# '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: $$\nu(s)$$.

Thank you for your help.

• 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) 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). 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 Jun 27 '15 at 14:11
• Well cf. my first paragraph... have you got a reference or proof of this? Jun 27 '15 at 16:58
• OK, I've cobbled something together. Apologies for any earlier confusion 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 $(*)$.