Tensor products
In this lecture I'm going to take a step back and define tensor products
of vector spaces and their generalization: tensor product over rings
* Definition of tensor product @
* Hom(V,W) is "naturally" isomorphic to V^* @ W (as vector spaces)
* Tensor product of representations means?
* X(V @ W)=X(V)X(W)
* Irreps of the Cartesian product of groups
* Tensor algebra 1.11 of Etingof for example
* Extension of scalars
------------------------------------------
* Definition of tensor product @
First let's define the tensor product of two vector spaces V,W over a
field k.
We give the universal definition:
DEFINITION: (Tensor product of vector spaces over a field)
V @_k W is the universal object among vector spaces such that
f bilinear
V x W -----------> V@W
| /
| ---------
g| / h
| /
| /
v /
U <
if f and g are bilinear maps and U is a vector space then there exists
a unique LINEAR map h:V@W->U such that the above diagram commutes.
UNIQUENESS: *IF* V@W exists then it is unique up to linear isomorphism
(Let U be (V@W)' be another tensor product of V,W.)
EXISTENCE: Define V@W to be the vector space consisting of finite linear
combinations of v@w where v in V and w in W.
Impose the conditions that force the map
V x W --> V@W
(v,w) |-> v@w
to be bilinear, i.e., (av+bv')@w = av@w+bv'@w, av@w=v@(aw) etc etc.
MEANS?
Let D=V x W be a vector space consisting of pairs (v,w).
Consider the subspace S of D spanned by the vectors
(v+v',w)-(v,w)-(v',w)
(av,w)-(v,aw)
etc
Then V@W is defined as the quotient vector space D/S.
Thus v@w MEANS [(v,w)] (equivalence class).
Hence (v+v')@w = v@w+v'@w MEANS
(v+v',w)-(v,w)-(v',w) in S which is true.
Exer: Now you need to define the claimed linear map by sending
v @ w to g(v,w) and extend linearly. One then has some well-definedness
to verify.
---------------------------------
Upgrade, now suppose V, W are R-modules. Then define V @_R W the same
way. For this lecture, we are only concerned about the tensor product as
vector spaces. However we will have reason to care about tensoring over
a ring in our discussion of induced representations.
---------------------------------
* Hom(V,W) is "naturally" isomorphic to V^* @ W (as vector spaces)
I assume here that V, W are finite dimensional.
Recall that Hom(V,W) is the vector space of linear transformations from
V->W.
Recall V^* is the dual vector space of V, consisting of linear
functionals from V->k = base field. That is, V^*=Hom(V,k).
Proof idea: Let T in Hom(V,W). Fix a basis {w1,...,wn} of W.
A linear transformation T:V->W is determined by its action on a basis
{v1,...,vm} of V.
Now, write
T(v)=f1(v)w1+...+fn(v)wn
The fi's vary with v in V, and in fact we claim that linearity of T
implies each of the fi's are linear, i.e., they are linear functionals.
Hence we can map T in Hom(V,W) to f1@w1+....+fn@wn.
This map is clearly injective. It is also surjective because if
v1^*,...,vm^* is the basis of V^* dual to the basis v1,...,vm of V
then vi^* @ wj corresponds to the transformation T:V->W such that
T(vi)=wj
and T(vk)=0 for all k \neq i. QED
Remark: If you want a more "formal" proof see, e.g.,
https://math.stackexchange.com/questions/679584/why-is-texthomv-w-the-same-thing-as-v-otimes-w
Observe: Suppose V and W are G-modules. Then V^* is a G-module.
Hence V^* @ W is a G-module. But that means Hom(V,W) is a G-module.
Exer: What is the action? (Ans: if g in G and F:V->W is a linear
transformation then g.F sends v to g.F(g^{-1}.v).
Application: The Hom(V,W) iso V^* @ W has been applied to study the
complexity of matrix multiplication. Here's the basic idea (see the
papers/book of Joseph Landsberg).
