Relative homology of $M^k$ and the fat diagonal

Question

For every topological space $M$ and some $k \geq 2$, recall the fat diagonal $M^k_{k-1} \subset M^k$, defined via

$$ M^k_{k-1} := \{(x_1,\dots,x_k) \in M^k : \exists i \neq j \text{ with } x_i = x_j\}.$$

I would like to understand the relative homology $H_\bullet(M^k, M^k_{k-1})$ for $M = S^1$ and $k$ arbitrary. Since this seems like a good pair to me, we can identify $$H_\bullet(M^k, M^k_{k-1}) \cong \tilde{H}_\bullet(M^k / M^k_{k-1}),$$

but now I've already run out of quick and dirty ideas. Any hints or suggestions? And is there something even more general one can say outside of the case $M = S^1$?

Answer 1

I will describe the steps of a solution:

1. Let $N$ denote an (open) regular neighborhood of the fat diagonal in $M^k=(S^1)^k$ and let $X$ denote the complement $M^k \setminus N$. The space $X$ is a manifold with boundary, its interior is homeomorphic to the complement $Z=M^k\setminus M^k_{k-1}$. Using the excision, you get: $$ H_*(X,\partial X)\cong H_*(M^k, M^k_{k-1}). $$

2. The space $X$ is disconnected, consists of $(k-1)!$ homeomorphic components; the components $X_\sigma$ are indexed by permutations $\sigma\in Sym_{k-1}$. See below. The manifold $X$ is obviously oriented. Using Poincare duality, we get:
   $$ H_*(X_\sigma,\partial X_\sigma)\cong H^{k-*}(X_\sigma)\cong H^{k-*}(Z_\sigma), $$ where $Z_\sigma$ is the unique component of $Z$ containing $X_\sigma$.

3. The space $Z$ is homeomorphic to the disjoint union of $(k-1)!$ copies $Z_\sigma$ of $$ S^1\times {\mathbb R}^{k-1}. $$ In order to see this, fix the location (at $i=\sqrt{-1}\in {\mathbb C}$) of the first point in the tuple $(z_1,...,z_k)$ of distinct points in $S^1$. Using the stereographic projection, identify $S^1 \setminus \{i\}$ with the real line. The rest of the points $(z_2,...,z_k)$ are then free to move on the real line as long as they do not collide. Fixing their order in ${\mathbb R}$ amounts to picking a permutation $\sigma\in Sym_{k-1}$. Once this is done, the configuration space is given by the inequalities $$ x_{i_2}< x_{i_3}<...<x_{i_k} $$ where $(i_2,...,i_k)$ is the permutation $\sigma$. Compare my answer here. (For some reason, I made it more complicated than necessary dividing out by the conformal group instead of $S^1$.)

4. To conclude, $$ H_i(M^k, M^k_{k-1})\cong \oplus_{s=1}^{(k-1)!} H^{k-i}(Z_s), $$ $H^{k-i}(Z_s)=0$, unless $i=k$ or $i=k-1$. In these special cases, you get ${\mathbb Z}$ as the cohomology group. Thus, $H_i(M^k, M^k_{k-1})=0$ unless $i=k$ or $i=k-1$; for $i=k$ or $i=k-1$ you get $$ H_i(M^k, M^k_{k-1})\cong {\mathbb Z}^{(k-1)!}. $$