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An interesting thing happened the other day. I was computing the Stiefel-Whitney numbers for $\mathbb{C}P^2$ connect sum $\mathbb{C}P^2$ to show that it was a boundary of another manifold. Of course, one can calculate the signature, check that it is non-zero and conclude that it can't be the boundary of an oriented manifold. I decided it might be interesting to calculate the first and only Pontrjagin number to check that it doesn't vanish. I believe Hirzebruch's Signature Theorem can be used to show that it is 6, but I was interested in relating the Stiefel-Whitney classes to the Pontrjagin classes.

I believe one relation is

$p_i (\mathrm{mod} 2) \equiv w_{2i}^2$ (pg. 181 Milnor-Stasheff)

So I went ahead and did a silly thing. I took my first Chern classes of the original connect sum pieces say 3a and 3b, used the fact that the inclusion should restrict my 2nd second "Stiefel-Whitney Class" (scare quotes because we haven't reduced mod 2) on each piece to these two to get $w_2(connect sum)=(3\bar{a},3\bar{b})$. I can use the intersection form to square this and get $3\bar{a}^2+3\bar{b}^2=6c$ since the top dimensional elements in a connect sum are identified. Evaluating this against the fundamental class gives us exactly the first Pontrjagin number! This is false. Of course this is wrong because it should be 9+9=18 as pointed out below. This does away with my supposed miracle example. My Apologies!

This brings me to a broader question, namely of defining Stiefel-Whitney Classes over the integers. This was hinted at in Ilya Grigoriev's response to Solbap's question when he says

On thing that confuses me: why are the pullbacks of the integer cohomology of the real Grassmanian never called characteristic classes?

Of course the natural reason to restrict to $\mathbb{Z}/2$ coefficients is to get around orientability concerns. But it seems like if we restrict our orientation to orientable bundles we could use a construction analogous to those of the Chern classes where Milnor-Stasheff inductively declare the top class to be the Euler class, then look at the orthogonal complement bundle to the total space minus its zero section and continue. I suppose the induction might break down because the complex structure is being used, but I don't see where explicitly. If someone could tell me where the complex structure is being used directly, I'd appreciate it. Note the Euler class on odd dimensional fibers will be 2-torsion so this might produce interesting behavior in this proposed S-W class extension.

Another way of extending Stiefel-Whitney classes would be to use Steenrod squares. Bredon does use Steenrod powers with coefficient groups other than $\mathbb{Z}/2$ (generally $\mathbb{Z}/p$ $p\neq 2$), but this creates awkward constraints on the cohomology groups. Is this an obstruction to extending it to $\mathbb{Z}$ coefficients? It would be interesting to see what these two proposed extensions of S-W classes do and how they are related.

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Stiefel-Whitney classes were originally defined as obstruction classes to sections of Stiefel-bundles of a manifold. If you take the pull back of integral homology you no longer get these kinds of obstructions. Your questions seems to be more aimed towards "why do we call the pull-back of certain cohomology classes characteristic, and others not?" –  Ryan Budney Feb 27 '10 at 23:08
    
That's funny - I was just about to ask this question when I saw yours. –  Ilya Grigoriev Feb 28 '10 at 2:26
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5 Answers

up vote 14 down vote accepted

I'm grateful to Allen Hatcher, who pointed out that this answer was incorrect. My apologies to readers and upvoters. I thought it more helpful to correct it than delete outright, but read critically.

If $X$ and $Y$ are cell complexes, finite in each degree, and two maps $f_0$ and $f_1\colon X\to Y$ induce the same map on cohomology with coefficients in $\mathbb{Q}$ and in $\mathbb{Z}/(p^l)$ for all primes $p$ and natural numbers $l$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. To see this, write $H^n(Y;\mathbb{Z})$ as a direct sum of $\mathbb{Z}^{r}$ and various primary summands $\mathbb{Z}/(p^k)$, and note that the summand $\mathbb{Z}/(p^k)$ restricts injectively to the mod $p^l$ cohomology when $l\geq k$. One can take only those $p^l$ such that there is $p^l$-torsion in $H^\ast(Y;\mathbb{Z})$. (I previously claimed that one could take $l=1$, which on reflection is pretty implausible, and is indeed wrong.)

We can try to apply this to $Y=BG$, for $G$ a compact Lie group. For example, $H^{\ast}(BU(n))$ is torsion-free (and Chern classes generate the integer cohomology), and so rational characteristic classes suffice. In $H^{\ast}(BO(n))$ and $H^{\ast}(BSO(n))$ there's only 2-primary torsion. That leaves the possibility that the mod 4 cohomology contains sharper information than the mod 2 cohomology. It does not, because, as Allen Hatcher has pointed out in this recent answer, all the torsion is actually 2-torsion.

It's sometimes worthwhile to consider the integral Stiefel-Whitney classes $W_{i+1}=\beta_2(w_i)\in H^{i+1}(X;\mathbb{Z})$, the Bockstein images of the usual ones. These classes are 2-torsion, and measure the obstruction to lifting $w_i$ to an integer class. For instance, an oriented vector bundle has a $\mathrm{Spin}^c$-structure iff $W_3=0$.

