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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.

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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.

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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 $nk-(n+1)$.

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  • $\begingroup$ 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

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