Schur multiplier of the symmetric group

Question

It is known that $H^2(S_n, \mathbb{C}^\times) \cong \mathbb{Z}/2\mathbb{Z}$ for $n \geq 4$. Is an explicit formula as a function $S_n \times S_n \to \mathbb{C}^\times$ for a representative of the nontrivial element known? Thanks.

Answer 1

I can't answer your question as framed, but there is an alternative way of looking at the Schur Multiplicator which could be relevant and which goes back to a paper

Miller, Clair, `The second homology of a group', Proc. American Math. Soc. 3 (1952) 588-595.

For a group $G$ she defines a group $G \wedge G$ as generated by elements $g \wedge h, g,h \in G$ factored by relations

$$gg' \wedge h = (gg'g^{-1}\wedge {ghg^{-1}})(g \wedge h)$$ $$g \wedge hh' =(g \wedge h)(hgh^{-1}\wedge hh'h^{-1})$$ $$ g \wedge g =1$$ for all $g,g',h,h' \in G$. Then the commutator map $[\;,\; ]: G \times G \to G$ factors through $\kappa: G \wedge G \to G $ and Miller proves that the kernel of $\kappa$ is isomorphic to $H_2(G)$. This work was subsumed into work on a nonabelian tensor product of groups $G,H$ which act on each other and on themselves by conjugation, and a bibliography on this, starting with Miller's paper, is given at http://groupoids.org.uk/nonabtens.html . Paper [7] there gives some information on $S_n$ in these terms.