frac{\partial x'^{\beta}}{\partial x^{\nu}}g'_{\alpha\beta}(x')=\frac{\partial x'^{\alpha}}{ \partial x^{\mu}}\frac{\partial x'^{\beta}}{\partial x^{\nu}}g_{\alpha\beta}(y)$

and because $x'(p)=y(f(p))$, we have finally:

$g_{\mu\nu}(x)=\frac{\partial y^{\alpha}}{ \partial x^{\mu}}\frac{\partial y^{\beta}}{\partial x^{\nu}}g_{\alpha\beta}(y)$

 

[WannabeNewton]

$$g'_{\varphi(p)}(\varphi_{*}\partial_{i},\varphi_{*}\partial_{j}) = g'_{\varphi(p)}(\frac{\partial x^{l'}}{\partial x^i}\partial_{l'}, \frac{\partial x^{k'}}{\partial x^j}\partial_{k'}) = \frac{\partial x^{l'}}{\partial x^i} \frac{\partial x^{k'}}{\partial x^j}g'_{lk} = g_{ij}$$

EDIT: So yes, I agree with what you said for the case of coordinate transformations.

 

[maxverywell]

$$g'_{\varphi(p)}(\varphi_{*}\partial_{i},\varphi_{*}\partial_{j}) = g'_{\varphi(p)}(\frac{\partial x^{l'}}{\partial x^i}\partial_{l'}, \frac{\partial x^{k'}}{\partial x^j}\partial_{k'}) = \frac{\partial x^{l'}}{\partial x^i} \frac{\partial x^{k'}}{\partial x^j}g'_{lk} = g_{ij}$$

Are the coordinates x and x' the coordinates of p and f(p) respectively?

Also, why $\varphi_{*}\partial_{i}=\frac{\partial x^{l'}}{\partial x^i}\partial_{l'}$?

 

[WannabeNewton]

That is what the pushforward does under a change of coordinates to the coordinate representation of the original coordinate basis.

 

[maxverywell]

So $\phi_* \partial_i= \partial_i$?
This would mean that under isometries the basis vectors don't change.

 

[WannabeNewton]

So $\phi_* \partial_i= \partial_i$?
This would mean that under isometries the basis vectors don't change.

Oops, let me be more precise. Let $(U,\varphi)$ and $(V,\psi)$ be two charts on a smooth manifold $M$ and let $p\in U\cap V$. Say $\varphi$ has coordinate functions $(x^{i})$ and that $\psi$ has coordinate functions $(x^{i'})$. The transition map $(\psi\circ \varphi^{-1}):\varphi(U\cap V)\rightarrow \psi(U\cap V)$ is given by $(\psi\circ \varphi^{-1})(x) = (x^{1'}(x),...,x^{n'}(x))$. The pushforward of the transition map then gives $(\psi\circ \varphi^{-1})_{*}\partial_{i}|_{\varphi(p)} = \frac{\partial x^{j'}}{\partial x^{i}}(\varphi(p))\partial_{j'}|_{\psi(p)}$.

 

[maxverywell]

Oops, let me be more precise. Let $(U,\varphi)$ and $(V,\psi)$ be two charts on a smooth manifold $M$ and let $p\in U\cap V$. Say $\varphi$ has coordinate functions $(x^{i})$ and that $\psi$ has coordinate functions $(x^{i'})$. The transition map $(\psi\circ \varphi^{-1}):\varphi(U\cap V)\rightarrow \psi(U\cap V)$ is given by $(\psi\circ \varphi^{-1})(x) = (x^{1'}(x),...,x^{n'}(x))$. The pushforward of the transition map then gives $(\psi\circ \varphi^{-1})_{*}\partial_{i}|_{\varphi(p)} = \frac{\partial x^{j'}}{\partial x^{i}}(\varphi(p))\partial_{j'}|_{\psi(p)}$.

Actually I am more confused now.
At first you said that $\varphi$ is diffeomorphism $\varphi:(M,g)\to (N,g')$ and now it's the local coordinate map $\varphi:M\to\mathbb{R}^n$. We want the push-forward $\varphi_*$.

 

[WannabeNewton]

If we are just talking about a change of coordinates, then all we care about is the transition map that goes from one chart to the other, at some point on the manifold; the diffeomorphism is the transition map, not the coordinate map itself (transition maps are automatically diffeomorphisms by definition of a smooth atlas). I was using a single symbol to denote the transition map before, which was certainly my mistake since I did the opposite in post #2, so I cleaned that up in post #9. See also Lee "Introduction to Smooth Manifolds" page 72.

EDIT: Let me rewrite the line here in a clearer fashion. $g_{ij} = g(\partial_{i},\partial_{j}) = \frac{\partial x^{k'}}{\partial x^{i}}\frac{\partial x^{l'}}{\partial x^{j}}g(\partial_{k'}, \partial_{l'}) = \frac{\partial x^{k'}}{\partial x^{i}}\frac{\partial x^{l'}}{\partial x^{j}}g_{k'l'}$. The change of coordinate is taken into account implicitly from the way the coordinate basis transforms.

Now, for a general endomorphic isometry $f:M\rightarrow M$, we will have the following: $g_{ij}(p) = g_{f(p)}(f_{*}\partial_{i},f_{*}\partial_{j}) = g_{f(p)}(\frac{\partial f^{k}}{\partial x^{i}}\partial_{k}, \frac{\partial f^{l}}{\partial x^{j}}\partial_{l}) = \frac{\partial f^{k}}{\partial x^{i}}\frac{\partial f^{l}}{\partial x^{j}}g_{kl}(f(p))$

What was confusing me before with regards to your notation is that you kept mixing up coordinate transformation notation with that of endomorphic isometries so I couldn't tell what exactly it was you wanted.

 

[maxverywell]

Thnx WannabeNewton, especially for mentioning the Lee's book, it's very nice. and I started reading it.

 

[WannabeNewton]

and I started reading it.

Good, good. I'm sure you'll like it! Don't forget to have fun with it :). I should warn you though that Lee's book assumes you have done topology (mostly point-set topology and some algebraic for the later parts) so if you haven't done topology before then you might want to brush up on that first. Cheers!