**Taken from An Invitation to Quantum Groups and Duality by Timmermann.**

Let be a quantum group with a comultiplication . We make the following definitions. A *corepresentation *of on a complex vector space is a linear map that dualises representations with the coassociativity and counit properties:

, and

.

Now we wish to dualise the terms *invariant*, *irreducible, unitary, intertwiner, equivalent, matrix elements.* As we want a theory dual to group representation we won’t use Timmermann’s definitions at face value but instead construct them from their group representation counterparts. This might prove difficult. In attempting to dualise group representation theory as quantum group corepresentation theory is ‘everything’ dualised? Our first term here provides a problem. Do we need an invariant corepresentation or a ‘coinvariant representation’?

### Invariant

An invariant subspace of a group representation is a subspace such that

for all and .

This means that for the family of linear maps , is stable subspace. How this is dualised is important in how we may hope to write a corepresentation as a direct sum of irreducible corepresentations so we need the right definition. Timmermann calls *invariant *if . If we could view the co-representation as a family of endomorphisms on then we might be able to write down a definition of co-invariant or maybe say that Timmermann’s definition is what we need.

As an example of what we might need to do let be the regular action of a group and let be the subspace of constant functions. This set is invariant. Thinking about this yields a definition of *co-invariant. *

A subspace is *co-invariant *for if .

Timmermann’s definition makes good sense. One is confused between invariant co-representation and co-invariant co-representation. I am guessing that Timmermann’s definition will allow us to do what we want.

### Irreducible

We call a representation *irreducible *if it has no non-trivial subspaces. We can do the exactly the same for co-representations.

### Unitary

No problem here we want where we might take the Haar inner product on : , where is the Haar state on …I forgot this.

### Intertwiners & Equivalent Corepresentations

In the group representation picture, an intertwiner of and is a bijective linear map which makes the following diagram commute.

We can find an appropriate definition of an* intertwiner* of two corepresentations and of a quantum group by reversing arrows:

Two representations and if there exists an intertwiner for them. The intertwiner by definition is bijective. Two corepresentations can now be said to be *equivalent *if there they have an intertwiner.

## Schur’s Lemma

*Suppose that and are two irreducible corepresentations of a quantum group and let an intertwiner. *

*If and are not equivalent then**If is complex and is an intertwiner of with itself then .*

First we show that the kernel and image of an intertwiner are invariant subspaces. The kernel is invariant by the following diagram:

Similarly the image is invariant by the following diagram:

Hence by irreducibility the kernel and image are either trivial or the whole space. The rest of the theorem follows through from the classical argument

Now we want to work towards some orthogonality relations.

From this point we follow Timmermann. Timmermann works with compact algebraic quantum groups of which finite quantum groups are.

### Theorem 3.2.1

*Let be a corepresentation of over a vector space . *

*If is a algebraic compact quantum group, then is equivalent to a unitary corepresentation.*

*If is unitary and is an invariant subspace, then the complement is also invariant.**Every element**is contained in some finite-dimensional invariant subspace of**. In particular,**has finite dimension if**is irreducible.**If for every finite-dimensional invariant subspace**, the restriction**is equivalent to a unitary corepresentation, in particular, if**is an algebraic compact quantum group, then**is equivalent to a direct sum of finite-dimensional irreducible unitary corepresentations.*

*Proof*

1. Choose some Hermitian inner product on . Since the Haar stare of is positive and faithful (proof in here), defines a Hermitian inner product on . By a standard argument defines a Hermitian product on . As is injective, we can define a second Hermitian inner product on by the formula

for all ,

and expressed in Sweedler’s notation. An associated -valued product

for all . Timmermann goes on to show that is unitary with respect the inner product

2. This is a standard argument according to Timmermann.

3. Denote by the representation

.

Note that is the space of intertwiners from to .

Since , by definition, the subspace contains . Evidently, is invariant for , so by a previous proposition of Timmermann (3.1.7 (vi)) also for . If , where and , then is contained in the linear span of the . Therefore has finite dimension.

4. Follows from 3. & 4. and Zorn’s Lemma.

### Proposition 3.2.3

*Two irreducible unitary co-representations are equivalent if and only if they admit a unitary intertwiner.*

## Schur’s Orthogonality Relations

The matrix elements of irreducible co-representations satisfy an analogue of Schur’s orthogonality relations known from the representation theory of compact groups. It is precisely this kind of a result that we need to write a quantised Diaconis Upper Bound Lemma.

### Lemma

*Let be a Hopf *-algebra with normalised integral , let and be co-representations on finite-dimensional vector spaces and , respectively, and let . Denote by and the co-representation operators corresponding to and , respectively, and define by *

* and (Y(R\otimes 1)X^{-1}).*

*Then* *. If , then *

### Proposition 3.2.6

*Let be a Hopf *-algebra with a normalised integral , and let and be inequivalent irreducible co-representations of on vector spaces and , respectively. Then for all and ,*

.

*If and are unitary, then for all and .*

*Proof *: As the co-representations are irreducible we can take the vector spaces and to be finite dimensional. Let and be the* corepresentation operators *defined by

and .

Let be the matrix element of given by

and be the matrix element of given by

.

…I am completely lost now!

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