## Posts Tagged ‘combinatorics’

### A Bosuk-Ulam-kind theorem for simplexes

October 17, 2011

This little note came out of a lunch discussion with NYU grad student Alfredo Hubard earlier this week. I think the problem-solving process was quite amusing hence worth shearing.

Back in kindergarten, we all learned this theorem called ‘at any given time, there are two opposite places on Earth having exactly the same temperature and air pressure’. Yes that’s the Bosuk-Ulam theorem. I remember at some point I came across a much less famous theorem in some kind of discrete/combinatorial geometry, saying:

Theorem: Any map from a n-dimensional simplex to $\mathbb{R}^{n-1}$ must have a pair of intersecting opposite faces.

Note: each k-dimensional face of the n-dimensional simplex has a unique, well-defined $(n-1)-k$ dimensional opposite face, as shown: Some examples of maps from the 3-simplex to $\mathbb{R}^2$: i.e. in general they can be quite a mess. I think it can be proved by Thurston’s simplex straightening argument. (haven’t checked carefully)

To me this is like Bosuk-Ulam except for instead of considering a large amount of antipodal pairs, we consider only finitely many such pairs. Hence a discrete analoge.

However, one should note that although they are of the same nature, neither follows from the other.

So somehow this theorem came up during the lunch and Alfredo mentioned to me that professor Guth wondered whether the theorem can be proved for mappings from the simplex to lower dimensional simplicial complexes (instead of $\mathbb{R}^n$). i.e.

Question: Given $f: \Delta^{n+1} \rightarrow S$ where $\Delta^{n+1}$ denote the $(n+1)$-dimensional simplex and $S$ is a $n$-dimensional simplicical complex. Then must there be a pair of opposite faces with intersecting image?

So we started to throw out random ideas.

First of all, although only boundary faces of the simplex has non-empty opposite faces (hence can possibly be intersecting pair), it is important that $f: \Delta^n \rightarrow S$ is defined on the solid simplex. (i.e. if one just map the boundary, then we may let $S$ be topologically a sphere and make the map a homeomorphism!) So the moral is, we kind of need to ‘claps’ the simplex to a simply connected lower dimensional thing first, then map it to our $S$, hence $S$ having non-trivial topology won’t be of much help. Looks almost like the $\mathbb{R}^n$ case, doesn’t it? Perhaps the image on $S$ can be complicated and has non-trivial topology, but this is merely ‘wrapping’ a contractible, $n$-dimensional thing around. But wrapping around can only cause more overlapping hence making the faces intersecting more, not less.

The above line of thoughts give an immediate proof in the case when $S$ is a surface and $n=2$: Lift the map to the universal cover and apply the theorem for $\mathbb{R}^2$. (It’s a little more tricky when the surface is $\mathbb{S}^2$ but you can work it out~ the map restricted to $\partial \Delta$ must be of even degree) Note this won’t generalize to higher dimensions (even for manifolds) since universal covers are no longer that similar to $\mathbb{R}^n$.

So what’s the main difference between complexes and manifolds? well, one can have more than two $n$ dimensional faces attached to the same $n-1$ dimensional face. I decided to first think of whether Bosuk-Ulam is still true if we further assume that the map extends to the solid ball (as seen above, it is true in the surface case).

After trying to generalize that ‘lifting’ proof for a while, we realized the Bosuk-Ulam does not work for simplicial complexes in all dimensions! For very simple reason, i.e. if we map the $n-sphere$ to $\mathbb{R}^n$ by projecting down, the pre-image of the center point is the only pair of antipodal points that’s mapped together. Hence if instead of $\mathbb{R}^n$ we have three $n$-simplicies attached along a single $(n-1)$-simplex (think of it as a piece of $\mathbb{R}^n$ with a vertical simplex attached in the middle): Now we can still project to that piece of $\mathbb{R}^n$, with the image of antipodal point lying on the middle $(n-1)$ simplex, all we need to do is to separate this pair while not creating new pairs! But this is easy, just ‘drug’ the upper sheet into the third $n$-simplex a little bit: (the region outside of the red neibourhood is unchanged) It’s easy to see that no new antipodal pair is moved together.

So now we turned back to the simplex and realized it’s even easier to argue: project in orthogonal direction to a $n$-face, originally the only intersecting pair of opposite faces was the vertex and that $n$-face. Now if we lift the vertex int the third sheet, nothing can intersect~ OK~ perhaps not a useful answer but problem solved!

So far I do not know the answer to the following:

Questions:

1. Is this intersecting faces theorem true for mapping $n$-simplexes to $n$-dimensional manifolds?

2. Is the Bosuk-Ulam true if we consider maps from spheres to $n$-dimensional manifolds which extends to the ball?

The later might be well-known. But so far I can only find a theorem by Conner and Floyd, stating that any map from $S^n$ to a lower dimensional manifold must have a pair of antipodal points mapped together.

### Kissing numbers of sphere packings

September 26, 2011

Enough of the ‘trying to understand recent big theorems’ on this blog, let’s do something light and fun this week!

As I was browsing the ArXiv one day, one thing led to anther and I eventually arrived in a very short and cute paper of Greg Kuperberg and Oded Schramm.

