About a year ago, Azita Mayeli, Jonathan Pakianathan and I published a paper proving the Fuglede Conjecture in , prime. In this context, the conjecture says that the following. We say that tiles if there exists such that

for all , where is the indicator function of $E$.

We say that is spectral if there exists such that

, , is an orthogonal basis for in the sense that every function can be written as a finite sum

and

, for all .

In dimension and higher this result is known to be false. In five dimensions this was demonstrated by Terry Tao in one direction and by Kolountzakis and Matolcsi in the other. In dimension four the conjecture was disproved, in both direction, by Kolountzakis and Matolcsi. But the two and three dimensional cases remained open and it was widely speculated that the higher dimensional counter-examples will eventually apply there as well, However it turns out that something special is going on in lower dimensions and this is where we now turn out attention.

Theorem (Iosevich, Mayeli and Pakianathan) The Fuglede Conjecture holds in , prime.

The paper was published in Analysis and PDE last year (https://projecteuclid.org/euclid.apde/1508432238)

I will sketch the proof below and explain the basic ideas involved. We shall focus on the case because this allows us to avoid the technicalities involving non-zero vectors that are orthogonal to themselves.

Basic notation: if , define

. It is not difficult to check that

If , denotes its indicator function.

The set of directions determined by a set is the set of difference , , under the equivalence relation , , if for some .

Lemma 1 (number theory) Let . Suppose that there exists such that . Then

i) for all .

ii) is equi-distributed on lines perpendicular to in the sense that

To prove the lemma, we write

where .

It follows that

Since the minimal polynomial of is

we conclude that

This proves part ii) and part i) follows at once.

Lemma 2 (combinatorics) Let of size . Then determines every possible direction in .

To prove this note that if a direction is missing, is a subset of a graph

, so ,

With the two lemmas in tow, we are ready to prove the theorem above. Suppose that is spectral. Then is an orthogonal basis and by elementary linear algebra, . The orthogonality hypothesis implies that Lemma 1 applies and we conclude that .

Suppose that .Then determines every possible direction since . Invoking orthogonality, we see that

for all , . By Lemma 1, for all and since determines every possible direction, for all . This implies that .

Thus we may assume that . The case is not very interesting, so we consider the case . Invoking Lemma 1 again, we see that for each , has exactly one point on the line , where is a fixed vector such . As we noted above, such a vector must exist since we are assuming that is spectral, so for all , .

By rotation, we may assume that has exactly one point on the lines given by the equation . But then it is not difficult to see that that tiles with the tiling set .

Let’s now assume that tiles by translation with a tiling set . Then must be a divisor of . The only interesting case is . Tiling means that

for all . Taking the Fourier transform of both sides, we see that

for all . We conclude from this that there exists such that . If not, then for all . It would follow that , which is not possible since , which implies that .

Since there exists such that , we can invoke Lemma 1 again to see that contains exactly one point on the lines perpendicular to . By rotation, we may assume that has exactly one point on the lines given by the equation . But then one can check by a direct calculation that the set is an orthogonal basis for , where .

This completes the proof. Here are some exercises for the interested reader.

Exercise: Work out the details of the case .

Exercise: Prove that if and tiles by translation, then is spectral.

It is not known whether spectral implies tiling in .