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I am still having a hard time with tensors...

The answer is probably obvious, but is it always the case (for an arbitrary metric tensor [tex]g_{\mu \nu}[/tex] that [tex]g_{ab}g_{cd}=g_{cd}g_{ab}[/tex] ?

I was trying to find a formal proof for that, and was wondering if we could use the relations:

(1) [tex]g_{ab}=\frac{\partial x^{c}}{\partial x^{a}} \frac{\partial x^{d}}{\partial x^{b}} g_{cd}[/tex]

(2) [tex]g_{cd}=\frac{\partial x^{a}}{\partial x^{c}} \frac{\partial x^{b}}{\partial x^{d}} g_{ab}[/tex]

And then multiply the left-hand side of (1) and (2) together and use the fact that the fractions of partial derivatives commute with the metric tensor and cancel each other to get:

[tex]g_{ab}g_{cd}

=(\frac{\partial x^{c}}{\partial x^{a}} \frac{\partial x^{d}}{\partial x^{b}} g_{cd}) (\frac{\partial x^{a}}{\partial x^{c}} \frac{\partial x^{b}}{\partial x^{d}} g_{ab})

= \frac{\partial x^{c}}{\partial x^{a}}\frac{\partial x^{a}}{\partial x^{c}} \frac{\partial x^{d}}{\partial x^{b}} \frac{\partial x^{b}}{\partial x^{d}} g_{cd} g_{ab}

= g_{cd}g_{ab}[/tex]

Does that make sense ?!

Thanks for your help...

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# Commutation of metric tensor

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