3
$\begingroup$

Let $(A, \Delta)$ be a Hopf $^*$-algebra. Assume that $\{u^\alpha\}_{\alpha \in I}$ is a maximal collection of pairwise inequivalent irreducible unitary corepresentation matrices. I want to show that $$\mathfrak{C}(u^\beta) \cap \sum_{\alpha \ne \beta} \mathfrak{C}(u^\alpha) = 0.$$

This is claimed in Timmerman's book "An invitation to quantum groups and duality" in the proof of theorem 3.2.12 (which characterises algebraic compact quantum groups among the Hopf $^*$-algebras) in the implication (iii) $\implies$ (v). As a hint, Timmerman says to use the following facts:

Proposition 3.2.11

Proposition 3.2.2

Can someone explain how to use these facts to reach the desired conclusion?

We can associate to every corepresentation matrix $u^\alpha$ a corepresentation $(V_\alpha, \delta_\alpha)$. These are mutually pairwise inequivalent and irreducible. Then note that we have to show that $$\mathfrak{C}(\delta_\beta) \cap \mathfrak{C}(\boxplus_{\alpha \ne \beta} \delta_\alpha) = 0$$ but I can't see why this holds.

$\endgroup$
4
  • $\begingroup$ What is $\mathfrak C(u^\alpha)$? $\endgroup$
    – LSpice
    Oct 27, 2021 at 14:39
  • $\begingroup$ Anyway, it seems that the point is that $\Delta\rvert_{\mathfrak C(\delta_\beta)}$ decomposes into copies of $\delta_\beta$, whereas $\Delta\vert_{\mathfrak C(\boxplus_{\alpha \ne \beta} \delta_\alpha)}$ decomposes into sums of copies of the various $\delta_\alpha$, and these don't intertwine. $\endgroup$
    – LSpice
    Oct 27, 2021 at 14:43
  • $\begingroup$ @LSpice Thanks for your comments. $\mathfrak{C}(u^\alpha)$ is the space of matrix coefficients of the corepresentation matrix $u^\alpha$. In other words, it is the linear span of the matrix entries. $\endgroup$
    – Andromeda
    Oct 27, 2021 at 18:33
  • $\begingroup$ @LSpice I was also thinking in this direction, which inspired my other question: mathoverflow.net/questions/407196/… $\endgroup$
    – Andromeda
    Oct 27, 2021 at 18:34

1 Answer 1

2
+200
$\begingroup$

I will try to use notations compatible with those in Theorem 3.2.12 from the mentioned Timmerman's book "An invitation to quantum groups and duality".

So for each $\alpha$ i will denote by $V_\alpha$ the space $\mathfrak C(u^\alpha)$, by $\delta_\alpha$ the corepresentation of $A$ on $V_\alpha$ defined by the restriction of the (right regular corepresentation) $\Delta$, by $V$ the whole space $A$ (seen as a vector space after applying the forgetful functor).

Assume there is some $v\ne 0$ in the intersection $$ V_\alpha\cap\underbrace{\left(\sum_{\beta\ :\ \beta\ne \alpha}V_\beta\right)}_{=:W^\alpha} \subseteq V\ . $$ In the lattice of invariant subspaces of $V$ (with $\min$ given by intersection) let us consider the smallest invariant subspace $U\ne 0$ containing $v$. (The family of such spaces is not empty, contains at least $V$.) Then $U\subseteq V_\alpha$, (else consider $U\cap V_\alpha$, and contradict the assumed minimality,) and $U\subseteq W^\alpha$.

For the convenience of the reader, i am roughly citing from loc. cit.

Proposition 3.2.11 Let $\delta$ be an irreducible corepresentation of a Hopf $*$-algebra $(A,\Delta)$. Then the corresponding corepresentation $\Delta\Big|_{\mathfrak C(\delta)}$ is equivalent to a direct sum correpresentation $\delta^{\boxplus n}$ involving $n$ copies of $\delta$ for some suitable $n\in\Bbb N$.

Applied for $\delta_\alpha$ instead of $\delta$, this gives: $ \displaystyle \Delta\Big|_{V_\alpha} = \Delta\Big|_{\mathfrak C(\delta_\alpha)} \cong \delta_\alpha^{\boxplus n_\alpha} $ for some $n_\alpha\in\Bbb N$.

On the other side, let us build $U^\perp$ inside $V_\alpha$ (not inside $V$), it is also an invariant subspace, since $U$ is, and we split $\Delta$ restricted on $V_\alpha$ into the pieces. We obtain: $$ \delta_\alpha^{\boxplus n_\alpha} \cong \Delta\Big|_{V_\alpha} = \Delta\Big|_U \boxplus \Delta\Big|_{U^\perp} \ . $$ The projectors on the $\boxplus$-summands are intertwiners. Further splitting the two summands on the R.H.S. above, and applying Schur' Lemma for corepresentations, we see that only $\delta_\alpha$ components may occur in $\Delta\Big|_U$.

