Is the theory of categories decidable? - MathOverflow

Is the theory of categories decidable?

2010-01-23T06:49:55Z

There are a lot of theorems in basic homological algebra, such as the five lemma or the snake lemma, that seem like they'd be more easily proven by computer than by hand. This led me to consider the following question: is the theory of categories decidable?

More specifically, I was wondering whether or not statements about abelian categories can be determined true or false in finite time. Also, if they can be determined to be false, is it possible to explicitly describe a counterexample? If it is known to be decidable, is anything known about the complexity? (Other decidable theories often have multiply-exponential time complexities.) If it is known to be undecidable, say by embedding the halting problem, then can I change my assumptions a bit and make it decidable? (For example, maybe I shouldn't be looking at abelian categories after all.)

Thanks in advance.

Edit: It appears a clarification is needed. My goal was to consider the minimal theory that could state things like the five lemma, but not necessarily prove them. For example, I want to say:

If in an abelian category, you have a bunch of maps $0\to A \to B \to C\to 0$ and $0\to A' \to B' \to C'\to 0$ which make up two short exact sequences and some more maps $a:A\to A'$, $b:B\to B'$, $c:C\to C'$ which commute with the previous maps, and $a$ and $c$ are isomorphisms, then $b$ is an isomorphism, too.

Sentences of this form would be inputs to a program, which decides if this statement is in fact true in ZFC (or your other favorite axiomatization of category theory). The point here is that I am restricting the sentences one can input into the program, but keeping ZFC or whatnot as my framework.

I hoped (perhaps naively) that if I restricted the class of sentences, it might be decidable whether or not these statements were true. For example, I imagined that every such theorem is either proven by diagram chasing, or it is possible to find a concrete example of maps among, say, R-modules that contradict the result.

Answer by Joel David Hamkins for Is the theory of categories decidable?

2010-01-23T18:38:06Z

Thanks for clarifying your question. The formulation that you and Dorais give seems perfectly reasonable. You have a first order language for category theory, where you can quantify over objects and morphisms, you can compose morphisms appropriately and you can express that a given object is the initial or terminal object of a given morphism. In this language, one can describe various finite diagrams, express whether or not they are commutative, and so on. In particular, one can express that composition is associative, etc. and describe what it means to be a category in this way.

The question now becomes: is this theory decidable? In other words, is there a computable procedure to determine, given an assertion in this language, whether it holds in all categories?

The answer is No.

One way to see this is to show even more: one cannot even decide whether a given statement is true is true in all categories having only one object. The reason is that group theory is not a decidable theory. There is no computable procedure to determine whether a given statement in the first order language of group theory is true in all groups. But the one-point categories naturally include all the groups (and we can define in a single statement in the category-theoretic language exactly what it takes for the collection of morphisms on that object to be a group). Thus, if we could decide category theory, then we could decide the translations of the group theory questions into category theory, and we would be able to decide group theory, which we can't. Contradiction.

The fundamental obstacle to decidability here, as I mentioned in my previous answer (see edit history), it the ability to encode arithmetic. The notion of a strongly undecidable structure is key for proving various theories are undecidable. A strongly undecidable theory is a finitely axiomatizable theory, such that any theory consistent with it is undecidable. Robinson proved that there is a strongly undecidable theory of arithmetic, known as Robinson's Q. A strongly undecidable structure is a structure modeling a strongly undecidable theory. These structures are amazing, for any theory true in a strongly undecidable structure is undecidable. For example, the standard model of arithmetic, which satisfies Q, is strongly undecidable. If A is strongly undecidable and interpreted in B, then it follows that B is also strongly undecidable. Thus, we can prove that graph theory is undecidable, that ring theory is undecidable and that group theory is undecidable, merely by finding a graph, a ring or a group in which the natural numbers is interpreted. Tarski found a strongly undecidable group, namely, the group G of permutations of the integers Z. It is strongly undecidable because the natural numbers can be interpreted in this group. Basically, the number n is represented by translation-by-n. One can identify the collection of translations, as exactly those that commute with s = translation-by-1. Then, one can define addition as composition (i.e. addition of exponents) and the divides relation is definable by: i divides j iff anything that commutes with sⁱ also commutes with s^j. And so on.

I claim similarly that there is a strongly undecidable category. This is almost immediate, since every group can be viewed as the morphisms of a one-object category, and the group is interpreted as the morphisms of this category. Thus, the category interprets the strongly undecidable group, and so the category is also strongly undecidable. In particular, any theory true in the category is also undecidable. So category theory itself is undecidable.

Answer by François G. Dorais for Is the theory of categories decidable?

2010-01-23T20:39:48Z

The theory of categories is undecidable. By the theory of categories I mean the theory with two types Ob (objects) and Ar (arrows) together with operations dom:Ar → Ob, cod:Ar → Ob, 1:Ob → Ar, and o:Ar×Ar→Ar (possibly partial composition), and the obvious axioms.

One way to see this is to interpret the theory of groups — which is undecidable by a beautiful theorem of Trakhtenbrot — within the theory of pointed categories, that is categories with a distinguished object * (which is an inessential extension). Indeed, the definable set of invertible arrows from * to * form a group, and every group can be interpreted as the set of arrows in a category with * as its only object. I suspect that the theory of Abelian categories is not decidable either, but I haven't tried to prove that (yet).

Answer by Bjorn Poonen for Is the theory of categories decidable?

2010-01-24T00:25:29Z

This answer builds on those of F. G. Dorais and Joel David Hamkins to answer your "specific question", the question left open by them, namely whether the theory of abelian categories is decidable.

The answer is still no.

Even the following more limited family of problems is undecidable:

Given words $r, r_1,\ldots,r_m$ in $x_1,\ldots,x_n$ (i.e., each $r_i$ is a finite product of the $x_i$ and their inverses), decide whether it is true that whenever the $x_i$ are interpreted as automorphisms of an object $M$ in an abelian category, $r_1=\cdots=r_m=1_M$ implies $r=1_M$.

If the answer to the corresponding instance of the word problem for finitely presented groups is yes, then the answer to this abelian category question is yes. Conversely if the answer to the word problem instance is no, then we can construct the finitely presented group $G = \langle x_1,\ldots,x_n | r_1,\ldots,r_m \rangle$, form the group ring $\mathbb{Z}G$, and let $M$ be $\mathbb{Z}G$ as a module over itself, which shows that the answer to the abelian category question is no too.

So if there were an algorithm to decide this family of abelian category problems, there would also be an algorithm to decide the word problem for finitely presented groups. But P. S. Novikov proved in 1955 that the latter algorithm does not exist.