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Background: I'm a logic student with very little background in cohomology etc., so this question is fairly naive.


Although mathematical logic is generally perceived as sitting off on its own, there are some striking applications of algebraic/geometric/combinatorial ideas to logic. In general, I'm very interested in the following broad question:

"How should I go about looking for pieces of mathematics far from mathematical logic, which have bearing on some piece of mathematical logic?"

Right now, I'm specifically interested in the following:

"When should I think 'cohomology!'?"

The specific example I'm motivated by is a pair of papers by Dan Talayco (http://arxiv.org/pdf/math/9311205.pdf, http://www.sciencedirect.com/science/article/pii/0168007295000240) in which he develops cohomology theories for two purely set-theoretic objects: Hausdorff gaps, and particularly weird infinite trees ("Todorcevic trees").

At the beginning of his paper on Hausdorff gaps, Talayco mentions that

"the original observation that gaps are cohomological in nature is due to Blass."

This is something I want to be able to do! I can tell that e.g. Hausdorff gaps are all about "not being able to fill something in," but it's a long way between that vague statement and the intuition that there should be a cohomology theory around it, let alone coming up with the specifics. So my question is:

Question. When should I suspect that some piece of mathematics (ideally far from algebra/geometry) has a cohomological interpretation, and how should I go about figuring out what the specifics should be?

To clarify: although 'useful' is always good, I'm just asking how I can tell that cohomology can be attached to some piece of mathematics (especially logic), regardless of whether it yields new results.

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  • $\begingroup$ "ideally far from algebra/geometry": I hope you do understand that this is where cohomology comes from, and that most major applications of cohomology still belong to algebra/geometry(topology). anyway if you have (as you say) very little background in cohomology, you should learn proper cohomology first, and this belongs to algebra/geometry. I doubt you will be able to use cohomology otherwise. You look like someone who tries to drive a car but still struggles to walk $\endgroup$ – Mister Benjamin Dover Jan 11 '15 at 12:32
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    $\begingroup$ I think learning a subject like obstruction theory is very informative on this kind of question that you have. Obstruction theory in a sense was the reason people took cohomology seriously -- that it wasn't just some formal trinket. It has to do with the problem of assembling types of "global objects" from things like local data -- or any kind of intermediate data -- stuff that's smaller than the entire space. $\endgroup$ – Ryan Budney Jan 11 '15 at 19:46
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    $\begingroup$ @Laters I disagree with the last analogy; to me it is more like someone who can pilot an airliner Boeing and tries to pilot a fighter jet. Different specialties, no harm in trying, needs to be aware of the differences. But this is why the OP posted the question in the first place. $\endgroup$ – guest Jan 11 '15 at 20:05
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    $\begingroup$ @RyanBudney if you could expand this comment into an answer with references, I think it would be a great start for the OP. Cohomology (and more general derived functors) as a measure of obstructions is, I think, the best way for the OP to approach the topic. $\endgroup$ – guest Jan 11 '15 at 20:07
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I guess I will take a crack at this.

First of all, it is probably worthwhile for you to learn some cohomology in its original home so that you have some intuition for it, and some knowledge of what theorems there are, how to compute it, etc. You do not give much indication in your answer as to how much knowledge of cohomology you have currently.

Generally the intuition behind cohomology groups is that they measure the failure of "locally consistent" things to be "globally consistent".

Examples:

The first de Rham cohomology group of the punctured plane is 1 dimensional since there is (up to the gradient of a global function) only 1 vector field on this space which is locally a gradient of a function but not globally the gradient of a function.

The Penrose triangle represents a nontrivial cohomology class over the multiplicative group of positive reals, since it "locally" looks like a perspective drawing, but there is no "global" object realizing that.

If local exchange rates between countries allow arbitrage, then there is no globally consistent exchange rate, so current exchange rates give a nontrivial cohomology class.

The axiom of choice says every surjection splits. In fact, even without the axiom of choice, every surjection splits locally (for every point in the codomain, I can find an inverse image), and so the axiom of choice is a a local to global statement: these local inverses can be assembled into a global section. Blass has written a bit about this in Blass - Cohomology detects failure of the axiom of choice, but there is still a lot more work to be done with this concept.

The moral is just to be on the lookout for situations where things seems to fit together in small bits, but somehow the whole does not work out. There is, more than likely, cohomology playing into this somehow.

I will mention that (from my perspective) sheaf cohomology probably formalizes this intuitive perspective the best, since you do not have to start with a sequence of maps with differentials (where do those come from?) just a notion of local objects and how to patch them. So I would recommend learning some sheaf cohomology if you are planning on looking for cohomology far from algebraic topology.

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  • $\begingroup$ First, “Penrose” is a proper noun and should be capitalized. Second, why do you think Penrose triangle is a perspective drawing rather than an orthographic drawing? $\endgroup$ – Incnis Mrsi Jan 14 '15 at 20:29
  • $\begingroup$ I changed "penrose" to "Penrose". Apologies to him. I am not sure about the distinction between perspective and orthographic: will have to read about it. I think the message is clear regardless. $\endgroup$ – Steven Gubkin Jan 14 '15 at 20:32
  • $\begingroup$ Lovely! The example about exchange rates is a beautiful illustration of cohomology in an unexpected context, and the last paragraph about sheaf cohomology helps me a lot to understand a topic that's always been mysterious to me. How does Galois cohomology (which, from my point of view, measures the failure of the rational-points functor to be right exact—at least $\mathrm H^1$) fit into this picture? $\endgroup$ – LSpice Jan 22 at 16:38
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As Steven Gubkin and Ryan Budney have already pointed out, cohomology is often used to measure how far a "locally consistent" object is from being "globally consistent". I thought I might describe another example of this in set theory, which it turns out contains a lot of open problems. As far as I know, this example hasn't been written down anywhere, so I'll have to be a bit verbose. I won't attempt an actual answer to the OP's question, but I hope that what I write down might be helpful.

Disclaimer: Justin Moore told me about this problem many years ago, when I was just beginning grad school. What I'm writing down is what I've been able to reconstruct from my memory of that conversation; it might not be the most up-to-date information on the problem, or the best description of it.

Given a subset $A$ of $\omega$, let $\Gamma(A) = \prod_{n\in A} \mathbb{Z} / \bigoplus_{n\in A} \mathbb{Z}$. Thus, an element of $\Gamma(A)$ can be described as the equivalence class of a function $f : A\to\mathbb{Z}$, where the equivalence is "$f$ and $g$ differ on at most finitely-many coordinates."

Now given a family $\mathcal{A}\subseteq\mathcal{P}(\omega)$, and an $n < \omega$, we define the groups $$ C_n(\mathcal{A}) = \prod_{A_1,\ldots,A_n} \Gamma(A_1\cap\cdots \cap A_n)$$

When $n = 0$, this definition is a little ambiguous, so we take the opportunity to identify $C_0(\mathcal{A})$ with $\Gamma(\omega)$. (One can argue that that's the correct way of doing things, but I'll leave it to the reader.) Now we can define coboundary maps $\delta_n : C_n\to C_{n+1}$ (for all $n$, including $n = 0$) by $$\delta_n(F)(A_1,\ldots,A_{n+1}) = \sum_{k=0}^n (-1)^k F(A_1,\ldots,\widehat{A_{k+1}},\ldots,A_{n+1})$$ where as usual $\widehat{A_{k+1}}$ means we drop $A_{k+1}$ from the list. One can prove as usual that $\delta_{n+1}\circ \delta_n = 0$, so $H_n(\mathcal{A}) = \ker{\delta_{n+1}} / \textrm{im}\;\delta_n$ makes sense.

If you work through the definitions, you can see that $\delta_0$ maps a function (or more accurately, its equivalence class) to its restrictions to elements of $\mathcal{A}$. $\delta_1$ takes a collection of functions defined on members of $\mathcal{A}$ to their differences (on the pairwise intersections).

Hence, a member of $\ker{\delta_1}$ is a family of functions $f_A : A\to \mathbb{Z}$ ($A\in\mathcal{A}$) such that $f_A\upharpoonright A\cap B$ and $f_B\upharpoonright A\cap B$ agree mod-finite for every $A,B$. Such families have been studied before by set theorists, and are called coherent families. The question of whether such a coherent family is in $\textrm{im}\;\delta_0$ is exactly what comes up in Dow, Simon and Vaughan's paper "Strong homology and the proper forcing axiom". They prove there the following (I'm paraphrasing a little bit):

Theorem 1: Assume $\mathfrak{d} = \omega_1$. Then there is a $P$-ideal $\mathcal{I}\subseteq\omega$ (in fact, $\mathcal{I}$ is just $\emptyset\times\textrm{fin}$) such that $H_0(\mathcal{I}) \neq 0$.

Theorem 2: Assume the Proper Forcing Axiom. Then $H_0(\mathcal{I}) = 0$ for every $P_{\aleph_1}$-ideal $\mathcal{I}$.

Actually, it's not hard to prove that $2^\omega < 2^{\omega_1}$ implies that $H_0(\mathcal{A})$ has size $2^{\omega_1}$, whenever $\mathcal{A}$ is a $\subset^*$-increasing $\omega_1$-sequence in $\mathcal{P}(\omega)$. Moreover, Velickovic proves Theorem 2 from just OCA in his paper "OCA and automorphisms of $\mathcal{P}(\omega)/\mathrm{fin}$".

Okay, so we have a lot of natural questions!

Question: What is the possible behavior of $H_n(\mathcal{A})$ for various sets $\mathcal{A}$, and $n\ge 1$? Does PFA (or MM, or whatever) imply that they're all trivial, whenever (say) $\mathcal{A}$ is a $P$-ideal? Can we consistently get $H_n(\mathcal{A}) \neq 0$ for all $n\ge 1$? All at the same time? Etc.

Here's another, entirely unrelated problem. It's known that for every $n$, there is a $\sigma$-$n$-linked poset of size $\mathfrak{b}$, which has no $n+1$-linked subset of size $\mathfrak{b}$. (See Todorcevic, "Remarks on cellularity in products.") Can you express this using cohomology (or maybe homology)?

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