Proving that a function's image contains (1/n,...,1/n) - MathOverflow

Proving that a function's image contains (1/n,...,1/n)

2011-07-12T23:35:06Z

This question is a follow-up to a previous question answered by Neil Strickland:

http://mathoverflow.net/questions/67318/map-from-simplex-to-itself-that-preserves-sub-simplices

Let $B$ denote the closed unit ball in $\mathbb{R}^2$ and let $\Delta_{n-1}$ denote the $(n-1)$-simplex. I have a continuous function $f(x_1,\dots,x_n):B^n \rightarrow \Delta_{n-1}$ defined for all subsets $\lbrace x_1,\dots,x_n\rbrace \subset B$ of size $n$ that satisfy $x_i \neq x_j$ for all pairs $i,j$ (in other words, the function is only defined if all of the $n$ arguments are distinct). This function has the property that, if $\sigma$ denotes a permutation, then $f(\sigma(x_1,\dots,x_n)) = \sigma(f(x_1,\dots,x_n))$. In other words, permuting the arguments of the function merely permutes the output. My question is: are there non-trivial sufficient conditions on $f$ under which the point $(1/n , \dots, 1/n)$ lies in the image of this map? (or, even better, is this always the case?)

Here's one property of the map $f$ that I can add regarding the requirement that arguments be distinct: if $\lbrace \mathbf{x}_k \rbrace$ is a sequence of $n$-tuples (with distinct entries) in $B$ that converges to an $n$-tuple $\bar{\mathbf{x}}$ with (possibly) non-distinct entries, then the limit of $f(\mathbf{x}_k)$ exists if and only if, for each pair of entries $x_i^k$ and $x_j^k$ in the $n$-tuple, the unit direction vector from $x_i^k$ to $x_j^k$ (i.e. $\frac{x_i^k - x_j^k}{||x_i^k - x_j^k||}$) has a limit.

Answer by Neil Strickland for Proving that a function's image contains (1/n,...,1/n)

2011-07-13T11:11:50Z

This turns out to be a remarkably interesting question. I don't have a complete answer but here is a start.

First, for any space $X$ I'll write $F_n(X)\subset X^n$ for the space distinct $n$-tuples. The question asks whether there exist $\Sigma_n$-equivariant maps $f:F_n(B^2)\to\Delta_{n-1}$ with the fixed point $b=(1/n,\dotsc,1/n)$ not in the image. If there is such a map then we can push it away from $b$ to the boundary of $\Delta_{n-1}$, which is homeomorphic to $S^{n-2}$. On the other side, there is an obvious embedding $i:B\to\mathbb{R}^2$ and one can choose an embedding $j:\mathbb{R}^2\to B$ such that $ij$ and $ji$ are isotopic to the respective identity maps; using this we see that $F_n(B^2)$ is equivariantly homotopy equivalent to the space $X=F_n(\mathbb{R}^2)$. This space is well-known: the following paper is one entry point to the literature:

\bib{MR1344842}{article}{
   author={Cohen, F. R.},
   title={On configuration spaces, their homology, and Lie algebras},
   journal={J. Pure Appl. Algebra},
   volume={100},
   date={1995},
   number={1-3},
   pages={19--42},
   issn={0022-4049},
   review={\MR{1344842 (96d:55005)}},
   doi={10.1016/0022-4049(95)00054-Z},
}

In particular:

$\pi_1(X)$ is the pure braid group $Br_n$ on $n$ strings. Moreover, the higher homotopy groups are trivial, so $X$ is the classifying space $BBr_n$.
$X$ has an equivariant deformation retract $X_0$ that is a finite simplicial complex of dimension $n-1$. This means that $H^k(X)=0$ for $k>n-1$.
The cohomology of $X$ is completely known, together with the action of $\Sigma_n$. In particular, the top group $H^{n-1}(X)$ is the module known as $\text{Lie}(n)$ (or maybe the dual of that?). As a $\mathbb{Z}[\Sigma_{n-1}]$-module this is free of rank one, but the $\Sigma_n$-action is harder to describe. The standard description also implies that all Steenrod operations in $H^*(X_0;\mathbb{Z}/p)$ are trivial.

