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Stream: deprecated: algebra & CT

Topic: Subalgebras of dual Hopf algebra

Alonso Perez-Lona (Aug 27 2023 at 17:22):

Hi all. What is the relation between the Hopf subalgebras of a finite-dimensional Hopf algebra H, and the Hopf subalgebras of the dual Hopf algebra H^*? Is it as straightforward as "A is a Hopf subalgebra of H iff A^* is a Hopf subalgebra of H^*" ?

Jean-Baptiste Vienney (Aug 27 2023 at 18:08):

Well, we can just try and see

Jean-Baptiste Vienney (Aug 27 2023 at 18:11):

I guess that the definition of a Hopf subalgebra of a Hopf $k$-algebra $(H,\eta,\nabla,\epsilon,\Delta,S)$ is a sub-$k$-vector space $A \subseteq H$ such that:

• if $\lambda \in k$, then $\eta(\lambda) \in A$
• if $a_1,a_2 \in A$, then $\nabla(a_1 \otimes a_2) \in A$
• if $a \in A$, then $\Delta(a) \in A \otimes A$
• if $a \in A$, then $S(a) \in A$

Jean-Baptiste Vienney (Aug 27 2023 at 18:18):

Now for every Hopf algebra $H$, I obtain by duality these maps:

• $\eta^*: H^* \rightarrow k^*$
• $\Delta^*: (H \otimes H)^* \rightarrow H^*$
• $\epsilon^*: k^* \rightarrow H^*$
• $\nabla^*: H^* \rightarrow (H \otimes H)^*$
• $S^*:H^* \rightarrow H^*$

Jean-Baptiste Vienney (Aug 27 2023 at 18:21):

If $H$ is finite-dimensional, I know by the compact closed structure of the category of finite-dimensional vector spaces, that we have this isomorphism of $k$-vector spaces:

• $(H \otimes H)^* \cong H^* \otimes H^*$

Moreover, we also have:

• $k^* \cong k$

Jean-Baptiste Vienney (Aug 27 2023 at 18:27):

Let's denote as below the maps that we obtain by combining the dual maps with these isomorphisms:

• $\eta^\vee:H^* \rightarrow k$
• $\nabla^\vee:H^* \otimes H^* \rightarrow H^*$
• $\epsilon^\vee:K \rightarrow H^*$
• $\Delta^\vee:H^* \rightarrow H^* \otimes H^*$
• $S^\vee:H^* \rightarrow H^*$

Jean-Baptiste Vienney (Aug 27 2023 at 18:28):

I guess that we call the dual of a finite-dimensional Hopf algebra $H$ is $(H^*,\eta^\vee,\nabla^\vee,\epsilon^\vee,\Delta^\vee,S^\vee)$

Jean-Baptiste Vienney (Aug 27 2023 at 18:46):

-> If $A$ is a Hopf sub-algebra of the finite-dimensional Hopf algebra $H$, then is $A^*$ a Hopf subalgebra of $H^*$?

Jean-Baptiste Vienney (Aug 27 2023 at 19:04):

I guess that you can't go without using the sacrilegious isomorphisms $A \cong A^*$ and $H \cong H^*$ ie. choosing basis.

Jean-Baptiste Vienney (Aug 27 2023 at 19:06):

I don't see how you would see $A^*$ as a subspace of $H^*$ without that

Jean-Baptiste Vienney (Aug 27 2023 at 19:32):

Let's choose a basis $(e_1,...,e_n)$ of $A$ and complete it into a basis $(e_1,...,e_{n+p})$ of $H$.