With the same argument applied for the restriction of $\Delta$ to $W^\alpha$, (and with an orthocomplement of $U$, built inside $W^\alpha$,) we see that only $\delta_\beta$ components for one or more values of $\beta\ne \alpha$ may occur in $\Delta\Big|_U$.

Contradiction. (The identity of $U$ is inside a sum of homomorphisms from the $n_\alpha$ pieces $ \delta_\alpha$ to corresponding copies of pieces $\delta_\beta$, but $\operatorname{Hom}(\delta_\alpha,\delta_\beta)=0$ for $\beta\ne\alpha$.)

$\square$



Later edit.

I'm very thankful for the comments, and try here to say more on the way to use the "same argument" as presented for $\Delta|_{V_\alpha}$ also for the restriction $\Delta|_{W_\beta}$. There is a constraint in doing so. The question addresses the situation from the implication (iii) $\Rightarrow$ (iv) from Theorem 3.2.12 in the cited book of Timmerman. So the arguments have to be given in this in-between situation. (For this reason a Haar functional $h$ will not appear.)

Some references for these details / discussion:

([T] is target oriented, algebraic CQG's are in focus, although it collects many results not involving a Haar state $h$. Koornwinder and Dijkhuizen are extract the linear algebra for the first time, [K] §2 is a survey. The $C^*$-algebraic framework of CGQ as presented for instance in [MvD] is isolated first a the Hopf-algebraic level. So [K] is well suited for this answer. We use only this reference below.)

Relevant results:

  • [K] Proposition 1.28, showing that matrix elements of mutually inequivalent corepresentations are linearly independent, i think this is the missing piece in the puzzle, claim and proof are added below.

  • [K] Lemma 1.25, that will be cited here with slightly changed notations to match those in [T]:

    Let $(A,\Delta)$ be a coalgebra. Let $\delta:V\to V\otimes A$ be a corepresentation of $A$ on a finite dimensional vector space $V$. Suppose that $V$ is a direct sum of subspaces $V_i$, $i=1,2,\dots , n$, and that each $V_i$ is a direct sum of subspaces $W_{ij}$, $j= 1, 2,\dots , m_i$, and that there are mutually inequivalent irreducible corepresentations $\delta_1,\delta_2, \dots , \delta_n$ such that each subspace $W_{ij}$ is invariant, and $\delta$ restricted to $W_{ij}$ is equivalent to $\delta_i$. Let $U$ be a nonzero invariant subspace of $V$ such that $\delta$ restricted to $U$ is an irreducible corepresentation $\delta'$. Then, for some $i$, $U\subseteq V_i$, and $\delta'$ is equivalent to $\delta_i$.

    The proof of Lemma 1.25 uses adapted Schur Lemmas ([K], Lemma 1.24 (b), (c)), or [T], Proposition 3.2.2 (ii)) for corepresentations, and the following facts.

    • [K] Lemma 1.24 (a) from [K], Proposition 3.2.2 (i) from [T] - image and kernel of an intertwiner $T:\delta_V\to\delta_W$ are invariant subspaces in $W$, respectively $V$.

    • [K] Proposition 1.23 from [K], or 3.2.1 (ii) [T]:

      Let $\delta:V\to V\otimes A$ be a unitary corepresentation of a Hopf $*$-algebra $(A,\Delta)$ on a finite dimensional Hilbert space $V$.

      • (a) Let $W$ be an invariant subspace of $V$. Then the orthogonal complement of $W$ in $V$ is also invariant.
      • (b) $V$ is a direct sum of invariant subspaces on each of which the restriction of $\delta$ is an irreducible corepresentation of $A$.

      Proof: (a) Consider $w\in W$. Let $W^\perp\subseteq V$ be the orthogonal complement. We fix $v\in W^\perp$, show $\delta v\in W^\perp\otimes A$. We write: $$ \begin{aligned} \delta v &= \sum v_0\otimes a_1\ ,\\ \delta w &= \sum w_0\otimes b_1\ ,\\ \end{aligned} $$ with corresponding components $(v_0)$ in $V$, $(w_0)$ in $W\subseteq V$, and $(a_1)$, $(b_1)$ chosen respectively linearly independent in $A$. The compatibility of the scalar product on $A$ with $\delta$ (making $\delta$ unital), $$ (v,w)\;1_A = \underbrace{\sum(v_0,w_0)\; b_1^*a_1}_{=:(\delta v,\delta w)} $$ is rewritten as ([K] (1.42)) : $$ \sum (v_0,w)\; Sa_1 =\sum \underbrace{(v,w_0)}_{=0}\; b_1^*=0\ .$$ So the components $(v_0,w)$ are each equal to zero. Since $w\in W$ is arbitrary, each component $v_0$ is in $W^\perp$, so $\delta v\in W^\perp\otimes A$.

