# What does the curvature of a surface have to do with tensor products? (Feb 1st, 2022)

The question I would like to address in this article is: what is a tensor? This question has two answers. If you ask an algebraist, he (she) will tell you it is an element of the tensor product of two modules. If you ask a geometer, she (he) will ramble about “global constructions that only depend on point-wise values” for hours on end. We should note that we will primarily focus on what a tensor is from the perspective of a geometer, the intuition behind it and how we get from that to the usual formalism.

Frankly, I feel like there isn’t much to explain, yet I never had this explained to me and I always felt it was difficult reconcile my intuition with the formalism most commonly adopted. This is the primary reason I wrote this article: I would have loved to read it in the past.

In differential geometry and related fields, information can often be obtained by passing from the non-linear to the linear via infinitesimal approximations. Often times this comes in form of ${C}^{\infty }$-linear functions between the spaces of smooth sections of two fiber bundles. Specifically, if $M$ is a smooth manifold and $E\to M$, $F\to M$ are vector bundles over $M$, then the sets $\Gamma \left(E\right)$ and $\Gamma \left(F\right)$ of smooth (global) sections of $E$ and $F$, respectively, have a natural structure of a ${C}^{\infty }\left(M\right)$-modules, and sometimes linear maps $\Gamma \left(E\right)\to \Gamma \left(F\right)$ show up. If such a map $\tau :\Gamma \left(E\right)\to \Gamma \left(F\right)$ satisfies the condition that $\tau {\left(\xi \right)}_{p}$ depends only on ${\xi }_{p}$ — and not $\xi$ on its entirety — then $\tau$ is called a tensor.

Often times it is convenient to also consider multilinear maps $\tau :\Gamma \left({E}_{1}\right)×\Gamma \left({E}_{2}\right)×\cdots ×\Gamma \left({E}_{n}\right)\to F$ — i.e. maps that are linear in each coordinate. Again, if $\tau {\left({\xi }^{1},{\xi }^{2},\dots ,{\xi }^{n}\right)}_{p}$ is determined by ${\xi }_{p}^{i}$ then $\tau$ is called a tensor. The classical examples of tensors are differential forms. A perhaps more interesting example is a Riemannian metric: for each point $p\in M$ we fix a positive-definite bilinear form ${\text{g}}_{p}:{T}_{p}M×{T}_{p}M\to ℝ$ which “varies smoothly with $p$”. This construction induces a tensor

where $\left(\text{g}\left({V}^{1},{V}^{2}\right)\right)\left(p\right)={\text{g}}_{p}\left({V}_{p}^{1},{V}_{p}^{2}\right)$.

This is what a tensor is supposed to be: for each $p\in M$ we fix some multilinear function between the $p$-fibers of some vector bundles that “varies smoothly with $p$”. The meaning of “varies smoothly with $p$” is still imprecise, dare I not say unclear. We should point out that often times it is more convenient to define tensors in terms of global sections rather than defining the fiber-wise transformations, such as in the case of the curvature tensor $R\left(X,Y\right)Z={\nabla }_{X}{\nabla }_{Y}Z-{\nabla }_{Y}{\nabla }_{X}Z-{\nabla }_{\left[X,Y\right]}Z$ of a connection $\nabla$ or the Nijenhuis tensor $N\left(X,Y\right)=\left[X,Y\right]+J\left[JX,Y\right]+J\left[X,JY\right]-\left[JX,JY\right]$ of an almost complex structure $J$.

Hence the need to consider tensors in geometry. Working with multilinear maps can be a bit of an annoyance, however. It would be convenient if we could somehow look at a tensor as a straight linear map — instead of a multilinear map. This brings us to the algebraic answer to our initial question. Given a ring $R$ and two $R$-modules $M$ and $N$, their tensor product $M{\otimes }_{R}N$ is the $R$-module which enjoys the universal property that

where $\text{Bil}\left(M×N,L\right)$ is the module of $R$-bilinear maps $M×N\to L$.

