# Intrinsic, coordinate-free interpretation of "inverse" of a metric tensor

I am taking a course on general relativity. There is some part where I think they are abusing terminology and I would like to interpret more properly.

Given an indefinite metric tensor $g$ on a smooth manifold $M$, write $g$ in component form in local coordinates as $g_{\mu\nu}$. We define the following contravariant tensor in component form in local coordinates as $g^{\mu\nu}$, where $g^{\mu\nu}g_{\nu\lambda}=g_{\lambda\nu}g^{\nu\mu}=\delta_\lambda^\mu$, i.e. $g^{\mu\nu}$ is (the components of) the "inverse" of the metric tensor.

The above is how they defined $g^{\mu\nu}$. I find it problematic to me. Firstly, the definition relies on coordinates. Secondly, metric tensors are type $(0,2)$ tensors instead of type $(1,1)$, i.e. they are not linear transformations. Without reference to coordinates, there is no such thing as "inverse" of type $(0,2)$ tensor.

I propose two ways to define and interpret $g^{\mu\nu}$ in an as-coordinate-free-as-possible way:

1. While $g$ is nondegenerate, it induces an isomorphism between a tangent space $T_pM$ and its dual $T_p^*M$ by $v\mapsto g_p(v,\cdot)$. This isomorphism transfers the indefinite inner product $g_p$ from $T_pM$ to $T_p^*M$. This inner product on $T_p^*M$ is type $(2,0)$ on the manifold, and $g^{\mu\nu}$ would be the local coordinates expression of this inner product.

2. While $g$ is nondegenerate, hopefully there exists a unique type $(2,0)$ tensor on $M$, denoted $g^{-1}$ (this is NOT an inverse of $g$. It is just a symbol) for which forming the tensor products $g^{-1}\otimes g$ and $g\otimes g^{-1}$ and then do a tensor contraction on each of them would both give the identity linear transformation on each tangent space. Then $g^{\mu\nu}$ would be the local expression of $g^{-1}$.

Are the above interpretations of $g^{\mu\nu}$ correct, i.e. can we recover the equation $g^{\mu\nu}g_{\nu\lambda}=g_{\lambda\nu}g^{\nu\mu}=\delta_\lambda^\mu$? I am quite sure interpretation 2 is correct, because multiplying the two components is corresponds to tensor product, and making an upper index and a lower index equal and sum over that index corresponds to tensor contraction. But I am not very sure about interpretation 1.

• It is definitely true that $g^{\mu\nu}$ are the metric components of $T^*M$ in local coordinates. Approach number 1 seems to me very natural and healthy. Mar 12, 2018 at 20:22
• I think you answered your own question in Interpretation 1 Mar 13, 2018 at 14:42

Let's start with your definition 1, and prove $$g_{\lambda\nu}g^{\nu\mu}=\delta^\mu_\lambda$$ ($g^{\mu\nu}g_{\nu\lambda}=\delta^\mu_\lambda$ will then follow by symmetry of $g$ and $g^{-1}$).
Let's call the isomorphism $v\mapsto g_p(v,-)$ by the name $k$. Then our definition of $g^{-1}$ is $$g^{-1}_p(f,h)=g_p(k^{-1}f,k^{-1}h)$$ (note that $g^{-1}$ is clearly symmetric).
To prove that $$g_{\lambda\nu}g^{\nu\mu}=\delta^\mu_\lambda$$ it suffices to show that $$x^\lambda g_{\lambda\nu}g^{\nu\mu}f_\mu=x^\mu f_\mu$$ for all $x$ and $f$. The left hand side is equal to $$g^{-1}_p(g_p(x,-),f)$$ which by definition of $k$ is $$g^{-1}_p(kx,f).$$ Now applying the definition of $g^{-1}$ gives $$g_p(k^{-1}kx,k^{-1}f)$$ which simplifies to $$g_p(x,k^{-1}f).$$ Using symmetry we have $$g_p(k^{-1}f,x)$$ which is just $$(kk^{-1}f)x$$ which simplifies to $$fx.$$
You can indeed say that $g_{ik}g^{kj}=\delta_i^j$. This is because the definition of $g^{ij}$ is that in each basis its components are such that it is the inverse of $g_{jk}$.