Check if Matrix Root of Inverse Metric Tensor is a Tensor

Question

According to the Square Root Theorem for tensors (e.g. see here, here, or here), any positive definite symmetric tensor $A$ has a unique positive definite symmetric tensor $B$ such that $A=B^2$. (This appears to be a matrix-oriented view, just talking about rank-2 tensors.)

My question is specifically about the inverse metric tensor $g^{ij}$, which is a symmetric positive definite (assuming $g$ is SPD, so too is $g^{-1}$) type-(2,0) contravariant tensor.

Thus, there exists $\sigma$ such that $\sigma^2=g^{-1}$. How can I prove $\sigma$ is a tensor (i.e. its transformation law) if it indeed is one?

I know that we have the following transformation law: $$ \bar{g}^{ij} = \frac{\partial \bar{x}^i}{\partial x^a} \frac{\partial \bar{x}^j}{\partial x^b} g^{ab} $$ or $\bar{g}^{-1} = J^{-1}gJ^{-T}$.

In matrix terms, I can say $\bar{\sigma}^2 = J^{-1}\sigma\sigma J^{-T}$. Then because $\sigma$ is symmetric I can write $\bar{\sigma}\bar{\sigma}^T=J^{-1}\sigma(J^{-1}\sigma)^T$ and so maybe conclude $\bar{\sigma}=J^{-1}\sigma$.

Can anyone shed some light on this? In other words, if $\sigma$ is the matrix field computed from the matrix square root of the (matrix) components of $g^{-1}$, do the components of $\sigma$ form a tensor?

Answer 1

Since I'm not sure how to interpret your question, let me discuss the linear algebra picture first. Given a finite dimensional real vector space $V$, a $(1,1)$ tensor is an element of $V^{*} \otimes V \cong \operatorname{Hom}(V,V)$, i.e a linear map. It makes sense to ask whether a linear map $T$ has a square root (a map $S$ such that $S^2 = T$) but it doesn't make any sense to ask whether $T$ is symmetric without an additional structure so let us put an inner product on $V$. Then we define what is a symmetric $(1,1)$ tensor and the references you quote show that a $(1,1)$ positive-definite tensor has a unique positive-definite square root (which is also a $(1,1)$ tensor).

Now, a $(2,0)$ tensor is an element of $V^{*} \otimes V^{*}$ which can be identified with a bilinear form on $V^{*}$. Now, it makes sense to talk about a symmetric and positive-definite $(2,0)$ bilinear form without any additional structure but it doesn't make any sense to talk about the square root of a bilinear form.

Returning to the manifold setting, if you have a metric, you can use it to convert any $(0,2)$ (or $(2,0)$) tensor field to a $(1,1)$ tensor field by raising (or lowering) indices and then ask whether the resulting tensor field has a square root. However, if you do it to the metric (the metric inverse), you'll get the identity tensor which is the square root of itself.