TITLE: Analogy between Stiefel-Whitney and Chern classes
QUESTION [16 upvotes]: There is a clear similarity between Stiefel-Whitney and Chern classes, if one replaces base field $\mathbb R$ with $\mathbb C$, coefficient ring $\mathbb Z/2$ with $\mathbb Z$ and scales the grading by a factor of $2$. For instance, both can be defined by the same axioms (functoriality, dimension, Whitney sum, value on the tautological bundle over $\mathbb P^1$).
Is there a deep reason behind this correspondence? The best explanation I have so far is the structure of classifying spaces. Being Grassmanians, they admit Schubert cell decomposition (which is essentially an algebraic fact). For cohomology of complex Grassmanians, differentials vanish for dimensionality reasons, and for real ones they vanish when reduced mod $2$.
There is a number of similar phenomena, for instance, $BO(1,\mathbb R) = K(\mathbb Z/2, 1)$ and $BU(1) = K(\mathbb Z, 2)$ which says that in both cases, topological line bundles are completely determined by their first characteristic class.
Also, I have been told that Thom polynomials for Thom-Boardman singularities of maps between real or complex manifolds have the same coefficients when expressed in $w_i$ and $c_i$. Can these facts be explained in a similar way?

REPLY [7 votes]: Any rank $n$ real vector bundle $E\to X$, $X$ compact $CW$-complex is $\newcommand{\bZ}{\mathbb{Z}}$ is $\bZ/2$-oriented and, as such it has a $\bZ/2$-Thom class $\tau_E\in H^n_{cpt}(E,\bZ/2)$. Then 
$$w_n(E)=\zeta^*\tau_E\in H^n(X,\bZ/2),$$ where $\zeta:X\to E$ is the zero-section
Any complex vector bundle $E\to X$, $X$ compact $CW$-complex of complex rank $n$  is  $\bZ$-oriented and, as such it has a $\bZ$-Thom class $\tau_E\in H^{2n}_{cpt}(E,\bZ)$. (Note that $2n$ is the real rank of $E$.) Then 
$$c_n(E)=\zeta^*\tau_E\in H^{2n}(X,\bZ).$$
Thus in both cases the top  Stieffel-Whitney class and the top Chern class are Euler classes, with different choices of coefficients. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bP}{\mathbb{P}}$
To get the rest  of the   Stieffel-Whitney/Chern classes on then needs to rely  on some basic facts
$$ H^\bullet(\bR\bP^n,\bZ/2)\cong\bZ/2[w]/(w^n+1),\;\; H^\bullet(\bC\bP^n,\bZ)\cong\bZ[c]/(c^n+1),\tag{1}
$$
where $w\in H^1(\bR\bP^n,\bZ/2)$, $c\in H^2(\bC\bP^n,\bZ)$ are the Euler classes of the (duals) of the tautological line bundles.
These results  suffice to construct the   Stieffel-Whitney/Chern classes. This is the approach pioneered by Gronthendieck for the construction of Chern classes. For details see Chapter 5 of these notes.. As explained in Example 4.3.5. of these notes  duality, under the guise  of Thom isomorphism is also responsible for  the isomorphisms (1).   
One could claim that duality or Thom isomorphism is what makes things work. What is behind Thom isomorphism? As  described in Bott-Tu,  this follows from two basic facts about cohomology. The first is the Poincare lemma with compact supports
$$H^k_{cpt}(\bR^n, G)=\begin{cases} 0, & k\neq n,\\
G, &k=n.
\end{cases}
$$
and the second is the Mayer-Vietoris principle which, roughly specking says that one can recover the cohomology of an union from the cohomology of its parts. View this as a local-to-global principle, a way a patching local data to obtain  global information.  The orientability condition is the one that allows the local-to-global transition.

REPLY [4 votes]: Let me try a high-brow answer using equivariant stable homotopy theory. 
By the stable Thom isomorphism, the integral (co)homology of $BU$ agrees with that of $MU$; likewise the $\mathbb{Z}/2$-(co)homology of $BO$ agrees with that of $MO$. 
Let $H\underline{\mathbb{Z}}$ be the $C_2$-equivariant Eilenberg--Mac Lane spectrum for the constant Mackey functor and let $M\mathbb{R}$ be the Real cobordism spectrum. Using the theory of Hu and Kriz, we see that $H\underline{\mathbb{Z}}$ is Real oriented and thus the $RO(C_2)$-graded groups $H\underline{\mathbb{Z}}^{\bigstar}M\mathbb{R}$ agree with $H\underline{\mathbb{Z}}^{\bigstar}[[ \overline{c}_1, \overline{c}_2, \dots ]]$, where $\overline{c}_i \in H\underline{\mathbb{Z}}^{i+i\sigma}M\mathbb{R}$. The $\overline{c}_i$ define thus maps $M\mathbb{R} \to \Sigma^{i+i\sigma}H\underline{\mathbb{Z}}$, which forget to the usual Chern classes $c_i\colon MU \to H\mathbb{Z}$. 
The geometric fixed points of $M\mathbb{R}$ are $MO$, while those of $H\underline{\mathbb{Z}}$ are $\prod_{k\geq 0}\Sigma^{2k}H\mathbb{Z}/2$ and thus come with a projection $p\colon \Phi^{C_2}H\underline{\mathbb{Z}} \to H\mathbb{Z}/2$. Thus combining $\Phi^{C_2}$ with $p$, the $\overline{c}_i$ define maps $MO \to \Sigma^iH\mathbb{Z}/2$, which we claim to be the $w_i$. 
I have not thought through this, but I guess one way to show a thing like this is to use homology instead. We have $\pi_{\bigstar}H\underline{\mathbb{Z}} \otimes M\mathbb{R} \cong \pi_{\bigstar}H\underline{\mathbb{Z}}[\overline{b}_1, \dots]$. Write $\pi_*\Phi^{C_2}H\underline{\mathbb{Z}} = \mathbb{Z}/2[u]$. Then 
$$H_*(MO; \mathbb{Z}/2)[u]\cong (\Phi^{C_2}H\mathbb{Z}/2)_*(MO) \cong \pi_* \Phi^{C_2}(H\underline{\mathbb{Z}} \otimes M\mathbb{R}) \cong \mathbb{Z}/2[u][\Phi^{C_2}\overline{b}_1, \dots] $$
Here, we use that we can get the geometric fixed points by inverting one element $a_{\sigma}$ and that $M\mathbb{R}$ is of finite type. Killing the $u$ on both sides, gives that we obtain indeed the whole homology of $MO$ by this geometric fixed points construction. Thus, I expect that by playing around with dual (co)homology classes, one should get indeed that one obtains the $w_i$ from the $\overline{c}_i$. 
(Why was it natural to use the projection $p$? Indeed, when we apply geometric fixed points to the $\overline{c}_i$ all other components are $0$. This we can see by lifting the $\overline{c}_i$ to $BP\mathbb{R}^{\bigstar}M\mathbb{R}$ and observing that the geometric fixed points of $BP\mathbb{R}$ are just $H\mathbb{Z}/2$.)