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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?

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  • $\begingroup$ что касается предпоследнего абзаца -- может быть, просто повезло, что у $S^0$ и $S^1$ нет высших гомотопических групп?.. $\endgroup$ – Andrey Ryabichev May 12 at 7:50
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Here is one way I like to think of the analogy.

The maximal torus of diagonal matrices $T^{n} \subset U(n)$ gives a map $BT^n \to BU(n)$ which on integral cohomology gives an isomorphism from $H^{\ast}(BU(n))$, which is a polynomial algebra on $n$ generators of degrees $2i$, $1\leq i\leq n$, to $H{\ast}(BT^n)^{\Sigma_n}$, which is the polynomial algebra on the symmetric polynomials $\sigma_i$ in the $n$ standard degree 2 generators. The Chern class $c_i$ is the element of the domain that maps to $\sigma_i$.

The maximal $2$-torus of diagonal matrices $(C_2)^{n} \subset O(n)$ gives a map $BC_2^{n} \to BO(n)$ which on mod $2$ cohomology gives an isomorphism from $H^{\ast}(BO(n))$, which is a polynomial algebra on $n$ generators of degrees $i$, $1\leq i\leq n$, to $H^{\ast}(BC_2^n)^{\Sigma_n}$, which is the polynomial algebra on the symmetric polynomials $\sigma_i$ in the $n$ standard degree $1$ generators. The Stiefel-Whitney class $w_i$ is the element of the domain that maps to $\sigma_i$.

Thinking of $BT^{n}$ as $(CP^{\infty})^{n}$ and $BC_2^{n}$ as $(RP^{\infty})^{n}$ may help.

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    $\begingroup$ I like to imagine that these maps are part of the explanation of splitting principle, since $BT^n$ and $B(C_2)^n$ are classifying spaces for split bundles. This makes generators of their cohomology a sort of "universal Chern roots". $\endgroup$ – Troshkin Michael May 13 at 0:47
  • $\begingroup$ I agree. I gave a new proof of the splitting priniciple in math.uchicago.edu/~may/PAPERS/Split.pdf that emphasizes that point of view. $\endgroup$ – Peter May May 13 at 17:39
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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.

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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 $H\mathbb{Z}/2$.)

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  • $\begingroup$ Could you please add few words to mention what $\Phi$ is? Hu and Kriz do explain it referring to Lewis-May-Steinberger, but... $\endgroup$ – მამუკა ჯიბლაძე May 11 at 11:09
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    $\begingroup$ Sure. $\Phi = \Phi^{C_2}: \mathrm{Sp}^{C_2} \to \mathrm{Sp}$ is the geometric fixed point functor. It is characterized by the following properties: 1) $\Phi^{C_2}\Sigma^{\infty}X = \Sigma^{\infty}X^{C_2}$; 2) $\Phi^{C_2}$ is symmetric monoidal; 3) $\Phi^{C_2}$ commutes with homotopy colimits. If you want to compute the geometric fixed points of $M\mathbb{R}$ you can argue like follows: $M\mathbb{R}_{k+k\sigma}$ is the Thom space of the universal bundle on $BU(n)$ with the complex conjugation action. Its fixed points are the corresponding Thom space over $BO(n)$... $\endgroup$ – Lennart Meier May 11 at 11:43
  • $\begingroup$ ...further $M\mathbb{R} \simeq \mathrm{hocolim} \Sigma^{-k-k\sigma} M\mathbb{R}_{k+k\sigma}$ (this is a general fact, see e.g. Section 2 of Hill--Hopkins--Ravenel's Kervaire paper). As $\Phi^{C_2}$ is symmetric monoidal, it also preserves duals and thus takes $\Sigma^{-k-k\sigma}(-) = D(S^{k+k\sigma}) \otimes -$ to $\Sigma^{-k}$ as the $C_2$-fixed points of $S^{k+k\sigma}$ are $S^k$. Thus, we see that the geometric fixed points of $M\mathbb{R}$ are the homotopy colimit over $k$-fold desuspensions of Thom spaces over $BO(n)$, i.e. exactly $MO$. $\endgroup$ – Lennart Meier May 11 at 11:47
  • $\begingroup$ Is there a particular reason why you use $M\mathbb{R}$ instead of $\Sigma^\infty BU_{\mathbb{R}}$ or is that only for convenience of computation? $\endgroup$ – Denis Nardin May 11 at 16:07
  • $\begingroup$ @DenisNardin You're right that it would be actually more natural to use the latter because one doesn't have to use the Thom isomorphism. It was more out of habit. $\endgroup$ – Lennart Meier May 11 at 17:42
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$\newcommand{\Z}{\mathbf{Z}}$The Chern classes may be viewed as a map $c: \mathrm{BU}(n) \to \prod_{i=1}^n K(\Z,2i)$. Call the target $X_n$. Complex conjugation defines a $\Z/2$-action on $\mathrm{BU}(n)$, whose fixed points are $\mathrm{BO}(n)$. Give $X_n$ the $\Z/2$-action defined by viewing $K(\Z, 2n)$ as $\Omega^\infty(\Sigma^{2n,n} \mathrm{H}\Z)$, where $S^{2n,n} = S^{n + n\sigma}$ (here, $S^\sigma$ is the one-point compactification of the sign representation of $\Z/2$), and $\mathrm{H}\Z$ is the Eilenberg-Maclane spectrum associated to the constant Mackey functor. I forgot to write this in the previous version of this answer, but (as in Lennart's answer) there is a projection map $\Phi^{C_2} \mathrm{H}\Z\to \mathrm{H}\mathbf{F}_2$. The map $c$ is $\Z/2$-equivariant, and taking $\Z/2$-fixed points and composing with the above projection yields the map $\mathrm{BO}(n) \to \prod_{i=1}^n K(\Z/2, i)$ given by the Stiefel-Whitney classes. I haven't done this explicitly, but to check that these are in fact the Stiefel-Whitney classes, one reduces to the case of line bundles (by the splitting principle); then, it follows from the fact that $\mathrm{BU}(1)$ is equivariantly equivalent to $\Omega^\infty \Sigma^{2,1} \mathrm{H}\Z$ (and this identification is given by the first Chern class), and taking fixed points and projecting gives the identification of $\mathrm{BO}(1)$ with $\Omega^\infty \Sigma \mathrm{H}\mathbf{F}_2$ (where this identification is given by the first Stiefel-Whitney class). [This is essentially the same as Lennart's answer.]

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    $\begingroup$ It's not true that $\Sigma^nH\mathbb{Z}/2$ is the homotopy fixed points nor the genuine fixed points, nor the geometric fixed points of $\Sigma^{(1+\sigma)n}H\mathbb{Z}$. In which sense did you want to take the fixed points? $\endgroup$ – Denis Nardin May 11 at 8:56
  • $\begingroup$ @DenisNardin You're right. Edited. (Had excluded the step made explicit in Lennart's reply.) $\endgroup$ – skd May 11 at 15:36

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