Let $G$, $H$ be finite groups. Consider the group algebra $\mathbb{C}G$ acting on $L^2(G)$, making $\mathbb{C}G$ into a C* algebra, and the resulting positive elements, say $P_G\subset \mathbb{C}G$. Similarly we have $P_H\subset \mathbb{C}H$. Now take any function $f:G\to H$, and this induces a linear map which we also denote $f:\mathbb{C}G\to \mathbb{C}H$.

Conjecture. If $f(P_G)\subset P_H$ and $f$ maps the identity to the identity, then $f$ is a group homomorphism.

I would be grateful if anyone could shed any light on this problem. This conjecture is related (is dual to) to one from the PhD thesis of Fatemah Al Ghamdi.

  • 3
    $\begingroup$ The hypothesis implies that $f\colon {\mathbb C}G\to{\mathbb C}H$ is a Jordan homomorphism. Hence by Stormer's theorem, there is a central projection $p$ in ${\mathbb C}H$ such that $f(\,\cdot\,)p$ is a homomorphism and $f(\,\cdot\,)(1-p)$ is an anti-homomorphism (as suggested in Yemon's answer). Probably, this would lead to that $f$ itself is either a homomorphism or an anti-homomorphism? $\endgroup$ – Narutaka OZAWA Feb 18 at 0:51
  • $\begingroup$ @NarutakaOZAWA Indeed something about the original question reminded me of M. Walter's theorem concerning isometric isomorphism of Fourier algebras, which would lead to the kind of conclusion that you suggest. $\endgroup$ – Yemon Choi Feb 18 at 0:57
  • $\begingroup$ How does the hypothesis imply that we have a Jordan homomorphism? This is very interesting, and this sounds like the critical point! $\endgroup$ – Edwin Beggs Feb 18 at 19:57
  • $\begingroup$ Sorry that is the paper: On Positive Linear Maps Preserving Invertibility - just found that. $\endgroup$ – Edwin Beggs Feb 18 at 21:05
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    $\begingroup$ @Edwin Beggs: I learned it from Kirchberg's paper (Invent Math 1993) and recorded in my QWEP survey (IJM 2004). A correction on the previous comment: $p$ is in the center of the C*-algebra generated by the range of $f$. In any case, by the Jordan multiplicativity, $f(xy)$ equals to either $f(x)f(y)$ or $f(y)f(x)$ for each $x,y \in G$. The rest is an algebra/combinatorics problem... $\endgroup$ – Narutaka OZAWA Feb 19 at 1:35

I think the "modified conjecture" (that is, one gets a homomorphism or anti-homomorphism) can be proved by elementary calculations, in this setting. (Walter's theorem, which Yemon alludes to, uses only that we have an isometry, not that the map has the rather special form). However, my tolerance for such calculations is low at the moment (and I cannot see how to organise the data in a "clever" way) so I'll just indicate the technique and some partial results.

The idea is that $a\in\mathbb CH$ is positive if and only if $(a\xi|\xi)\geq 0$ for all $\xi\in\ell^2(H)$. We aim to choose $\xi$ intelligently, but first we make a general observation. Let $(\delta_h)_{h\in H}$ be the standard orthonormal basis of $\ell^2(H)$ and let $\lambda_x$ be the left-multiplication operator given by $x\in H$. Then $$ (\lambda_x\xi|\xi) = \sum_{h\in H} \xi_{x^{-1}h} \overline{\xi_h} = \sum_h \overline{\xi_h} \check\xi_{h^{-1}x} = \overline\xi \star \check\xi = \varphi(x), $$ say. I call such $\varphi$ positive definite. Here $\check\xi_h = \xi_{h^{-1}}$ for each $h$, and $\star$ is convolution.

Choose $\xi = \delta_e + u\delta_h$ for $u\in\mathbb C$, so $\check\xi = \delta_e + u\delta_{h^{-1}}$ and thus $$ \varphi(x) = (\overline{\xi} \star \check\xi)(x) = (1 + |u|^2)\delta_{x,e} + u\delta_{x,h^{-1}} + \overline{u} \delta_{x,h}. $$

Claim: $f(x^{-1}) = f(x)^{-1}$ for each $x\in G$.

I write $1$ for $\lambda_e$. To show this, consider $a = (1+z\lambda_x)^\ast (1+z\lambda_x) \geq 0$ in $\mathbb CG$, where $z\in\mathbb C$. Then $$ f(a) = (1+|z|^2) + z\lambda_{f(x)} + \overline{z} \lambda_{f(x^{-1})}. $$ By hypothesis, $f(a)\geq 0$ so $\langle f(a), \varphi \rangle \geq 0$. Towards a contradiction, suppose $f(x^{-1})\not=f(x)^{-1}$.

  • If $f(x)=e$ then choose $h=e$ so $\varphi(x) = t\delta_{x,e}$ for some $t\geq 0$, so $$ \langle f(a), \varphi \rangle = t(1+|z|^2+z) $$ which is not positive for all $z\in\mathbb C$.
  • If $f(x)\not=e$ but $f(x^{-1})=e$ then argue similarly.
  • So $f(x)\not=e$ and $f(x^{-1})\not=e$. Set $h=f(x)$ so then $$ \langle f(a), \varphi \rangle = (1+|u|^2)(1+|z|^2) + \overline{u}z +\overline{uz}\delta_{f(x), f(x^{-1})} + uz\delta_{f(x)^{-1},f(x)}. $$
  • Regardless of the values of $\delta_{f(x), f(x^{-1})}$ and $\delta_{f(x)^{-1},f(x)}$, this can never be positive for all $z,u$.
  • So the claim holds.

The function $\sigma:x\to x^{-1}$ induces a function on the ${\rm C}^*$-algebra ${\mathbb C}H$ that sends positive elements to positive elements; indeed, this function is akin to the usual involution on the group ${\rm C}^*$-algebra except that it is linear rather than conjugate linear. Consequently, composing any homomorphism $f:G\to H$ with $\sigma$ yields a function $\sigma f$ whose extension to ${\mathbb C} G\to {\mathbb C}H$ sends positive elements to positive elements, yet which will usually not be a homomorphism (unless $H$ is abelian).

P.S. This example suggests replacing the positivity property in your question by "complete positivity", but I haven't thought this through.

  • $\begingroup$ Many thanks for correcting this! $\endgroup$ – Edwin Beggs Feb 20 at 17:44

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