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.
 A: 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] An invitation to Quantum Groups and Duality, Thomas Timmermann, EMS/AMS, 2008


*[K] General Compact Quantum Groups, a Tutorial, Tom H. Koornwinder, 1994


*[DK] CQG Algebras: A Direct Algebraic Approach to Compact Quantum Groups, Mathijs S. Dijkhuizen, Tom H. Koornwinder, Lett. Math. Phys. 32 (1994), no. 4, 315–330.


*[MvD] Notes on Compact Quantum Groups, Ann Maes, Alfons Van Daele, 1998
([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].)
