I am puzzled by the possible non-zero $w_3(G)$ of a principal $SO(3)$-$G$ bundle over the 4-manifold $X_4$, and how can $w_3(G)$ be related to the lower dimensional $w_2(G)$. Here $w_i$ is the Stiefel Whitney classes over the principal $SO(3)$ bundle.

May we consider a few cases of examples, say $X_4=S^4,\mathbb{P}^4(\mathbb{R}),\mathbb{P}^2(\mathbb{C}),$ connected sum of them, etc.

Can we construct some examples of non-zero $w_3(G)$ of a principal $SO(3)$-$G$ bundle over some compact 4-manifold $X_4$? How explicitly can they be related to the lower dimensional $w_2(G)$ and other characteristic class constraint?


There are two relevant relations in the cohomology of $BSO(2n+1)$: first, the Euler class is 2-torsion, and in fact $e = \beta w_{2n}$; and second, the Euler class in the cohomology of $BSO(k)$ reduces to $w_k$ mod 2. Therefore, for a rank $2n + 1$ vector bundle $E$, $w_{2n+1}(E) = \beta w_{2n}(E) \mod 2$.

This operation, taking the integral Bockstein $H^{k}(X;\Bbb Z/2) \to H^{k+1}(X; \Bbb Z)$, and composing with reduction mod 2 $H^{k+1}(X;\Bbb Z) \to H^{k+1}(X;\Bbb Z/2)$ is actually the first Steenrod square operation, written $\text{Sq}^1: H^*(X;\Bbb Z/2) \to H^{*+1}(X;\Bbb Z/2)$.

Now the Dold-Whitney theorem says that on a 4-complex $X$, $SO(3)$-bundles are classified up to isomorphism by $(w_2, p_1) \in H^2(X;\Bbb Z/2) \times H^4(X;\Bbb Z)$, where these satisfy the constraint $w_2^2 = p_1 \pmod 4$ - note that here we are using the Pontryagin square to square a $\Bbb Z/2$ class into a $\Bbb Z/4$-class. In addition, there is always an $SO(3)$-bundle with a given pair $(w_2, p_1)$ as long as it satisfies this constraint.

In any case, on an oriented closed manifold, $H^4(X;\Bbb Z) = \Bbb Z$, so there are $\Bbb Z$-many $SO(3)$-bundles with fixed $w_2$.

Knowing this relationship and the Dold-Whitney classification, your question reduces to finding oriented examples where $\text{Sq}^1: H^2(X;\Bbb Z/2) \to H^3(X;\Bbb Z/2)$ is nonzero. Certainly this is false for $S^4, \Bbb{CP}^2$, since those have no third cohomology. Unfortunately, $\text{Sq}^1: H^2(\Bbb{RP}^4;\Bbb Z/2) \to H^{3}(\Bbb{RP}^4;\Bbb Z/2)$ is zero, even though both of these groups are nonzero. (Easiest way to see this is probably a by hand chain level calculation.)

(It is worth knowing that $e(E) = \beta w_2(E) \in H^3(X;\Bbb Z)$ is precisely the obstruction to writing $E \cong \Bbb R \oplus \lambda$, where $\Bbb R$ is the trivial line bundle and $\lambda$ an oriented plane bundle, and $2e(E) = 0$ is the obstruction to writing $E \cong \xi \oplus \lambda$ where now $\xi$ is a real line bundle and $\lambda$ is a 2-plane bundle; that is, it is always possible to split in this way.)

As an example of an oriented manifold where this is true, try $S^1 \times \Bbb{RP}^3$. If $x$ denotes the generator of $H^1(S^1;\Bbb Z/2)$ and $y$ the generator of $H^1(\Bbb{RP}^3;\Bbb Z/2)$, then $\text{Sq}^1(xy) = xy^2$, which is nonzero; even better, the cup-square of $xy$ is zero (since $x^2 = 0$). So pick the bundle with $w_2 = xy$ and $p_1 = 0$; this has nonzero $\beta w_2$. There is probably an explicit description of this bundle but I haven't expended any effort to think of it.

I suspect the Enriques surface provides another example.

  • $\begingroup$ thanks! +1, where can I find relevant info/Refs on the things you said? They are a lot infos here. :) $\endgroup$ – annie heart Jul 14 '18 at 3:15
  • $\begingroup$ Can $\beta w_2$ be in $\mathbb{Z}$ class? If so, what is the allowed $\mathbb{Z}$ valued, in your last $S^1 \times \mathbb{RP}^3$ example? Is that any integer $\mathbb{Z}$ as $\beta w_2=\beta (xy)=xy^2 \in \mathbb{Z}$? $\endgroup$ – wonderich Jul 14 '18 at 3:20
  • $\begingroup$ @annieheart For characteristic classes of a general vector bundle, see Milnor-Stasheff's book. The specific relation $e = \beta w_{2k}$ here is a formula that holds in the ring $H^*(BSO(2n+1);\Bbb Z)$. I knew this from the formulas in Brown's calculation (it is illuminating when reading that to consider the special case $BSO(3)$). I learned about Steenrod squares from a bit in Hatcher's algebraic topology book, in particular, some things about $\text{Sq}^1$ and the Bockstein operation. $\endgroup$ – user98602 Jul 14 '18 at 6:50
  • $\begingroup$ The bit about splitting off a line bundle (which was INCORRECT as written at first, and has been fixed) is from interpreting these classes via obstruction theory to finding a nonvanishing section of the unit sphere bundle (resp. projective bundle); this is also covered somewhat briefly in Milnor-Stasheff. $\endgroup$ – user98602 Jul 14 '18 at 6:51
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    $\begingroup$ Calculating some cohomology groups for the Enriques surface $X$: $H^2(X;Z) = Z^{10}\oplus Z_2$, $H^2(X;Z_2) = Z_2^{10}\oplus Z_2 \oplus Z_2$, $H^3(X;Z) = Z_2$, $H^3(X;Z_2) = Z_2$, we see that there is a class $\alpha$ in $H^2(X;Z_2)$ with no integral lift. Therefore the Bockstein takes $\alpha$ to the nonzero element in $H^3(X;Z)$, which reduces mod 2 to the nonzero element in $H^3(X;Z_2)$. So, $Sq^1\alpha \neq 0$, and one can take the $SO(3)$ bundle determined by $w_2 = \alpha$ and $p_1 = $ (any integral lift of the Pontryagin square of $\alpha$) as an example with $w_3 = Sq^1w_2 \neq 0$. $\endgroup$ – Aleksandar Milivojevic Jul 18 '18 at 23:50

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