9
$\begingroup$

The wikipedia page Covering groups of the alternating and symmetric groups gives explicit presentations for the double covers of the symmetric group Sn (n ≥ 4). Can someone provide a similar presentation, or better yet an explicit combinatorial description, of the double cover of the alternating group An? (I really only care about the case of large n, in case it matters.)

$\endgroup$

1 Answer 1

12
$\begingroup$

Yeah, Schur did this a long time ago. Let $\tilde \Sigma_n \to \Sigma_n$ be a double cover (there are two) -- lets denote them $\tilde \Sigma_n = \Sigma_n^\epsilon$ where $\epsilon \in \{+1, -1\}$.

Schur uses the notation $[a_1 a_2 \cdots a_k]$ for a specific lift of the cycle $(a_1 a_2 \cdots a_k) \in \Sigma_n$ to $\Sigma_n^\epsilon$ -- might as well call these $k$-cycles. Then his presentation goes like this:

$$[a_1 a_2 \cdots a_k] = [a_1 a_2 \cdots a_i][a_i a_{i+1} \cdots a_k] \ \ \forall 1 < i< k$$

and all $k$-cycles, $k>1$.

$$[a_1 a_2 \cdots a_k]^{[b_1 b_2 \cdots b_j]} = (-1)^{j-1}[\phi(a_1) \phi(a_2) \cdots \phi(a_k)]$$

where $\phi$ is the cycle $(b_1 b_2 \cdots b_j)$

$$[a_1 a_2 \cdots a_k]^k = \epsilon$$

for all $k$-cycles -- ie this is always $+1$ or $-1$ depending on which extension of $\Sigma_n$ you're interested in. And:

$$[a_1 a_2 \cdots a_k][b_1 b_2 \cdots b_j] = (-1)^{(k-1)(j-1)}[b_1b_2 \cdots b_j][a_1 a_2 \cdots a_k]$$

provided the cycles $(a_1 a_2 \cdots a_k)$ and $(b_1 b_2 \cdots b_j)$ are disjoint.

The map $\tilde \Sigma_n \to \Sigma_n$ sends $[a_1 \cdots a_k]$ to $(a_1 \cdots a_k)$. So this gives you a corresponding presentation of the double of $A_n$ -- take your favourite presentation of $A_n$, lift the relators and see what happens using the above relations.

A small extra tidbit -- think of $\Sigma_n$ as being the group of orientation-preserving isometries of $\mathbb R^{n}$ that preserves a regular (n-1)-simplex. Then if you lift this group to $Spin(n)$, the extension you want is the one where $[a_1 a_2 \cdots a_k]^k = -1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .