# cohomology groups and the pontryagin construction

I'm just beginning to learn about cohomology groups, and have been told that at this point, it definitely behooves me to step away from the crutch of geometric intuition, but I'm going to try to lean on it for a little longer anyway.

When learning about higher homotopy groups, the only way that I was able to make sense of $\pi_{n+k}(S^n)$ at first was by using the Pontryagin construction to think of elements of the group as framed cobordism classes of $k$-folds in $\mathbb{R}^{n}$, and once I has some vastly oversimplified picture in my head, I was able to make sense of many other proofs and statements with greater ease.

Now, I understand that one may think of $H^n(X;G)$ as $[X,K(G,n)]$. This leads me to believe that there should be something similar to the Pontryagin construction for visualizing these things. Is this known, and is there a reference? Or is it wrong, because there's not really a good way to visualize $K(G,n)$, or because for $X$ not a sphere the whole thing fails, or for any other reason?

Thanks for any insight!

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Though I cannot really answer your question, let me explain why not, and what is related to your question.

Let us first recall how to go form algebra to geometry using the Pontryagin-Thom construction. The general statement of Pontryagin-Thom is that (vaguely speaking) homotopy classes of maps into a Thom space correspond to "enriched" bordism groups. The passage from the homotopy classes to bordism classes is obtained by approximating your map by a smooth one, then make it transversal to the zero section in your Thom space and form a transversal inverse image. In your example you just replace your map $S^{n+k} \to S^n$ by a map regular at $p \in S^n$ and take the inverse image of this point. This will be a $k$-dimensional submanifold of $S^{n+k}$ and it is even a framed manifold.

What did we use in order to make such a construction work?

So suppose we have a space $Y$ and want to get geometry into the set $\lbrack X, Y \rbrack$. In your case we want to choose $Y$ to be an Eilenberg-McLane space $K(A,n)$.

Firstly we needed the notion of smoothness and transvers inverse images, respectively suitable subspaces of the target $Y$ of which we want to take this transverse inverse image.

Now note, that in the general Pontryagin-Thom construction the input is some smooth bundle $E$ over a smooth manifold $M$. Then you take the Thom-space of this bundle which still has a natural zero section $M \to Th(E)$ and in a neighbourhood of this zero section $Th(E)$ is a manifold, where we do have notions of smooth maps and transversal inverse images.

Moreover note that also better the source (i.e. $X$) should be a smooth manifold, such that we can talk about smooth and transverse maps after all.

So if we want to have a similar construction as Pontryagin-Thom in the case $\lbrack X, K(A,n) \rbrack$ we would at least want $X$ to be a manifold, and also a model of $K(A,n)$ which is a manifold.

Let us look at this last obstruction, namely that $K(A,n)$ has the homotopy type of a closed manifold. For the case $n=1$ at least one has the necessary condition on $A$ to be torsionfree (because if $K(A,1)$ is a closed manifold then it is in particular a finite $CW$-complex and thus has homology bounded above by the dimension. But if a group contains torsion, then one knows that the group homology is not bounded, and group homology is isomorphic to singular homology of $BA = K(A,1)$). For $n\geq 2$ I do not know of any criterion which must be satisfied such that $K(A,n)$ can be a smooth closed manifold, but I suspect it seldomly is. Consider for example the easiest case $K(\mathbb{Z},2) \simeq \mathbb{C}P^\infty$ which is not a manifold on the nose.

One could weaken this condition in saying that you want $K(A,n)$ to be a union of finite sub complexes which all have the homotopy type of closed manifolds (like $\mathbb{C}P^\infty$), unfortunately I still do not know when this can happen in general. But at least then if $X$ where a closed manifold (of a finite $CW$-complex) any map would factor over some finite stage, and hence it would map into a smooth manifold, as we wanted.

I hope this helped you a little bit in understanding why I do not think there is something similar to the Ponrtyagin-Thom construction for singular cohomology. (in any case, the Pontryagin-Thom construction is a method of translating bordism theories into stable homotopy theory, the instance being that $\pi_{n+k}(S^k)$ is independent of $k$ for large $n$ compared to $k$ and is called the $n^{th}$ stable homotopy group of spheres).

Let me nevertheless try to explain, why in the case of $X$ a manifold, there is gemeotry in $H^k(X;A)$ i.e. in $\lbrack X,K(A,k)\rbrack$.

Firstly let us assume the manifold $X$ has dimension $n$ and is oriented (for simplicity). Then by Poincaré duality we have that $H^k(X;A) \cong H_{n-k}(X;A)$. Now one knows that every closed oriented manifold $N$ of dimension $n-k$ together with a map $f: N \to X$ gives rise to a class $f_*(\lbrack N \rbrack) \in H_{n-k}(X;A)$ and these classes should be considered to be "geometric", I think. Note that this in particular holds for oriented submanifolds of $X$

So a good question is, given a cohomology class in $H^k(X;A)$ can it be represented by some manifold of dimension $n-k$ (maybe even by a sub manifold)?

These questions have been studied a lot, let me mention some result concerning these.

For the case $A = \mathbb{Z}/2\mathbb{Z}$ the answer is always yes, i.e. every mod $2$ homology class of a manifold $X$ is represented by some manifold that maps to $X$. This follows from a theorem of Thom which (in modern terms) states that the unoriented bordism spectrum $MO$ splits as a wedge of spectra, one of which is the mod $2$ Eilenberg-McLane spectrum, and the map being the mod $2$ orientation.

For integer coefficients the same statement is true only up to dimension $6$ (I am not really sure about this number). But it is wrong in general, Peter Teichner for example explicitely constructs a manifold and a homology class which is not represented in this sense.

But geometric means allow you to prove for example that ever codimension $2$ homology class is represented. Precisely: Let $X$ be an $n$-dimensional oriented closed manifold. Then every class $\alpha \in H_{n-2}(X;\mathbb{Z})$ is represented by a submanifold of $X$.

Let me finish this answer by giving a sketch of a proof. You know that $$H_{n-2}(X;\mathbb{Z}) \cong H^2(X;\mathbb{Z}) \cong \lbrack X,\mathbb{C}P^\infty\rbrack.$$

The last isomorphism is explicitely given by pulling back the tautological bundle and then taking the first chern class of this vector bundle over $X$. Now since $X$ is compact indeed any map to $\mathbb{C}P^\infty$ factors over $\mathbb{C}P^N$ for some $N$. You can make this map transverse to $\mathbb{C}P^{N-1}$ and take the inverse image. This will be a codimension $2$ submanifold of $X$ and it will represent the Poincaré dual of the cohomology class you started with.

In Bredon's book "Geometry and Topology" these questions are also addressed to some extend.

To sum things up, to my knowledge, in general a Pontryagin-Thom construction is not applicable for singular cohomology. But there are geometric interpretations of homology classes (and hence for manifold also for cohomology classes by Poincaré duality) using the theory of spectra and how certain bordism spectra split up as wedge sums containing Eilenberg-McLane spectra.

I hope this helps a little.

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Thank you. It certainly does help; I should have immediately realized that $K(G,n)$ not usually being a manifold would be the chief difficulty. I expect that I'll be coming back to this answer as I learn more about generalized cohomology theories. –  gmoss Feb 12 '13 at 3:02