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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