# How to understand all characteristic classes as generators of classifying space cohomologies

Context: I'm somewhere in the middle of my study of differential geometry and starting to learn about characteristic classes. I like to have a general intuitive understanding of a concept before diving into details, and I've been struggling to assemble that understanding for characteristic classes. Getting an answer to the following couple of questions would be a huge help.

To set the stage, let's restrict to the smooth category and define characteristic classes as natural transformations between

• the (contravariant) functor $$A$$ from the category of smooth manifolds $$\mathrm{Man}^\infty$$ to $$\mathrm{Set}$$ which maps a manifold $$M$$ to its associated set of isomorphism classes of vector bundles with base $$M$$ (and mapping morphisms to pullbacks), and
• a cohomology functor $$B$$ from $$\mathrm{Man}^\infty$$ to $$\mathrm{Set}$$ which maps a manifold to its set of cohomology groups.

We can develop an alternative characterization as follows. Exploiting a bijection between isomorphisms classes of real $$n$$-vector bundles and isomorphism classes of $$GL_n(\mathbb{R})$$-principal bundles, one can construct the classifying space $$BO(n)$$ and show that all real bundles over $$M$$ are pullbacks of the universal bundle $$\gamma_n$$ over this space. This is useful because it means that $$BO(n)$$ represents the functor $$A$$, so by the Yoneda lemma, we can think about characteristic classes for $$n$$-vector bundles as choices of elements in the cohomology of $$BO(n)$$. A similar result holds for complex vector bundles if we replace $$BO(n)$$ with $$BU(n)$$. These ideas are all nicely summarized, for example, in this REU paper.

This can in turn be made more tangible by computing some cohomology rings for these classifying spaces. For example,

• $$H^*(BO(n), \mathbb{Z}_2) \cong \mathbb{Z}_2[x_1, \dots, x_n]$$
• $$H^*(BU(n), \mathbb{Z}) \cong \mathbb{Z}[y_1, \dots, y_n]$$

and you can pick out canonical manifestations of these isomorphisms. Putting together all of the pieces, characteristic classes for real vector bundles with coefficients in $$\mathbb{Z}_2$$ are generated by the $$x_i$$ (the Stiefel-Whitney classes), while for complex vector bundles with coefficients in $$\mathbb{Z}$$ they are generated by the $$y_i$$ (the Chern classes).

Now, there are only so many characteristic classes discussed in the literature: Stiefel-Whitney, Chern, Thom, Euler, Pontryagin, Todd. My questions are as follows:

1. Is there a concise way to see how Thom, Euler, Pontryagin, and Todd classes fit in this framework of looking at the generators of $$H^*(BG,R)$$ for $$G = O(n), U(n)$$ and some choice of ring $$R$$?
2. How come the list of characteristic classes stops there, given the number of rings one can consider? That is, why does it suffice to look at some subset of all of the possible rings, and why do we choose the ones we do?
• Thom class is not a characteristic class. Characteristic classes are cohomology classes in the base of the bundle, whereas Thom class is defined on the total space, it is the pullback of the Euler class by the zero section. Commented Dec 18, 2023 at 18:26
• @MarianoSuárez-Álvarez Thank you for this reference recommendation! It looks like Ch. 20 adopts the perspective I describe, and Th. 20.4.3 and Th. 20.5.5 perform some kind of classification for (semi-)arbitrary rings, but I still don't fully understand what I'm supposed to take away from those results... Commented Dec 19, 2023 at 17:44
• @feynhat Thanks for this clarification: it seems to me that Euler should be the zero-section pullback of Thom, though, not the other way around? (see discussions on other posts like here and here) Commented Dec 19, 2023 at 17:49
• @sheride Yes, you are correct, that's what I meant. Commented Dec 24, 2023 at 20:57

This is a deep and interesting question. Let me be brave and stab at it, also to prompt me to read more on this topic. I may keep editing the post, and hopefully there will be a flow of of discussion going.

So characteristic classes are cohomology classes defined for vector bundles on manifolds, and the bundles have structure groups, that is, the group of linear transformations allowed for the transition functions of the bundle.

There are basically real bundles with structure group as $$GL(n, R)$$ or complex bundles with structure group as $$GL(n, C)$$ (say rank is $$n$$).

Somehow the topology of a Lie group is captured by its maximal compact subgroup. So instead, we consider $$U(n)\subset GL(n, C),\quad O(n)\subset GL(n, R).$$ This is called reduction of the structure group, and geometrically, we are putting a Hermitian metric on a complex bundle and a Riemannian metric on a real bundle (they always exist by partition of unity and local triviality of the bundle).

Therefore the classifying spaces $$BU(n)$$ and $$BO(n)$$ come into play, as mentioned in the original post. These geometrically are some Grassmannians of $$n$$-planes in infinite dimensional spaces. Their cohomology classes become the universal characteristic classes.

Note that there is a close relationship between the real cohomology ring of the classifying space with the ring of adjoint-invariant functions on the corresponding Lie algebras. Geometrically, this is the Chern-Weil theory, where you can plug in the curvature form, which is a Lie-algebra valued 2-form, into the invariant function, to produce a real cohomology class, whose degree is twice the degree of the invariant function. (There could be a more topological approach through the work of Borel, Hirzebruch, etc.)

