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Suppose $G$ is a finitely generated group, with given generating set $S={g_1, \dots, g_n}$. (Assume that if $g\in S$, then $g^{-1}\notin S$. (EDIT: Also assume that $S$ is minimal in the sense that no proper subset of $S$ is generating.)) Given complex numbers $a_1, \dots , a_n$, we can form the element $\sum a_j g_j$ of the group ring $\mathbb C[G]$ and consider its norm as an element of the full group C*-algebra $C^\ast(G)$: $$ \|\sum a_j g_j \| =\sup_\pi \|\sum a_j \pi(g_j)\| $$ where the supremum is over all unitary representation $\pi$ of $G$ on Hilbert space. I am interested in the quantity $$ \alpha(S) = \inf_{\|a\|_1=1} \|\sum a_j g_j\|. $$ (Here $\|a\|_1 =\sum |a_j|$).

Question: Are there known examples of groups $G$ and generating sets $S$ for which $\alpha(S)<1$?

(I am mostly interested in the case where all the generators have infinite order, so I haven't spent much time looking at finite groups. So, in case there are finite examples, a follow-up question would be for examples with all $g$ of infinite order.) (EDIT: Since the C*-norm will be strictly smaller than $\ell^1$ norm, there should be lots of examples, see answers & comments below. We can still ask for examples when the generating set is minimal, and then there is the harder question of just how small $\alpha(S)$ can be.)

Some background and discussion: The question arose from an attempt to analyze the convex set of all points in $\mathbb C^n$ of the form $$ (\langle \pi(g_1)x,y\rangle, \dots \langle \pi(g_n)x,y\rangle ) $$ where $\pi$ ranges over all representations of $G$ on Hilbert spaces $\mathcal H$ and $x,y$ range over all vectors of the unit ball of $\mathcal H$. I am interested in how large a polydisk centered at the origin such a set can contain; estimating the norm of $\sum a_j g_j$ is essentially the dual problem to this one. In this setting, cases where $\alpha(S)$ is small are the "enemy", so I'm trying to understand more about when this happens.

It is not hard to prove the bounds $$ \frac{1}{\sqrt{n}} \leq \alpha(S) \leq 1 $$ (indeed the upper bound is trivial, and the lower bound comes from considering the regular representation of $G$ on $\ell^2(G)$. ) As simple examples, it can be shown that the free abelian group $\mathbb Z^n$ and the free group $\mathbb F^n$ (with their standard $n$-element generating sets) both have $\alpha(S)=1$. Now, one can try to get better lower bounds on $\alpha(S)$ in particular cases by considering particular representations $\pi$ (such as the regular representation), since then one is dealing with particular concrete unitaries, but it seems much harder to improve the upper bound (when it can be improved) unless one has a handle on a faithful representation of $C^\ast(G)$. Of course this is the case when $G$ is amenable, though at the same time it's not clear that amenability is relevant, given the free (non-amenable) and free abelian (amenable) examples just mentioned.

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I think that there is a rather simple iff condition for $\alpha(S)=1$~: if $S=\{g_1,g_2,\dots,g_n\}$, $\alpha(S)=1$ iff $g_1^{-1} g_2,\dots g_1^{-1} g_n$ generate a free abelian subgroup in the abelianization of $G$. I do not have time right now to write a proof, but if you are interested I can do it when I come back. –  Mikael de la Salle May 12 '12 at 7:13

3 Answers 3

up vote 6 down vote accepted

Here is an algebraic characterization of $\alpha(S)=1$: if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=1$ if and only if $\{g_1^{-1}g_k,k=2\dots n\}$ generate a free abelian group in the abelianization $G^{ab}$ of $G$.

In fact, since $\alpha(S)$ is unchanged under the action of $G$ by left (or right) multiplication, it is natural to request for example that $1$ belongs to $S$. We therefore consider, for a subset $S \subset G$, $\widetilde \alpha(S) = \alpha (S \cup \{1\})$. Studying $\alpha$ and $\widetilde \alpha$ are clearly the same problem, since for example if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=\widetilde \alpha(g_1^{-1} g_2,\dots,g_1^{-1} g_n)$.

The condition $\widetilde \alpha(S)=1$ can be expressed in terms of the image $q(S)$ of $S$ in the abelianization $G^{ab}$ of $G$. Namely the claim is that $\alpha(S)=1$ if and only if the group generated by $q(S)$ in $G^{ab}$ is $\mathbf Z^S$ (i.e. it is free abelian).

One direction is clear: $\|a_0+\sum_{g \in S} a_g g\|_{C^*(G)} \geq \|a_0+\sum_{g \in S} a_g g\|_{C^*(G^{ab})}$, so that if $q(S)$ generates a free abelian group, $\widetilde \alpha(S)=1$.

For the other direction, assume that $q(S)$ does not generate a free abelian group. Equivalently there exists a word $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ with $g_i \in S$, $\varepsilon_i \in \{-1,1\}$ that represents $1$ in $G$ but such that $\sum_{i, g_i=g} \varepsilon_i \neq 0$ for at least one $g \in S$ (in other words, this word is an element of the free group on $S$ generators $F_S$ that is trivial in $G$ but nontrivial in $F_S^{ab}=\mathbf Z^S$). In particular, one can choose $a_g \in \mathbf C$ for $g \in S$ such that $\prod a_{g_i}^{(\varepsilon_i)} <0$ where I use the notation $z^{(1)}=z$ and $z^{(-1)}=\bar z$.

