The Lovasz number of the plane is about 3.48287

As seen in this comment on Polymath and explicated further in Fernando de Oliveira Filho’s thesis, section 4.4.

I actually spent much of today thinking about this so let me try to explain it in a down-to-earth way, because it involved me thinking about Bessel functions for the first time ever, surely a life event worthy of recording.

So here’s what we’re going to do.  As I mentioned last week, you can express this problem as follows:  suppose you have a map h: R^2 -> V, for some normed vector space V, which is a unit-distance embedding; that is, if |x-x’|_{R^2} = 1, then |h(x)-h(x’)|_V = 1.  (We don’t ask that h is an isometry, only that it preserves the distance-1 set.)

Then let t be the radius of the smallest hypersphere in V containing h(R^2).

Then any graph embeddable in R^2 with all edges of length 1 is sent to a unit-distance graph in V contained in the hyperplane of radius t; this turns out to be equivalent to saying the Lovasz number of G (ok, really I mean the Lovasz number of the complement of G) is at most 1/(1-2t).  So we want to show that t is bounded below 1, is the point.  Or rather:  we can find a V and a map from R^2 to V to make this the case.

So here’s one!  Let V be the space of L^2 functions on R^2 with the usual inner product.  Choose a square-integrable function F on R^2 — in fact let’s normalize to make F^2 integrate to 1 — and for each a in R^2 we let h(a) be the function F(x-a).

We want the distance between F(x-a) and F(x-b) to be the same for every pair of points at distance 1 from each other; the easiest way to arrange that is to insist that F(x) be a radially symmetric function F(x) = f(|x|); then it’s easy to see that the distance between F(x-a) and F(x-b) in V is a function G(a-b) which depends only on |a-b|.  We write

g(r) = \int_{\mathbf{R}^2} F(x)F(x-r) dx

so that the squared distance between F(x) and F(x-r) is

\int F(x)^2 dx - \int F(x)F(x-r) dx + \int F(x-r)^2 dx = 2(1-g(r)).

In particular, if two points in R^2 are at distance 1, the squared distance between their images in V is 2(1-g(1)).  Note also that g(0) is the square integral of F, which is 1.

What kind of hypersphere encloses all the points F(x-a) in V?  We can just go ahead and take the “center” of our hypersphere to be 0; since |F| = 1, every point in h(R^2) lies in (indeed, lies on) the sphere of radius 1 around the origin.

Hey but remember:  we want to study a unit-distance embedding of R^2 in V.  Right now, h sends unit distances to the distance 2(1-g(1)), whatever that is.  We can fix that by scaling h by the square root of that number.  So now h sends unit distances to unit distances, and its image is enclosed in a hypersphere of radius

2(1-g(1))^{-1}

The more negative g(1) is, the smaller this sphere is, which means the more we can “fold” R^2 into a small space.  Remember, the relationship between hypersphere number and Lovasz theta is

2t + 1/\theta = 1

and plugging in the above bound for the hypersphere number, we find that the Lovasz theta number of R^2, and thus the Lovasz theta number of any unit-distance graph in R^2, is at most

1-1/g(1).

So the only question is — what is g(1)?

Well, that depends on what g is.

Which depends on what F is.

Which depends on what f is.

And of course we get to choose what f is, in order to make g(1) as negative as possible.

How do we do this?  Well, here’s the trick.  The function G is not arbitrary; if it were, we could make g(1) whatever we wanted.  It’s not hard to see that G is what’s called a positive definite function on R^2.  And moreover, if G is positive definite, there exists some f giving rise to it.  (Roughly speaking, this is the fact that a positive definite symmetric matrix has a square root.)  So we ask:  if G is a positive definite (radially symmetric) function on R^2, and g(0) = 1, how small can g(1) be?

And now there’s an old theorem of (Wisconsin’s own!) Isaac Schoenberg which helpfully classifies the positive definite functions on R^2; they are precisely the functions G(x) = g(|x|) where g is a mixture of scalings of the Bessel function $J_0$:

g(r) = \int_0^\infty J_0(ur) A(u)

for some everywhere nonnegative A(u).  (Actually it’s more correct to say that A is a distribution and we are integrating J_0(ur) against a non-decreasing measure.)

So g(1) can be no smaller than the minimum value of J_0 on [0,infty], and in fact can be exactly that small if you let A become narrowly supported around the minimum argument.  This is basically just taking g to be a rescaled version of J_0 which achieves its minimum at 1.  That minimum value is about -0.4, and so the Lovasz theta for any unit-distance subgraph on the plane is bounded above by a number that’s about 1 + 1/0.4 = 3.5.

To sum up:  I give you a set of points in the plane, I connect every pair that’s at distance 1, and I ask how you can embed that graph in a small hypersphere keeping all the distances 1.  And you say:  “Oh, I know what to do, just assign to each point a the radially symmetrized Bessel function J_0(|x-a|) on R^2, the embedding of your graph in the finite-dimensional space of functions spanned by those Bessel translates will do the trick!”

That is cool!

Remark: Oliveira’s thesis does this for Euclidean space of every dimension (it gets more complicated.)  And I think (using analysis I haven’t really tried to understand) he doesn’t just give an upper bound for the Lovasz number of the plane as I do in this post, he really computes that number on the nose.

Update:  DeCorte, Oliveira, and Vallentin just posted a relevant paper on the arXiv this morning!

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4 thoughts on “The Lovasz number of the plane is about 3.48287

  1. Emmanuel Kowalski says:

    The J_0 certainly appears here because it is the Fourier transform of the uniform measure on the unit circle (up to a multiplicative factor)…

  2. voloch says:

    Re: “Filho’s thesis”. Filho means Son in Portuguese. In the context of someone’s name it has the same meaning as Junior does in English. His last name is Oliveira. You wouldn’t refer to Tate’s thesis as “Junior’s thesis”. Sorry, pet peeve.

  3. JSE says:

    Thanks, Felipe, I fixed it!

  4. JSE says:

    But I am totally referring to Tate’s thesis as “Junior’s thesis” from now on.

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