Classification of $SU(2)$ principal fibre bundles over four-dimensional manifolds I would like to find a pedagogical reference where the classification, up to isomorphism, of principal $SU(2)$ bundles over a four-dimensional compact, oriented manifold is explained. In particular I am interested in the torus $T^4$ case. Is there a similar classification when the base manifold is non-compact, in particular in the case when it is $\mathbb{R}^4$ or $\mathbb{R}^4-\left\{ 0\right\}$? A pedagogical answer explaining this problem would be also very welcome.
Thanks.
 A: Let $X$ be an $(n-1)$-connected CW complex with $\pi_n(X) = G$. By attaching cells of dimensions at least $n+2$, we obtain a CW complex $Y$ the same $(n+1)$-skeleton as $X$, but with $\pi_i(Y) = 0$ for $i > n$. For $i \leq n$, we have, by cellular approximation, 
$$\pi_i(Y) = \pi_i(Y^{(n+1)}) = \pi_i(X^{(n+1)}) = \pi_i(X),$$ so $Y$ is a $K(G, n)$ with $X$ as a subcomplex.
If $M$ is a CW complex of dimension at most $n$, then 
$$[M, X] = [M, X^{(n+1)}] = [M, Y^{(n+1)}] = [M, Y] = [M, K(G, n)] = H^n(M, G).$$
Now consider the case $X = BSU(2)$. As $\pi_i(BSU(2)) = \pi_{i-1}(SU(2)) = \pi_{i-1}(S^3)$, $BSU(2)$ is $3$-connected and $\pi_4(BSU(2)) = \mathbb{Z}$, so for a CW complex $M$ of dimension at most four
$$\operatorname{Prin}_{SU(2)}(M) = [M, BSU(2)] = [M, K(\mathbb{Z}, 4)] = H^4(M, \mathbb{Z}).$$
The isomorphism $[M , BSU(2)] \to [M, K(\mathbb{Z}, 4)]$ is given by $[f] \mapsto [\iota \circ f]$ where $\iota : BSU(2) \to K(\mathbb{Z}, 4)$ is the inclusion map.
As $\mathbb{Z} \cong \pi_4(K(\mathbb{Z}, 4)) \cong H_4(K(\mathbb{Z}, 4); \mathbb{Z})$, the identity map $\operatorname{id} : \mathbb{Z} \to \mathbb{Z}$ gives rise to an element of $\operatorname{Hom}(H_4(K(\mathbb{Z}, 4); \mathbb{Z}), \mathbb{Z})$ and hence an element $\alpha$ of $H^4(K(\mathbb{Z}, 4); \mathbb{Z})$. The isomorphism $[M, K(\mathbb{Z}, 4)] \cong H^4(M; \mathbb{Z})$ is given by $[g] \mapsto g^*\alpha$.
The composition of these two isomorphisms is an isomorphism $[M, BSU(2)] \to H^4(M; \mathbb{Z})$ given by $[f] \mapsto (\iota\circ f)^*\alpha = f^*(\iota^*\alpha)$. The map $\iota^* : H^4(K(\mathbb{Z}, 4); \mathbb{Z}) \to H^4(BSU(2); \mathbb{Z})$ is an isomorphism because $BSU(2)^{(5)} = K(\mathbb{Z}, 4)^{(5)}$. As $\alpha$ is a generator of $H^4(K(\mathbb{Z}, 4); \mathbb{Z})$, $\iota^*\alpha$ is a generator of $H^4(BSU(2); \mathbb{Z}) \cong \mathbb{Z}c_2$, so $\iota^*\alpha = \pm c_2$.
Therefore the isomorphism $\operatorname{Prin}_{SU(2)}(M) \to H^4(M, \mathbb{Z})$ constructed above is either $P \mapsto c_2(P)$ or $P \mapsto -c_2(P)$. Either way, we see that for a CW complex $M$ of dimension at most four,  principal $SU(2)$-bundles over $M$ are completely determined by their second Chern class; moreover, every element of $H^4(M; \mathbb{Z})$ arises as the second Chern class of some $SU(2)$-principal bundle on $M$.
A: Here's what I have so far. Let $X$ be a $4$-manifold (I don't need any other assumptions on $X$, although I will really be thinking of $X$ as a $4$-dimensional CW complex). A principal $SU(2)$-bundle over $X$ has a second Chern class $c_2 \in H^4(X, \mathbb{Z})$. 

Claim: Such a bundle is trivializable iff $c_2$ vanishes. 

Note that if $X$ is connected, then $H^4(X, \mathbb{Z})$ vanishes unless $X$ is closed, so the classification is only interesting in this case; in particular, every principal $SU(2)$-bundle over $\mathbb{R}^4$ or $\mathbb{R}^4 \setminus \{ 0 \}$ is trivializable.
