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Let $G$ be a locally compact Hausdorff group. Is the group algebra $L^1(G)$ separable? Would you please help me or introduce references that can help me. Thanks

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No, even if the group is compact, Abelian and separable. For instance, the Cantor group $\{0,1\}^{\omega_1}$, where $\omega_1$ is the first uncountable cardinal, is a good counter-example. This group is obviously compact and Abelian. It is also separable by the Hewitt–Marczewski–Pondiczery theorem. However $L_1(\{0,1\}^{\omega_1})$ is non-separable.

Indeed, if it were separable, the group C*-algebra $C^*_r(\{0,1\}^{\omega_1})$ would be separable as it contains $L_1(\{0,1\}^{\omega_1})$ as a dense subspace. This is not true because $$\widehat{\{0,1\}^{\omega_1} } = \bigoplus_{\alpha<\omega_1} \{0,1\}, $$ so you have uncountably pair-wise orthogonal non-zero projections in $C^*_r(\{0,1\}^{\omega_1})$.

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Not to refute “… even if the group is compact, Abelian…”, but how could a topological space be “separable”? – Incnis Mrsi Aug 21 '14 at 17:55
And what is the criterion on the group? At least, the converse is true: if L1(G) is separable, then G also is? – Yulia Kuznetsova Oct 3 '14 at 13:37
Yulia, yes. $L_1(G)$ is separable if and only if $C_r(G)$ is separable (as the former is dense in the latter). $L_1(G)$ is separable if and only if $G$ is second-countable. – Tomek Kania Oct 3 '14 at 13:44
Dear Tomek, I am seeking without success a reference for this: $L^1(G)$ separable iff $G$ second countable. Can you please give me one? – Yulia Kuznetsova Oct 3 '14 at 14:58
Yulia, for compact groups this is a result of Grekas (Theorem 3.1 in; see also the corrigendum). – Tomek Kania Oct 3 '14 at 17:09

No, because any discrete group is locally compact and admits a Haar measure that assigns the measure 1 to any element. $L^1$ on a non-countable set with such measure isn’t separable.

BTW I doubt that we necessarily have separability even for compact groups. Compactness is not a very strong condition to infer “nice” properties of function spaces. There are spaces that are compact and Hausdorff albeit very thick, in the sense there are damn much functions on them (although I am not sure it is the case for $L^1$).

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