How does one go from Chern--Weil to cohomology classes on BGL(n,C)?

Given a manifold $M$ and a complex vector bundle $V$ over $M$, we can equip $V$ with a $\mathfrak g\mathfrak l_n(\mathbb C)$ connection and from this connection compute a closed differential $2k$-form (from the curvature of the connection) which thus determines an element $c_k(V)\in H^{2k}(M,\mathbb C)$ (by deRham theory). This value is independent of the connection chosen. If $f:N\to M$ is a smooth map, then $c_k(f^\ast V)=f^\ast c_k(V)$.

I've often heard that Chern--Weil theory gives cohomology classes $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$. However, if we take Chern--Weil theory to mean the above boxed summary, then this does not seem obvious to me unless we use the fact that we can choose $B\operatorname{GL}_n(\mathbb C)$ to be a direct limit of manifolds (namely Grassmannians).

My question is whether there is more abstract way of constructing the classes $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$ from Chern--Weil theory, using only the fact that $B\operatorname{GL}_n(\mathbb C)$ is the classifying space of complex vector bundles. I am fine with assuming that $B\operatorname{GL}_n(\mathbb C)$ is a direct limit of finite CW complexes, but of course, Chern--Weil theory does not obviously define Chern classes for vector bundles over CW complexes.

• Every CW complex is homotopic to a manifold. Since $BGL_n(\mathbb C)$ is defined only up to homotopy, those statements are equivalent. The issue seems to me to be showing that viewing $BGL_n(\mathbb C)$ as a direct limit of manifolds in different ways doesn't change the result. Nov 10 '12 at 21:30
• You said "I've often heard that Chern--Weil theory gives cohomology classes $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$".. I think there is some confusion (either I am confused or you are confused).. Chern-Weil theory gives cohomology classes in $H^{2k}(M,\mathbb{C})$ or $H^{2k}(M,\mathbb{C})$... Given topological/Lie group $G$ we have classifying space $BG$ and what is called universal bundle $EG\rightarrow BG$... Given a bundle $P\rightarrow M$, there exists a unique map (upto homotopy) $M\rightarrow BG$ that determines $P\rightarrow M$ (continued) Jan 3 '19 at 4:10
• This map $M\rightarrow BG$ then gives a map on cohomology $H^*(BG)\rightarrow H^*(M)$.. Images of cohomology classes in $H^*(BG)$ in $H^*(M)$ are called (among other ways to define) Characteristic classes... I think this is not called as Chern-Weil theory.. Jan 3 '19 at 4:12

Why are you not happy with using Grassmanians as in:

A related approach that might interest you is given in Dupont's book

http://www.amazon.com/Curvature-Characteristic-Classes-Lecture-Mathematics/dp/3540086633

using simplicial manifolds. A simplicial manifold $X$ is a sequence of manifolds $\lbrace X_n \rbrace$ and various maps between them. From a simplicial manifold you can construct a topological space called its realisation. This is how you define $EG \to BG$. Although it isn't a manifold you can realise its topology using the finite dimensional spaces $X_n$ which is how you tie things back to the statement of Chern-Weil theory given in the question. In this example the simplicial space is also the one arising in the bar construction and Milnor's join construction of $EG \to BG$.

Expanding on Will Sawin's comment:

Every finite CW complex is homotopy equivalent to a finite simplicial complex (by an approximation argument) and a simplicial complex is homotopy equivalent to a manifold (take a regular neighborhood of the natural geometric realization in $\mathbb R^{\#\text{vertices}}$). The naturality of the "very fine approximation of a CW complex by a simplicial complex" means that we get well-defined classes $c_k(V)\in H^{2k}(X,\mathbb C)$ for any complex vector bundle $V$ over a finite CW complex $X$ (alternatively, take two small manifold "extensions" of $X$; we can embed these in a third manifold extension, so they give the same cohomology class $c_k(V)$). These classes satisfy $f^\ast c_k(V)=c_k(f^\ast V)$ for $f:Y\to X$ where $Y$ and $X$ are both finite CW complexes because we can extend the map $f$ to the approximation/regular neighborhood.

Now we have $c_k$ as characteristic classes of complex vector bundles over finite CW complexes which satisfy naturality (pullback), and so abstract nonsense about representability implies they come from unique $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$ (which we can construct by filtering $B\operatorname{GL}_n(\mathbb C)$ by finite CW complexes in any way we like).

I can think of several versions, besides those that have been mentioned in the earlier answers:

1. You can in fact construct $B GL_n (\mathbb{C})$ as a manifold, but of course an infinite-dimensional one. Start with a countably dimensional Hilbert space $H$. Look at the Stiefel manifold $V_n (H)$ of linear embeddings $\mathbb{C}^n \to H$. Being an open subset of $H^n$, it is a secound-countable Hilbert manifold. It can be proven directly that $V_n (H)$ is contractible and that the quotient $V_n (H) \to V_n (H)/CL_n (\mathbb{C})$ is a principal bundle. The proof for the second fact is more or less the same as in the finite-dimensional case, the first fact in proven in an Eilenberg-swindly way.

Now second-countable Hilbert manifolds are a particularly simple type of infinite dimensional manifolds. They have smooth partitions of unity, and as a consequence the proof of the de Rham theorem (for example the one given in Bredon' book) can be carried out without any substantial change.

The theory of connections on principal bundle works in the same way for Hilbert manifolds as base space (if the fibre is a finite-dimensional Lie group). So you get a Chern-Weil homomorphism in the universal case.

If you replace $GL_n (\mathbb{C})$ by any closed subgroup $G$, then $V_n (H) \to V_n (H)/G$ is a Hilbert manifold model for $BG$; and the same arguments as before work.

1. There exist a simplicial set model for $BG$, classifying $G$-bundles with connection. The set of $p$-simplices is the set of all triples $(P,\pi,\omega)$, where $\pi:P \to \Delta^p$ is a smooth $G$ principal bundle and $\omega$ a connection $1$-form on $P$. To turn it into a set (and to make the simplicial structure precise), you take those $P$ with $P \subset \Delta^p \times \mathbb{R}^{\infty}$ (as a manifold).

By the ordinary Chern-Weil construction, you get a simplicial differential form on this simplicial set. What do I mean by this? Observe that forms on the standard simplices assemble to a simplicial d.g.a: $q \mapsto \mathcal{A}^{\ast} (\Delta^q)$. For a simplicial set $X_{\bullet}$, you look at the set of simplicial set maps $X_{\bullet} \to \mathcal{A}^q (\Delta^{\bullet})$; which is a vector space, and for varying $q$ gives a d.g.a.; which by definition is the simplicial de Rham complex.

There are two things to be proven here: that the simplicial set I described is indeed $BG$ and that the simplicial de Rham complex computes the real cohomology. The second one you find in the book ''Rational homotopy theory'' by Felix, Halperin, Thomas. For the first part, I do not have a reference; this is folklore.