Let $V$ be a finite-dimensional vector space and let $V^{*}$ denote its dual vector space. A tensor, $T,$ of type $(k,l)$ over $V$ is a multilinear map $$T: V^{*} \times ... V^{*} \times V \times ... \times V \to \mathbb{R}$$ with $k$ copies of $V^{*}$ and $l$ copies of $V.$
The metric tensor $g$ is a tensor of type $(0,2)$ denoted $g_{ab}$. The inverse of $g_{ab}$ (which exists on account of the nondegeneracy of $g_{ab}$) is a tensor of type $(2,0)$ and is denoted simply as $g^{ab}.$ Thus, by definition, $g^{ab}g_{bc}=\delta^{a}_{c}$ where $\delta^{a}_{c}$ is the identity map (viewed as a map from $V_p$ into $V_p$).
In general, raised or lowered indices on any tensor denote application of the metric or inverse metric to that slot. Thus, for example, if $T^{abc}_{de}$ is a tensor of type $(3,2),$ then $T^{a}\ _{b}\ ^{cde}$ denotes the tensor $g_{bf}g^{dh} g_{ej} T^{afc}_{hj}.$
Reference: General Relativity; Robert M. Wald; Chapter 2
I have a few questions...
- What does it mean to be the "inverse" of the metric? I assume it is the following sense: A tensor of type $(0,2)$ can be viewed as inducing two maps from $V$ to $V^{*}$. Because $g$ is symmetric, these maps are the same. So the inverse of $g$ is the corresponding map from $V^{*}$ to $V.$ Is this correct?
In the above would it have meant the same thing if Wald wrote $T^{acde}_{b}$ instead of $T^{a}\ _{b}\ ^{cde}$? Or is there a distinction between the two?
In general, $g_{bf}T^{af} = T_{a}^{b}.$ But now if we let $T$ be $g$ itself this implies $g_{bf}g^{af} = g_{a}^{b} = \delta_{a}^{b}.$ But it does not seem right that $g_{a}^{b} = \delta_{a}^{b}$ is always true. So what is the issue here?