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Is it true that for every group $G$ and $f\in \mathbb C[G]$ it holds that $$\dim(\mathbb C[G]*f)\mathop{supp}(f)\geq |G| ?$$

Here, $\mathbb C[G]$ is the group algebra, and by $\mathbb C[G]*f$ I mean left ideal of the group algebra $\mathbb C[G]$ generated by $f$.

Essentially this is uncertainty principle for non-commutative groups. Since $supp \hat {f} = dim C[G]*f$ in case $G$ is abelian.

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It might help to explain your notation more carefully. – Simon Wadsley Feb 19 '11 at 21:16
Check out…. – Michael Feb 19 '11 at 23:29
up vote 8 down vote accepted

The answer is yes, this always holds.

Note that

$$\dim(im(f)) \cdot \|f\|^2 \cdot | {\rm supp}(f)| \geq \tau(f^*f) \cdot |{\rm supp}(f)| \geq |G| \cdot \|f\|^2_1.$$

Here, $\tau \colon \mathbb C[G] \to \mathbb C$ is the non-normalized trace on $\mathbb C[G]$, coming from the inclusion $\mathbb C[G] \subset M_{|G|} \mathbb C$. It is best decribed by $$\tau(\sum_{g \in G} a_g g) = |G| \cdot a_e.$$ Also, $f \mapsto f^*$ denotes the usual involution, i.e. $$(\sum_{g \in G}a_g g)^{*} = \sum_{g \in G} \bar a_g g^{-1}.$$ The first inequality, follows since $\tau(f^*f)$ is just the sum of the eigenvalues of $f^*f$, which is obviously bounded by the number of non-zero eigenvalues (which equals the dimension of image of $f$) times the size of the largest eigenvalue (which equals $\|f\|^2$). Note that by direct computation $$\tau\left((\sum_{g \in G} a_g g)^* (\sum_{g \in G} a_g g)\right)=|G| \cdot \sum_{g \in G} |a_g|^2.$$ The second inequality follows from this observation and the Cauchy-Schwarz inequality applied to $f\cdot \chi_{{\rm supp} f}$, where the product is here the pointwise product of coefficients and $\|f\|_1$ denotes the usual 1-norm on $\mathbb C[G]$.

Now, since each group element acts as a unitary (and hence with operator norm $1$) on $M_{|G|} \mathbb C$, we get $\|f\|_1 \geq \|f\|$ and hence

$$\dim(im(f)) \cdot | {\rm supp}(f)| \geq |G|.$$

This even has an extension to all (possibly infinite) groups with essentially the same proof. The appropriate statement is then that for the normalized Murray-von Neumann dimension (with respect to the group von Neumann algebra $LG$) of the closure of the image of $\lambda(f)$ acting on $\ell^2 G$ via the left-regular representation $\lambda$, we have

$$\dim_G \left (\overline{im(\lambda(f))} \right) \cdot |{\rm supp}(f)| \geq 1.$$

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Your proof shows also that $1$ on the right-hand side can be replaced by $|f|_1/|f|$, where $|f|_1$ is the $l^1$-norm of $f$ and $|f|$ is the operator norm of $f$. Which can be interpreted as saying that non-amenable groups are more "Heisenberg-uncertain" than ameanable groups :-). – Łukasz Grabowski Mar 1 '11 at 13:11
You are right, I noticed this as well. Anyhow, this does not seem to be a strong inequality. – Andreas Thom Mar 1 '11 at 14:00

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