5
$\begingroup$

Yesterday I understood that I can't live without this construction:

Let $G$ be a group, $A$ an associative algebra over $\mathbb R$ and $n\in{\mathbb N}$. We consider a sequence of maps $\varphi_k:G\to A$, $k=0,...,n$, satisfying the following three identities: $$ \varphi_k(1_G)=\begin{cases}1_A,& k=0\\ 0,& k\ne 0\end{cases} $$ $$ \varphi_k(a\cdot b)=\sum_{0\le l\le k}\begin{pmatrix}k \\ l\end{pmatrix}\cdot \varphi_l(a)\cdot \varphi_{k-l}(b),\qquad a,b\in G, $$ $$ \forall k\ge 1\qquad \varphi_0(a)\cdot\varphi_k(b)=\varphi_k(b)\cdot\varphi_0(a),\qquad a,b\in G, $$

In particular, this means that $\varphi_0:G\to A$ is a homomorphism (of $G$ into the group of invertible elements in $A$): $$ \varphi_0(1_G)=1_A,\qquad \varphi_0(a\cdot b)=\varphi_0(a)\cdot\varphi_0(b),\qquad a,b\in G. $$ And in the case of $\varphi_0(a)=1$ ($a\in G$), the map $\varphi_1:G\to A$ becomes a generalization of what is called real character on $G$: $$ \varphi_1(1_G)=0,\qquad \varphi_1(a\cdot b)=\varphi_1(a)+\varphi_1(b),\qquad a,b\in G. $$

In what I consider $G$ is a locally compact group, $A$ is a $C^*$-algebra, and the maps $\varphi_k:G\to A$ are continuous and satisfy the following supplementary identities: $$ \varphi_0(a)^*=\varphi_0(a^{-1}),\qquad a\in G. $$ $$ \varphi_1(a)^*=-\varphi_0(a)^{-2}\cdot\varphi_1(a),\qquad a\in G. $$ $$ \varphi_2(a)^*=-\varphi_0(a)^{-2}\cdot\varphi_2(a)+\varphi_0(a)^{-3}\cdot\varphi_1(a),\qquad a\in G. $$ $$ ... $$

I never saw something like this before, so my main question is

What is this?

Did anybody consider the sequences of maps $\varphi_k:G\to A$ like these? If yes, what is known about them? I am curious, for example, in

How long can these sequences $\varphi_k:G\to A$ (with non-zero $\varphi_k$) be?

It is easy to see, for example, that if $G$ is a compact group, and $A$ is a $C^*$-algebra, then only $\varphi_0$ can be non-zero: $$ \forall k\ge 1\qquad \varphi_k=0. $$ This follows from the fact that $\varphi_0$ in this case acts into the set of unitary elements, therefore $$ \|\varphi_0(a)\|=1, $$ and $\varphi_1$ satisfies the identity $$ \varphi_1(a^m)=m\cdot\varphi_0(a)^{m-1}\cdot\varphi_1(a) $$ So if $\varphi_1(a)\ne 0$, then $\|\varphi_1(a^m)\|\to\infty$ as $m\to\infty$, but this is impossible, since $\varphi_1:G\to A$ is a continuous map of a compact space $G$. As a corollary, $\varphi_1=0$, and the same reasoning for other $\varphi_k$ with $k>0$.

Any references, thoughts, feelings will be appreciated.

$\endgroup$
  • $\begingroup$ If you extend by linearity to the group algebra, you get a sequence of linear maps between two associative algebras satisfying the same conditions. So these axioms make sense in this context, with no group on the left side. Maybe "ra.rings-and-algebras" would be a natural tag then. $\endgroup$ – YCor Sep 24 '14 at 15:31
  • $\begingroup$ @YCor, yes initially $G$ and $A$ both were algebras. I reformulated this for the case when $G$ is a group. $\endgroup$ – Sergei Akbarov Sep 24 '14 at 15:36
  • 4
    $\begingroup$ +1 for the first sentence. $\endgroup$ – Nick Gill Sep 24 '14 at 16:30
  • $\begingroup$ I think you mean $\varphi_n(a.b)$ rather than $\varphi_k(a.b)$ in the second condition. $\endgroup$ – Neil Strickland Sep 24 '14 at 17:17
  • 4
    $\begingroup$ The first two conditions just mean that the map $\psi(u)=\sum_k\varphi_k(u)t^k/k!$ gives a ring homomorphism $\psi\colon\mathbb{R}[G]\to A[t]$. $\endgroup$ – Neil Strickland Sep 24 '14 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.