2 edited body

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 $V_n (H) \to V:n V_n (H)/G$ 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.

1

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.