Mathematical Embarrassments



Mathematical embarrassments are problems that should be solved already



Terry Tao is one of the greatest mathematicians of our time. Tao has already solved many long-standing open problems, which earned him a Fields Medal. What I cannot understand about him is how he can be so productive. He is the author of one of the best blogs in the world, he writes original articles that solve hard open problems, he writes great survey articles on a wide range of topics, and he is the author of terrific books on topics from additive combinatorics to partial differential equations. He is impressive.

Today I want to talk about a topic that he has discussed: mathematical embarrassments—the name is mine, but I hope he will agree with spirit of this term.



There is an old story about Godfrey Hardy and John Littlewood, who together, of course, published many great papers. Steven Krantz tells a version of the story in his fun book, Mathematical apocrypha:

It is said that Landau thought “Littlewood” was a pseudonym for Hardy so that it would not seem like he wrote all the papers.

I think Tao might consider using this trick and invent a co-author or two. It would help make the rest of us feel better.

Let’s now turn to mathematical embarrassments.

Mathematical Embarrassments

A mathematical embarrassment (ME) is a problem that should have been solved by now. An ME usually is easy to state, seems approachable, and yet resists all attempts at an attack. There may be many reasons that they are yet to be solved, but they “feel” like they should have been solved already.

Some Examples

The and problem. The problem is to prove that and are both transcendental numbers. We know that one of these must be transcendental. For if both were algebraic, then so would,

which contradicts the known fact that is transcendental.

Similarly, at least one of and must be transcendental, for otherwise

would be an algebraic polynomial with transcendental roots. It seems ridiculous that we are able to show the transcendence of and , but not these numbers.

The linear recursion zero problem. An ME due to Tao is a problem about linear recurrences. Given a linear recurrence over the integers,

with initial conditions, does there exist an so that ? He only asks for a decision procedure: he says that it is faintly outrageous that this problem is still open; it is saying that we do not know how to decide the halting problem even for “linear” automata!

Mersenne composites. The problem is to prove that for an infinite number of primes ,

is composite. It is an open question whether there are infinite number of Mersenne primes. Currently, only 47 Mersenne primes are known. It seems more likely that there are an infinite number of Mersenne composites. Leonhard Euler has shown that

Theorem: If and is prime, then is prime if and only if .

Thus, if and are both primes, then is a composite number. So an approach to the problem is to show there are infinitely many ‘s of the above form.

The Jacobian conjecture. This is one of my favorite problems outside theory proper. Consider two polynomials and . The Jacobian Conjecture asks when is the mapping:

one to one? The values of are allowed to be any complex numbers. Clearly, a simple necessary condition is that the mapping be locally one to one: this requires that the determinant of the Jacobian of the mapping to be everywhere non-zero. That is that for all and ,

for some non-zero constant . If the value of is not a constant, it is a non-trival polynomial, which must take the value zero for some values . So for those values, the mapping will not even be one to one locally.

Note, is the value of the partial derivative of with respect to , and the same for . The Jacobian Conjecture is that this necessary condition is sufficient.

This is the conjecture for two dimensions; there is a similar one for higher dimensions, but this case is already open. Note, the maps can be messy; for example, the mapping

is one to one. Then, compose this with some linear invertible transformation, and the mapping can start to look very strange.

The Jacobian Conjecture is related to another famous theorem called the Ax—Grothendieck theorem. In fact, the Jacobian conjecture implies the Ax—Grothendieck theorem.

Theorem: Suppose that is a polynomial function. Then, if is one-to-one, then is onto.

The proof is based on the following trivial observation: if a map from a finite set to itself is one-to-one, then it is onto. Amazing. The full proof is based on a simple compactness argument and this simple observation. See this for a nice exposition by Michael O’Connor.

The existence of explicit non-linear lower bounds. One of the great ME for theory is that we have no circuit lower bounds that are non-linear on any explicit problem. The natural proof and other barriers do not seem to be relevant here. I do not believe that they stop a lower bound of the form:

The BPP hierarchy problem. The problem is to show that is more powerful than . Even with respect to oracles, we do not know if this is true. It is hard to believe that such a basic question is still open. There is a partial result due to Jin-Yi Cai, Ajay Nerurkar, and D. Sivakumar.

The number of planar graphs. How many labeled planar graphs are there, with a given number of vertices ? If is the number of such graphs, then the best known bounds are of the form:

where . How can we not know the number of planar graphs?

Well it turns out that I am embarrassed—Omer Giménez and Marc Noy know. They have proved—in 2008—an asymptotic estimate for the number of labeled planar graphs:

where and are explicit computable constants. This a beautiful result, which solves a long standing ME.

There more known about 2-connected planar graphs—see the paper by Edward Bender, Zhicheng Gao, and Nicholas Wormald for some pretty results. One that I like very much is that they can show that a random 2-connected planar graph is highly likely to contain many copies of any fixed planar graph:

Theorem: For any fixed planar graph , there exist positive constants and such that the probability that a random labeled 2-connected planar graph G with vertices has less than vertex disjoint copies of is .

The Barrington problem. David Barrington’s famous theorem shows that polynomial size bounded width computations can compute all of . He does this by using computations over simple groups. An open question is can simple groups be replaced by solvable groups? I think the conventional wisdom is that this must be impossible. But there seems to be no progress on this problem. The reason it is interesting is that the construction of Barrington fails easily for solvable groups, but that is not a proof.

See this set of lecture notes and the blog post by Lance Fortnow for more discussion on Barrington’s theorem.

The problem. This problem is given a series of integers

is there a subset of size that sums to ? How can order be the best possible running time?

This algorithm is essentially the same as the exponential algorithm for the knapsack problem that I talked about before. Recall, that the knapsack problem is to find a 0-1 vector so that

The method re-writes the problem as

where —we can assume that is even. Then, all the left hand sums are computed, and all the right hand sums are computed too. There is a solution to the original problem if and only if these sets have a value in common. Since there are of them, this method takes time times a polynomial in .

The Hilbert subspace problem. Given a Hilbert space and a linear operator is there a subspace so that it is non-trivial, invariant under , and closed? Invariant under means that for each in , is also in . This is a classic result for finite Hilbert spaces, but is has long been open in general. Several special cases of this problem have been resolved, but not the general case.

Open Problems

What are your favorite ME? Let me know, please.