# 'Test Functions' to Lower Bound the Norm of Elements of Dual Quantum Group

There may well be an answer to this question in a simpler category than that of finite dimensional quantum groups and in that case this question is more suitable to math.stack and I apologise in advance if this is the case.

Consider a finite dimensional quantum group $A=F(\mathbb{G})$ with a Haar state $h:A\rightarrow\mathbb{C}$. In the finite dimensional case (and in more generality in fact), this allows us to define '$p$-norms' on $A$ via

$$\|a\|_p:=\sqrt[p]{h((a^*a)^{p/2})}.$$

In particular, $\|a\|_1=h((a^*a)^{1/2})$ and $\|a\|_2=\sqrt{h(a^*a)}$. There is a Cauchy-Schwarz Inequality:

$$\|ab\|_1\leq\|a\|_2\|b\|_2.$$

We can define a dual quantum group $\hat{A}$ via the map $\mathcal{F}:A\mapsto A'$, $\mathcal{F}(a)(b)=h(ba)$. In this, finite dimensional case, $\hat{A}=A'$, and the multiplication in $\hat{A}$ is given by the convolution:

$$\nu\star\mu=(\nu\otimes\mu)\Delta,$$

the Haar state $\hat{h}:\hat{A}\rightarrow\mathbb{C}$ is given by

$$\hat{h}(\mathcal{F}(a))=\varepsilon(a),$$

and the involution is

$$\nu^*(a)=\overline{\nu(S(a)^*)}.$$

I am interested in finding bounds for

$$\|\nu\|_1=\hat{h}((\nu^*\nu)^{1/2}).$$

Via $\varepsilon\star \nu=\nu$, we have the following upper bound on $\|\nu\|_1$:

$$\|\nu\|_1=\|\varepsilon\star \nu\|_1\leq\|\varepsilon\|_2\|\nu\|_2=\sqrt{\dim A}\cdot\|\nu\|_2.$$

For my application, I have a way of calculating and bounding $\|\nu\|_2$ above but I am also interested in bounding below:

$$?\leq \|\nu\|_1\leq \sqrt{\dim A}\|\nu\|_2.$$

In the classical case where $G$ is a finite group, a 1-norm on $F(G)$ might be given by

$$\|f\|_1'=\sum_{t\in G}|f(t)|,$$

and this fits quite well into this framework:

$$\|f\|_1=h\left((f^*f)^{1/2}\right)=h(|f|)=\frac{1}{|G|}\sum_{t\in G}|f(t)|.$$

Now I am more interesting in looking at $\|\cdot\|_1:\hat{A}\rightarrow[0,\infty)$. Now classically one might define the 1-norm on $\mathbb{C}G\supset M_p(G)$ as

$$\|\nu\|_1'=\sum_{t\in G}|\nu(\delta_t)|.$$

An advantage of working with this norm is that we have

$$\|\nu\|_1'=\sup_{\underset{f\in F(G)}{\|f\|_{\infty}\leq1}}|\nu(f)|,$$

so that we can generate lower bounds by looking at test functions $\phi\in F(G)^1$ and so we have

$$|\nu(\phi)|\leq \|\nu\|_1'\leq \sqrt{|G|}\|\nu\|_2'.$$

The problem with using $\|\cdot\|_1:\hat{A}\rightarrow[0,\infty)$ via the Haar state $\hat{h}$:

$$\|\nu\|_1=\hat{h}((\nu^*\nu)^{1/2}),$$

is that even in the classical case I don't quite have something like

$$\|\nu\|_1=\frac{1}{|G|}\sum_{t\in G}|\nu(\delta_t)|.$$

When we are in $A=F(G)$ the involution is simply

$$f^*(s)=\overline{f(s)},$$ and with pointwise multiplication and positivity in the C*-algebra equivalent to positivity of the coefficients, we have

$$(f^*f)^{1/2}(s)=|f(s)|,$$

so the 1-norm works quite nicely in there.

Things are more complicated however in $\hat{A}=\mathbb{C}G$ (as is alluded to in this question of mine). In general, even for symmetric probability measures, we don't have

$$(\nu^*\nu)^{1/2}=\nu.$$

Now in the classical case I can just use $\|\cdot\|_1'$ and the ordinary C-S to get my upper bounds. However in the truly non-commutative case I want to use $\|\cdot\|_1$... if I can get myself some lower bounds! Otherwise I can just use and bound $\|\cdot \|_2$ above.

In the not-necessarily-commutative, quantum group case, is there a way to generate lower bounds on $\|\nu\|_1=\hat{h}((\nu^*\nu)^{1/2})$ via 'test functions': $$\|\nu\|_1\geq\sup_{s\in S}F(s,\nu)?$$ Perhaps $F(s,\nu)$ involves convolving $\nu$ with some element of $s\in S\subset\hat{A}$ with a small norm or maybe something like hitting $\nu$ with an element of $s\in S\subset A$ with small norm: $\nu(s)$.

• I find your question somewhat unclear. Are you merely asking for a way to express the noncommutative L^1-norm given by a tracial state as a supremum using the natural pairing? (Your definition of the 1-norm seems to be the usual one if the Haar state is tracial, as it is for Kac examples, but I am not sure it is the correct definition in the non-tracial case) – Yemon Choi Jun 27 '15 at 13:58
• Yes to expressing it as a supremum --- or even greater than a supremum. You can assume Kac (and if this isn't enough tracial also). – JP McCarthy Jun 27 '15 at 14:01
• Hmm, well perhaps I have misunderstood your question, but if $\tau$ is a faithful normal trace on a von Neumann algebra $M$, then IIRC $\tau((x^*x)^{1/2})$ is equal to the supremum of $|\tau(xy)|$ as $y$ runs over all elements in unit ball of M – Yemon Choi Jun 27 '15 at 14:11
• Well cf. my first paragraph... have you got a reference or proof of this? – JP McCarthy Jun 27 '15 at 16:58
• OK, I've cobbled something together. Apologies for any earlier confusion – Yemon Choi Jun 29 '15 at 11:55

In a more general setting than that of the original question: suppose we have a faithful normal state $h$ on a von Neumann algebra $M$. Suppose furthermore that $h$ is tracial, meaning that $h(xy)=h(yx)$ for all $x,y\in M$. (Warning! there are important examples of compact quantum groups where the Haar state is faithful but not tracial.)
In this setting we may define $\Vert x\Vert_{L^1(M,h)}$ to be $h(|x|)$. With this definition it is not clear that we have a norm; however, it is known that one has $$h(|x|) = \sup\{ | h(xy) | \colon y \in M, \Vert y\Vert_M \leq 1 \} \tag{*}$$ and the proof has been given by Martin Argerami on MathStackExchange. The motivating example to keep in mind is the commutative case $M=L^\infty[0,1]$ with usual ess.sup norm and usual weak-star topology, with $h(x) = \int_0^1 x(t)\,dt$.
one sees that Segal took the RHS of $(*)$ as the definition of the $L^1$-norm on $M$ given by $h$. He then, in part (d) of Corollary 10.1, shows that the RHS of $(*)$ is equal to the LHS of $(*)$.