7
$\begingroup$

Consider the collection of symmetric groups $\{\Sigma_n\}_{n\geq1}$ as a semi-simplicial set (i.e. a simplicial set without degeneracies) as follows. Consider $i\in\{1,\dots,n+1\}$ and $\pi\in\Sigma_{n+1}$ represented as a sequence $(\pi(1),\dots,\pi(n+1))$, then $$d_{i-1}(\pi)=(\pi(1)-\epsilon_1,\dots, \widehat{\pi(i)},\dots,\pi(n+1)-\epsilon_{n+1})$$ where for every $j\in\{1,\dots,n+1\}$ $$\epsilon_j=\begin{cases} 0 & if\ \ \pi(j)<\pi(i)\\ 1 & if\ \ \pi(j)>\pi(i). \end{cases}$$

I am wondering what it's known about the homotopy type of the geometric realization of this semi-simplicial set?

$\endgroup$

1 Answer 1

11
$\begingroup$

It is contractible. To see this first observe that it is simply connected. It has 2 arcs $a = (1,2)$ and $b = (2,1)$, however the tirangle $(3,1,2)$ gives us the relation that $ab = b$ and symmetrically $ba = a$ so $\pi_1 = \{1\}$ for this space. To see that the homology vanishes you can use the homotopy operator on the homology complex given by $H((i_1,...,i_{n+1})) = (i_1,...,i_{n+1},n+2)$. It is pretty clear that $dH \pm Hd = Id$ in this case, just note that all the first face maps just turn formally $n+2$ into $n+1$ (giving precisely the Hd term) while the last face map eliminate the $n+2$-st term, giving back the original chain.

So, this simplicial complex is a simply connected acyclic space, hence contractible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.