Convergence of coeffiecients of element in infinite symmetric group algebra

Question

Let $\mathbb{C}S_\infty$ - infinite symmetric group algebra (generated by all finite permutations, i.e. $\mathbb{C} S_\infty = \bigcup_{n=1}^\infty \mathbb{C} S_n$). Turn it into $C^*$-algebra by identity $\sigma^* = \sigma^{-1}$ for every permutation $\sigma \in S_\infty$. Take completion of $\mathbb{C}S_\infty$ by spectral norm $||x||_{spec}^2=r(x^*x)$ where $r(x)=\sup\{|\lambda| :\lambda \in \mathbb{C}, x-\lambda 1$ - not invertible$\}$ - spectral radius of element $x$. Let $x \in \mathbb{C} S_\infty$ - some element of $(\mathbb{C}S_\infty, ||\cdot||_{spec})$, imagine it like $x=\sum_{\sigma \in S_\infty} \sigma \lambda_\sigma $ it is true that $\sum_{\sigma \in S_\infty} |\lambda_\sigma|^2$ convergence?

Answer 1

You cannot use that technique to construct a C$^*$-algebra. The spectral radius does not give a norm in an arbitrary $*$-algebra.

For an easy example, consider $\mathbb C[x]$ with complex conjugation as an involution. The only invertible polynomials are the constants, so for any polynomial $p$ of degree at least 1 $$ \sigma(p^*p)=\mathbb C. $$ The spectrum is not very useful when you have almost no inverses.

When you have a group algebra, to get a C$^*$-algebra you need to induce the norm via a representation. The two canonical ones are the left regular representation (thus giving you the reduced C$^*$-algebra of the group), and the universal representation (giving the universal group algebra).

In the case of the reduced C$^*$-algebra, one can show that $C^*_r(\Gamma)$ can be identified with some convolution operators: $$ C^*_r(\Gamma)\subset\{x\in\ell^2(\Gamma):\ \forall y\in\ell^2(\Gamma),\ x*y\in\ell^2(\Gamma)\}. $$ So in that case the elements of the group algebra are indeed given by properly chosen $\ell^2$-sequences, but not all.