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
g:X(M)ΓX(M)βCβ(M)β Ξ(MΓR)
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
HomRβ(MβRβN,L)β BilRβ(MΓN,L),
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
Ξ(E1ββE2βββ―βEnβ)βΞ(F)
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
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