**Main question:** How can we describe the double covers $(2\cdot\mathfrak{S}_n)^+$ and $(2\cdot\mathfrak{S}_n)^-$ of the symmetric group $\mathfrak{S}_n$ as permutation groups? (I.e., what sort of set with combinatorial structure do they act faithfully upon?)

I am not necessarily asking for the *minimal* permutation degree (I think this is an open problem; also see PS at bottom), but for some explicit combinatorial construction, reasonably uniform in $n$, that is hopefully reasonably small (perhaps merely exponential in $n$).

For a long time, I wrongly believed that one could form such a permutation representation of degree $2^{\lfloor n/2\rfloor}$ (or perhaps twice or four times that, or thereabouts) by taking an appropriate basis in the basic spin representation (and their negatives, and perhaps also times the imaginary unit), but I now realize that this does not work (at least, not as I thought it would). So I wonder if something else can be done.

The following question may is closely related to the above (trying to lift some $\mathfrak{S}_n$-set $X$ to a $(2\cdot \mathfrak{S}_n)$-set $\tilde X$ consisting of "signed" elements of $X$):

**Alternate question:** Given a set $X$ on which $\mathfrak{S}_n$ acts faithfully and transitively, with stabilizer $H$, is there some way to detect (merely by looking at $X$) whether $H$ lifts in $2\cdot \mathfrak{S}_n$ to a direct product $\{\pm 1\}\times H$, or equivalently (see below), whether the wreath product $1 \to \{\pm 1\}^X \to \{\pm 1\}\wr_X\mathfrak{S}_n \to \mathfrak{S}_n\to 1$ contains a non-split extension $1 \to \{\pm 1\} \to 2\cdot \mathfrak{S}_n \to \mathfrak{S}_n\to 1$?

Indeed, if $2\cdot \mathfrak{S}_n$ acts transitively on a set $\tilde X$, with stabilizer $H$, and the action does not factor through $\mathfrak{S}_n$, then $H$ intersects $\{\pm 1\}$ (center of $2\cdot\mathfrak{S}_n$) trivially, so maps isomorphically to a subgroup of $\mathfrak{S}_n$, which we can also call $H$, and the latter lifts to $2\cdot\mathfrak{S}_n$ as a direct product $\{\pm 1\}\times H$ (and we can see $\tilde X = (2\cdot\mathfrak{S}_n)/H$ as a double covering of $X = \mathfrak{S}_n/H$). But then, it follows from Derek Holt's paper "Embeddings of Group Extensions into Wreath Products" (*Quarter. J. Math. Oxford* **29** (1978) 463–468), theorem 2, that the wreath product contains our $2\cdot \mathfrak{S}_n$. Conversely, when this is the case, $2\cdot \mathfrak{S}_n$ acts faithfully and transitively (through the wreath product) on $\tilde X := \{\pm 1\}\times X$.

For example, if $H$ is generated by an $n$-cycle, and either $n$ is odd or we $n\equiv 2\pmod{4}$ if we are considering $(2\cdot\mathfrak{S}_n)^+$, we get a permutation action of degree $2(n-1)!$ on the set $\mathfrak{S}_n/H$ of cyclic orders with a sign added. But this isn't a great improvement over the regular action ($2n!$) and it's not very explicit, so I'm hoping we can do better.

**PS:** If my computations with Gap are correct, the minimal permutation representation for (Gap's choice) $(2\cdot\mathfrak{S}_n)^-$ is $16,48,80,240,240,480$ for $n=4,5,6,7,8,9$. This is not in the OEIS, I wonder if it's worth adding.

**PPS:** The corresponding values for $(2\cdot\mathfrak{S}_n)^+$ are $8,40,80,240,480,480$ (also not in the OEIS), so apparently they're not the same. (The trick I found to represent $(2\cdot\mathfrak{S}_n)^+$ under Gap is to multiply by a fourth root of unity the generators of $(2\cdot\mathfrak{S}_n)^-$ that lift an odd permutation.)