For $k \geq 2, n \geq 1$ let $$M^{n,k} = \{(x_1,\dots,x_k) \in S^n \times \dots \times S^n \ | \ x_1 + \dots + x_k = 0\}$$ This is a compact CW-complex and almost, but not quite, a manifold. I generally want to understand this space better. More concretely, I am interested in the following questions:

(1) What is the dimension of $M^{n,k}$ in a reasonable sense?

(2) For which $n,k$ is $M^{n,k}$ simply connected?

(3) For $k = p$ prime, the space admits a free $\mathbb Z/p$ action by cyclically permuting the coordinates. What can be said about the cohomology of $(\mathbb Z/p)\backslash M^{n.k}$?

For small $k$ and $n$, it is possible to give a "hands on" description of $M^{n,k}$: $M^{n,2}$ is obviously homeomorphic to $S^n$. It is easy to see that $M^{n,3}$ is homeomorphic to the unit sphere bundle of the tangent bundle of $S^n$. The space $M^{1,4}$ is homeomorphic to $S^1 \times X$ where $X$ is the 1-dimensional CW-complex with three 0-cells and six 1-cells, such that for any pair of two different 0-cells there are two 1-cells joining them. $M^{2,4}$ is already pretty hard to describe, but it is definitely of dimension 5.


Your space $M^{n,k}$ is a moduli space of closed $k$-gons in $\mathbb{R}^{n+1}$ with sides of length 1, viewed up to translation. As such it is a fairly well-studied object in the literature, see for example

Farber, Michael; Fromm, Viktor The topology of spaces of polygons, Trans. Amer. Math. Soc. 365 (2013), no. 6, 3097–3114.

In particular, it is a closed smooth manifold of dimension $n(k-1)-1$ when $k$ is odd, and has singularities when $k$ is even.


Let me attempt to offer a thought on your first question. While $M^{n,k}$ is not a manifold in general, it is a real affine algebraic variety. Indeed, $S^n$ is a real affine variety, so that the same is true of the product $(S^n)^k$. Taking the vanishing locus of the algebraic map $(S^n)^k\rightarrow\mathbb{R}^{n+1}$, $(x_1,\ldots,x_k)\mapsto x_1+\ldots+x_k$, we obtain the closed subvariety $M^{n,k}$.

As a real variety, $M^{n,k}$ has a dimension, namely $nk-(n+1)$.

  • $\begingroup$ I think the algebraic map actually goes to $\mathbb{R}^{n+1}$, hence the discrepancy in our dimensions. $\endgroup$ – Mark Grant Nov 20 '14 at 14:24
  • $\begingroup$ Absolutely, yes! $\endgroup$ – Peter Crooks Nov 20 '14 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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