      (b) Apply the previous result inductively w.r.t. the dimension of $V$.

      $\square$


[K] Proposition 1.28:

Let $(A,\Delta)$ be a Hopf algebra with invertible antipode. Let $(\delta_\alpha)$, $\alpha\in \Lambda$ ($\Lambda$ being some index set) be a collection of mutually inequivalent irreducible matrix corepresentations of $A$. Then the set of all matrix elements $a^\alpha_{ij}$ is a set of linearly independent elements.

(See also the comments on this Proposition in [DK] after Theorem 2.1 in §2, CQG Algebras. There is no scalar product required.)

Proof:

  • (i)


Let us work first with only one corepresentation $\delta_\alpha$, so we rename it to $\delta:V\to V\otimes A$. We fix a basis of $V$ and obtain the matrix coefficients $(a_{ij})$ of $\delta$, spanning a subspace $\mathfrak C(\delta)$.

Recall that associated to $A$ we have two regular corepresentations induced by $\Delta$: $$ \begin{aligned} \delta^A_r &:A\to A\otimes A\ , & \delta_r &= \Delta\ , & \delta_r a_{ij} &= \sum_k a_{ik}\otimes a_{kj}\ ,\\ \delta^A_l &:A\to A\otimes A\ , & \delta_l &= \tau(S\otimes\operatorname{id})\Delta\ , & \delta_r a_{ij} &= \tau(S\otimes\operatorname{id})\sum_k a_{ik}\otimes a_{kj}\\ &&&&& = \tau\sum_k S a_{ik}\otimes a_{kj}\\ &&&&& = \sum_k a_{kj} \otimes S a_{ik}\\ &&&&& = \sum_k a_{kj} \otimes a'_{ki}\ . \end{aligned} $$ $\tau=\tau_{12}$ is the switch of corresponding $\otimes$-factors (on the positions $1$ and $2$). Here we use the notation $$ a'_{ij} := Sa_{ji}\ . $$ The matrix (coefficients) $(a'_{ij})_{ij}$ also correspond to a corepresentation, the contragredient corepresentation: $$ \sum_k a'_{ik}\otimes a'_{kj} = \sum_k Sa_{ki}\otimes Sa_{jk} = \sum_k (S\otimes S)\tau a_{jk}\otimes a_{ki} = (S\otimes S)\tau\sum_k a_{jk}\otimes a_{ki} = (S\otimes S)\tau\Delta a_{ji} = \Delta S a_{ji} = \Delta a'_{ij}\ . $$ Now consider the corepresentation $\tilde \Delta$ of the Hopf algebra $B:=A\otimes A$ on the space $A$. It is defined as a composition $$ \begin{aligned} A&\overset{\delta_r}\longrightarrow A\otimes A\overset{\delta_l \otimes\operatorname{id}}\longrightarrow (A\otimes A)\otimes A\cong A\otimes (A\otimes A)=A\otimes B\ ,\\ \\ &\qquad\text{ and explicitly on elements $a\in A$:} \\ a&\overset{\delta_r}\longrightarrow \Delta a = \sum a_1\otimes a_2\\ &\qquad \overset{\delta_l \otimes\operatorname{id}}\longrightarrow \sum \delta_l a_1\otimes a_2 \\ &\qquad\qquad = \sum \tau(S\otimes \operatorname{id})\Delta a_1\otimes a_2\\ &\qquad\qquad = \sum (\tau(S\otimes \operatorname{id}) a_{11}\otimes a_{12})\otimes a_2\\ &\qquad\qquad = \sum a_{12}\otimes S a_{11}\otimes a_2= \sum a_2\otimes S a_1\otimes a_3\ ,\\ &\qquad\text{ so its action on the matrix element $a_{ij}$ is:}\\ \Delta^{(2)}\underbrace{a_{ij}}_a &=(\Delta\otimes\operatorname{id})\Delta a_{ij}=\sum_{k,j} \underbrace{a_{ik}}_{a_1}\otimes\underbrace{a_{kl}}_{a_2}\otimes\underbrace{a_{lj}}_{a_3} \\ a_{ij} &\overset{\tilde\Delta}\longrightarrow \sum_{k,l}a_{kl}\otimes S a_{ik}\otimes a_{lj} \\ &\qquad = \sum_{k,l}\underbrace{a_{kl}}_{\in\mathfrak C(\delta)}\otimes \underbrace{a'_{ki}\otimes a_{lj}}_{\in B=A\otimes A} \in\mathfrak C(\delta)\otimes B\ . \end{aligned} $$ So we can consider the restriction $\tilde\Delta|=\tilde\Delta|_{\mathfrak C(\delta)}:\mathfrak C(\delta)\to\mathfrak C(\delta)\otimes B$.

