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.
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 Cβ-linear functions between the
spaces of smooth sections of two fiber bundles. Specifically, if M
is a smooth manifold and EβM, FβM are vector
bundles over M, then the sets Ξ(E) and
Ξ(F) of smooth (global) sections of E and
F, respectively, have a natural structure of a
Cβ(M)-modules, and sometimes linear maps
Ξ(E)βΞ(F) show up. If such a map
Ο:Ξ(E)βΞ(F) satisfies the condition that
Ο(ΞΎ)pβ depends only on ΞΎpβ — and not
ΞΎ on its entirety — then Ο is called a tensor.
Often times it is convenient to also consider multilinear maps
Ο:Ξ(E1β)ΓΞ(E2β)Γβ―ΓΞ(Enβ)βF — i.e.
maps that are linear in each coordinate. Again, if
Ο(ΞΎ1,ΞΎ2,β¦,ΞΎn)pβ is determined by
ΞΎpiβ then Ο is called a tensor. The classical
examples of tensors are differential forms. A perhaps more interesting example
is a Riemannian metric: for each point pβM we fix a
positive-definite bilinear form gpβ:TpβMΓTpβMβR which
βvaries smoothly with pβ. This construction induces a tensor
where (g(V1,V2))(p)=gpβ(Vp1β,Vp2β).
This is what a tensor is supposed to be: for each pβM we fix
some multilinear function between the p-fibers of some vector
bundles that βvaries smoothly with pβ. The meaning of βvaries
smoothly with pβ 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=βXββYβZββYββXβZββ[X,Y]βZ
of a connection β or the Nijenhuis tensor
N(X,Y)=[X,Y]+J[JX,Y]+J[X,JY]β[JX,JY]
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 MβRβN is the
R-module which enjoys the universal property that
where BilRβ(MΓN,L) is the module of R-bilinear
maps MΓNβL.
In other words, R-multilinear maps
M1βΓM2βΓβ―ΓMnββN
naturally correspond to R-linear maps
M1ββM2βββ―βMnββN. 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 EiββM
are bundles over M, there is a vector bundle
E1ββE2βββ―βEnββM whose fibers are
The relationship between these two notions of tensor should now be
clear: tensorsΞ(E1β)ΓΞ(E2β)Γβ―ΓΞ(Enβ)βΞ(F) are
called tensors because they correspond to Cβ(M)-linear maps
which are in turn canonically identified with Cβ(M)-linear maps
In fact, there’s a natural isomorphism of sheaves of
Cβ-modules
Ξ(β,E1β)βCββΞ(β,E2β)βCβββ―βCββΞ(β,Enβ)β Ξ(β,E1ββE2βββ―βEnβ) π€‘
To recap: we’ve just shown that a tensor
Ο:Ξ(E1β)Γβ―Ξ(Enβ)βΞ(F) can be naturally
identified with some
ΟβHomCβ(M)β(Ξ(E1βββ―βEnβ),Ξ(F)). A
natural question to ask ourselves at this point is: does Ο
correspond to some ΟβΞ(Hom(E1βββ―Enβ,F))? First
of all, why does this make sense? Recall that given two vector spaces
V and W, the set Hom(V,W) of linear
transformations VβW is again a vector space. Hence we can
consider the vector bundle Hom(E1βββ―βEnβ,F)βM whose
fibers are
The previously mentioned example of Riemannian metrics does hint at an
inclusion
which takes Ξ·βΞ(Hom(E1βββ―βEnβ,F)) to
iΞ·:Ξ(E1βββ―βEnβ)βΞ(F) with
iΞ·(ΞΎ)pβ=Ξ·pβ(ΞΎpβ) — notice this is precisely what we did
to get from βa bilinear form in TpβM for each
pβMβ to a Riemannian metric seen as a tensor. The meaning of
βa transformation at each fiber p that varies smoothly with
pβ is now much clearer too: this is a smooth section of
Hom(E1βββ―βEnβ,F). The inclusion i is not
surjective. This is because in general if
Ο:Ξ(E1βββ―βEnβ)βΞ(F) is a homomorphism
the value of Ο(ΞΎ1,β¦,ΞΎn)pβ may very well depend
on ΞΎi in their entirety, and not only on ΞΎpiβ.
We claim, however, that the image of i consists precisely of the
multilinear functions E1βΓβ―ΓEnββF that are tensors — i.e. such that Ο(ΞΎ1,β¦,ΞΎn)pβ is determined by
ΞΎpiβ. Indeed, if we consider the map
given by
sΟpβ(v1β,β¦,vnβ)=Ο(ΞΎ1,β¦,ΞΎn)pβ, where
T(E1βΓβ―ΓEnβ,F)βHomCβ(M)β(E1βββ―βEnβ,F)
is the subspace of tensors and ΞΎiβΞ(Eiβ) are such that
ΞΎpiβ=viβ, we can very quickly check that
i=sβ1, establishing an isomorphism of
Cβ(M)-modules
The definition of sΟpβ(v1β,β¦,vnβ) does not depend on the
choice of ΞΎi precisely because the value of
Ο(ΞΎ1,β¦,ΞΎn)pβ depends only on
ΞΎpiβ=viβ! In conclusion, a tensor
Ο:E1βΓβ―ΓEnββF is called a tensor because it
corresponds to a smooth section of Hom(E1βββ―βEnβ,F).
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 Hom(E1βββ―βEnβ,F) as
E1ββββ―βEnβββF.
This is because given two vector spaces V and W, the
space Hom(V,W) is canonically isomorphic to
VββW. Taking V=E1β|pβββ―βEnβ|pβ
and W=F|pβ, this translates to an isomorphism of vector bundles
Hom(E1βββ―βEnβ,F)βE1ββββ―βEnβββF.
In fact, usually the differential structure of
Hom(E1βββ―βEnβ,F) is defined via the identification
with E1ββββ―βEnβββF. 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
ΟβΞ(E1ββββ―βEnβββF)β.
Also, if FβM is the trivial line bundle MΓR, one
usually refers to E1ββββ―βEnβββF by simply
E1ββββ―βEnββ, because tensoring by MΓR is the same as doing nothing. For instance, a Riemmanian metric is most
often defined as a tensor gβΞ(TβMβTβM)
satisfying special conditions. That about wraps it up. I hope this helped
someone π