Sketch. A principal $SU(2)$-bundle over $X$ is classified by a map $X \to BSU(2)$. The universal second Chern class determines a map $c_2 : BSU(2) \to B^4 \mathbb{Z}$, where $B^4 \mathbb{Z}$ denotes the Eilenberg-MacLane space $K(\mathbb{Z}, 4)$, inducing an isomorphism on $\pi_4$. In fact $BSU(2)$ is $3$-connected and
$$\pi_4 BSU(2) \cong \pi_3 SU(2) \cong \mathbb{Z}$$
is its first nontrivial homotopy group. It follows that the homotopy fiber of $c_2$ is the $4$-connected cover of $BSU(2)$. Hence if the pullback of $c_2$ to $X$ vanishes, then the classifying map $f : X \to BSU(2)$ lifts to the $4$-connected cover of $BSU(2)$. But $X$ is a $4$-manifold, so all maps from $X$ to a $4$-connected space are nullhomotopic. Hence the pullback of $c_2$ vanishes iff $f$ admits such a lift iff $f$ is nullhomotopic iff the bundle it classifies is trivializable.
A: Isomorphism classes of principal $SU(2)$-bundles $P\to X$ over a closed 4-manifold $X$ are classified by $H^4(X;\mathbb{Z})$. One assigns to $P$ the second Chern class (a.k.a. Euler class) of the associated $\mathbb{C}^2$-bundle. (In the oriented case, the number $c_2(P)[X]$ is also called the instanton number.) Over open 4-manifolds, all $SU(2)$-bundles are trivial. 
$SU(2)$ is 2-connected and its classifying space is 3-connected. It follows from this fact and the standard obstruction-theory argument (climbing inductively from the 0-skeleton up to the 3-skeleton) that maps from a 3-dimensional CW complex to $BSU(2)$ are null-homotopic. Since an open 4-manifold $Y$ has the homotopy type of such a complex, say by Morse theory, $SU(2)$-bundles over $Y$ are trivial. 
Over a closed, connected 4-manifold $X$, we can trivialize $P$ in the complement of a 4-ball $B$, and also over a neighborhood of $B$. The isomorphism class of $P$ is then determined by the homotopy class of the clutching function $f\colon \partial B\to SU(2)$ obtained by comparing the trivializations over $\partial B$. The map $f$ is determined by the integer $\deg f$ (whose sign is fixed by an orientation of $X$). But $\deg f$ equals the Euler number of the associated $\mathbb{R}^4$-bundle $V$, i.e., the signed count of (transverse) zeros of a section of $V$: the basic case to consider is of the identity map $B^4 \to B^4$, which has a single non-degenerate zero, and extends the identity map $S^3\to S^3$.
[Reverted after an over-hast edit; thanks to the commenters for ironing things out. I've added some explanation.] It is still true in the closed non-orientable that the Euler class classifies $SU(2)$-bundles. This follows from obstruction theory, which interprets the Euler class (as computed via $\deg f$) as the class in $H^4(X;\pi_3(SU(2))$ that precisely obstructs trivializing $P$ over the 4-skeleton. When $X$ is closed, connected, and non-orientable, reduction $H^4(X;\mathbb{Z}) \to H^4(X;\mathbb{Z}/2)$ is an isomorphism. 
(The Poincare dual statement, involving reduction of coefficients for $H_0$ twisted by the orientation line bundle, is easier to visualize.) One concludes that the mod 2 Euler class $w_4(V)\in H^4(X;\mathbb{Z}/2)=\mathbb{Z}/2$ exactly classifies $P$.
A: Thanks for the answers. Correct me if I am wrong, but so far I have understood the following:
1) Every principal $SU(2)$-bundle over $M = \mathbb{R}^{4}$ or $M = \mathbb{R}^{4}-\left\{ 0\right\}$ is trivial, namely it is of the form $M\times SU(2)$. In fact, Every principal $G$-bundle, where $G$ is a Lie group, over $\mathbb{R}^{4}-\left\{ 0\right\}$ is trivial.
2) Let $M$ be a four-manifold. Then a principal $SU(2)$-bundle over $M$ is trivial if and only if $c_{2} = 0$, its second Chern class, vanishes. Since if $M$ is not closed then $c_{2} = 0$ we have that every $SU(2)$ bundle over a non-compact base is trivial. What happens if the base is compact but it has boundary?
3) Let us restrict then to the case where the base $M$ is closed. Suppose that I have computed $c_{2}\in H^{4}(M,\mathbb{Z})$ in a local patch on the base, because all I know is the expression of the connection $A$ on some local patch on the base. How can I obtain what is the topology of the bundle?
4) When it is said that $SU(2)$ bundles are topologically classified by elements in $H^{4}(M,\mathbb{Z})$, what does this really mean? The classification is up to homotopy equivalence, isomorphism of principal bundles...? I read that if $M$ is closed then $H^{4}(M,\mathbb{Z})\simeq \mathbb{Z}$. Does this mean that the different topologies of $SU(2)$-bundles over $M$ are in bijection with $\mathbb{Z}$? I found this a very funny and interesting result. Is there a more explicit way to see this isomorphism? Aside from the fact that the zero class corresponds to the trivial bundle.
5) This is a very trivial question, but I want to confirm it. If I have a trivial principal bundle, then any connection that I can consistently define in that trivial bundle will have zero curvature? 
Thank you very much.