Let V=Mat_{2 x 2}
(in general V=Mat_{n x n} works fine for what we say but it's already
interesting for n=2)
Think about the matrix multiplication map
V x V -> V
M N -> MN
Notice this map is BILINEAR.
HENCE one can equivalently think of a LINEAR map from V@V->V
A LINEAR map from V@V->V is a POINT in Hom(V@V,V).
But Hom(V @ V, V) is isomorphic to (V @ V)^* @ V
But (V @ V)^* is iso to V^* @ V^*
[Why? Fix a basis of V, say v1,...,vm with dual basis
v1^*,...,vn^*. A basis of V@V is vi @ vj which has dual
basis (vi @ vj)^*. Then for the "<-" map send
vi^* @ vj^* to (vi @ vj)^* and extend linearly. This map is
injective and since the dimensions agree, surjective.]
Hence
Hom(V@V, V) iso (V @ V)^* @ V
iso V^* @ V^* @ V
iso V @ V @ V
That is, Matrix multiplication can be interpreted as a POINT in the
vector space V @ V @ V. Stratten, Landsberg and others start from there
to use rep theory and algebraic geometry to analyze the complexity of
matrix mult (look up "border rank").
--------------------------------
* Tensor product of representations means?
Suppose V and W are G-modules. Look at V@W AS VECTOR SPACES and define
an action of G on V@W by g.(v@w)=(g.v)@(g.w) ["diagonal action"].
Definition: The *representation ring* Rep(G) of G (also called a
Grothendieck ring of G is defined as follows: the elements are given by
isomorphism classes [V] of G-modules V. The addition is by direct sum
[V]\oplus [W]=[V\oplus W] and product is by tensor product
[V]*[W]=[V@W].
Exer: Prove this is well-defined.
In class exer: What is the multiplicative identity for Rep(G)? (Ans: Let
I be the 1-dimensional trivial representation, i.e., G->C^* by sending
g->1.
Observe that if V is any Vector space then V@C is isomorphic to V. (why?)
Hence [V]@[C]=[V@C]=[V].)
In class exer: How about the additive 0?
(Ans: *formally* for each [V] there is -[V].)
--------------
In class exer: If V is irreducible, and W is one dimensional (and hence
irreducible) G-modules, then V@W is irreducible. ANS: Pick a basis
v1,v2,..,vn of V and a basis w of W. Look at the basis v1@w, ...,vn@w of
V@W. If U inside V@W is G-invariant, it has a basis v1'@w,...vk'@w, then
by the action of G, v1',...,vk' spans a G-invariant subspace of V of the
same dimension.
-------------
* X(V @ W)=X(V)X(W)
Now suppose G is a finite group and V and W are two complex G-modules
associated to f:G->GL(V) and g:G->GL(W).
Exer: If f(x)=M(x) [an n x n matrix; n=dim V]
g(x)=N(x) [an m x m matrix; m=dim W]
then
the map h:G->GL(W) that describes the G-module structure of V@W is
M@N (Kronecker product of matrices).
Solution: Pick a basis v1,...,vn of V and w1,...,wm of W.
Thus an mn size basis of V@W is vi@wj.
g.vi -> \sum_{r=1}^n m_{ri} vr
g.wj -> \sum_{s=1}^m n_{sj} wj
Hence
g.(vi@wj) -> (\sum_{r=1}^n m_{ri} vr) @ (\sum_{s=1}^m n_{sj} wj)
Now order the basis of vi@wj by
v1@w1, v1@w2,..., v1@wm, v2@w1, v2@w2, ... v2@wm, ....
e.g,
g.(v1@w1) -> (\sum_{r=1}^n m_{r1} vr) @ (\sum_{s=1}^m n_{s1} ws)
and the coefficient of v1@w2 is m11 n11
and the coefficient of v1@w2 is m11 n21
etc which agrees with the first column of M@N. etc. QED
Exer: Tr(M@N) = Tr(M)Tr(N)
Ans: you check.
