Raising and lowering indices using metric tensor

[cianfa72]

TL;DR

How to use metric tensor to raise and lower indices of the inverse metric tensor

I'm in trouble with the notations used to rise or lower indices in tensor calculus.
Consider the following:

$$g_{\mu \nu} g^{\nu \rho} \text{ vs. } g_{\mu \nu} g^{\rho \nu}$$
Pick the former: there two way to see it: in the first we can say $g_{\mu \nu}$ acts on $g^{\nu \rho}$ as $g_{\mu \nu} (g^{\nu \rho})$ lowering its 1st index $\nu$ substituting $\mu$ into it to get $\delta_{\mu}\text{ }^{\rho}$. Likewise we can equally say $g^{\nu \rho}$ acts on $g_{\mu \nu}$ as $g^{\nu \rho} (g_{\mu \nu})$ raising its 2nd index $\nu$ substituting $\rho$ into it to get $\delta_{\mu}\text{ }^{\rho}$. So far so good.

Now pick the latter: seen as $g_{\mu \nu} (g^{\rho \nu})$ the metric tensor acts on the inverse lowering its 2nd index $\nu$ to get $\delta^{\rho}\text{ }_{\mu}$. On the other hand we can equally say the inverse metric tensor acts ont the metric tensor as $g^{\rho \nu} (g_{\mu \nu})$ raising its 2nd index to get $\delta_{\mu}\text{ }^{\rho}$.

What's going on, what I am missing ?

 

[anuttarasammyak]

They are the same because g is a symmetric tensor.

 

[cianfa72]

They are the same because g is a symmetric tensor.

Yes, it doesn't matter at all since $g$ is symmetric. However, from the formal viewpoint of tensor calculus, the results in the second case look really different.

 

[Dale]

Yes, it doesn't matter at all since $g$ is symmetric. However, from the formal viewpoint of tensor calculus, the results in the second case look really different.

But only $g$ can be used to raise and lower indices, and $g$ is symmetric.

 

[cianfa72]

But only $g$ can be used to raise and lower indices, and $g$ is symmetric.

So your point is: in the second case $g_{\mu \nu} g^{\rho \nu}$ seen as $g_{\mu \nu} (g^{\rho \nu})$, before calculating the contraction, swap the indices of the inverse metric getting $g_{\mu \nu} (g^{\nu \rho})$, therefore $$g_{\mu \nu} g^{\nu \rho} = \delta_{\mu}\text{ }^{\rho}$$ as in the first case.

 

[SiennaTheGr8]

To be clear, the Kronecker delta is symmetric too.