# Intuitive explanation for metric inverse definition

Let $$V$$ be a tangent space at a point of a manifold, with an associated metric $$g_{ab}: V \times V \mapsto \mathbb{R}$$. The inverse $$g^{ab}$$ of the metric $$g_{ab}$$ is defined as:

$$g^{ab}g_{bc} = \delta^a_c$$

With $$\delta^a_c: V^* \times V \mapsto \mathbb{R}$$ being interpreted as the identity map $$V \mapsto V$$, i.e. given a $$v \in V$$, we're left with $$\delta^a_c: V^* \mapsto \mathbb{R}$$ which can be associated with a vector from $$V$$, which accordingly should be the same vector as $$v$$.

This is all written using abstract index notation, hence the latin letters do not refer to individual components in a specified basis but rather slots of the tensors, with the two operations:

1. Contraction $$CT = T^{ab \dots k \dots}_{a'b' \dots k \dots}$$ being a repetition of indices to denote a sum over the tensor with basis vectors inserted into corresponding slots
2. Outer product $$T_1 \otimes T_2 = (T_1)^{ab \dots}_{a'b' \dots}(T_2)^{a''b'' \dots}_{a'''b''' \dots}$$ being tensors written next to each other to denote tensor product defined in the usual sense.

I tried seeing whether this definition makes sense by taking the outer product of $$g^{ab}$$ and $$g_{bc}$$, then contracting, obtaining the following result:

$$g^{ab}g_{bc} = g^{-1}(-, v^{*b})g(v_b, -)$$ Where $$\{v^{*b}\}$$ is an (arbitrary) basis of $$V^*$$ and $$\{v_b\}$$ an (arbitrary) basis of $$V$$. Seeing this result, I figured that since tensors are multilinear, and $$g(v_b, -)$$, gives a real number, I can move $$g$$ into the second slot:

$$g^{-1}(-, v^{*b})g(v_b, -) = g^{-1}(-, v^{*b}g(v_b, -))$$

Now I do see how this new tensor $$g^{ab}g_{bc}$$ takes a dual vector and a vector and gives a real number (just like $$\delta^a_c$$), however I'm finding it difficult to see how it is the identity map. Surely, using the summation convention makes it easy to see, if the metrics are viewed as matrices, but I'm trying to see how this works out without translating it into tensor components.

Moreover, I don't find it intuitive to use this identity map in tensor relations. For example (from Wald's book on General Relativity):

$$g^{ab} = g^{ac}g^{bd}g_{cd}$$

By evaluating the product on the RHS from right to left, and raising/lowering indices, I see how this becomes true. But, assuming that the order of the product doesn't matter, wouldn't the following also be true:

$$g^{ab} = (g^{ac}g_{cd})g^{bd} = \delta^a_d g^{bd}$$

I don't find it intuitive how $$\delta^a_d g^{bd} = g^{ab}$$.

To summarize my questions:

1. Is there any intuitive explanation for why $$g^{ab}g_{bc} = \delta^a_c$$? I understand that this is a definition, but could it become intuitively clear using outer products and contraction?
2. In the example from Wald's book, is there any explanation for how $$\delta^a_d g^{bd} = g^{ab}$$? More generally, is there a simple relation involving the $$\delta^a_b$$-tensor, when acting on other tensors, such as $$\delta{v} = v$$ for an ordinary identity map?

Let $$M$$ be an riemannian manifold, $$p \in M$$ a point and $$T_pM$$ the tangent space at $$p$$. The metric tensor $$g$$ at a point $$p$$ is an inner product $$g_p : T_pM \times T_pM \to \mathbb{R}$$. Since it is an inner product in $$T_pM$$, we can define a linear transformation $$\hat{g}_p : T_pM \to T_pM^*$$, where $$T_pM^*$$ is the dual vector space of $$T_pM$$, by $$\hat{g}_p(v) = g_p(v,.)$$. The non-degeneracy of $$g_p$$ implies that $$\hat{g}_p$$ is an isomorphism between vector spaces.

In coordinates, you can check that $$g_{ab}$$ at the point $$p$$ is just the matrix that represents the map $$\hat{g}_p$$. Therefore $$g^{ab}$$ is the matrix representation of the inverse map $$\hat{g}^{-1}_p$$. When we use the metric tensor to raise or lower indices, we are basically applying the maps $$\hat{g}$$ and $$\hat{g}^{-1}$$ to vector and covector fields.

For last, when you write $$g_{ab}$$, the label of the index don't really matter as long as it is consistent inside a summation. Since the symbol $$\delta^{a}_b$$ is just the identity map in coordinates, $$g^{db} \delta^{a}_d$$ is just changing the label of the index from $$d$$ to $$a$$. This is really useful for tracking how the indices are changing after a big sum.

Also, you copied the example wrong, it's $$g^{ab} = g^{ac}g^{bd}g_{cd}$$, otherwise, we would have $$g_{ab} = g^{ab}$$ which is rarely the case.

• Thanks! One note is that I'm using abstract index notation, so there is no assumed summation in my post (even though the equations look the same). The labels don't represent components in a particular basis but are rather symbolic. I do understand how $g^{ab}g_{bc} = \delta^a_c$ if you think of it in terms of matrices, but I was wondering if it can make intuitive sense thinking in terms of coordinate-independent maps? Aside from that, thinking of $\delta^a_b$ as changing the label of the index is a great explanation and helped me understand it intuitively!
– Max
Commented Jan 5, 2022 at 17:01
• I forgot that Wald uses a weird notation, but in my notation, the coordinate independent way to express $g^{ab}g_{bc}= \delta^a_b$ at the point $p$ is by $\hat{g}^{-1} \circ \hat{g} = Id$. The metric tensor itself doesn't have an "inverse", so you need to see it as an linear transformation to define $g^{ab}$ in coordinates, then you can define a tensor $\tilde{g}$ that is an inner product in the dual space such that its coordinate expression is given by $g^{ab}$. Commented Jan 5, 2022 at 20:26
• Wald's notation is not really different from the normal coordinate notation. He only wants to get extra attention to the fact that tensors are coordinate-free objects and acts on coordinate free objects, and if it is needed, you can always write your equations using coordinates when it is relevant. With his notation you can just change the labels from roman letters to greek letters and assume summation with repeated up-down indices. Commented Jan 5, 2022 at 20:29