Hence combining the above with the definition of character f:G->GL(V)
as Tr(f) gives:
Theorem: X(V @ W)=X(V)X(W) [for finite groups]
Remark 1: although we are proving this in the finite group case,
characters more generally have this property. For example for compact
connected Lie groups. See for example
Kamnitzer's notes "Representation theory of compact groups and complex
reductive groups, Winter 2011"
The point is that for such groups one has Haar measure which allows one to
turn sums into integrals and run the SAME finite group arguments.
Remark 2: Suppose V and W are irreducible representations of G, in general
V@W will not be irreducible. Now we have proved that characters behave well
with respect to direct sum and tensor product. Thus to decompose
V@W=n1 W1 \oplus ...\oplus nh Wh
where Wi are irreducibles and ni are their multiplicities
(= dim Hom(Wi, V@W))
It suffices to compute instead
X(V@W)=X(V)X(W) (a class function) and expand it in the basis of
irreducible characters. In other words, once one computes the character
table, the tensor product problem is just linear algebra: multiply two rows
of the character table and expand in the rows.
On the other hand, we know each ni is a nonnegative integer.
FAMOUS OPEN PROBLEM: Give a combinatorial rule for these multiplicities
in the case G=S_n (the symmetric group). Very few cases are known. Probably
the most general case solved is by J. Blasiak ("the one hook case").
Exercise: Show that in the case of S_n, the tensor product multiplicities
are symmetric in the arguments (lambda,mu,nu).
[Solution sketch: In general, suppose V is a G-rep and let V^* be the dual
representation (see the Introductory notes). One can check that X_{V^*}=(X_V)^*.
But in the case of the symmetric group, all characters are real (and in fact,
integer valued). Hence in this case X_{V^*}=X_V, that is the representation is
self dual.
The significance of this is that by Schur's lemma, the tensor multiplicity
t_{lambda,\mu}^{nu} = dim Hom(V_{\nu}, V_{\lambda}@V_{\mu})
= dim V^{\nu}^* @ V_\lambda @ V_{\mu}
= dim V^{\nu} @ V_{\lambda} @ V_{\mu}
= dim V^{\nu} @ V_{\lambda}^* @ V_{\mu}
= dim V_{\lambda}^* @ V_{\nu} @ V_{\mu}
= dim Hom (V_{lambda, V_{\nu} @ V_{\mu})
= t_{nu,mu}^{\lambda}
etc. QED]
--------
* Irreps of the Cartesian product of groups
Suppose (G,*_G) and (H,*_H) are finite groups.
Def: The product G x H means the group with set G x H such that
(g1,h1)*_{G x H} (g2,h2) = (g1 *_G g_2, h1 *_H h2)
Exer: Check this is indeed a group structure.
Now suppose p:G->GL(V) and q:H->GL(W) are representations of G and H.
We can build a representation
p @ q: G x H -> GL(V @ W)
by the action
(g,h).v@w = gv @ hw.
Exer: The character X_{p @ q} satisfies
X_{p @ q}(g,h) = X_p(g) X_q(h).
Theorem 10 (of Serre Section 3.2)
(i) If p and q above are irreducible representations of G and H
respectively, then p @ q is an irrep of G x H.
(ii) Conversely, every irreducible representation of G x H is
isomorphic to a representation of the form p @ q.
Proof:
(i) Since p and q are irreducible, (p|p)=1 and (q|q)=1. That is
(1/|G|) \sum_{g in G} |X_p(g)|^2 = 1
and
(1/|H|) \sum_{h in G} |X_q(h)|^2 = 1.
Multiplying gives
(1/|G||H|) \sum_{(g,h) in G x H} |X_{p @ q}(g,h)|^2 = 1
and hence (X_{p @ q}|X_{p @ q})=1 which we know is equivalent to saying
X_{p @ q} is an irreducible character.
(ii) Given (i), we know that such X_p X_q = X_{p @ q} are irreducible
characters and thus an orthonormal set. As in the proof that the
set of irreducible characters of G form a basis of Class(G), it suffices
to show that if
f(g,h) in Class(G x H)
is orthonormal to all such characters is 0.
