Chern classes, classification of bundles, and Bockstein morphism

I'm doing a work on Chern classes and I have the following doubts, I do not know if anyone could support me with the doubts, or with bibliography / references

1. Given a $$n \in \mathbb{Z}$$, is it possible to construct a bundle of dimension 1, $$E \rightarrow \mathbb{S}^1 \times \mathbb{S}^1$$, such that $$C_1 (E) = n$$, with $$C_1$$ the first Chern class of the bundle?

2. How can you see that $$C_1 (E \otimes \mathbb{C}) = \beta \circ w_1 (E)$$ with $$C_1$$ the first Chern class of the bundle, $$\beta$$ the Bockstein morphism and $$w_1$$ the first Stiefel-Whitney class?

3. Is it possible to classify real bundles with an almost-complex structure over a manifold? (In particular, I'm interested in BSO (3), BSU (2) and BSO (8).)

Thanks

• The first question is relatively easy: note that $BU(1)$, the classifying space for complex line bundles, is also a $K(\mathbb{Z},2)$. In particular for any space $X$ the isomorphism $[X, BU(1)] \cong H^2(X;\mathbb{Z})$ sends a complex line bundle to its first Chern class. May 4, 2019 at 3:53
• Can you give a bit more elaboration for question 2? Is $E$ meant to be an arbitrary real bundle, or is it rank $1$ as in the first question? Are you finding this statement in a source, or are you trying to determine wether it's true or not? May 4, 2019 at 15:22
• @William Thank you for requesting details In fact, $E$ is an arbitrary real boundle of rank 1 May 6, 2019 at 6:00

I came up with a sketch of something that looks like it works for question 2. It uses the main idea I mentioned in my comment to answer your first question: namely real and complex line bundles correspond to elements of cohomology groups via the isomorphisms

$$w_1 \colon Prin_{O(1)}(X) \cong H^1(X;\mathbb{Z}/2)\ \ \text{ and }\ \ c_1 \colon Prin_{U(1)}(X) \cong H^1(X;\mathbb{Z})$$

resulting from the homotopy equivalences $$BO(1)\simeq K(\mathbb{Z}/2, 1)$$ and $$BU(1) \simeq K(\mathbb{Z},2)$$.

(I haven't solved 3 yet)

To answer question $$2$$, the meat is in showing that for any $$X$$ the morphism $$\beta\colon H^1(X;\mathbb{Z}/2) \to H^2(X;\mathbb{Z})$$ is induced by complexification of bundles.

Edit: Faster proof

The Bockstein $$\beta\colon H^1(-;\mathbb{Z}/2) \to H^2(-;\mathbb{Z})$$ corresponds to a homotopy class of functions $$K(\mathbb{Z}/2,1) \to K(\mathbb{Z}, 2)$$, and since this set of homotopy classes is $$H^2(\mathbb{RP}^\infty;\mathbb{Z}) \cong \mathbb{Z}/2$$, this map is either null-homotopic or in the unique class of maps which are not. To show it's not null-homotopic it suffices to come up with a space for which $$\beta$$ is non-zero. Indeed you can show that $$H^1(\mathbb{RP}^\infty;\mathbb{Z}) = 0$$ and $$H^1(\mathbb{RP}^\infty;\mathbb{Z}/2)\cong \mathbb{Z}/2$$ so $$\beta \colon H^1(\mathbb{RP}^\infty;\mathbb{Z}/2) \to H^2(\mathbb{RP}^\infty;\mathbb{Z})$$ is non-zero by the Bockstein exact sequence for $$\mathbb{Z} \stackrel{\cdot 2}{\rightarrow} \mathbb{Z} \to \mathbb{Z}/2$$. Considering that $$K(\mathbb{Z}/2,1) \simeq BO(1)$$ and $$K(\mathbb{Z}, 2) \simeq BU(1)$$, and that the complexification map $$c\colon BO(1) \to BU(1)$$ is not null-homotopic, it follows that $$c$$ represents $$\beta$$. See the bottom of my post for the conclusion of the argument.

My original proof with more theory and constructions and is more informative:

First we need some

General theory.

Given a short exact sequence of topological groups $$H\to G \to G/H$$ such that the quotient map is a fibre bundle, it can be extended to a "delooping" sequence

$$H \to G \to G/H \stackrel{\delta}{\to} BH \to BG$$

where each sequence of three spaces is equivalent to a fibration; moreover if $$H$$ is normal then the delooping sequence also extends one more step with a map $$BG \to B(G/H)$$. The map which is not given by functoriality is the connecting map $$\delta\colon G/H \to BH$$. To describe this map we use the fact that there is a model of $$BH$$ given by

$$BH \simeq EG/H \cong EG \times_G (G/H)$$

where $$X\times_G Y$$ is the "Borel Construction" or "Balanced Product" (see for example these notes, in particular Section 3, Proposition 8.3, and Theorem 11.3.) In words, there is a model of $$BH$$ constructed by starting with a model of $$EG$$ and replacing the fibres with copies of $$G/H$$; the result is a fibre bundle $$BH \to BG$$ and the connecting map $$\delta\colon G/H \to BH$$ is represented by the inclusion of one of these fibres.

