The first topic I’ll be covering in the Algebra and Set Theory seminar is Whitehead’s Conjecture, and I want to take some time to sketch out how those lectures are going to go.

In order to even talk about Whitehead’s conjecture, I’m going to need to talk about Ext. So given any abelian group $A$, we say that a free resolution of $A$ is a short exact sequence of the form

$0\to F_1 \to F_0 \to A \to 0$

where $F_1$ and $F_0$ are both free groups. Now, every abelian group has a free resolution as we can take $F_0$ to be the free abelian group generated by using $A$ as an alphabet, and $F_1$ to be the kernel of the surjection induced by $a\mapsto a$. Hitting this complex with the $Hom(*, \mathbb Z)$ functor yields the following cochain complex:

$0\to Hom(F_0, \mathbb Z)\to Hom(F_1, \mathbb{Z})\to 0$.

Now that we have a cochain complex, we can ask about the cohomology groups of this complex, which are denoted by $Ext^n_\mathbb{Z} (A, \mathbb Z)$. So what on earth does this tell us about my original group $A$? Let’s borrow some (misleading) intuition from algebraic topology. One nice fact is that if my fundamental group is trivial, and all cohomology groups are trivial, then my space (provided it’s a CW complex) is contractible. So maybe if $Ext^1_\mathbb{Z}(A,\mathbb{Z})$ is trivial I can say something about my group.

It turns out that we can! This cohomology group being trivial is equivalent to the statement that the only group extension of $A$ by $\mathbb Z$ is just the direct sum $A\oplus \mathbb{Z}$. This is coming from the equivalence of the “derived functor” presentation of Ext and the “classical” version of Ext in terms of extension classes (hence ext). That’s pretty cool, but what about that topology bit? Obviously I can’t say that $A$ is contractible, but can I say that $A$ is “nice” in some other way? This leads us to Whitehead’s conjecture.

Whitehead’s Conjecture: For any abelian group $A$, if $Ext^1_\mathbb{Z}(A,\mathbb{Z})=0$, then $A$ is free.

The converse is actually a really easy theorem since Ext is invariant with respect to which free resolutions we take, and we can always resolve a free group in the dumbest way possible:

$0\to 0\to F \to F\to 0$

Of course, the first cohomology group of the resulting cochain complex is trivial. What’s interesting about Whitehead’s conjecture is that it’s independent of ZFC, so the resolution comes in two parts. Here’s my plan for how we’re going to tackle that:

1. We’re going to start off by proving some facts about Ext, and showing that Whitehead’s conjecture is true for countable abelian groups.
2. From here, we can use the proof of the countable case to see what we would need to “turn on induction” and prove this for higher cardinalities. This naturally leads us to Jensen’s $\diamondsuit$ principle.
3. At this point, we’re going to break from the algebra a bit to review some facts about stationary sets, introduce $\diamondsuit$, and prove some facts about it. In particular, we’ll need an equivalent definition that will be a bit more useful to us.
4. We now show that under $\mathrm{ZFC+\diamondsuit}$, we can prove that Whitehead’s conjecture holds for groups of size $\aleph_1$. In particular, we see how to continue going provided we can get diamond for other cardinals.
5. Again we take a break from our regularly scheduled algebra program to do some set theory. In particular we want to introduce Gödel’s constructible universe, $L$. Here we have quite a bit of structure, but important to us is the fact that $\diamondsuit_\kappa$ holds for every regular uncountable $\kappa$. I don’t know how much I’ll prove here about $L$, but at the very least we’ll need this model again later.
6. Now in $L$ we can push through the induction at regular stages, but singular stages seem to be problematic. Here we will prove Shelah’s singular compactness theorem which is stronger than what we need, and a result of ZFC. This final piece will finish the proof that if $V=L$, then Whitehead’s conjecture holds.
7. For the other half of the independence result, we’ll need to introduce Martin’s Axiom. Now this axiom does technically refer to forcing, but we can easily give a forcing free exposition provided we take the consistency of $\mathrm {ZFC+MA+\neg CH}$ for granted. On the other hand, I will take this opportunity to use Martin’s Axiom to motivate forcing and intuitively discuss how one might force to violate the continuum hypothesis.
8. With Martin’s Axiom, we assume that $\mathrm{MA+\neg CH}$ holds and construct an non-free abelian group $A$ of cardinality $\aleph_1$ such that $Ext^1_\mathbb{Z}(A,\mathbb{Z})=0$. This will finish up proof that Whitehead’s conjecture is independent of ZFC.

One thing that I find neat about this proof is that we get to touch on quite a bit of set theory as we go along, and it’s rather instructive as to how one can employ some powerful machinery to prove things about abelian group. The main reference I’ll be using for this part is Paul Eklof’s paper “Whitehead’s Problem is Undecidable”.