Schur cover of alternating groups

Question

Wilson's book "The finite simple groups" gives (in section 2.7) a description of the double cover of the alternating groups. First, one constructs a double cover $2S_n$ of the symmetric group $S_n$. By definition, a permutation $\pi$ has two lifts to $2S_n$, denoted by $+\pi$ and $-\pi$. Elements of $2S_n$ will be denoted by bracket notation as opposed to parentheses, i.e. $\pm[a_1,\ldots,a_k]$ are the lifts of the $k$-cycle $(a_1,\ldots,a_k)\in S_n$. He then defines multiplication in $2S_n$ by:

1. $[i,j]^{\pm\pi} = -[i^\pi,j^\pi]$, where $\pi$ is an odd permutation
2. A $k$-cycle can be written as $[a_1,\ldots,a_k] = [a_1,a_2][a_1,a_3],\cdots[a_1,a_k]$
3. A general element of $2S_n$ is a product of disjoint cycles

Then he says that you must be careful not to permute the cycles, or start a cycle at a different point.

His description is a bit confusing to me, mostly because he doesn't completely specify how to put elements of $2S_n$ into a ``canonical form''. I want to be certain that I'm understanding it correctly. Here are some questions:

(a) How do minus signs work when you multiply? For example, is $(-\pi)(-\sigma) = \pi\sigma$? (I would guess this is the case, as negation should presumably be viewed as multiplication by the nontrivial central element).

(b) What exactly is the difference between $[a_1,\ldots,a_k]$ and $[a_{k-i},\ldots,a_k,a_1,\ldots,a_{i-1}]$? (When are they the same?)

(c) Do disjoint cycles still commute in $2S_n$? (i.e., is $[1,2][3,4] = [3,4][1,2]$)? (In example 2.11, he shows that $[1,2][3,4] = -[3,4][1,2]$, but this doesn't answer the question)

(d) Are there other references for a similar "concrete" description of $2S_n$ that is more precise?

Answer 1

If you continue to the end of Section 2.7.2, there are actually two different double covers of $S_n$ (abstractly this comes from the fact that $S_n$ is not perfect, both of these double covers restrict to the same one on $A_n$.)

$2S_n^+$ is characterized by the rules you copied, plus $[i,j]^2=1$, while $2S_n^-$ has $[i,j]^2=-1$ instead.

To answer your questions:

(a): either of the double covers $2S_n^\pm$ fits into a central extension $$ 1 \to \{\pm 1\} \to 2S_n^\pm \to S_n \to 1, $$ so yes, $-[a_1,\ldots,a_k]$ is just notation for $(-1)\cdot [a_1,\ldots,a_k]$, and the element $(-1)$ commutes with everything.

(b): They have the same image in $S_n$, so they are equal or differ by $-1$. To determine which, observe that a more general form of the first property holds, where $$ [a_1,\ldots,a_k]^\pi = ([a_1,a_2] \cdots [a_1,a_k])^\pi = \operatorname{sgn}(\pi)^{k-1} [a_1^\pi,a_2^\pi]\cdots [a_1^\pi,a_k^\pi] = \operatorname{sgn}(\pi)^{k-1}[a_1^\pi,\ldots,a_k^\pi]. $$ In this case, you can take $\pi$ itself to be a lift of the permutation $(a_1,a_2,\ldots,a_k)^{k-i-1}$, so $[a_1,\ldots,a_k]$ and $[a_{k-i},\ldots,a_k,a_1,\ldots,a_{i-1}]$ differ by the sign $(-1)^{(k-1)(k-i-1)}$, I think.

(c): If $[a_1,\ldots,a_k]$ and $[b_1,\ldots,b_l]$ are disjoint, we have $$ [a_1,\ldots,a_k]^{[b_1,\ldots,b_l]} = \operatorname{sgn}((b_1,\ldots,b_l))^{k-1} [a_1,\ldots,a_k] = (-1)^{(l-1)(k-1)} [a_1,\ldots,a_k], $$ by the same observation as in the previous discussion. So pairs of disjoint odd cycles anticommute.

Multiplying elements

To understand the multiplication better, let us first solve the following problem: Given a sequence of transpositions $[i,j]$ which multiply to the trivial permutation in $S_n$, how can we find their product in $2S_n^{\pm}$? This product is an element of $\{\pm 1\}$, i.e. a single sign.

Starting with $\prod_i [a_i,b_i]$ with the property that their product is trivial in $S_n$, trace how the element $1\in \{1,\ldots,n\}$ gets moved, and mark all transpositions that contribute to this. For example, from left to right, there might be a bunch of transpositions that do not move $1$, then a $[1,3]$, then a bunch of transpositions that do not move $3$, then say $[3,2]$, then a bunch of transpositions that do not move $2$, and finally a $[2,1]$, followed by a bunch of transpositions that do not move $1$. We now perform the following move: Find the first marked transposition, say $[1,i]$ (otherwise rewrite $[i,1]=-[1,i]$), and let $[k,l]$ be the transposition immediately following it.

• If $[k,l]$ is unmarked, it means it does not contain $i$. We may use $[1,i][k,l] = [k,l]^{[1,i]} [1,i]$ to rewrite our sequence, where the first transposition now does not contain $1$, so it is still unmarked. So now the first marked transposition occurs further to the right.
• If they are both marked, either $k=i$ or $l=i$. Using $[k,l]=-[l,k]$, we may assume $l=i$. If we have $k=1$, then $[1,i][1,i] = \pm 1$ depending on whether we are in $2S_n^+$ or $2S_n^-$, and we may rewrite our sequence with two fewer transpositions. Otherwise, we have $[1,i][k,i] = -[k,1][1,i] = [k,i][k,1]$, and now again the first marked transposition occurs one further to the right.

Iterating this, we may rewrite our whole sequence by one without marked transpositions, i.e. $1$ does not occur. Repeating this with $2$, etc., we reduce our sequence to a single sign.

How does this help calculating the multiplication in $2S_n^\pm$? Well, given any expression $\sigma$ in terms of cycles in $2S_n^\pm$, we may first multiply them in $S_n$, and express their product in terms of disjoint cycles. Lifting this normal form to $2S_n^\pm$, you have found a product $\rho$ of disjoint cycles in $2S_n^\pm$ with the same image in $S_n$. So $\sigma=\pm \rho$. To determine the sign, write $\sigma\rho^{-1}$ as product of cycles and apply the above algorithm.