# 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 $E \rightarrow M$, $F \rightarrow M$ are vector bundles over $M$, then the sets $\Gamma ( E )$ and $\Gamma ( F )$ of smooth (global) sections of $E$ and $F$, respectively, have a natural structure of a $C^\infty ( M )$-modules, and sometimes linear maps $\Gamma ( E ) \rightarrow \Gamma ( F )$ show up. If such a map $\tau : \Gamma ( E ) \rightarrow \Gamma ( F )$ 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 $\tau : \Gamma ( E_1 ) \times \Gamma ( E_2 ) \times \cdots \times \Gamma ( E_n ) \rightarrow F$ — i.e. maps that are linear in each coordinate. Again, if $\tau(\xi^1, \xi^2, \ldots, \xi^n)_p$ 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 $\text{g}_p : T_p M \times T_p M \rightarrow \mathbb{R}$ which “varies smoothly with $p$”. This construction induces a tensor

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

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 $R ( X , Y ) Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[ X , Y ]} Z$ of a connection $\nabla$ or the Nijenhuis tensor $N ( X , Y ) = [ X , Y ] + J [ J X , Y ] + J [ X , J Y ] - [ J X , J Y ]$ 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 $\text{Bil} \left ( M \times N , L \right )$ is the module of $R$-bilinear maps $M \times N \rightarrow L$.

In other words, $R$-multilinear maps $M_1 \times M_2 \times \cdots \times M_n \rightarrow N$ naturally correspond to $R$-linear maps $M_1 \otimes M_2 \otimes \cdots \otimes M_n \rightarrow N$. 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 $R = \mathbb{R}$, this construction induces a construction in the category of vector bundles over some fixed manifold $M$: if $E_i \rightarrow M$ are bundles over $M$, there is a vector bundle $E_1 \otimes E_2 \otimes \cdots \otimes E_n \rightarrow M$ whose fibers are

The relationship between these two notions of tensor should now be clear: tensors $\Gamma ( E_1 ) \times \Gamma ( E_2 ) \times \cdots \times \Gamma ( E_n ) \rightarrow \Gamma ( F )$ are called tensors because they correspond to $C^\infty ( M )$-linear maps

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

In fact, there’s a natural isomorphism of sheaves of $C^\infty$-modules $\Gamma ( - , E_1 ) \otimes_{C^\infty} \Gamma ( - , E_2 ) \otimes_{C^\infty} \cdots \otimes_{C^\infty} \Gamma ( - , E_n ) \cong \Gamma \left ( - , E_1 \otimes E_2 \otimes \cdots \otimes E_n \right )$ 🤡

To recap: we’ve just shown that a tensor $\tau : \Gamma ( E_1 ) \times \cdots \Gamma ( E_n ) \rightarrow \Gamma ( F )$ can be naturally identified with some $\tau \in \text{Hom}_{C^\infty ( M )} \left ( \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) , \Gamma ( F ) \right )$. 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 $\text{Hom} ( V , W )$ of linear transformations $V \rightarrow W$ is again a vector space. Hence we can consider the vector bundle $\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \rightarrow M$ 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 $i \eta : \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) \rightarrow \Gamma ( F )$ 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 $\varphi : \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) \rightarrow \Gamma ( F )$ is a homomorphism the value of $\varphi(\xi^1, \ldots, \xi^n)_p$ 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 $E_1 \times \cdots \times E_n \rightarrow F$ that are tensors — i.e. such that $\tau(\xi^1, \ldots, \xi^n)_p$ is determined by $\xi_p^i$. Indeed, if we consider the map

given by $s \tau_p (v_1, \ldots, v_n) = \tau(\xi^1, \ldots, \xi^n)_p$, where $\mathcal{T}(E_1 \times \cdots \times E_n, F) \subset \operatorname{Hom}_{C^\infty(M)}(E_1 \otimes \cdots \otimes E_n, F)$ is the subspace of tensors and $\xi^i \in \Gamma ( E_i )$ are such that $\xi_p^i = v_i$, we can very quickly check that $i = s^{- 1}$, establishing an isomorphism of $C^\infty ( M )$-modules

The definition of $s \tau_p (v_1, \ldots, v_n)$ does not depend on the choice of $\xi^i$ precisely because the value of $\tau(\xi^1, \ldots, \xi^n)_p$ depends only on $\xi_p^i = v_i$! In conclusion, a tensor $\tau : E_1 \times \cdots \times E_n \rightarrow F$ 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 $\text{Hom} ( V , W )$ 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 $\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \rightarrow E_1^* \otimes \cdots \otimes E_n^* \otimes F$. 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 $F \rightarrow M$ is the trivial line bundle $M \times \mathbb{R}$, 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 $M \times \mathbb{R}$ 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 😛