4
$\begingroup$

Divide the 1-dimensional sphere $S^1$ into arcs. Label each vertex between two arcs as either 1 or 2. Count the number of "full arcs" - arcs with both 1 and 2 at their endpoints. Divide these full arcs to two kinds: "positive" (1-2 going clockwise) and "negative" (2-1). Then, it is clear that the number of positive full arcs equals the number of negative full arcs: each time we go from 1 to 2, we must at some point go back from 2 to 1, since we are on a circle.

Is the following generalization to multi-dimensional spheres correct?

Triangulate the $n$-dimensional sphere $S^n$. Label each vertex of the triangulation by a label from $\{1,\ldots,n+1\}$. Count the number of "full simplices" - simplices with all $n+1$ different labels at their vertices. Divide labeled simplices into two equivalence classes: "positive" and "negative", such that for each two simplices in the same class, there is an orientation-preserving map from one to the other agreeing on the labels. To relabel a simplex keeping it in the same class, we can perform any even permutation on the labels (for example, if a simplex labeled 1-2-3-4 is positive, then the simplex labeled 2-3-1-4 is also positive, but the simplex labeled 2-1-3-4 is negative).

Conjecture: the number of positive simplices equals the number of negative simplices.

Is this conjecture correct?

Note: asked in math.SE with no replies

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ A bit of precision is needed. You assume that there is an inherently natural order of listing the vertices of a simplex. In general there isn't. However, for an $n$-simplex on $S^n$ there is such way it has to do with a choice of orientation on the sphere. In any case, Sperner's lemma shows that the number of full simplices is even. The cohomological proof of Sperner's lemma (www3.nd.edu/~lnicolae/Local-global.pdf) seems like a promising approach. $\endgroup$
    – Liviu Nicolaescu
    Commented Jan 29, 2017 at 12:59
  • $\begingroup$ See Theorem 2.25 in Prasolov's book Elements of Combinatorial and Differential Topology, Amer. Math. Soc., 2006. It answers your question if you take into account my previous comment on orientations. $\endgroup$
    – Liviu Nicolaescu
    Commented Jan 29, 2017 at 13:09
  • $\begingroup$ I tried to clarify "positive" and "negative." $\endgroup$
    – Douglas Zare
    Commented Jan 29, 2017 at 13:24

2 Answers 2

Reset to default
9
$\begingroup$

The labeling gives a piecewise linear map from $S^n$ to a simplex $\Delta$ with $n+1$ vertices, where label $k$ means the point is sent to vertex $v_k$, and we extend linearly on simplices. The map must have degree $0$ because $\Delta$ is contractible. The degree can be computed from the preimage of any point in the interior of $\Delta$, and it is the difference between the number of positive simplices and the number of negative simplices.

Improve this answer
$\endgroup$
5
  • $\begingroup$ So my conjecture is correct? $\endgroup$
    – Erel Segal-Halevi
    Commented Jan 29, 2017 at 13:31
  • 1
    $\begingroup$ This argument needs a homological (or homotopical) definition of degree. The usual situation is for maps $S^n\to S^n$ but here the map is to a simplex. One can fix this small problem by taking the simplex to be a simplex in a standard triangulation of $S^n$, so one has a map between spheres. Alternatively, one can regard $S^n$ and the simplex as compact connected orientable manifolds with (possibly empty) boundary and then use relative homology groups for the pairs consisting of a manifold and its boundary. $\endgroup$
    – Allen Hatcher
    Commented Jan 29, 2017 at 14:28
  • $\begingroup$ @Erel: Yes, your conjecture is correct. $\endgroup$
    – Kevin Walker
    Commented Jan 29, 2017 at 15:50
  • $\begingroup$ @Allen: Can't one use differential topology instead of homological degree? Extend the map to $B^{n+1}$, make the extended map transverse to your favorite point in the interior of the target simplex, and obtain an oriented 1-manifold pairing up the non-degenerate simplices in $S^n$. $\endgroup$
    – Kevin Walker
    Commented Jan 29, 2017 at 15:57
  • $\begingroup$ @Kevin: Sure, that works too, and perhaps involves less machinery. $\endgroup$
    – Allen Hatcher
    Commented Jan 29, 2017 at 17:47
4
$\begingroup$

Let me point out that this also admits a purely combinatorial proof. Consider your triangulated $S^n$ as the boundary of a triangulated ball $B^{n+1}$ (for example, but not necessarily, cone your triangulated sphere to a vertex in the interior of the ball) and extend your labeling of the vertices to the interior arbitrarily (in the example, simply choose a label in $\{1,\dots,n+1\}$ for the interior point).

Now, each $(n+1)$-simplex $\sigma$ in your triangulated ball that uses all $n+1$ labels has exactly two $n$-dimensional faces $\tau_1$ and $\tau_2$ that use all labels, and $\tau_1$ and $\tau_2$ have opposite orientations when seen from $\sigma$. If you now count the pairs $(\sigma,\tau)$ that arise this way (with their orientations), the $\tau$'s intersecting the interior of the ball cancel out (they belong to two $\sigma$'s, with opposite orientations) so that in the boundary there must be the same number of them with one and the other orientation.

(This proof is very much inspired in "Sperner's Lemma", which Sperner used to give a purely combinatorial proof of Brower's fixed point theorem)

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .