Can Group Theory Save Complexity Theory?



Can solvable groups be used in Barrington’s theorem?



Eric Allender is one of world’s experts on computational complexity. He is especially a master at the care and feeding of the complexity classes that are at the weaker end of the spectrum. These include classes like or among many others. Just because these classes seem to be weak does not mean they give out their secrets easily. Proving almost anything new about them seems well beyond all the techniques that are available to us.

Today I want to talk about a pretty result of Eric’s and how it connects to our last discussion on partial results.



I have never had the honor of working on a project with Eric, even though he was at Rutgers and I was nearby at Princeton for many years. We did however have an interesting situation arise where Eric and his colleagues wiped out Anastasios Viglas and me.

It was back in 1999 and for some reason the question of what is the computational cost of deciding the primality of a number occurred to both groups: one at Rutgers and one at Princeton. Proving lower bounds on general Turing machines is too hard, so both groups asked: could it be proved that testing whether a number is prime or not at least cannot be too easy. In particular, could we prove that primality did not lie in some low level complexity class.

The answer from Rutgers was “yes;” the answer from Princeton was “maybe.” Eric along with Michael Saks, and Igor Shparlinski proved that primality could not be in for any prime . Viglas and I proved a slightly weaker result, and worse our result needed an assumption about the structure of primes. It was one of those assumptions that must be right, but is probably hard to prove.

Unfortunately for us both papers were submitted to exactly the same conference: the 14 Annual IEEE Conference on Computational Complexity. Theirs got in—we did not. Clearly the right decision, but I always wondered what would have happened if we had submitted alone. Oh well.

Allender On Permanent

Now I will talk about the particular pretty result, an even more notable theorem of Eric’s from the late 1990’s. The actual results proved in his paper are stronger, but for our purposes the following simple statement is sufficient:

Theorem: The permanent is not in uniform .

Recall that is a constant depth circuit model that allows arbitrary threshold gates. In the theorem the circuits must be uniform: this means roughly that there is a simpler-than-logspace computation that tells which wires connect to which gates. There is some care needed in getting the model just right, so please look at Allender’s paper for all the details.

The proof uses a number of facts from complexity theory, but shows that a contradiction would follow if the permanent was in the threshold class. A very pretty and clever argument. I think it is well known, but I do not think Eric gets as much credit as he should for this result. It is definitely one of the strong results about the complexity of permanent.

A Result To Dream About

I decided that after the last discussion on partial results I would put out a partial result. The goal is to state a conjecture—actually a family of conjectures—so that if one is proved, then we can prove the following theorem:

Theorem?: The complexity class is properly contained in the class

The high-level issue is whether there is a loophole in the work by Barrington and others that an efficient group-algebra simulation of Boolean gates requires a non-solvable group. The issue involves the so-called commutator subgroup of a subgroup of a group . This is the group generated by the commutators where . The commutators themselves need not form a group—the question is, what group do they generate, and is it all of ? Well, if is an abelian subgroup then , the trivial identity subgroup, and we’ve figuratively “hit bottom.” If is not and not all of , then one can iterate and ask the same question about , and so on. A group is solvable provided you always hit bottom. Put another way, is non-solvable provided it has a subgroup such that . Barrington’s proof employs such a subgrou , indeed where is simple.

Of course, most of us probably believe that solvable groups cannot replace the simple groups in Barrington’s theorem, but that is an open problem. Suppose they can. Then the later work by Barrington and others on the solvable-group case would make equal to (a subclass of) . Allender’s theorem would then make properly contained in (which is equivalent for our purposes to the language class ). The partial work has thus been to establish a connection between a family of existential questions about solvable groups and results that would then fit together to prove an open separation theorem.

Moreover, if you don’t believe , i.e. if you believe Barrington’s characterization is tight, then you can read the connection the other way: results in complexity theory imply the impossibility of some pure statements in group theory. Maybe you can refute the particular statements we’ll propose by direct algebra, but if they stay open then we gain a meaningful dialogue between complexity and algebra.

I will state one of the questions in the family . They all imply as I stated that

The following is the question. Some group theory notation is needed. If are in a group, then

is the conjugate of by . Also denotes the group generated by the elements and . Finally, the order of an element in is denoted by .

Is there some solvable with four elements in the group so that the following are true? Each of the elements is not the identity.

Let be the subgroup . Then we require that some conjugate of and some conjugate of lie in . Let be the subgroup . Then we require that some conjugate of and some conjugate of lie in . And the following two numbers are relatively prime: and .

The last clause may force that must equal its commutator subgroup, which would violate being a solvable group. That would shoot down this member of , though we have others that might survive…

Open Problems

Prove . Or disprove it. If you do that then I will put out some even weaker conditions. Let me know.

Finally, I wish to thank Ken Regan for his special help with this post—as usual any mistakes are mine.

[fixed typo]