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I am trying to understand why for any oriented Riemannian four manifold $X$, we have following equality:

$$w_2(X)= w_2(\Lambda^{+}_{2}(TX))$$

Where $\Lambda^{+}_{2}(TX)$ is the bundle of self-dual two forms.

I am looking at Milnor-Stasheff's book but couldn't figure a theorem or a formula that might be related to this problem.

muratguner
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1 Answers1

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As usual, the splitting principle comes to the rescue. Below I am going to write direct sums as sums and tensor products as products for readability, which will unfortunately introduce some ambiguity between sums and products of vector bundles and of cohomology classes.

Write the tangent bundle formally as the sum of its Stiefel-Whitney roots

$$T = \sum_{i=1}^4 \alpha_i$$

so that its total Stiefel-Whitney class is

$$w(T) = \prod_{i=1}^4 (1 + w_1(\alpha_i))$$

and in particular

$$w_2(T) = \sum_{i < j} w_1(\alpha_i) w_1(\alpha_j).$$

The cotangent bundle is isomorphic to the tangent bundle, and the bundle of $2$-forms is its second exterior power, which gives

$$\Lambda^2(T^{\ast}) = \sum_{i < j} \alpha_i \alpha_j.$$

Now we need to use the hypothesis that $T$ is oriented, which means that

$$\prod_{i=1}^4 \alpha_i = 1$$

This lets us identify some of the six terms in $\Lambda^2(T^{\ast})$: we now have

$$\Lambda^2(T^{\ast}) = 2 (\alpha_1 \alpha_2 + \alpha_1 \alpha_3 + \alpha_1 \alpha_4)$$

because $\alpha_2 \alpha_3 = \alpha_1 \alpha_4$ and so forth. The bundles of self-dual and anti-self-dual $2$-forms come from splitting this sum up in a way that's invariant under even permutations, and I believe the only splitting that works is that they are each half the above, so

$$\Lambda^2(T^{\ast})^{+} = \alpha_1 \alpha_2 + \alpha_1 \alpha_3 + \alpha_1 \alpha_4.$$

Using the multiplicativity of the total Stiefel-Whitney class with respect to sums, we can now compute that the second Stiefel-Whitney class of this bundle is

$$w_1(\alpha_1 \alpha_2) w_1(\alpha_1 \alpha_3) + w_1 (\alpha_1 \alpha_2) w_1(\alpha_1 \alpha_4) + w_1(\alpha_1 \alpha_3) w_1(\alpha_1 \alpha_4).$$

At this point I could not find a nice way to check that this is equal to $w_2(T)$, so I did the terrible thing, which is to use the identity $\sum w_1(\alpha_i) = 0$ to eliminate $\alpha_4$ from both expressions. It checks out but it's not particularly worth writing down.

Qiaochu Yuan
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  • Isn't $\bigwedge^2(\alpha_1\oplus\alpha_2\oplus\alpha_3\oplus\alpha_4) \cong \bigoplus_{i < j}\bigwedge^1\alpha_i\otimes\bigwedge^1\alpha_j \cong \bigoplus_{i<j}\alpha_i\otimes\alpha_j$? As you've written $\bigwedge^2(T^)$ (i.e. just $i \neq j$) both $\alpha_i\alpha_j$ and $\alpha_j\alpha_i$ appear, so it seems that it has 12 direct summands, and $\bigwedge^2(T^)^+$ has 6 (you didn't include any of the terms with $\alpha_4$). Am I missing something? – Michael Albanese Mar 14 '17 at 17:20
  • @Michael: my bad, I got confused and dropped $\alpha_4$ halfway through for some reason. – Qiaochu Yuan Mar 14 '17 at 19:33
  • Thanks a lot Qiaochu for your answer, and Michael for your comment. – muratguner Mar 16 '17 at 23:57
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    @QiaochuYuan Do you have references for your statements about the orientability of $T$ in terms of the $\alpha_i$, and about the SD/ASD bundles coming from an even-permutation-invariant splitting of the sum? – Brian Klatt Mar 08 '18 at 21:38