So, we have sphere packings, which is simply a bunch of spheres (say in $\mathbb{R}^3$) with disjoint interiors, some of them might be tangent to others, in which case we say the two sphere ‘kiss‘, like this: (points of tangency are marked with a red cross) We can associate a graph to a sphere packing with vertices representing spheres and join two vertices with an edge if two sphere kisses: Question: What kind of graph can appear this way?

Now if we go back to two-dimensions, it’s a classical theorem that any planar graph can appear as the nerve of a circle packing. (In fact, as mentioned at the end of this earlier post, something much stronger is true. i.e. it’s a theorem of Schramm that one can ‘kiss’ pretty much any given set of planar objects with a given nerve.)

In light of this, it’s nature to wonder whether every graph can be realized as a nerve of some sphere packing (since all graph embeds in some oriented surface which then embeds in $\mathbb{R}^3$). Turns out, no.

However I would imagine it’s not easy to show things such as a given graph cannot be realized. In the paper they came up with what’s perhaps the first restriction that shows not all graphs can be realized in such packing, namely:

The average kissing number is, as one might expect, in average how many kisses does each sphere get, i.e.

2 * number of tangency points / number of balls $= 2|E|/|V|$

If one thinks about it, how might one attempt to construct a packing with super large average kissing number? Well, perhaps we start with a single sphere, put a huge amount of small spheres around it (because one cannot fit many large spheres), then this middle one does get a lot of kisses, but what about the smaller ones? If they are equal sized then all the small spheres only gets roughly 7 kisses…now we need to put even smaller spheres around each of those…If we stop at any stage, there would be more smallers spheres that’s not taken care of than the larger ones with lots of kisses, so it’s not clear at all if the average can blow up.

As usual, things are much simpler in two dimensions: Since the Euler characteristic of any planar graph is 2 ( $|V|+|F|-|E|=2$), and $|F| \leq 2|E|/3$ (any face needs at least 3 edges around them, precisely two faces share an edge) we have $|V| \geq |E|/3+2$, hence $2|E|/|V| \leq 6 - 12/|V|$ < $6$

On the other hand, for our all-time favorite hexagonal packing of congruent circles, if we take more and more layers, the number of balls with six kisses grow like $n^2$, the number of balls on the boundary (hence with less than six kisses) is like $n$. Hence by adding enough layers we can make the average as close to 6 as we want~ Hexagonals are the best, as expected~

Now comes what’s in the paper: they showed that for any sphere packing, the average kissing number is less than $8+4\sqrt{3}$. Hence any graph with average degree larger than 15 cannot be realized. They also provided an example of sphere packings with average kissing number larger than 12.

For simplicity I’ll only reproduce the ‘warm-up case’ in the paper which proves the kissing number is no more than 24. (well, since our goal here is only to say that not all graphs can be realized.) I find this observation (due to Schramm) is very simple, cute, and works in any dimensions.

Let’s first define some numbers: $k_c(n) =$ the maximum number of congruent sphere kissing one sphere (i.e. how many unit spheres we can place around a unit sphere) $k_c$ can be easily computed, for example $k_c(2) = 6$ and $k_c(3)=12$, etc.

Let $k(n) = \sup \{$ average kissing number of sphere packings in dimension $n \}$. As we have seen $k(2) = k_c(2) = 6$ and we are interested in giving upper bounds to $k(3)$.

Theorem: $k(n) \leq 2 k_c(n)$.

Given a sphere packing $P$, let $E$ be the set of kissing pairs and $V$ be the set of spheres.

Define map $f: E \rightarrow V$ where $f$ maps each pair to the sphere in the pair with smaller radius (if the two spheres has the same radius, then just randomly choose one).

Since each sphere can only be surrounded by $k_c(n)$ spheres of its size (of course larger than its size would give us fewer), we conclude that $f$ is at most $k_c$ to $1$. Therefore, we have $|E| \leq k_c(n) |V|$, i.e. $k = 2|E|/|V| \leq 2k_c(n)$

How simple!

Now going from 24 to $8+4\sqrt{3}$ requires some estimates relying on us being in 3-dimensions. Essentially, given any sphere in the packing, if we enlarge it by a ratio $\lambda$ w.r.t its center, then this ‘shell’ intersect some other spheres in disjoint circles. Those circles of course cuts out a fraction of the area that’s < 1 on the shell. Summing this fraction over all $\lambda$ shells results a number that’s less than $|V|$.

Now if we look at each pair of balls, say take pair (V_1, V_2), it turns out that if they kiss, the fraction of shell around $V_1$ that’s cut out by $V_2$ plus the fraction of shell around $V_2$ cut out by $V_1$ is a constant depending only on $\lambda$.

Hence we get a relation $|V| \geq c(\lambda) \times |E|$. Choose an optimum $\lambda$ gives $8+4 \sqrt{3}$ which is a bit less than 15.

So we know $k(3) \leq 8+4 \sqrt{3}$ but larger than 12, the problem of finding the exact value of $k(3)$ remains open, so is the problem of giving other restrictions on the graph for being the nerve of a sphere packing.