If $$G$$ and $$H$$ are moreover abelian, then $$BG$$, $$BH$$ and $$B(G/H)$$ are all groups as well, and even abelian themselves, so they too have classifying spaces which are also abelian groups. In this case the delooping sequence extends arbitrarily far:

$$H \to G \to G/H \stackrel{\delta}{\to} BH \to BG \to B(G/H) \stackrel{B\delta}{\to} B^2H \to B^2G \to B^2(G/H) \stackrel{B^2\delta}{\to} \dots$$

Suppose finally that we are in the extra-special case that $$G$$ and $$H$$ are even discrete. Since $$B^nA \simeq K(A, n)$$ for any discrete abelian group $$A$$, given any $$X$$ the homotopy classes of maps into the delooping sequence give the usual long exact sequence of cohomology groups

$$H^0(X;H) \to H^0(X;G) \to H^0(X;G/H) \stackrel{\beta}{\to} H^1(X;H) \to H^1(X;G) \to H^1(X;G/H) \stackrel{\beta}{\to} H^2(X;H) \to \dots$$

Our case.

We are considering $$\beta\colon H^1(X;\mathbb{Z}/2) \to H^2(X;\mathbb{Z})$$ coming from the short exact sequence $$\mathbb{Z} \stackrel{\cdot 2}{\rightarrow} \mathbb{Z} \to \mathbb{Z}/2$$. Our standard model of $$E\mathbb{Z}$$ will be $$\mathbb{R}$$ with the usual $$\mathbb{Z}$$ action, so that $$B\mathbb{Z} \cong S^1 \cong U(1)$$. To model the map $$B(\cdot 2)$$ we construct a model of $$B(2\mathbb{Z})$$ as above to get $$\mathbb{R} \times_\mathbb{Z}(\mathbb{Z}/2)$$, which takes all of the $$\mathbb{Z}$$ fibres in the quotient map $$\mathbb{R} \to S^1$$ and replaces them with copies of $$\mathbb{Z}/2$$. This results in a connected double cover of $$S^1$$, since for any $$[r,\epsilon]\in \mathbb{R} \times_\mathbb{Z}(\mathbb{Z}/2)$$ the path sending $$t$$ to $$[r + t,\epsilon]$$ is a path from $$[r,\epsilon]$$ to $$[r+1,\epsilon] = [r,\epsilon + 1]$$. Therefore it is equivalent to the unique non-trivial double-cover $$S^1 \to S^1$$, and it follows that the connecting map $$\mathbb{Z}/2 \to B\mathbb{Z}$$ in the delooping sequence is actually modelled by the standard inclusion $$\mathbb{Z}/2 \hookrightarrow S^1$$, which is also known as the "complexification" map $$\kappa\colon O(1) \to U(1)$$. Therefore the operation $$\beta\colon H^n(X;\mathbb{Z}/2) \to H^{n+1}(X; \mathbb{Z})$$ is induced by $$B^n(\kappa)$$. In particular when $$n=1$$ we get the map $$BO(1) \to BU(1)$$ which sends a real line bundle to its complexification, or more specifically if $$V \to X$$ is classified by $$f\colon X\to BO(1)$$ then $$B\kappa \circ f$$ classifies $$V\otimes \mathbb{C}$$ (I think this is worked out in Mirura-Toda somewhere). We end up with the following diagram for all $$X$$

$$\require{AMScd}$$ $$\begin{CD} Prin_{O(1)}(X) @>\otimes\mathbb{C}>> Prin_{U(1)}(X)\\ @V w_1 V \cong V= @V c_1 V \cong V\\ H^1(X;\mathbb{Z}/2) @>>\beta> H^2(X;\mathbb{Z}) \end{CD}$$

which expresses what you're trying to show for line bundles.

If $$E$$ is not necessarily a line bundle, but instead has rank $$n\geq 1$$, replace $$E$$ with its determinant line bundle $$\Lambda^n E$$. Then it's known that $$w_1(\Lambda^n E) = w_1(E)$$ and $$c_1(\Lambda^n E\otimes \mathbb{C}) = c_1(\Lambda^n (E\otimes \mathbb{C})) = c_1(E\otimes \mathbb{C})$$ (the proof I know uses the splitting principle to come up with a formula in terms of elementary symmetric polynomials, I'm not sure of a source off the top of my head).