# On quadratic forms, Pontryagin Squares, and H^4(K(pi,2),U(1))

I am trying to get a concrete handle on the isomorphism $H^4(K(\pi_2,2),U(1)) \simeq \{$quadratic forms $\pi_2 \to U(1) \}$. This is explained in Eilenberg and Maclane's http://www.jstor.org/stable/1969702 and its companion but I am having a hard time getting just what this 4-cocycle should assign to a 4-simplex in $K(\pi_2,2)$. I am primarily interested in understanding the map from the right to the left.

I have a guess at something which may be close, which is there is a canonical closed 2-form on $K(\pi_2,2)$ valued in $\pi_2$. Using the associated bilinear form of the given quadratic form, I can wedge this form with itself to obtain a closed 4-form valued in $U(1)$. I worry that instead of the ordinary square, I need to be doing some factoring, perhaps using the Pontryagin square instead.

Any help, especially with some intuition, would be much appreciated.

• How does the tag 'characteristic-classes' apply? Perhaps the tag 'cohomology' would be more appropriate instead? Also, which article by Eilenberg and MacLane are you specifically referring to? They have several articles in the Annals of Mathematics. – Ricardo Andrade Aug 1 '13 at 0:02
• I suppose you're right about the tags. I had this paper and its companion in mind. jstor.org/stable/1969702 – Ryan Thorngren Aug 1 '13 at 0:07
• There was a brief blog discussion at: sbseminar.wordpress.com/2008/04/10/reference-hunt-i – S. Carnahan Aug 1 '13 at 4:07
• Thanks, Scott. That clarifies some of Whiteheads notation considerably. – Ryan Thorngren Aug 1 '13 at 6:59
• If you have a quadratic form, you get a unique cohomology class in $H^4$, but if you want an explicit cocycle, you will have to make a noncanonical choice in the coboundary coset. – S. Carnahan Aug 1 '13 at 13:44

## 1 Answer

To a quadratic form $q: \pi_2 \to U(1)$, we get a corresponding Pontryagin square operation $H^2(-; \pi_2) \to H^4(-; U(1))$, and such cohomology operations are given by elements of $H^4(K(\pi_2, 2); U(1))$. Unfortunately, I don't know if it's possible to get an explicit cochain-level description of the Pontryagin square operator just from the quadratic form. In Proposition 7.3 of my paper Extensions of groups by braided 2-groups, I write down a (group) cochain-level description of the Pontryagin square corresponding to an abelian 3-cocycle, but I don't know if there's a way in general to get an abelian 3-cocycle from a quadratic form. One good reference to look at is Baues's Combinatorial Homotopy and 4-Dimensional Complexes.

One intuitive way to think of the appearance of the Pontryagin square is that we are describing braided 2-groups with homotopy groups $\pi_2$ and $U(1)$, and the Pontryagin square operation $H^2(G; \pi_2) \to H^4(G; U(1))$ on a group $G$ is the obstruction to lifting an extension of $G$ by $\pi_2$ to an extension of $G$ by the entire braided 2-group (in the sense I describe in my paper).

EDIT: I should mention that you can avoid referring to the Pontryagin square directly. Per Baues, $H_4(K(\pi_2, 2); \mathbb{Z}) = \Gamma(\pi_2)$, the receptor of the universal quadratic map from $\pi_2$, so by universal coefficients, $H^4(K(\pi_2, 2); U(1)) = \operatorname{Hom}_{\mathbb{Z}}(\Gamma(\pi_2), U(1))$, which is the same as quadratic forms on $\pi_2$ valued in $U(1)$.

• Thanks for your answer. Abelian 3-cocycles are indeed the same as quadratic forms. I don't know the isomorphism there either, though, so I think it might take me a while to get my answer out of your paper. – Ryan Thorngren Aug 1 '13 at 7:07
• Actually I may have been mislead by the blog post Scott linked above, so don't take my word for that isomorphism. – Ryan Thorngren Aug 1 '13 at 8:15