In other words, $R$-multilinear maps ${M}_{1}×{M}_{2}×\cdots ×{M}_{n}\to N$ naturally correspond to $R$-linear maps ${M}_{1}\otimes {M}_{2}\otimes \cdots \otimes {M}_{n}\to N$. We should point out that the tensor product of modules can always be shown to exist by means of an explicit construction — whose elements are usually called tensors. If we fix $R=ℝ$, this construction induces a construction in the category of vector bundles over some fixed manifold $M$: if ${E}_{i}\to M$ are bundles over $M$, there is a vector bundle ${E}_{1}\otimes {E}_{2}\otimes \cdots \otimes {E}_{n}\to M$ whose fibers are

The relationship between these two notions of tensor should now be clear: tensors $\Gamma \left({E}_{1}\right)×\Gamma \left({E}_{2}\right)×\cdots ×\Gamma \left({E}_{n}\right)\to \Gamma \left(F\right)$ are called tensors because they correspond to ${C}^{\infty }\left(M\right)$-linear maps

which are in turn canonically identified with ${C}^{\infty }\left(M\right)$-linear maps

In fact, there’s a natural isomorphism of sheaves of ${C}^{\infty }$-modules $\Gamma \left(-,{E}_{1}\right){\otimes }_{{C}^{\infty }}\Gamma \left(-,{E}_{2}\right){\otimes }_{{C}^{\infty }}\cdots {\otimes }_{{C}^{\infty }}\Gamma \left(-,{E}_{n}\right)\cong \Gamma \left(-,{E}_{1}\otimes {E}_{2}\otimes \cdots \otimes {E}_{n}\right)$ 🤡

To recap: we’ve just shown that a tensor $\tau :\Gamma \left({E}_{1}\right)×\cdots \Gamma \left({E}_{n}\right)\to \Gamma \left(F\right)$ can be naturally identified with some $\tau \in {\text{Hom}}_{{C}^{\infty }\left(M\right)}\left(\Gamma \left({E}_{1}\otimes \cdots \otimes {E}_{n}\right),\Gamma \left(F\right)\right)$. A natural question to ask ourselves at this point is: does $\tau$ correspond to some $\tau \in \Gamma \left(\text{Hom}\left({E}_{1}\otimes \cdots {E}_{n},F\right)\right)$? First of all, why does this make sense? Recall that given two vector spaces $V$ and $W$, the set $\text{Hom}\left(V,W\right)$ of linear transformations $V\to W$ is again a vector space. Hence we can consider the vector bundle $\text{Hom}\left({E}_{1}\otimes \cdots \otimes {E}_{n},F\right)\to M$ whose fibers are

The previously mentioned example of Riemannian metrics does hint at an inclusion

which takes $\eta \in \Gamma \left(\text{Hom}\left({E}_{1}\otimes \cdots \otimes {E}_{n},F\right)\right)$ to $i\eta :\Gamma \left({E}_{1}\otimes \cdots \otimes {E}_{n}\right)\to \Gamma \left(F\right)$ with $i\eta {\left(\xi \right)}_{p}={\eta }_{p}\left({\xi }_{p}\right)$ — notice this is precisely what we did to get from “a bilinear form in ${T}_{p}M$ for each $p\in M$” to a Riemannian metric seen as a tensor. The meaning of “a transformation at each fiber $p$ that varies smoothly with $p$” is now much clearer too: this is a smooth section of $\text{Hom}\left({E}_{1}\otimes \cdots \otimes {E}_{n},F\right)$. The inclusion $i$ is not surjective. This is because in general if $\phi :\Gamma \left({E}_{1}\otimes \cdots \otimes {E}_{n}\right)\to \Gamma \left(F\right)$ is a homomorphism the value of $\phi {\left({\xi }^{1},\dots ,{\xi }^{n}\right)}_{p}$ may very well depend on ${\xi }^{i}$ in their entirety, and not only on ${\xi }_{p}^{i}$.

