Representation of a Hopf algebra

From Academic Kids

In mathematics, there is a concept of a representation of a Hopf algebra.

Let's say we have an unital associative algebra H. We know what a representation of H is. It's just a module. But let's just say for the moment we didn't know that but we'd like, for some plain algebra H, for some reason or other




(note that this relation is not satisfied, for let's say reps of a Lie algebra)

for all X in (the underlying vector space of) the rep and any A, B in H.

Then, it's obvious




for all X in the rep and all A, B, C in H.

A sufficient condition for this to hold is if H is unital and associative.

Now let's say we wish for any two reps V1 and V2 that there's a <math>\otimes<math> operation defined on the reps such that <math>V_1\otimes V_2<math> is also a rep AND the underlying vector space of <math>V_1\otimes V_2<math> is just the product of the underlying vector spaces of V1 and V2. This would not hold in general for an arbitrary algebra, but let's just say we insist upon it.

And let's also assume this product operation for reps is functorial. In other words, there is a linear binary operation <math>\Delta:H\rightarrow H\otimes H<math> such that for any X in <math>V_1\otimes V_2<math> and any A in H,

<math>A[X]=\Delta A[X_{(1)}\otimes X_{(2)}]=A_{(1)}[X_{(1)}]\otimes A_{(2)}[X_{(2)}]<math>.

Here, we are using Sweedler's notation, which is kind of like an index free form of Einstein's summation convention.

Well, what does it mean when we say <math>V_1\otimes V_2<math> is also a rep?


<math>\Delta 1[X_{(1)}\otimes X_{(2)}]=1[X]=X=X_{(1)}\otimes X_{(2)}=1[X_{(1)}]\otimes 1[X_{(2)}]=(1 \otimes 1)[X_{(1)}\otimes X_{(2)}]<math>


<math>\Delta(AB)[X_{(1)}\otimes X_{(2)}]=(AB)[X]=A[B[X]]=\Delta A[\Delta B[X_{(1)}\otimes X_{(2)}]]=(\Delta A )(\Delta B)[X_{(1)}\otimes X_{(2)}]<math>

So, it would be sufficient if <math>\Delta 1=1\otimes 1<math> and <math>\Delta(AB)=(\Delta A)(\Delta B)<math>.

Let's say we'd like to turn the collection of all reps into a monoidal category with respect to <math>\otimes<math>. Let's assume it's not weak for the moment, which means, in this stronger form, that <math>V_1\otimes(V_2\otimes V_3)<math> and <math>(V_1\otimes V_2)\otimes V_3<math> are equivalent instead of being merely isomorphic and that there's an identity rep <math>\epsilon_H<math>, called the trivial rep such that <math>\epsilon_H\otimes V<math>, V and <math>V\otimes \epsilon_H<math> are all equivalent (isomorphic for the corresponding weak version, which is slightly more trickier).

It's obvious εH would have to be 1 dimensional. What's more, because the three reps above are assumed to be equivalent (isomorphic in the weak version), if we assume the equivalence is functorial (over all reps), then there's an element of εH, which we'll just call 1 such that <math>1\otimes X<math>, X and <math>X \otimes 1<math> are all equal for any X in V. In fact, as a matter of convenience, we can just assume the underlying vector space of εH is just the field F that H is over.

Now, we can define a linear map <math>\epsilon:H\rightarrow F<math> given by ε[A]≡A[1].

Obviously, for any X in <math>V_1\otimes(V_2\otimes V_3)=(V_1\otimes V_2)\otimes V_3<math> (<math>\simeq<math> for the weak version),

<math>((id\otimes \Delta)\Delta A)[X_{(1)}\otimes X_{(2)}\otimes X_{(3)}]=A_{(1)}[X_{(1)}]\otimes A_{(2)(1)}[X_{(2)}]\otimes A_{(2)(2)}[X_{(3)}]=A_{(1)}[X_{(1)}]\otimes A_{(2)}[X_{(2)}\otimes X_{(3)}]=A[X]=((\Delta\otimes id)\Delta A)[X_{(1)}\otimes X_{(2)}\otimes X_{(3)}]<math>

Insisting that <math>(id\otimes \Delta)\Delta A=(\Delta \otimes id)\Delta A<math> is sufficient.






<math>(\epsilon(A_{(1)})A_{(2)})[X]=A_{(1)}[1][A_{(2)}[X]]=A_{(1)}[1]\otimes A_{(2)}[X]=A[1\otimes X]=A[X]=(A_{(1)}\epsilon(A_{(2)}))[X]<math>

Insisting that ε(A(1))A(2)=A=A(1)ε(A(2)) is sufficient.

Note that I've just motivated the whole concept of a bialgebra and what it means to have a rep of a bialgebra!

Now we all know about dual vector spaces. For the case where the field is R</sub>, it's just the dual vector space. For the case where the field is <b>C, there's a complex conjugation involved. To save space on notational issues, we'll just work with the real case for the moment. Extending it to C is fairly straightforward.

Now let's say we'd like to have a dual representation V* for every rep V such that the underlying vector spaces of both reps are dual to each other. However, we needn't insist V**=V. Let's also assume this * here is functorial over the monoidal category of reps. In other words, there is a linear map <math>S:H\rightarrow H<math> such that for any A in H, X in V and Y in V*,


where <.,.> is just the contraction for dual vector spaces.

We also require that there is a H-intertwiner (as maps between bialgebra reps, because we haven't defined a Hopf algebra rep yet) from εH to <math>V^*\otimes V<math> which maps 1 to <math>1_{V^*\otimes V}<math> where <math>1_{V^*\otimes V}<math> is the unique element of <math>V^*\otimes V<math> which satisfies <math><1_{V^*\otimes V},X>=X<math> and <math>=Y<math> for all X and Y.

Not so obviously, this implies

<math>A[1_{V^*\otimes V}]=\epsilon(A)[1_{V^*\otimes V}]<math>

which is satisfied if the sufficient requirement that



This is merely one justification of a Hopf algebra. It turns out the same few definitions are motivated in many other topics, which is why the concept of a Hopf algebra is not merely abstract nonsense.

If V is a rep of a Hopf algebra H, then X in V is said to be invariant under H if for all A in H, A[X]=ε(A)X. Note that the subset of all invariant elements of V forms a subrep of V.

This article describes a vector space rep of a Hopf algebra. The interesting thing about Hopf algebra is that it also has algebra reps with an additional structure over and above the module structure of an ordinary linear rep. See algebra representation of a Hopf algebra.

See also Tannaka-Krein reconstruction theorem.


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