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...

  1. 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?
  1. 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?

  2. 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?


2 Answers 2


Keep in mind that the dual space $V^*$ can be viewed as the space of all linear maps from $V \to R$. For clarity, I'll use a functional notation for the various maps.

  1. The metric $g_{ab}$ is a map from $V \times V \to R$, but can equally well (as you say) be viewed as a map from $V \to V^*$ sending (with an abuse of notation for clarity) $v$ to the unique linear map $g_{ab}(v) = h \in V^*$ which sends $w \mapsto h(w) \equiv g_{ab}(v,w)$. In this view, the inverse of $g_{ab}$ is a map $g^{ab}$ from $V^* \to V$ such that $g^{ab} \circ g_{ab} = \mathrm{id}_{V}$ and $g_{ab} \circ g^{ab} = \mathrm{id}_{V^*}$.

  2. No, the two expressions $T^{acde}_{b}$ and $T^{a}{} _{b}{} ^{cde}$ are not equal. When the metric tensor is used to raise or lower an index, the resulting objects might belong to spaces which maps the same number of copies of $V^*$ and $V$ to $R$, no matter which index is raised or lowered, but when you pay attention to the order of the parameters passed to the resulting map, they are in fact different multilinear maps with distinct input spaces. In a functional notation, you can think of the function $T$ being composed with the function $g$ on one of its inputs or outputs. To get a more visual or intuitive sense of this, you can look at Penrose's graphical tensor notation -- though I wouldn't recommend calculating with it.

  3. $g_a^b$ is just notational sugar for $g_{ac} g^{bc}$, and the fact that the components of this are equal to the Kronecker delta tensor is exactly because $g^{ab}$ is the inverse of $g_{ab}$, in the sense described in point 1.

(My answer probably does not add much beyond J. Murray's, but I'd begun writing it and didn't want to throw the work away)

  • $\begingroup$ I just edited in the tensors in the second part: I assume this is what you meant? $\endgroup$ Commented Jan 27, 2021 at 4:44
  • $\begingroup$ @NiharKarve It is, thanks! It also reminded me that I mean to say something about the difference between the objects from the view of function composition, and recommend at least looking at how Penrose's graphical notation treats them. $\endgroup$
    – jwimberley
    Commented Jan 27, 2021 at 13:45
  1. Yes. A $(0,2)$-tensor can be thought of as a map from $V$ to $V^*$, and if the tensor is non-degenerate (as a Lorentzian metric must be), then this map is invertible. The inverse metric is indeed the inverse map from $V^*$ to $V$.
  2. In general, the order of the indices matters, which means that they should be written as you have done. Otherwise the meaning of an expression like $T^a_b$ is ambiguous.
  3. It follows immediately from point 1 that in component notation, $g^{ab}g_{bc}= \delta^a_{\ \ c}$, as you have shown. This is true simply by definition of what it means to be an inverse transformation. Given a tensor with components $T^{ab}$, the expression $T^a_{\ \ b}$ is by definition shorthand for the longer expression $g_{bc}T^{ac}$. Applying this shorthand to the metric yields the expression you wrote.

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