Commutation of metric tensor

1. Feb 13, 2009

emma83

Hello,

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 $$g_{\mu \nu}$$ that $$g_{ab}g_{cd}=g_{cd}g_{ab}$$ ?

I was trying to find a formal proof for that, and was wondering if we could use the relations:
(1) $$g_{ab}=\frac{\partial x^{c}}{\partial x^{a}} \frac{\partial x^{d}}{\partial x^{b}} g_{cd}$$
(2) $$g_{cd}=\frac{\partial x^{a}}{\partial x^{c}} \frac{\partial x^{b}}{\partial x^{d}} g_{ab}$$

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

Does that make sense ?!

2. Feb 13, 2009

malawi_glenn

in

$$g_{ab}g_{cd}=g_{cd}g_{ab}$$

$g_{ab}$ is just a number, and so is $g_{cd}$, and numbers commute.

for instance, $g_{02} = 0$, $g_{00} = 1$ (or $g_{00} = -1$) depending on which your original definition of metric is)

3. Feb 13, 2009

emma83

Argh! Thanks a lot, again I mixed up "tensors" with "tensor components"...
So the commutation relation I wrote holds for the tensor components of any tensor (not only the metric), doesn't it ?

On the other hand, the tensor (objects) do not necessarily commute with each other, right ?

4. Feb 13, 2009

malawi_glenn

no the tensor objects can be non-commuting things like matricies, e.g.

$$\sigma ^{\mu \nu} = \frac{i}{2}[\gamma ^\mu, \gamma^\nu]$$
where the gamma's are dirac 4x4 matricies