If we can show that there is no $\Sigma_n$-equivariant map from $X_0$ to $S^{n-2}$ then we will be done.

In the case $n=2$ we just have $X_0=S^1$ and $S^{n-2}=S^0$ with $\Sigma_2$ acting antipodally on both sides: it is clear that there is no equivariant map, as required.

In the case $n=3$ we have $S^{n-2}=S^1=K(\mathbb{Z},1)$, so the nonequivariant mapping set is $[X_0,S^1]=H^1(X_0)$, and one can check that this is just $\mathbb{Z}^3$ with the action given by permuting the coordinates and multiplying by the signature. The only fixed point for this action is zero, so any map $X_0\to S^1$ that is equivariant-up-to-homotopy is nonequivariantly homotopic to a constant map. Here the action of $\Sigma_3$ on $S^1$ is generated by a reflection and a rotation through $2\pi/3$, so there are no fixed points. This means that constant maps $X_0\to S^1$, although equivariant-up-to-homotopy, cannot be equivariant on the nose. I suspect that there are no equivariant maps, and it should be possible to prove this by equivariant obstruction theory (ie working up the skeleta of $X_0$) but I do not see the details at the moment.

For $n>3$ we still have an evident map $[X_0,S^{n-2}]\to H^{n-2}(X_0)$, but it need not be bijective. We can compare $S^{n-2}$ with the fibre of the map $Sq^2:K(\mathbb{Z},n-2)\to K(\mathbb{Z}/2,n)$, recalling that $Sq^2$ acts trivially on $H^*(X_0)$, which should give an explicit description of $[X_0,S^{n-2}]$. With a bit of representation theory we should be able to calculate the group of equivariant-up-to-homotopy maps $X_0\to S^{n-2}$. We would then need some equivariant obstruction theory to improve this to understand whether there are any equivariant maps. Because $\Sigma_n$ acts freely on $X_0$ and $S^{n-2}$ is nonequivariantly $(n-3)$-connected and $X_0$ is $(n-1)$-dimensional, this obstruction theory will only involve the last two or three skeleta of $X_0$, so it should hopefully be tractable.

Answer by gowers for Proving that a function's image contains (1/n,...,1/n)

2011-07-13T14:00:18Z

This isn't an answer to your question but rather a comment that is too long for a comment. I once tried to solve what was then an open problem called the Knaster hypothesis. It's a beautiful and highly plausible statement: that if $f$ is a continuous function defined on the sphere $S^n$ and $x_1,\dots,x_{n+1}$ are points in $S^n$, then there is an orthogonal map $T\in SO(n+1)$ such that $f(Tx_1)=\dots=f(Tx_{n+1})$. One reason for being interested in the conjecture was that, as observed by Milman, it easily implies Dvoretzky's theorem with an essentially optimal bound in all parameters.

When $n=1$ we have two points in the circle and the result can be proved by an easy application of the intermediate value theorem. When $n=2$ the result is true but nothing like so easy. In that case we can think about the map from $SO(3)$ to $\mathbb{R}^3$ that takes $T$ to the triple $(f(Tx_1),f(Tx_2),f(Tx_3))$. This map is continuous, and if $f$ is a counterexample to the Knaster hypothesis then the image of this map does not intersect the line $x=y=z$. So one can hope to try to prove the result by contradiction, showing that the image somehow "surrounds" the line $x=y=z$ enough to be forced to contain it (just as a map from the closed disc to itself that preserves the boundary must map something to the centre of the disc). It is this that I am reminded of by your question, though I don't claim that there is a close connection.

If I remember correctly a proof along those lines can be given when $n=2$, but the conjecture in general turned out to be false. This interesting (but slightly disappointing since the hypothesis would have been such a nice theorem) result was proved by Kashin and Szarek. Their function $f$ was just the $\ell_\infty$ norm. The set of points was slightly more complicated but not too bad.