Jean-Baptiste Vienney (Aug 27 2023 at 19:35):

Then we have the dual basis $(e_1^*,...,e_{n+p}^*)$ of $H^*$. For every $1 \le k \le n$, write $e_i^\sharp = e_i^*|_{A}$

Jean-Baptiste Vienney (Aug 27 2023 at 19:56):

$(e_1^\sharp,...,e_n^\sharp)$ is a basis of $A^*$. Indeed it's a free family of $n=dim(A^*)$ vectors:

Jean-Baptiste Vienney (Aug 27 2023 at 19:58):

$e_i^\sharp(e_j)=e_i^*(e_j)=\delta_{i,j}$ and thus $\sum \lambda_i e_i^\sharp = 0 \Rightarrow \forall 1 \le j \le n, (\sum \lambda_i e_i^\sharp)(e_j) = \lambda_j = 0$

Jean-Baptiste Vienney (Aug 27 2023 at 20:03):

Thus, we have an isomorphism $\sum \lambda_i e_i^\sharp \mapsto \sum \lambda_i e_i^*$ from $A^*$ to $Span(e_1^*,...,e_n^*)$

Jean-Baptiste Vienney (Aug 27 2023 at 20:04):

With this identification, you can see $A^*$ as a subspace of $H^*$

Jean-Baptiste Vienney (Aug 27 2023 at 20:05):

And you can ask whether $A^*$, ie. $Span(e_1^*,...,e_n^*)$ is not only a sub-vector space of $H^*$ but moreover a Hopf sub-algebra of $H^*$.

Jean-Baptiste Vienney (Aug 27 2023 at 20:08):

With these coordinates, it's going to be a hassle...

Jean-Baptiste Vienney (Aug 27 2023 at 20:16):

By definition, we want to know whether:

• if $\lambda \in k$, then $\eta^\vee(\lambda) \in A^*$
• if $b_1,b_2 \in A^*$, then $\nabla^\vee(b_1 \otimes b_2) \in A^*$
• if $b \in A^*$, then $\Delta^\vee(b) \in A^* \otimes A^*$
• if $b \in A^*$, then $S^\vee(b) \in A^*$
  where $A^*$ must be understood as $Span(e_1^*,...,e_n^*)$

Jean-Baptiste Vienney (Aug 27 2023 at 20:17):

Pff, I think there is probably a better approach that this coordinate nightmare...

John Baez (Aug 27 2023 at 21:59):

Alonso Perez-Lona said:

What is the relation between the Hopf subalgebras of a finite-dimensional Hopf algebra H, and the Hopf subalgebras of the dual Hopf algebra H^*? Is it as straightforward as "A is a Hopf subalgebra of H iff A^* is a Hopf subalgebra of H^*" ?

Subalgebras of a finite-dimensional Hopf algebra correspond to quotient algebras of its dual.

Jean-Baptiste Vienney (Aug 27 2023 at 22:01):

How do you know this?

John Baez (Aug 27 2023 at 23:13):

For starters, taking duals is a contravariant functor sending inclusions $i: V \hookrightarrow W$ to quotient maps $i^\ast: W^\ast \to V^\ast$, algebras to coalgebras and coalgebras to algebras.

Jean-Baptiste Vienney (Aug 27 2023 at 23:14):

What are you quotienting by in your quotient map?

John Baez (Aug 27 2023 at 23:27):

The subspace of $W^\ast$ consisting of functionals that vanish on all of $im(V) \subseteq W$.

Jean-Baptiste Vienney (Aug 27 2023 at 23:28):

Ok

Jean-Baptiste Vienney (Aug 27 2023 at 23:39):

That was the main point I think ahah. I didn't know that. The rest doesn't look too difficult.

Jean-Baptiste Vienney (Aug 27 2023 at 23:43):

I will definitely remember that

Jean-Baptiste Vienney (Aug 27 2023 at 23:52):

Could you give more details on this point?

Jean-Baptiste Vienney (Aug 27 2023 at 23:54):

You say that if $i:V \rightarrow W$ is an (injective?) linear map between two (finite-dimensional?) vector spaces and $E=\{\phi \in W^*~s.t.~\forall v \in V~\phi(i(v)) = 0\}$, then $V^{*} \cong W^*/E$

Jean-Baptiste Vienney (Aug 27 2023 at 23:55):

What's the isomorphism exactly?