[I'm sceptical of your example in $2\mathbb{CP}^2$. So far as I can see, $3a+3b$ squares to 18, not 6, and indeed, $p_1$ is not a square.]

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Thank you for pointing out that! I guess I wanted to see 6 and I saw 6. –  Justin Curry Feb 27 '10 at 23:13
    
This is a very interesting answer; I really enjoyed reading it. Also, it also made quite a few more questions appear in my mind. First, you imply that the cohomology of the Grassmanian with Q coefficients corresponds to the Pontryagin classes, is that true? Secondly, did people ever try to calculate the cohomology of the Grassmanian with Z/p coefficients where p is not 2? Are the resulting classes interesting? Are they related to Steenrod powers the same way as the Stiefel-Whitney classes related to Steenrod Squares? Oh, and can you see them geometrically (e.g. like obstructions) in any way? –  Ilya Grigoriev Feb 28 '10 at 3:01
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@Ilya: Thanks - the cohomology mod 2-torsion of a real (unoriented) Grassmannian is polynomial in the Pontraygin classes; see Hatcher, "Vector bundles and K-theory", Theorem 3.16. Over Z/p for p odd, you'll therefore just get the polynomials in the mod p Pontryagin classes. As you say, one should get examples of mod p char classes by applying the Steenrod p-powers in the Thom space; maybe it would be fun to work out a formula for them. To say that a char class is zero mod p (i.e. divisible by p) is the sort of thing that the signature theorem or other index theorems sometimes tell us. –  Tim Perutz Feb 28 '10 at 4:33
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Of course any cohomology class of $BG$, with any coefficients, serves as a characteristic class of $G$-principal bundles ($G$ is an arbitrary group). This is more or less the definition of characteristic classes. However, if $G=O(n)$ or $SO(n)$, it is quite difficult to get a hand on integral coefficients. John Klein gave a link here:

What characteristic class information comes from the 2-torsion of $H^*(BSO(n);Z)$?

To see the essential ingredients for the definition of Stiefel-Whitney classes for real vector bundles (and similar series), it is helpful to ignore Milnor-Stasheff and forget about the cell decompositions of Grassmannians for a moment. (I learnt the following definition from Matthias Kreck) Let $V \to X$ be a real $n$-dimensional vector bundle and $L \to RP^{\infty}$ be the universal line bundle. The external tensor product $V \boxtimes L$ is a bundle over $X \times RP^{\infty}$. It has an Euler class $e \in H^n (X \times RP^{\infty};Z/2)$. Use Kuenneth to write this group as $\oplus_{k=0,...,n} H^k (X) \otimes H^{n-k} (RP^{\infty})$. Under this isomorphism, $e$ becomes $\sum_k w_k(V) \otimes x^{n-k}$, $x$ the generator of $H^{\ast}(RP^{\infty})$.

The same construction yields the Chern classes, replacing $Z/2$ by $Z$ and $R$ by $C$ throughout.

What you see from this construction is that if you wish to have integral classes, you need the Euler class, i.e. orientability. But, no matter whether $V$ is oriented or not, the bundle $V \boxtimes L$ is not oriented.

What you can do is to replace $L \to RP^{\infty}$ by the universal $2$-dimensional oriented vector bundle $U \to BSO(2)=CP^{\infty}$. The point is that $U$ and hence $V \boxtimes U$ is a complex vector bundle and hence oriented. More precisely

$$V \boxtimes_R U \cong V \boxtimes_R (C \otimes_C U) = (V \otimes_R C) \boxtimes_C U.$$

You get the Pontrjagin classes! You can play the same game with the quaternions and the universal quaternionic line bundle $H \to HP^{\infty}$. Here it is important that for each quaternionic vector bundle $V \to X$, the bundle $V \boxtimes_H H$ is only real oriented and not complex. The classes obtained in this way are also called Pontrjagin classes.

Having defined these classes, one computes the cohomology of the classifying spaces $BG(n)$ ($G=U,O,SO,Sp$) with different coefficient rings $A$ with the help of the Gysin sequence of the sphere bundle $BG(n) \to BG(n+1)$ and induction on $n$. An important point is that the computation goes smoothly if (and only if!) the Euler numbers of the occuring spheres are either zero or invertible (in $A$). Of course, the two cases produce quite different looking results.

If $G=U$ or $G=Sp$, all spheres are odd-dimensional and thus have zero Euler number. Thus the compuation goes well for any $A$.

If $G=O,SO$, then there are even-dimensional spheres around, with Euler number $2$. Therefore the computation is smooth with $Z/2$-coefficients and also if $2$ is invertible in the coefficient ring. But the results are really different in characteristic $2$ and $\neq 2$! If $2$ is neither zero nor invertible in the coefficient ring, things become messy at this point.