Suppose f(g,h) is such a class function, then
0 = (f|X_p @ X_q) MEANS
0 = \sum_{g,h} f(g,h) X_p^*(g) X_q^*(h) (!)
Fix q and let
t(g) = \sum_{h in H} f(g,h) X_q(h)^*. (!!)
Then (!) is rewritten as
0 = \sum_{g} t(g) X_p(g)^*. (!!!)
But t is a class function on G since f is a class function on G x H.
Hence (!!!) MEANS
0 = g(t|X_p) for all irreducible p
This implies t = 0 (identically). (why? Because X_p's form an
orthonormal basis.)
Hence (!!) says
0= \sum_{h in H} f(g,h) X_q(h)^* for all q.
That means f(g,h), which for fixed g is a class function on H satisfies
(f(g,-),X_q)=0 for all q
This implies f(g,-)=0 identically on H and since g was arbitrary,
f(g,h)=0 for all (g,h) in G x H, as desired. QED
Remark: The Theorem is true more generally. See, e.g.,
https://math.stackexchange.com/questions/2502919/irreducible-representations-of-direct-product-of-compact-groups/2503141#2503141
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The Tensor Algebra, and Friends --
Often people consider V^{@n} and more generally V^{@n}@(V^*)^{@m},
the latter space being the tensors of type (m,n)
ex: tensors of type (0,1) are vectors
ex: tensors of type (1,0) are co-vectors (linear functionals)
ex: tensors of type (2,0) are bilinear forms (why?)
ex: tensors of type (1,1) are linear operators, i.e. T in Hom(V,V)
(why? General fact/exer: Hom(V,W) is CANONICALLY isomorphic to V^{*}@W.)
Recall:
Def: An *algebra* is a vector space A over a field k with a binary
operation *:A x A -> A.
Def: An *associative algebra* is a vector space A over a field k
with a binary operation *:A x A -> A such that (a*b)*c=a*(b*c).
(Non associative) Example: A Lie algebra. This is a vector space V with
a bracket operation [ , ]:V x V->V that is bilinear, [a,a]=0 (and hence
[ , ] is skew-symmetric [a,b]=-[b,a]) and satisfies the Jacobi identity
[[a,b],c]+[[b,c],a]+[[c,a],b]=0.
Ex 1. Any vector space with [,]=0 (abelian Lie algebra)
Ex 2. Any associative algebra with [a,b]=ab-ba, such as n x n matrices.
Ex 3. Derivations of an algebra, that is, linear maps D:A->A satisfying
the Leibniz rule
D(ab)=D(a)b+aD(b)
An example of a derivation is this: suppose A is a Lie algebra.
Fix x in A. The map ad_x:A->A defined by ad(x)y:=[x,y] is a derivation
(exercise).
Example: The *tensor algebra*. This is
T(V) = \oplus_{n>=0} V^{\otimes n}.
The multiplication is a*b=a\otimes b.
The multiplication is *graded* in the sense that if a in V^{@m} and b
in V^{@n} then a*b=a@b in V^{@(m+n)}
The tensor algebra is important because many important algebras are quotients
of it. It is universal in the following sense.
Suppose f:V->A is a linear map from a vector space V to an associative
algebra A over a field k. Then there is a UNIQUE algebra homomorphism g:
T(V)->A that makes this diagram commute.
i
V ------> T(V)
\ |
\ |
\ |
f \ |g
\ |
\ |
\ v
\> A
(Compare this to the free group in the category of groups. Thus
T(V) is a functor from vector spaces to associative algebras.)
In class Exer: what should the proper subobject of an associative
algebra be?
Ans: A two sided ideal I of A, which means a subspace I of A such that
aI is contained in I for all a in A (left ideal) AND
Ia is contained in I for all a in A (right ideal).
With this one can define the quotient algebra A/I.
