therightwomanwf
2022-07-08
Answered

I would like to know, under what condition on the group $G$ (abelian, compact or localement compact ...), the algebra ${L}^{1}(G)$ is commutative?

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Sariah Glover

Answered 2022-07-09
Author has **16** answers

For a locally compact group $G$, the convolution algebra ${L}^{1}(G)$ is commutative if and only if $G$ is Abelian.

Callum Dudley

Answered 2022-07-10
Author has **4** answers

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Consider $\stackrel{\u2015}{F}}_{p$ , the algebraic closure of $F}_{p$ . I want to see that: for every proper subfield $K\le \stackrel{\u2015}{{F}_{p}},\frac{{\stackrel{\u2015}{F}}_{p}}{K}$ is not a finite extension.

It is known that, and can be somewhat easily shown that$\stackrel{\u2015}{F}}_{p}={\cup}_{n\ge 1}{F}_{{p}^{n}$ .

Now, if any of the proper subfields have the form$F}_{{p}^{n}$ , it is easy enough to see that ${\stackrel{\u2015}{F}}_{p}\ne {F}_{{p}^{n}}({\alpha}_{1},\cdots ,{\alpha}_{m})$ for some $\alpha}_{i$ by going high up enough, i.e, to some big enough m such that $\alpha}_{i}\in {F}_{{p}^{m}}\subseteq \stackrel{\u2015}{{F}_{p}$

The problem is characterizing the proper subfields. Is every subfield of$\stackrel{\u2015}{{F}_{p}}$ going to have this form? Can we have an infinite intermediate subfield?

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