Errata on Tensors (Jun 2nd, 2022)

February this year I posted some comments on how tensors show up in geometry. However, having enrolled in Claudio Gorodski's course in Riemannian Geometry at IME-USP, I now know a thing or two about tensors I didn’t knew at the time. I’m proud to say that most of what I said stands up to the test of time, but I would still like to make some corrections.

If you’re interested in Riemannian geometry, I strongly recommend Claudio’s course notes.

First and foremost, in the previous post I explicitly stated that there are vector bundles $E_i\to M$ and $F\to M$ such that the map

that takes a tensor $T\in \Gamma \left(\text{Hom}\left(E_1\otimes \cdots \otimes E_n,F\right)\right)$ to the homomorphism $iT:\Gamma \left(E_1\otimes \cdots \otimes E_n\right)\to \Gamma \left(F\right)$ given by $\left(iT\right){\left({\xi }^1,\dots ,{\xi }^n\right)}_p=T_p\left({\xi }_p^1,\dots ,{\xi }_p^n\right)$ is not surjective. This is not the case: every $C^{\infty }\left(M\right)$-multilinear map $T:\Gamma \left(E_1\right)\times \cdots \times \Gamma \left(E_n\right)\to \Gamma \left(F\right)$ is the image of some tensor, which is to say, the value of $T{\left({\xi }^1,\dots ,{\xi }^n\right)}_p$ depends only on the values of ${\xi }_p^i$.

This result is known as the tensor field characterization lemma, and a proof can be found in here. This particular proof is specific to the case where each $E_i=TM$ and $F=M\times ℝ$ — which is to say, the proof is specific to the case where $T:X\left(M\right)\times \cdots \times X\left(M\right)\to C^{\infty }\left(M\right)$ — but I believe the same argument should work for arbitrary smooth vector fields. The proof, however, is heavily reliant on the existence of smooth partitions of unity, which hints at the fact that perhaps the lemma does not hold for holomorphic vector fields over complex manifolds — i.e. perhaps there some complex manifold $X$ with holomorphic vector fields $E_i\to X$ and an $O_X$-multilinear map $T:\Gamma \left(E_1\right)\times \cdots \times \Gamma \left(E_n\right)\to \Gamma \left(F\right)$ which is not the image of a holomorphic tensor.

The fact that this proof is restricted to the particular case where $E_i=TM$ and $F=M\times ℝ$ is not a coincidence: in most cases we are interested in tensors with the signature $X\left(M\right)\times \cdots \times X\left(M\right)\times \Gamma \left(T^{*}M\right)\times \cdots \times \Gamma \left(T^{*}M\right)\to C^{\infty }\left(M\right)$. The space of tensors which take $s$ fields and $r$ cofields is usually referred to as $T^{\left(r,s\right)}M$. In particular, if $M$ is a Riemannian manifold with metric $g$, the canonical isomorphism $T_{p}M\to T_p^{*}M$ given by the metric $g_p$ induces an isomorphism of $C^{\infty }\left(M\right)$-modules $X\left(M\right)\to \Gamma \left(T^{*}M\right)$.

Another way to put it is to say that $g$ induces an isomorphism $T^{\left(0,1\right)}M\to T^{\left(1,0\right)}M$, which in turn induces the so called musical isomorphisms $♯:T^{\left(s,r+1\right)}M\to T^{\left(s+1,r\right)}M$ and $♭:T^{\left(s+1,r\right)}M\to T^{\left(s,r+1\right)}M$. In particular, $T^{\left(r,s\right)}M$ is canonically isomorphic to $T^{\left(0,r+s\right)}M$ via ${♯}^s$, which is the space of tensors with signature $X\left(M\right)\times \cdots \times X\left(M\right)\to C^{\infty }\left(M\right)$ — so that the proof of the lemma for this particular case is sufficient. That about wraps it up. Hope this helped someone 😛