How does one go from Chern--Weil to cohomology classes on BGL(n,C)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:53:50Z http://mathoverflow.net/feeds/question/112024 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112024/how-does-one-go-from-chern-weil-to-cohomology-classes-on-bgln-c How does one go from Chern--Weil to cohomology classes on BGL(n,C)? unknown (google) 2012-11-10T21:06:34Z 2012-11-11T15:31:28Z <p>Let's assume we start with Chern--Weil theory in the following form:</p> <blockquote> <p>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 $c_k(f^\ast V)=f^\ast c_k(V)$.</p> </blockquote> <p>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).</p> <p>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.</p> http://mathoverflow.net/questions/112024/how-does-one-go-from-chern-weil-to-cohomology-classes-on-bgln-c/112038#112038 Answer by unknown (google) for How does one go from Chern--Weil to cohomology classes on BGL(n,C)? unknown (google) 2012-11-10T22:40:20Z 2012-11-10T22:40:20Z <p>Expanding on Will Sawin's comment:</p> <p>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 <code>$\mathbb R^{\#\text{vertices}}$</code>). 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.</p> <p>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).</p> http://mathoverflow.net/questions/112024/how-does-one-go-from-chern-weil-to-cohomology-classes-on-bgln-c/112066#112066 Answer by Michael Murray for How does one go from Chern--Weil to cohomology classes on BGL(n,C)? Michael Murray 2012-11-11T08:09:13Z 2012-11-11T12:24:42Z <p>Why are you not happy with using Grassmanians as in:</p> <p><a href="http://en.wikipedia.org/wiki/Classifying_space_for_U(n" rel="nofollow">http://en.wikipedia.org/wiki/Classifying_space_for_U(n</a>) ?</p> <p>A related approach that might interest you is given in Dupont's book</p> <p><a href="http://www.amazon.com/Curvature-Characteristic-Classes-Lecture-Mathematics/dp/3540086633" rel="nofollow">http://www.amazon.com/Curvature-Characteristic-Classes-Lecture-Mathematics/dp/3540086633</a></p> <p>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 $EG \to BG$.</p> http://mathoverflow.net/questions/112024/how-does-one-go-from-chern-weil-to-cohomology-classes-on-bgln-c/112077#112077 Answer by Johannes Ebert for How does one go from Chern--Weil to cohomology classes on BGL(n,C)? Johannes Ebert 2012-11-11T14:59:49Z 2012-11-11T15:31:28Z <p>I can think of several versions, besides those that have been mentioned in the earlier answers:</p> <ol> <li>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.</li> </ol> <p>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. </p> <p>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.</p> <p>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.</p> <ol> <li>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).</li> </ol> <p>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.</p> <p>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.</p>