Added in response To me, the most direct way to see the relation of $$H^*(BG)$$ to $$\text{inv}{\mathfrak g}$$ is through the maximal torus. Let $$T=(S^1)^n\subset G$$ be a maximal torus, where $$n$$ is the rank of $$G$$. Then $$BT = (BS^1)^n = ({\mathbb C}P^\infty)^n$$, and $$H^*(BT, {\mathbb R}) = {\mathbb R}[x_1, \dots, x_n],$$ where $$x_i\in H^2({\mathbb C}P^\infty)$$ is a generator. Furthermore, there is a natural action of the Weyl group $$W$$ of $$G$$ on $$T$$, so an induced action of $$W$$ on the polynomial algebra of cohomology.

We have a natural map $$BT\to BG$$, and so $$H^*(BG)\to H^*(BT)$$. The result is that this is an isomorphism to the invariant part under the Weyl group $$W$$, that is, $$H^*(BG) \cong (H^*(T))^W = ({\mathbb R}[x_1, \dots, x_n])^W = \operatorname{Inv}({\mathfrak g}),$$ since adjoint-invariant polynomials on $${\mathfrak g}$$ restrict to polynomials on $${\mathfrak t}$$ (the Lie algebra of $$T$$) invariant under the Weyl group $$W$$. Maybe see this paper of Borel, especially section 9. Addition ends.

For $$U(n)$$, the invariant functions have degrees $$1, 2, \dots, n,$$ and they are the coefficients of the characteristic polynomial $$\det(\lambda I - A) = \lambda^n + \sum_{i=1}^{n} (-1)^i c_i(A) \lambda^{n-i},\quad A\in {\mathfrak U}(n).$$ These correspond to the Chern classes for complex bundles. The corresponding simple Lie algebra type is $$A$$.

The simple Lie algebra types for $$O(n)$$ are $$B$$ (for $$n$$ odd) and $$D$$ (for $$n$$ even). The adjoint-invariant functions are still coefficients of the characteristic polynomials, but because of skew-symmetry, they only happen in the even degrees. Furthermore, for $$D_m=O(2m)$$, the final one, the determinant has a well-defined square root, called Pfaffian, which has degree $$m$$.

Therefore, the degrees of invariant functions for $$O(n)$$ are \begin{align} 2, 4, \cdots, 2m,\quad \text{if }n=2m+1,\\ 2, 4, \cdots, 2m-2, m, \quad\text{if }n=2m. \end{align}

The corresponding characteristic classes would have dimensions divisible by 4, and they correspond to the Pontryagin classes of a real bundle.

The particular one corresponding to the Pfaffian for $$SO(2m)$$ of degree $$2m$$ now is the Euler class of an even dimensional real bundle.

Therefore, there are relationships relating Chern and Pontryagin classes for a complex bundle regarded real or a real bundle complexified.

These are for real coefficients, and hence the torsion free part of the integral cohomology.

For $$BU(n)$$, there is no torsion. For $$BO(n)$$, there are only 2-torsions. The corresponding story would be the Stiefel-Whitney classes.

There does not seem to be a need for other coefficient rings, by the universal coefficient theorem.

At the end, I want to say that the Todd class is not a basic characteristic class, but rather a series in the cohomology ring, defined using the help of the splitting principle. It automatically terminates for a finite dimensional manifold, and is expressible in terms of Chern classes. Similarly, we have the Chern character.

I will stop now.

• Thanks for this really rich and thoughtful answer! I have a few follow-up questions, for you or anyone else: Commented Dec 25, 2023 at 19:38
• I added some words to the post and added a link to a paper of Borel to address the relationship to invariant functions. There is more about that Stiefel-Whitney class in that paper at the end of Sect 9. Commented Dec 25, 2023 at 23:21
• To me, the idea is that since $H^*(BO(n), {Z})$ has $Z/2$ torsion, why don't we study its cohomology with $Z/2$ coefficients? It would then have a simpler structure. Think of $n=1$, then $BO(1)=B(Z/2) = RP^\infty$ and $H^*(RP^\infty, Z) = Z/2$ for even dimensions except 0, but $H^*(RP^\infty, Z/2) = Z/2$ for all dimensions and it is a polynomial algebra $Z/2[x_1]$. Eventually the $H^*(BO(n), {Z}/2)$ is a polynomial algebra on the Stiefel-Whitney classes $w_1, \dots, w_n$. It is surprisingly hard to totally determine $H^*(BO(n), {Z})$ integrally, and only done later by E. Brown. Commented Dec 25, 2023 at 23:27
• It is not if the base manifold $M$ is orientable, it is if the bundle $E\to M$ is orientable. You can have an non-orientable bundle over an orientable manifold, say the Mobius band. If your bundle is non-orientable, then your classifying map only maps to $BO(n)$, but otherwise you lift to $BSO(n)$. (Again, $BSO(n)\to BO(n)$). The cohomology of $BSO(n)$ is richer, and has the Euler class. Otherwise, you don't have it as an integral class, and only the $Z/2$ shadow. Commented Dec 27, 2023 at 8:41
• See this paper of Edgar Brown for complete calculation. jstor.org/stable/2044298. In Thm 1.5, you see the integral cohomology of $BSO(n)$, and you have $X_n$, the Euler class. (Also note that if $n$ is even, $p_{n/2}=X_n^2$, that is what I said that the determinant is the square of the Pfaffian for matrices in $D_n$. But in Thm 1.6 for $BO(n)$, the Euler class is gone. Commented Dec 27, 2023 at 8:46