# Characterising algebraic compact quantum groups among Hopf $^*$-algebras

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:

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.

• What is $\mathfrak C(u^\alpha)$? Oct 27, 2021 at 14:39
• 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. Oct 27, 2021 at 14:43
• @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. Oct 27, 2021 at 18:33
• @LSpice I was also thinking in this direction, which inspired my other question: mathoverflow.net/questions/407196/… Oct 27, 2021 at 18:34

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].)

• 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 $U \subseteq V_\alpha$)? Nov 6, 2021 at 20:47
• 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. Nov 6, 2021 at 21:37
• 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 $\Delta\vert_U$). Nov 6, 2021 at 22:24
• Oh, that was an irritating typo, sorry, i wanted $U$ to contain the vector $v$, have to correct, and reread again. Nov 6, 2021 at 23:00
• 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! Nov 7, 2021 at 18:32