Consider now the element $X=1+\sum_{g \in S} a_{g} g$. I claim that $\|X\|_{C^*(G)}< 1+\sum |a_g|$. Indeed, $\|X\|^{2k} = \|(XX^*)^k\|$, and formally $(XX^*)^k$ is a sum of terms of the form $\sum_w c_w w$ where the sum is over all $|S+1|^{2k}$ words in $S\cup{1}$ and $S^{-1} \cup 1$ alternatively, with $\sum |c_w| = (1+\sum |a_g|)^{2k}$. But in this sum, there are two particular elements: $w_0$, the one corresponding to the word with all $1's$, for which $c_w=1$, and (at least) one word, $w_1$ that corresponds to $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ once we remove all the $1$'s, for which $c_w=\prod a_{g_i}^{(\varepsilon_i)} <0$. But both words correspond to $1$ in $G$, so that by the triangle inequality we can write $$\|(XX^*)^k\|_{C^*(G)} \leq |1+c_{w_1}| + \sum_{w \neq w_0,w_1} |c_w|< \sum_w |c_w|=(1+\sum_{g \in S} |a_g|)^{2k}.$$ QED


Some additional remark. The fact that $\widetilde \alpha(S)=\widetilde \alpha(S')$ if $S$ and $S'$ are the generators of $F_n$ and $\mathbf Z^n$ says that the linear space space spanned by $S$ and $S'$ in $C^*(G)$ can be isometrically isomorphic when the groups generated are not. Such phenomenon cannot happen if one considers the operator space version of the question. It is a theorem by Pisier that if a map $T:span(1,u_1,\dots,u_n) \to span(1,v_1,\dots,v_n)$ sending $1$ to $1$ and the unitary $u_i$ to the unitary $v_i$ is completely isometric, then $T$ extends to $*$-isomorphism of the $C^*$-algebras generated. When $u_i=g_i$ (resp $v_i=g_i'$) in $C^*(G)$ (resp. $C^*(G')$), this implies that $G$ and $G'$ are isomorphic.

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This is very helpful, thank you! –  Mike Jury May 12 '12 at 18:15

Hold on. The only restriction on $S$ is that $g$ and $g^{-1}$ are never both in $S$? It seems like all you're asking is whether the full group C*-algebra norm of an element of ${\bf C}[G]$ is ever strictly less than its $l^1$ norm, for elements satisfying the condition that the coefficients of $g$ and $g^{-1}$ never both nonzero.

Surely that already fails for $G = {\bf Z}$ and $S = {\bf N}$? It's just saying that the natural map from $l^1({\bf N})$ into the disk algebra isn't an isometry.

Edit: counterexamples verifying the minimality condition on $S$ are in the comments.

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Ah, duh. I hadn't thought very hard about the conditions on $S$. It is probably natural to impose the condition that $S$ be "minimal" in some sense. Say, no proper subset of $S$ is generating? (I think I've been tacitly assuming this in all the examples I've looked at.) I'll edit the question to include this condition. Even so, now that you say it seems obvious that there should be (lots of?) examples, just on the grounds that the C*-norm won't equal the $\ell^1$ norm. Thanks. –  Mike Jury May 12 '12 at 3:27
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Right. But actually checking that something is an example might still be hard if you have to compute norms. Here's an idea: the C* norm of $x = \sum a_jg_j$ squared equals the C* norm of $x^*x = \sum a_j\bar{a}_k g_jg_k^{-1}$. If the $l^1$ norm of the latter is less than the $l^1$ norm of $x$ squared, then we know the C* norm of $x$ is less than its $l^1$ norm. This can give us an easy way to generate examples where the C* norm doesn't equal the $l^1$ norm. –  Nik Weaver May 12 '12 at 4:16
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For example, let $G = ({\bf Z}/2)^2$ with standard generators $a$ and $b$ and let $S = \{a,b\}$. Then we can take $x = 1a + ib$, so the $l^1$ norm of $x$ is 2 but the $l^1$ norm of $x^*x = (1a - ib)(1a + ib) = 2e$ is 2, so that the C* norm of $x$ can be at most $\sqrt{2}$. –  Nik Weaver May 12 '12 at 4:36
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Or if you want an example with infinite order generators, take $G = \langle a,b: ab^{-1} = ba^{-1}\rangle$ and again let $S = \{a,b\}$ and $x = 1a + ib$. Again the $l^1$ norm of $x$ is 2 but its C* norm is at most $\sqrt{2}$. –  Nik Weaver May 12 '12 at 4:53

Hi, Mike. Take a look at the paper "Computing Norms in Group C*-Algebras" by Akemann and Ostrand, Amer. J. Math. 98 (1976), 1015–1047. For instance, if $g_1, \ldots, g_n$ are the standard generators of a free group, they calculate that $\|\sum_1^n g_i\| = 2\sqrt{n-1}$. It looks like that would (taking each $a_i = 1/n$, and with $n \geq 3$) already make your infimum less than one. Specifically, it would give an upper bound of $2\sqrt{n-1}/n$ for $\alpha(S)$ in this case.

Edit: I stand corrected, that is the reduced group C*-algebra norm.

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Hi NIk, If I'm not mistaken (I'll have to go back and look at it again), the Akemann-Ostrand result is about estimates only in the regular representation, and so applies only to the reduced group C*-algebra. It seems not hard to prove that $\alpha(S)=1$ in the free group case, since it would be bounded below by the corresponding $\alpha$ for the free abelian group, and it is easy to see that this $\alpha(S)$ is 1, since $C^\ast(\mathbb Z^n)$ is ismoporphic to $C(\mathbb T^n)$, and we can take the generating unitaries to be the coordinate functions $z_1, \dots z_n$. –  Mike Jury May 12 '12 at 1:58
    
Oh, you're right. Let me think a little more. –  Nik Weaver May 12 '12 at 2:09

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