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The definition you mention of the SW classes reminds me of the construction of the Steenrod operations given in Steenrod & Epstein. This answer is very enlightening! –  Sean Tilson Mar 1 '11 at 22:24
    
@Johannes: Do you know if a similar construction (taking the Euler class of exterior tensor product with the canonical line bundle) yields the SW-classes for unoriented cobordism? –  Mark Grant Sep 7 '11 at 11:22
    
@Mark: I am not sufficiently familiar with the computational aspects of bordism theory to answer this question quickly. –  Johannes Ebert Sep 7 '11 at 18:22
    
@Johannes: Thanks anyway. I think the answer is no, the coefficients of the formal group law are involved in the formula (this is what led me to ask mathoverflow.net/questions/74770/…) –  Mark Grant Sep 9 '11 at 14:52
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The integral cohomology rings of both $BO(n)$ and $BSO(n)$ were computed by E. H. Brown, Proceedings AMS, 85, 2, 1982, p. 283-288. These rings are generated by the Pontrjagin classes, Bocksteins of monomials in even Stiefel-Whitney classes and, in the case of $BSO(2k)$, the Euler class. The description is as follows. All torsion is 2-torsion. The subalgebra generated by the Pontrjagin classes (and the Euler class in the case of $BSO(2k)$) has no torsion and is subject to just one relation: the square of the Euler class is the corresponding Pontrjagin class in the $BSO(2k)$ case. The torsion ideal can be identified with the $A$-submodule of the mod 2 cohomology generated by the image of $Sq^1$ where $A$ is the subalgebra generated by the reductions of the Pontrjagin classes, and the reduction of the Euler class in the case of $BS(2k)$. The key observation is Lemma 2.2.

The cohomology of $BO(n)\times BO(m)$ and $BSO(n)\times BSO(m)$ can be described in a similar way. E.H. Brown also computes the images of the Pontrjagin and Euler classes under the Whitney sum maps $BO(n)\times BO(m)\to BO(n+m),BSO(n)\times BSO(m)\to BSO(n+m)$. The Euler classes behave as expected; the torsion component of the images of the Pontrjagin classes is a bit more complicated. Finally, the image of the Bockstein of a monomial in the Siefel-Whitney classes can be computed using Lemma 2.2 and the action of the Steenrod algebra on the mod 2 cohomology.

So ``integral characteristic classes'' do not give any new tools for distinguishing real vector bundles up to isomorphism. However, in principle these classes may give new obstructions to representing bundles as Whitney sums and, by the splitting principle, as tensor products, symmetric or exterior powers etc.

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The "orthogonal complement" construction of Chern classes works like this: if $V\to X$ is a $U(n)$-bundle, let $p:S(V)\to X$ be the sphere bundle, and write $p^\*V\approx W\oplus \mathbb{C}$ for the decomposition of the pullback. We want to define $c_{n-1}(V)$ to be $e(W)$, but this is in the wrong group; to obtain a well-defined class in $H^{2n-2}X$, we need to know that $$ p^\*:H^{2n-2}X \to H^{2n-2}S(V)$$ is an isomorphism. It is, because the fiber of the bundle $p$ is $S^{2n-1}$, which is $(2n-2)$-connected.

If we try to do this for an $SO(n)$-bundle, we'd want to know that $$p^\*: H^{n-1}X\to H^{n-1}S(V)$$ is an isomorphism. But now the fiber is $S^{n-1}$, which is only $(n-2)$-connected. So the map $p^*$ on cohomology can fail to be surjective. So you may not be able to lift your element to $H^{n-1}(X)$. This fails already for the universal $SO(3)$-bundle; in this case, the Euler class $e(W)\in H^2S(V)=H^2BSO(2)$ doesn't lift to $H^2BSO(3)$.

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I think (but I'm not sure) that you CAN define characteristic classes from $BSO(n)$ with $\mathbb{Z}$ coefficients, but it's not nearly as easy to work with them or compute them.

One of the big problems is that the integer cohomology ring of the infinite Grassmanians is quite nasty while with $\mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Q}$ coefficients, it's not so bad (giving rise to the Stiefel-Whitney classes and with a bit more work, the Pontrjagin classes and Euler class).

Another general issue is the following: For the standard inclusions $SO(k)\rightarrow SO(n)$ as a block form, one wants to know what the induced maps $BSO(k)\rightarrow BSO(n)$ look like, or at least the induced map $ H^{\*}(BSO(n))\rightarrow H^{\*}(BSO(k)) $ looks like. For $\mathbb{Z}/2\mathbb{Z}$ coefficients, it's easy: the map has kernel all of the $w_i$ with $i > k$ and is an isomorphism on the rest. It's equally easy for rational coefficients. (And, as an aside, one can repeat the same questions with $BU(n)$ and the Chern classes. Turns out the cohomology ring of $BU(n)$ with $\mathbb{Z}$ coefficients is fine, and these induced maps are also easy to compute.)

Not only are the integral cohomology rings of $BSO(n)$ messy, these induced maps on cohomology are not nearly as well understood.

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