See Section 1.4, 1.5 of Etingof:
https://math.mit.edu/~etingof/replect.pdf
Def: The *symmetric algebra* S(V) is the quotient of T(V) by the ideal
generated by v@w-w@v
Remark: Thus, if you fix a basis x1,x2,...,xn of V, S(V) is isomorphic
to k[x1,...,xn]. That is S(V) is the "coordinate-free" version of a
polynomial ring.
Def: The *exterior algebra* \wedge(V) is the quotient of T(V) by the
ideal I generated by v@v.
Notice that by definition, the ideal I defining the exterior algebra
contains v@w+w@v =(v+w)@(v+w)+v@v+w@w = (v+w)@(v+w)
-----------
Remark/Exer: Typically an element of \wedge(V) looks like linear
combinations of v1 ^ v2 ^ ... ^ vk. If V is n-dimensional then (exer)
any such expression is 0 when k>n). Now suppose T:V->V is a linear
operator ("a matrix"). Then this operator "induces a map"
T':Wedge^k->Wedge^k, namely
T(v1 ^ v2 ^ ... ^ vk):=T(v1)^T(v2)^...^T(vk).
When k=n, this map must act by a scalar d; this d is the determinant.
More generally, when kWedge^k
on a basis element e_{i1}^...^e_{ik}. The linear expansion of the image
is in e_{j1}^...^e_{jk}. Thus the associated matrix to T' is indexed by
k subsets of b and the I,J place is the I,J minor of the matrix
associated to T. Choice of basis is unnecessary in this discussion: specifically
for determinants, a linear operator T' from a one-dimensional space to itself
is determined by a scalar (the determinant).
See for instance:
https://math.stackexchange.com/questions/21614/is-there-a-definition-of-determinants-that-does-not-rely-on-how-they-are-calcula
---------------
Both the symmetric algebra and exterior algebra are graded
S(V)=\oplus_{n\geq 0} S^n(V)
\wedge(V) =\oplus_{n\geq 0} \wedge^n(V) (a finite direct sum if V is
finite dimensional).
remark: S^n(V) is the subspace of degree n polynomials in dim(V) many
variables. Confusingly, this is also denoted Sym^n(V). See below.
-------
Tensor square decomposition
(Related but different, in a notationally confusing way from the
concepts above)
Now suppose V is a G-module (no hypotheses on G).
Look at V@V as a G-module.
Let A be the automorphism on V@V such that A(x@y)=y@x (and extend
linearly).
Define Sym^2(V) to be z in V@V such that A(z)=z.
Alt^2(V) to be z in V@V such that A(z)=-z
In class exer: If e1,...,en is a basis of V then
{ei@ej+ej@ei}_{i\leq j} is a v.s. basis of Sym^2(V)
{e1@ej-ej@ei}_{iS is a ring homomorphism (think R subseteq S as a useful
special case, or even R=Reals contained in S=Complex numbers).
Then the module
M':=M @_R S
has an obvious right action by S: (m@s).s' = (m@ss').
-----
[In fact, the left action (by R) on M and the right action by S commute,
meaning that M' becomes a *(R,S)-bimodule*:
* S is a left R-module, i.e.,
r.s MEANS f(r)s
and
(r'r).s MEANS f(r')f(r)s [usual order of homomorphism].
* On the other hand, as we said: S is right S-module.
* the two actions commute (as we claim)
r.(s.s') = f(r)(ss')
= (f(r)s)s'
= (r.s).s'.]
------
[One can also define M":=S @_R M so that M" has a left S-action,
but to make things work out you need to treat M as a RIGHT R-module
via m.r MEANS f(r^{-1})m.]
^^^^^^
Ex) "Complexification". Let M=Reals[x] be the polynomial ring in x with
real coefficients. Here R=Reals and M is an R-module, i.e, a vector
space over Reals.
Then M':=M @_{Reals} Complex
extends the scalars and M' iso Complex[x].
Ex) "Complexification" II: Let V be a real vector space with basis
b_1,...,b_n. Then V @_{Reals} Complex is the same vector space spanned by
the same basis using complex coefficients, i.e., if V is iso Reals^n then
the complexification is iso Complex^n.