The above $\tilde\Delta|$ was built on $A$ using the Hopf algebra structure, mainly $\Delta$.

On the other hand, using the matrix coefficients $(a_{ij})$ and $(a'_{ij})$ and the corresponding corepresentations $\delta$ and $\delta'$ we obtain a corepresentation of $B=A\otimes A$. $$\tilde\delta:=\delta'\otimes\delta\ .$$ It is irreducible, since $\delta,\delta'$ are irreducible, [K] Lemma 1.26. It is given by: $$ \tilde\delta e_{ij} = \sum_{k,l} e_{kl}\otimes \underbrace{a'_{ki}\otimes a_{lj}}_{\in B=(A\otimes A)}\ . $$ From the match of the $B$-coefficients in the formulas for $\tilde \Delta a_{ij}|$ and $\tilde\delta e_{ij}$ [K] constructs the intertwiner $L:\tilde \delta\to\tilde\Delta|$ given by $L(e_{ij})=a_{ij}$.

The kernel of $L$ is an invariant space. If it is not $0$, then it is also proper since $\epsilon(a_{ii})=1\ne 0$, contradicting $\tilde\delta$ irreducible. So the kernel of $L$ is trivial, making it a bijection. Since $(e_{ij})$ is a linear independent system, the same holds for $(a_{ij})$.

  • (ii)


Let us consider now a list of inequivalent, irreducible corepresentations $\delta_\alpha$. Set $V_\alpha=\mathfrak C(\delta_\alpha)$ and the corresponding matrix elements $(a^\alpha_{ij})$. For each $\alpha$ we use a different index set $I(\alpha)$, so that the (disjoint) union of all $I(\alpha)$ is a set $I$.

Let $V$ be the span of all $V_\alpha$. Then $\Delta$ maps $\mathfrak C(\delta_\alpha)$ into $\mathfrak C(\delta_\alpha)\otimes \mathfrak C(\delta_\alpha)$. So $V$ is an invariant space of $\tilde\Delta$ over $B=A\otimes A$.

On the other side, consider elements $(e^\alpha_{ij})$ generating spaces $W_\alpha$ and set $W=\bigoplus W_\alpha$. $W$ comes naturally with a corepresentation $$ \boxplus_\alpha (\delta_\alpha' \boxtimes\delta_\alpha) $$ The pieces $(\delta_\alpha' \boxtimes\delta_\alpha)$ are inequivalent and irreducible (over $B=A\otimes A$), [K] Lemma 1.26.

Define again an intertwiner $L$ by setting $e^\alpha_{ij}\to a^\alpha_{ij}$. Its kernel is invariant. Suppose it is not trivial, so it has an invariant subspace $U\ne 0$. By [K] Lemma 1.25 $U$ is $W_\alpha$ for some suitable $\alpha$. For this index $\alpha$ we then have $a^\alpha_{ii}=0$ contradicting $\epsilon(a^\alpha_{ii})=1$.

$\square$

(All arguments are reproduced from [K].)

$\endgroup$
6
  • $\begingroup$ Thanks for your answer. Do you mean "... $U \ne 0$ CONTAINED in $V$..." as opposed to "... $U \ne 0$ containing $V$...". Also, how do you ensure that $U \cap V_\alpha \ne 0$ (so that we can conclude that $U \cap V_\alpha = U$ and thus $U \subseteq V_\alpha$)? $\endgroup$
    – Andromeda
    Nov 6, 2021 at 20:47
  • $\begingroup$ I think you can let $U$ be the intersection $V_\alpha \cap \bigcup_{\beta \ne \alpha} V_\beta$ and that the whole argument still goes through. $\endgroup$
    – Andromeda
    Nov 6, 2021 at 21:37
  • $\begingroup$ On hindsight, I think there is a problem when you try to apply the same argument for $W^\alpha$. You can't apply prop 3.2.11 anymore. So the question is: why can only $\delta_\beta$-components occur in $\Delta\vert_{W^\alpha}$ (and thus also in $\Delta\vert_U$). $\endgroup$
    – Andromeda
    Nov 6, 2021 at 22:24
  • $\begingroup$ Oh, that was an irritating typo, sorry, i wanted $U$ to contain the vector $v$, have to correct, and reread again. $\endgroup$
    – dan_fulea
    Nov 6, 2021 at 23:00
  • $\begingroup$ Can you elaborate why the paragraph "With the same argument applied for the restriction of $\Delta$ to $W^\alpha$, (and with an orthocomplement of $U$, built inside $W^\alpha$,) we see that only $\delta_\beta$ components for one or more values of $\beta\ne \alpha$ may occur in $\Delta\Big|_U$." is true? Note that proposition 3.2.11 only holds for irreducible coreps! $\endgroup$
    – Andromeda
    Nov 7, 2021 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.