# What does the curvature of a surface have to do with tensor products?

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 βglobal constructions that only depend on point-wise valuesβ 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.

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
βvaries smoothly with $p$β. 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 βvaries smoothly with $p$β. The meaning of βvaries
smoothly with $p$β 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 $\text{Bil}_R \left ( M \times N , L \right )$ 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 $\Gamma(-, E_1) \otimes_{C^\infty} \Gamma(-, E_2) \otimes_{C^\infty} \cdots \otimes_{C^\infty} \Gamma(-, E_n) \cong \Gamma(-, E_1 \otimes E_2 \otimes \cdots \otimes E_n)$ π€‘

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 βa bilinear form in $T_p M$ for each
$p \in M$β to a Riemannian metric seen as a tensor. The meaning of
βa transformation at each fiber $p$ that varies smoothly with
$p$β 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 $\xi_p^i$.

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 βa tensorβ in a formal sense he most likely means βsome $\tau \in \Gamma \left ( E_1^* \otimes \cdots \otimes E_n^* \otimes F \right )$β.

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 π