5
$\begingroup$

Let $G$ be a finite group, and consider the composition $X * Y$ on $\mathbb{C}G$ defined by $$(\sum_g \alpha_g u_g) * (\sum_g \beta_g u_g) = \sum_g \alpha_g \beta_g u_g.$$ This composition can be reformulated as a convolution using the Fourier transform.

Question: Let $X,Y \in \mathbb{C}G$ be positive elements. Is $X * Y$ also positive?

Remark: It is an application of Theorem 4.1 in http://dx.doi.org/10.1090/tran/6582 to finite groups theory, but we are here interested in a purely group-theoretic reference or proof.

It is like to Schur product Theorem because $X * Y$ is an entrywise product like the Schur product.

Bonus question: Can we extend to any locally compact group?

$\endgroup$
2
  • 3
    $\begingroup$ What means positive here? $\endgroup$ Aug 6, 2017 at 13:22
  • $\begingroup$ @AlexanderChervov: Positive means of the form $a^{\star}a$, with $(\sum_g \alpha_g u_g)^{\star} = \sum_g \overline{\alpha}_g u_{g^{-1}}$. $\endgroup$ Aug 6, 2017 at 13:26

1 Answer 1

4
$\begingroup$

The answer is positive. For ease of notation, let me write $a=\sum_g \alpha_g u_g$ and $b = \sum_g \beta_g u_g$.

Let's equip $\mathbb{C}G$ with the positive inner product induced by the standard basis. In terms of the usual normalized trace on a group algebra, this inner product is given by $(x,y)\mapsto\mathrm{tr}(x^*y)$. So when you decompose $\mathbb{C}G$ into a direct sum of matrix algebras, this is just the usual Hilbert-Schmidt inner product. In particular, the cone of positive elements is nicely self-dual: we have $y\geq 0$ if and only if $\mathrm{tr}(xy)\geq 0$ for all $x\geq 0$. Hence it is enough to show that $\mathrm{tr}\left(x(a \ast b)\right)\geq 0$ for all $x\geq 0$.

Now your composition operation $*:\mathbb{C}G\otimes\mathbb{C}G\to\mathbb{C}G$ is the adjoint of the usual comultiplication $\Delta:\mathbb{C}G\to\mathbb{C}G\otimes\mathbb{C}G$, meaning that $$ \mathrm{tr}\left(x(a \ast b)\right) = \mathrm{tr}(\Delta(x)(a\otimes b)). $$ Finally, we have $\Delta(x)\geq 0$ since $\Delta$ is a $*$-homomorphism, and $a\otimes b\geq 0$ since $a\geq 0$ and $b\geq 0$ by assumption.

I don't know about the generalization to the locally compact case.

$\endgroup$
4
  • $\begingroup$ Your nice proof extends immediately to any finite dimensional Kac algebra. $\endgroup$ Aug 6, 2017 at 15:34
  • $\begingroup$ So, $\forall a,b \in \mathbb{C}G$, $\exists c \in \mathbb{C}G$ such that $(a^{\star}a) * (b^{\star}b) = c^{\star}c$. Is there a formula for $c$? $\endgroup$ Aug 6, 2017 at 17:23
  • $\begingroup$ I agree that the argument works for any finite-dimensional Hopf C*-algebra. Are these the same as finite-dimensional Kac algebras? $\endgroup$ Aug 6, 2017 at 18:06
  • 1
    $\begingroup$ Yes, Kac algebra and Hopf C*-algebra are synonymous. $\endgroup$ Aug 6, 2017 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.