Obstruction Theory for Vector Bundles and Connections

I know that the property that a vector bundle on a manifold is flat is equivalent to it being a pull-back from the fundamental groupoid. Namely, there is a map $X \to P_{\le 1} X$ from $X$ to its first Postnikov truncation, and giving a flat connection on a bundle is essentially the same as writing it as a pull-back along this map.

Are there any known "higher equations" on a connection that guarantee that $V$ is a pull-back from a higher Postnikov truncation, say, that it is a pull-back from $P_{\le 2} X$? Intuitively that would mean that the vector bundle "depends only on the first 3 homotopies, including $\pi_0$", in some sense.

Edit: As mentioned in John Klein's answer and comments, the condition of being flat is stronger than being a pullback from the 1-st Postnikov filtra, since it is actually equivlent to lifting the map $X \to BG$ through the classifying space of the discrete group $G^\delta$. Thus, one can at most ask for a $\textbf{sufficient}$ differential geometric condition on a connection to guarantee that the vector bundle factor through $P_{\le 2}X$, say.

Thus, the remaining question is: Can we find such a sufficient condition on a connection, which is weaker than flatness?

• Already for line bundles and the $0$-th postinkov section this is nontrivial. The data of a line bundle is equivalent to giving the integral first chern class. Modulo torsion this is the curvature form. But if the chern class is torsion then the bundle is flat but potentially non trivial. Differential characters can help you here. For line bundles there's a cheeger simons character which classifies line bundles with connections modulo gauge equivalence. This solves the case of line bundles for the 0th postnikov section at least. – Saal Hardali May 4 '18 at 6:06
• Well, I guess you are arguing against the statement that flatness is equivalent to reduction to the foundamental groupoid. It sounds strange given that representations of the foundamental group and flat bundles are equivalent categories. Anyway, flatness do imply reduction to the foundamental groupoid, and my question is if we can impose weaker natural conditions on the connection that will ensure reduction to the second Postnikov piece. I don't require equivalence here. – S. carmeli May 4 '18 at 8:01
• Saal Hardali: Let $E\to \Bbb CP^1$ be the canonical line bundle. This is not flat. $\Bbb CP^1 = S^2$ and the bundle is classified by the inclusion map $S^2 \to BU(1) = \Bbb CP^\infty$. Note that this map is the map from $S^2$ to its second Postnikov section, so we get an example of what you want. – John Klein May 4 '18 at 21:37
• I think being a flat bundle is a stronger property than being pulled back from $P_{\leq 1}X$. For example take $X$ to be the torus (or any genus $g$ surface with $g\geq1$). Then $X = P_{\leq 1}X$ agrees with its Postnikov truncation. So every vector bundle is a pulled-back bundle. However line bundles corresponding to non-trivial elements of $H^2(X, \mathbb{Z}) =\mathbb{Z}$ will have non-trivial first Chern class, and hence are not flat. – Chris Schommer-Pries May 8 '18 at 14:44
• @ChrisSchommer-Pries You're 100% correct. The difference is this: if $G$ is the topological structure group of the bundle and $G^\delta$ is $G$ with the discrete topology, then flatness is the same as asserting the bundle, which is classified by a map $X\to BG$ lifts up to homotopy through $BG^\delta$. If it lifts, then a choice of lift $X\to BG^\delta$ factors through $B\pi= P_{\le 1}X$, where $\pi$ is the fundamental group of $X$. This is demanding more than having the map $X\to BG$ factor through $B\pi$. – John Klein May 9 '18 at 0:04

Correction: the definition below is wrong. It isn't true that a 1-flat reduction is the same as a flat reduction. One also needs to require that the map $Z \to BG$ factors through $BG^\delta$, where $G^\delta$ is $G$ with the discrete topology. (Also, one may as well replace $Z$ by the $k$-th Postnikov section in the definition of a $k$-structure).

The following is thoroughly useless general nonsense. Its main problem is that it lacks geometry. Even so, it kind of does the job and it is probably worth mentioning.

Let $G$ be the structure group of your vector bundle. For simplicity, I will assume that $G$ is connected (for example, $G$ could be $U(n)$ or $SO(n)$). Let our vector bundle be classified by a map $f: X\to BG$.

Definition: Let $k \ge 1$ be an integer. A $k$-structure is a pair $(Z,g)$ such that

$\bullet$ $Z$ is a path connected space.

$\bullet$ $Z$ has vanishing homotopy groups above dimension $k$, and

$\bullet$ $g: Z\to BG\,$ is a map.

[To such pairs $(Z_i,g_i)$ for $i=0,1$ and are equivalent there is a (finite chain of) weak homotopy equivalences over $BG$ from $Z_0$ to $Z_1$.]

Example: A 1-structure amounts to a discrete group $H$ and a map $BH\to BG$.

Definition: A $k$-flat reduction of $f$ consists of a $k$-structure $(Z,g)$ and a factorization of $f$ up to homotopy as $\require{AMScd}$ $$\begin{CD} X \to Z @>g>> BG\, . \\ \end{CD}$$

Examples:

(1). A 1-flat reduction of $f$ is the same thing as a flat reduction of the associated vector bundle.

(2). An oriented line bundle over $X$ automatically admits a $2$-flat reduction (this is a tautology, since $BSO(2) = K(\Bbb Z,2)$. In particular, the canonical line bundle over $\Bbb CP^1$ admits a 2-flat reduction but not a 1-flat reduction (since it isn't flat). More generally, a finite Whitney sum of oriented line bundles admits a $2$-flat reduction, since $BSO(2)^{\times j} = K(\Bbb Z^j,2)$.

(3). I don't know of examples in degrees $\ge 2$. However, if one is willing to work instead in the rational homotopy category, there are examples in higher degrees. Here's one: let $X= S^4$. This includes into $BS^3 = BSU(2)$. Rationally, this is a $K(\Bbb Q,4)$. Taking $G = SU(2)$ and $f: X\to BG$ to be the inclusion (which represents the quaternionic line bundle), we see that $f$ is rationally $4$-flat but not rationally $3$-flat.

• Thank you for your answer. I guess this is the correct framework to state correctly my question, even thought I don't see how it answers it. Can you give differential-geometric property that ensures 2-flatness? I see now that the question is harder then I thought, since the construction of $G \mapsto G^\delta$ does not trivially generalize, it don't seem to naturally seat in a tower of finer operations. This makes me less optimistic, but still... – S. carmeli May 9 '18 at 18:44
• @S.carmeli No I cannot give a differential geometric framework (I pointed that out in my post). – John Klein May 9 '18 at 19:45