This last statement is actually false! See the errata on this post.

We claim, however, that the image of $i$ consists precisely of the multilinear functions ${E}_{1}×\cdots ×{E}_{n}\to F$ that are tensors — i.e. such that $\tau {\left({\xi }^{1},\dots ,{\xi }^{n}\right)}_{p}$ is determined by ${\xi }_{p}^{i}$. Indeed, if we consider the map

given by $s{\tau }_{p}\left({v}_{1},\dots ,{v}_{n}\right)=\tau {\left({\xi }^{1},\dots ,{\xi }^{n}\right)}_{p}$, where $Τ\left({E}_{1}×\cdots {E}_{n},F\right)\subset {\text{Hom}}_{{C}^{\infty }\left(M\right)}\left({E}_{1}\otimes \cdots \otimes {E}_{n},F\right)$ is the subspace of tensors and ${\xi }^{i}\in \Gamma \left({E}_{i}\right)$ are such that ${\xi }_{p}^{i}={v}_{i}$, we can very quickly check that $i={s}^{-1}$, establishing an isomorphism of ${C}^{\infty }\left(M\right)$-modules

The definition of $s{\tau }_{p}\left({v}_{1},\dots ,{v}_{n}\right)$ does not depend on the choice of ${\xi }^{i}$ precisely because the value of $\tau {\left({\xi }^{1},\dots ,{\xi }^{n}\right)}_{p}$ depends only on ${\xi }_{p}^{i}={v}_{i}$! In conclusion, a tensor $\tau :{E}_{1}×\cdots ×{E}_{n}\to F$ is called a tensor because it corresponds to a smooth section of $\text{Hom}\left({E}_{1}\otimes \cdots \otimes {E}_{n},F\right)$. To finish things of, I would like to conclude our discussion by explaining a small notational quirk the reader will probably encounter in the literature: most people refer to $\text{Hom}\left({E}_{1}\otimes \cdots \otimes {E}_{n},F\right)$ as ${E}_{1}^{*}\otimes \cdots \otimes {E}_{n}^{*}\otimes F$.

This is because given two vector spaces $V$ and $W$, the space $\text{Hom}\left(V,W\right)$ is canonically isomorphic to ${V}^{*}\otimes W$. Taking $V={E}_{1}{\text{|}}_{p}\otimes \cdots \otimes {E}_{n}{\text{|}}_{p}$ and $W=F{\text{|}}_{p}$, this translates to an isomorphism of vector bundles $\text{Hom}\left({E}_{1}\otimes \cdots \otimes {E}_{n},F\right)\to {E}_{1}^{*}\otimes \cdots \otimes {E}_{n}^{*}\otimes F$. In fact, usually the differential structure of $\text{Hom}\left({E}_{1}\otimes \cdots \otimes {E}_{n},F\right)$ is defined via the identification with ${E}_{1}^{*}\otimes \cdots \otimes {E}_{n}^{*}\otimes F$. This is the formalism generally adopted, which is to say, when a geometer says “a tensor” in a formal sense he most likely means “some $\tau \in \Gamma \left({E}_{1}^{*}\otimes \cdots \otimes {E}_{n}^{*}\otimes F\right)$”.

Also, if $F\to M$ is the trivial line bundle $M×ℝ$, one usually refers to ${E}_{1}^{*}\otimes \cdots \otimes {E}_{n}^{*}\otimes F$ by simply ${E}_{1}^{*}\otimes \cdots \otimes {E}_{n}^{*}$, because tensoring by $M×ℝ$ is the same as doing nothing. For instance, a Riemmanian metric is most often defined as a tensor $\text{g}\in \Gamma \left({T}^{*}M\otimes {T}^{*}M\right)$ satisfying special conditions. That about wraps it up. I hope this helped someone 😛