Ex) "Induced representation" Suppose W is a right C[H]-module (i.e., M
is an H-representation where H acts by inverse). Then if we let R=C[H]
and S=C[G] there's an obvious ring homomorphism f:R->S (i.e.,
inclusion). Then
V=Ind_H^G(W) := C[G] @_{C[H]} W
has a left C[G] action, i.e., V is a C[G] module.
(One can also look at W @_{C[H]} C[G] which would have a RIGHT
C[G] action and then we'd have to make G act by inverse. Exer.)
This is THE induced representation.
Categorical perspective:
If H is a subgroup of G, we've already discussed *restriction* of
G-representations to H representations. This is a functor from the
The category of G-modules where the morphisms are G-equivariant maps.
to
The category of H-modules where the morphisms are H-equivariant maps.
In fact, it is a "forgetful" functor (forgets part of the G action).
[However being "forgetful" is not exactly a well-defined term (pun
intended).]
Recall
Def: a *functor* F from categories C->D MEANS a map that takes
objects X to objects F(X) and morphisms f:X->Y to morphisms F(f):F(X)->F(Y)
such that
F(id_X)=id_{F(X)}
F(g o f) = F(g) o F(g) [covariant].
Remark: if F(g o f) = F(f) o F(g) we call the functor contravariant.
-----
Definition: *Equivalence of categories* C and D means there is a functor
F:C->D such that the map Hom_C(c_1,c_2)->Hom_D(F(c_1),F(c_2)) is bijective
and "essentially surjective", i.e., each d in D is isomorphic to F(c) for
some c in C.)
(This is actually a characterization of equivalence of categories that
I find easier to absorb.)
-----
Def: An *adjunction* between categories C and D is a pair of covariant
functors
F: D->C and G: C->D
such that for all objects c in C and d in D we a bijection between
Hom_C(F(d),c) and Hom_D(d,G(c))
that is "natural" (exer: see technical definition
https://en.wikipedia.org/wiki/Natural_transformation)
F is called a "left adjoint" and G is called a "right adjoint".
This is a "somewhat akin weak form of equivalence"
in the sense that an actual equivalence provides an adjunction
(exer: clarify this but looking into weak equivalences in category theory!
Is adjunction actually a weak equivalence of categories?).
https://en.wikipedia.org/wiki/Adjoint_functors
EXAMPLE (tensor product): Consider the category C of R-modules. Then if M,N
are R-modules, M @_R N is an N module and Hom(M,N) iso M^* @_R N is an
R-module. If we fix N then
M @_R -
and
Hom(M,-)
are functors from C to itself.
Also, these two functors are left and right adjoint to one another:
Hom(M @_R N, P) is iso Hom(M, Hom(N,P)).
Separately, we also discussed short exact sequences
f1 f2 f3 f4
(0->A->B->C->0 where f1, f4 are the 0 maps and im(fi)=ker(f_{i+1})); this
is exactness at each "node".
A functor F is EXACT if it preserves these sequences. It is RIGHT
EXACT if
F(f2) F(f3) 0
F(A)->F(B)->F(C)->0
is exact. (Similar definition of left exact 0->F(A)->F(B)->F(C) is exact.)
The tensor product functor is RIGHT EXACT.
https://people.brandeis.edu/~igusa/Math205bS10/Math205b_S10_22.pdf
---------
ANOTHER EXAMPLE (induced representations): let H be a subgroup of G and let
D= category of H-modules with morphisms being H-equivariant maps
C= category of G-modules with morphisms being G-equivariant maps
and the left adjoint F:D->C.
In our context (finite groups), the bijection
Hom_C(F(d),c) \cong Hom_D(d,G(c))
is the catgorification of Frobenius reciprocity.
For finite groups Frobenius reciprocity states
_G = _H
----------
For finite groups H,G, there is a more "down to earth"(?) (isomorphic)
description of induced representation, which is useful for character
computation. This is discussed in the next lecture.