Uncertainty principle for non-commutative groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:49:46Z http://mathoverflow.net/feeds/question/55990 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55990/uncertainty-principle-for-non-commutative-groups Uncertainty principle for non-commutative groups Klim Efremenko 2011-02-19T17:02:43Z 2011-02-21T07:58:06Z <p>Is it true that for every group $G$ and $f\in \mathbb C[G]$ it holds that $$\dim(\mathbb C[G]*f)\mathop{supp}(f)\geq |G| ?$$ </p> <p>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 $f$.</p> <p>Essentially this is uncertainty principle for non-commutative groups. Since $supp \hat {f} = dim C[G]*f$ in case $G$ is abelian.</p> http://mathoverflow.net/questions/55990/uncertainty-principle-for-non-commutative-groups/56081#56081 Answer by Andreas Thom for Uncertainty principle for non-commutative groups Andreas Thom 2011-02-20T17:22:05Z 2011-02-21T07:54:52Z <p>The answer is yes, this always holds.</p> <p>Note that</p> <p>$$\dim(im(f)) \cdot \|f\|^2 \cdot | {\rm supp}(f)| \geq \tau(f^*f) \cdot |{\rm supp}(f)| \geq |G| \cdot \|f\|^2_1.$$</p> <p>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 $\mathbb C[G]$.</p> <p>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</p> <p>$$\dim(im(f)) \cdot | {\rm supp}(f)| \geq |G|.$$</p> <p>This even has an extension to <em>all</em> (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</p> <p>$$\dim_G \left (\overline{im(\lambda(f))} \right) \cdot |{\rm supp}(f)| \geq 1.$$</p>