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.

  • $\begingroup$ 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


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 $\beta\ne\alpha$.)


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 $V$.


[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.)


  • (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 $\epsilon(a^\alpha_{ii})=1$.


(All arguments are reproduced from [K].)

  • $\begingroup$ 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

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