# Components of the metric tensor and its inverse

Let $g$ denote a metric tensor. Then Wald writes (in his book on general relativity): "The inverse of $g_{ab}$...is a tensor of type $(2,0)$ and could be denoted as $(g^{-1})^{ab}$. It is convenient, however, to drop the inverse sign and denote it simply as $g^{ab}$. No confusion arises from this since the upper position of the indices distinguishes the inverse metric from the metric. Thus, by definition, $g^{ab}g_{bc} = \delta^{a}_c$ (viewed as a map from $V_p$ into $V_p$)".

I understand all of this fine. My only problem is with the notation. If we view g as a map from the tangent space into the cotangent space, then $g^{-1}$ is map from the cotangent space into the tangent space, and hence $g^{-1} \circ g$ is the identity map from the tangent space to the tangent space. Makes perfect sense. But, I don't see how the general component of $g^{-1} \circ g = \delta$ (the identity map on the tangent space) is given by $g^{ab}g_{bc}$.

• sorry, but I don't understand what your question is. Commented May 6, 2016 at 3:05
• $g^{-1} \circ g = 1_V$ Commented May 6, 2016 at 3:06
• Are you confused with the Einstein summation notation? $g^{ab} g_{bc}$ means $\sum_{b=1}^n g^{ab} g_{bc}$.
– user99914
Commented May 6, 2016 at 3:35
• This is really matrix multiplication
– user99914
Commented May 6, 2016 at 3:36
• I'm fine with the matrix multiplication. Is it sort of like how when you compose two linear transformations, then the matrix representation of the composition is the product of the matrices representing the original linear maps? Commented May 6, 2016 at 3:46

One can use the matrix for $g$ within the construction of the reciprocal basis $$\partial^k=\sum_sg^{ks}\partial_s,$$ or simply $\partial^k=g^{ks}\partial_s$.
It happens that $\partial_k\mapsto\partial^k$ is a change of basis in the tangent space.