Yeah, Schur did this a long time ago. Let tildeSigmantoSigman be a double cover (there are two) -- lets denote them tildeSigman=Sigmaenpsilon where epsilonin+1,−1.
Schur uses the notation [a1a2cdotsak] for a specific lift of the cycle (a1a2cdotsak)inSigman to Sigmaenpsilon -- might as well call these k-cycles. Then his presentation goes like this:
[a1a2cdotsak]=[a1a2cdotsai][aiai+1cdotsak]forall1<i<k
and all k-cycles, k>1.
[a1a2cdotsak][b1b2cdotsbj]=(−1)j−1[phi(a1)phi(a2)cdotsphi(ak)]
where phi is the cycle (b1b2cdotsbj)
[a1a2cdotsak]k=epsilon
for all k-cycles -- ie this is always +1 or −1 depending on which extension of Sigman you're interested in. And:
[a1a2cdotsak][b1b2cdotsbj]=(−1)(k−1)(j−1)[b1b2cdotsbj][a1a2cdotsak]
provided the cycles (a1a2cdotsak) and (b1b2cdotsbj) are disjoint.
The map tildeSigmantoSigman sends [a1cdotsak] to (a1cdotsak). So this gives you a corresponding presentation of the double of An -- take your favourite presentation of An, lift the relators and see what happens using the above relations.
A small extra tidbit -- think of Sigman as being the group of orientation-preserving isometries of mathbbRn that preserves a regular (n-1)-simplex. Then if you lift this group to Spin(n), the extension you want is the one where [a1a2cdotsak]k=−1.
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