Elements of length 0 in extended affine Weyl group for GL(n)

Question

As part of my research, I would like to understand the possible pairs of $(v,\sigma)\in \mathbb Z^n\times S_n$ satisfying the following condition: For $1\le i < j \le n$, we have $\sigma(i) < \sigma(j)$ iff $v(\sigma(i))=v(\sigma(j))$ and $\sigma(i) > \sigma(j)$ iff $v(\sigma(i)) = v(\sigma(j)) - 1$.

Equivalently, these are the length 0 elements of the extended affine Weyl group for $GL(n)$; I have seen the set of these elements denoted by $\Omega$.

Here are the observations I've made, and some other known facts.

If $\sigma$ is the identity, then the allowed vectors have all entries equal. Conversely if $v$ has all entries equal, then the only allowed permutation is the identity.

If $n>2$ and $\sigma$ is the permutation $i\mapsto n+1-i$, then there are no allowed vectors.

The length 0 elements form a subgroup, where multiplication is given by $(v,\sigma)(v',\sigma)=(v+\sigma\cdot v', \sigma\sigma')$, where $(\sigma\cdot v)(i) = v(\sigma^{-1}(i))$.

I would be happy to have pointers to literature where such things might be discussed.

Answer 1

Let $c$ be the cycle $i \mapsto i+ 1 \bmod n$. The pairs obeying your condition are of the form $$((\overbrace{x+1,x+1,x+1,\ldots,x+1}^{k},\overbrace{x,x,\ldots,x}^{n-k}), c^k).$$ Under the group multiplication that you write out, they are isomorphic to $\mathbb{Z}$, with generator $((0,0,0,\ldots,0,1), c)$.

It is easy to check that $((0,0,0,\ldots,0,1), c)$ obeys the given conditions, and hence its powers do. We now must check that any $(v, \sigma)$ obeying these condition is a power of $((0,0,0,\ldots,0,1), c)$.

Using your group operation, we can replace $\sigma$ by $c^k \sigma$ for any $k$, and therefore we can assume that $\sigma(1) = 1$. For any $j$, we then have $\sigma(1) < \sigma(j)$, so $v(1) = v(j)$, so $v$ is of the form $(x,x,\ldots, x)$, and we are done.

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The extended Weyl group of $GL_n$ is often described as the group of bijections $f:\mathbb{Z} \to \mathbb{Z}$ obeying the condition $f(i+n)=f(i)+n$. Then $\sigma$ is the reduction modulo $n$, and I'm not going to try to get the indices right to figure out what $v$ is. The length $0$ elements are the permutations of the form $i \mapsto i+k$.