Are Chern classes well defined up to contractible choice?

Question

The Chern classes are, by definition, cohomology classes. And cocycle representatives of the Chern classes are not unique.

But it might be the case that cocycle representatives of the Chern classes are unique up to contractible choice.

The following kind-of-answer is not what I'm looking for:
One obvious way to make Chern classes of vector bundles unique up to contractible choice is to pick, once and for all, cocycle representatives of the universal Chern classes $c_n\in H^{2n}(BU,\mathbb Z)$. But that's not a good answer, because this depends on making a choice of lifting someting in $H^{2n}(BU,\mathbb Z)$ to something in $Z^{2n}(BU,\mathbb Z)$. I don't want to have to make any arbitrary choices.

So, what I'm asking is:
is there a best way to pick (up to contractible choice) a cocycle representative of the universal Chern classes $c_n\in H^{2n}(BU,\mathbb Z)$?

Answer 1

I think that the best approach is to use the construction in the 1993 paper "Algebraic cycles and infinite loop spaces", which I learned about from this answer by Jeremy Hahn. To explain the context, recall that people now mostly use orthogonal spectra as a setting for stable homotopy theory (if they have any concrete foundation at all). One can compare orthogonal spectra with based spaces, but it is often more convenient to compare them with based orthogonal spaces, i.e. functors from $\mathcal{L}$ to based spaces, where $\mathcal{L}$ is the category of finite-dimensional real inner product spaces and linear isometric embeddings. There is a Quillen equivalence from orthogonal spaces to spaces given by $X\mapsto\text{colim}_nX(\mathbb{R}^n)$. This sends commutative monoids in based orthogonal spaces (which May used to call $\mathcal{J}_*$-functors) to infinite loop spaces. There is also a unitary version of all this based on finite-dimensional complex vector spaces equipped with hermitian inner products. The cited paper defines a morphism $c\colon P\to Q$ in the category of commutative monoids in based unitary spaces. Here $P$ corresponds to $\mathbb{Z}\times BU$, and $Q$ corresponds to $\prod_nK(\mathbb{Z},2n)$ and $c$ corresponds to the total Chern class. The definition of $Q$ involves Chow varieties, which involve formal $\mathbb{N}$-linear combinations of subvarieties of projective space. The definition of $P$ involves linear subvarieties and $c$ is just an inclusion.

Answer 2

Since the mapping space $map (BU(n);K(\mathbb{Z},2k))$ does not have contractible components, you need to make non-contractible choices somewhere. I understand Andre's question in the sense that you want as few choices as possible, and that the choices look more 'natural' than just picking cocycles for all Chern classes at random.

Here is an attempt for a construction for $\it{integral}$ coefficients. It does not quite work, but if you pass to $\it{rational}$ coefficients, the construction depends only on two noncontractible universal choices.

-(a) Fix an actual map $MU \to H \mathbb{Z}$ of spectra, implementing the Thom class of complex vector bundles.

-(b) Fix a cycle representing the generator of $H_2 (BU(1);\mathbb{Z})$ dual to $c_1 (L_1)$.

Choice (a) should give cocyle representatives for the Thom class of all of the universal bundles $L_n \to BU(n)$.

As for an arbitrary rank $n$ vector bundle $V \to X$, the space of bundle maps (fibrewise isomorphisms) $V \to L_n$ is contractible, the construction yields cocycle representatives for the Thom classes of arbitrary vector bundles, unique up to contractible choice once (a) is fixed.

Since the top Chern class $c_n (V)$ is given by pulling back the Thom class along the zero section, we obtain cocycle representatives for the top Chern class, unique up to contractible choice.

The lower Chern classes $c_k \in H^{2k}(BU(n);\mathbb{Z})$ can be defined by the following recipe: consider the external tensor product $L_n \boxtimes L_1 \to BU(n) \times BU(1)$. We can write $$c_n (L_n \boxtimes L_1) = \sum_{k=0}^{n} c_k (L_n) \boxtimes c_1 (L_1)^{n-k} $$ using the Kuenneth formula. Using the slant product, we can rewrite this formula as $$ c_k (L_n) := c_n (L_n \boxtimes L_1) / b_{n-k} $$ where $b_{n-k} \in H_{2n-2k}(BU(1))$ is the homology class with $\langle c_1 (L_1)^{n-k},b_{n-k} \rangle =+1$.

To turn this into something on the chain level, observe that the slant product is defined on the (co)chain level, depending on the (contractible) choice of an Eilenberg-Zilber map. Furthermore, we need cycles representing $b_{n-k}$.

With rational coefficients, such cycles can be defined using only the choice a cycle representing $b_1$, since you can recover $b_{n-k}$ as the $n-k$-fold Pontrjagin product of $b_1$, divided by $\frac{1}{(n-k)!}$.

Answer 3

In “Chern–Weil forms and abstract homotopy theory”, Freed and Hopkins compute the de Rham cohomology of $\def\B{{\sf B}}\B_∇ G$, the moduli stack (in groupoids) of principal $G$-bundles with connections on the site of smooth manifolds.

The de Rham cohomology of $\B_∇ G$ turns out to be isomorphic to the graded algebra of $G$-invariant polynomials (placed in even degrees) on the Lie algebra of $G$, with the zero differential.

This provides a completely canonical cocycle for Chern classes (taking $\def\U{{\rm U}} G=\U(n)$) as a morphism of stacks in groupoids on the site of smooth manifolds. This canonical cocycle can be used to produce cocycles in other models. For example, for a model using simplicial sets only, we can evaluate the above simplicial presheaves on smooth simplices $Δ^m$ and take the diagonal simplicial set, which yields a map of simplicial sets whose domain is weakly equivalent to $BG$ and the codomain is weakly equivalent to $\def\K{{\rm K}}\def\Z{{\bf Z}}\K(\Z,2k)$; the resulting simplicial map is a classifying map for the $k$th Chern class.