# Reduced compact quantum group and left and right multiplication

Let $$(A,\Delta)$$ be a compact quantum group in the sense of Woronowicz, and let $$A_0$$ be its dense Hopf subalgebra. We can construct from the Haar state $$h:A \to \mathbb{C}$$ an inner product $$\langle \cdot,\cdot\rangle: A_0 \times A_0 \to \mathbb{C}, ~~~~~ (a,b) \mapsto h(a^*b).$$ Let $$L^2(A_0)$$ be the completion of $$A_0$$ to a Hilbert space. Let $$\pi_L:A_0 \to \mathcal{B}(L^2(A_0))$$ denote the representation of $$A_0$$ on $$L^2(A_0)$$ given by left multiplication. This gives a norm on $$A_0$$ defined by $$\|a\| := \|\pi_L(a)\|_{op}$$, where $$\|\cdot\|_{op}$$ denotes the operator norm of $$\mathcal{B}(L^2(A_0))$$. The completion of $$A_0$$ with respect to $$\|\cdot\|_{op}$$ is called the reduced completion of $$A_0$$, and the coproduct of $$A_0$$ extends to this completion to give a compact quantum group. All this is explained in detail in Timmermann's book on quantum groups.

If we repeat the above construction but replace left multiplication by right multiplication do we get the same norm or something different? (note that right multiplication will give an anti-algebra map, but it's not clear that this does anything to the value of the norm.)

## 1 Answer

I think there are some subtle points here about what the "right action" even means.

For a general $$*$$-algebra $$A_0$$ and a functional $$\phi:A_0\rightarrow\mathbb C$$, we first of all have to decide what "positive" means for $$\phi$$. We could take this as being $$\phi(a^*a)\geq0$$ for all $$a$$. Then Cauchy-Schwarz holds and we can form $$L^2(A_0,\phi)$$. Why, however, do we have that left multiplication $$\pi_L:A_0\rightarrow L(A_0), \pi_L(a)(b) = ab$$ extends to a bounded operator on $$L^2(A_0,\phi)$$? For $$C^*$$-algebras, this is a basic but slightly subtle result: it follows from the inequality $$b^*a^*ab \leq \|a\|^2 b^*b$$.

For compact quantum group algebras $$A_0$$ I think you have to use the unitary corepresentation theory to show that $$\pi_L(a)$$ is a bounded operator, for each $$a\in A_0$$. Indeed, this is so: see Section 5.4.2 of Timmermann's book.

Thus, in general, there really is no notion of "right action", because again why need $$\pi_R(a):b\mapsto ba; A_0\rightarrow A_0$$ extends to a bounded operator on $$L^2(A_0)$$? Even for $$C^*$$-algebras, you do not have a right action. However, if $$h$$ is a trace (corresponding to $$A$$ being of Kac type) then a simple calculation shows that $$\pi_R$$ exists and gives the same norm: $$\|\pi_R(a)(b)\|^2_2 = h(a^*b^*ba) = h(baa^*b^*) = \|\pi_L(a^*)(b^*)\|^2_2 \leq \|\pi_L(a^*)\|^2 \|b^*\|^2_2$$ and $$\|b^*\|^2_2 = h(bb^*) = h(b^*b) = \|b\|^2_2$$.

Actually, for CQG algebras we can say more, because the state $$h$$ is KMS. This is not so clear in Timmermann's book I think, but compare Theorem 8.1.13 (ii) with Example 8.1.22. In particular, see the top of page 213. In short, there is an automorphism (not a $$*$$-automorphism) $$\sigma_{i/2}$$ of $$A_0$$ with $$h(a^*a) = h(\sigma_{i/2}(a) \sigma_{i/2}(a)^*) \qquad (a\in A_0).$$ Then consider $$\| \pi_R(a)(b)\|^2_2 = h(a^*b^*ba) = h(\sigma_{i/2}(ba)\sigma_{i/2}(ba)^*) = h(\sigma_{i/2}(b) \sigma_{i/2}(a) \sigma_{i/2}(a)^* \sigma_{i/2}(b)^*) = \|\pi_L(\sigma_{i/2}(a)^*)(\sigma_{i/2}(b)^*)\|_2^2,$$ and note that $$\|\sigma_{i/2}(b)^*\|_2^2 = h(\sigma_{i/2}(b)\sigma_{i/2}(b)^*) = h(b^*b) = \|b\|^2_2$$, here for $$a,b\in A_0$$. It follows that $$\pi_R(a)$$ is bounded with $$\| \pi_R(a) \| \leq \|\pi_L(\sigma_{i/2}(a)^*)\|$$ (actually, equal). So, $$\pi_R$$ is well-defined, but it does not induce the same norm, except when $$A$$ is Kac.

(My motivation for this argument was the following: One possible way to give some notion of a right action comes from von Neumann algebra theory, and the theory of "correspondences", see Takesaki, Volume 2, Chapter IX, Section 3. As $$h$$ is a KMS state, $$\pi_L(A_0)''$$ is in standard position on $$L^2(A_0)$$, and so there is a modular operator $$J$$. We can then define $$\pi_R(a) = J\pi_L(a)^*J$$. By definition, this gives the same norm on $$A_0$$, but it's not really what you would call the "right multiplication"...)

• So in short, the fact that we choose the inner product to be conjugate linear in the first variable makes left multiplication the "natural choice". Hence I would guess that if we were to choose the inner product to be conjugate linear in the second variable, that is $\langle a,b\rangle := \phi(ab^*)$ then right multiplication would be the natural choice? Mar 5, 2020 at 19:08
• Yes, that is completely right, and is (I think!) a very nice observation. But see also my edit to my answer. Mar 5, 2020 at 20:48