Jean-Baptiste Vienney (Aug 27 2023 at 23:59):

Oh no I see

Jean-Baptiste Vienney (Aug 28 2023 at 00:03):

No, I don't

John Baez (Aug 28 2023 at 07:38):

Assuming $i: V \to W$ is an injective map between finite-dimensional vector spaces, then we get $i^\ast \colon W^\ast \to V^\ast$ in the usual way:

$i^\ast(f) (v) = f(i(v))$

for all $f \in W^\ast$. My claim (a well-known fact) is that $i^\ast$ is surjective, with kernel equal to

$E=\{f \in W^*\textrm{ such that }~\forall v \in V~f(i(v)) = 0\}$,

John Baez (Aug 28 2023 at 07:43):

It then follows that $i^\ast$ gives an isomorphism from $W^\ast/E$ to $V^\ast$.

John Baez (Aug 28 2023 at 07:45):

(Whenever you have a surjective linear map between vector spaces $F: A \to B$ it gives an isomorphism $F^\prime: A/\mathrm{ker}(F) \to B$ defined by $F^\prime([a]) = F(a)$.)

John Baez (Aug 28 2023 at 07:47):

I just made a few claims here; if there's one that's hard to check let me know.

Alonso Perez-Lona (Aug 28 2023 at 13:42):

Makes sense, thanks everyone.

David Egolf (Aug 28 2023 at 15:53):

John Baez said:

Assuming $i: V \to W$ is an injective map between finite-dimensional vector spaces, then we get $i^\ast \colon W^\ast \to V^\ast$ in the usual way:

$i^\ast(f) (v) = f(i(v))$

for all $f \in W^\ast$. My claim (a well-known fact) is that $i^\ast$ is surjective...

Intuitively, this is interesting - it's like the ways in which $V$ relates to the underlying field are all "contained inside" of the ways in which the larger vector space $W$ relates to the underlying field. In symbols, I think the surjectivity of $i^\ast$ means that for any $g: V \to K$, where $K$ is the field our vector spaces are defined over (and viewed as a 1D vector space), then we can find a $f: W \to K$ so that $f \circ i = g$.

I wonder if a related more general statement is true (subject to some conditions to be determined): Let $i: V \to W$ be a representative of a subobject in some category, where we view subobjects as an isomorphism class of monomorphisms. So, $i: V \to W$ is a monomorphism. Then, let $g: V \to K$ be some morphism from $V$ to another object $K$. I wonder if we can always find a $f: W \to K$ so that $g$ factors through $W$ using $f$. That is, so that $g = f \circ i$.

I don't know what the minimum requirements are for this to be true. But let us consider the case where $W = V \coprod V'$ for some object $V'$. In this case, let $i: V \to V \coprod V'$ and $i': V' \to V \coprod V'$ be the two coprojection morphisms associated with $V \coprod V'$. Then let $g: V \to K$ be the morphism we wish to factor through $W$. Let $g': V' \to K$ be any morphism from $V'$ to $K$. Then, by the universal property of coproducts, there is a unique $f: V \coprod V' \to K$ so that $f \circ i = g$ and $f \circ i' = g'$.

So, in this case, there is always a way to factor $f: V \to K$ through a $g: W \to K$. Here are the two assumptions we made to achieve this, in addition to assuming that $i: V \to W$ is a monomorphism:

• $W$ is a coproduct of $V$ and some object $V'$, and $i: V \to W$ is one of the coprojections
• there is at least one morphism from $V'$ to $K$

Alonso Perez-Lona (Aug 30 2023 at 13:35):

Incidentally, what is the precise relation between Rep(H) and Rep(H^*) for H a finite-dimensional Hopf algebra? Can Rep(H^* ) be realized as some sort of Bimodule category of Rep(H)?

John Baez (Aug 30 2023 at 22:46):

I don't know a simple relation between Rep(H) and Rep(H*). Maybe there is one, but this is all I know:

If we start with, not just the category Rep(H) but also the forgetful functor

F: Rep(H) $\to$ Vect

we can recover H itself as the Hopf algebra of natural transformations from F to itself. This is called Tannaka duality. We can then form H* and then form Rep(H*) and its forgetful functor

G: Rep(H*) $\to$ Vect

But it would be nice to go from F: Rep(H) $\to$ Vect to G: Rep(H*) $\to$ Vect more directly!