7
$\begingroup$

Let $X$ be an oriented Riemannian 4-manifold. The bundle of 2-forms $\wedge^2 X$ can be decomposed into the bundle of self-dual and anti-self-dual forms, $\wedge^2_+ X \oplus \wedge^2_- X$, using the Hodge star. I would like to show that $$p_1(\wedge^2_+ X)=p_1(X)+2e(X)$$

This is part of Exercise 10.1.3(a) in Stipsicz and Gompf's '4-Manifolds and Kirby Calculus'. The exercise also asks the reader to show that $w_2(\wedge^2_+ X)=w_2(X)$, which I could do, using the splitting principle (see the answer to Second Stiefel-Whitney class of self-dual two forms of four manifolds). For reference, if $TX=E_1 \oplus E_2 \oplus E_3 \oplus E_4$, then $\wedge^2_+ X=(E_1 \otimes E_2) \oplus (E_1 \otimes E_3) \oplus (E_1 \otimes E_4)$.

I tried to use the same method for this part. Now the splitting principle does not hold for real bundles over $\mathbb{Z}$ coefficients, so I decompose $TX \otimes \mathbb{C}$ instead. Let $TX \otimes \mathbb{C}=E_1 \oplus E_2 \oplus E_3 \oplus E_4$, then $(\wedge^2_+ X) \otimes \mathbb{C}=\wedge^2_+ (T^*X \otimes \mathbb{C})=(E^*_1 \otimes E^*_2) \oplus (E^*_1 \otimes E^*_3) \oplus (E^*_1 \otimes E^*_4)$. Denote $a_i=c_1(E_i)$, we have$$p_1(X)=-c_2(TX \otimes \mathbb{C})=-\sum_{i<j} a_ia_j$$ $$p_1(\wedge^2_+ X)=-(a_1+a_2)(a_1+a_3)-(a_1+a_2)(a_1+a_4)-(a_1+a_2)(a_1+a_4)$$ Taking the difference, I must show that $-2e(X)=3a_1^2+a_1a_2+a_1a_3+a_1a_4$. In fact, since $X$ is orientable, $\wedge^4 TX \otimes \mathbb{C}=E_1 \otimes E_2 \otimes E_3 \otimes E_4$ is trivial, so $a_1+a_2+a_3+a_4=c_1(E_1 \otimes E_2 \otimes E_3 \otimes E_4)=0$. So what I really need to show is $$e(X)=-a_1^2=-c_1(E_1)^2$$ The problem is, I have no idea why this true. In fact, I am inclined to believe that I have made a mistake somewhere, due to the asymmetry of this formula.

Any help with the original question, following my approach or not, is welcomed!

EDIT: The result is proved using Chern-Weil theory on P.195 of Walschap's "Metric Structures in Differential Geometry". But I'm still curious for an algebraic topology proof.

$\endgroup$
  • 1
    $\begingroup$ You might like to check "Metric Structures in Differential Geometry" by Gerard Walschap. In section 6.5, around pg 195 he proves some theorems along these lines. See for example Propositions 5.3, 5.4 on pg 195, and the remarks proceeding them. $\endgroup$ – Tyrone Feb 25 '18 at 14:57
  • $\begingroup$ @Tyrone Thank you for the reference! The result is indeed proved in Walschap using (real) Chern-Weil theory. But I'm still curious for an algebraic topology proof. $\endgroup$ – Chi Cheuk Tsang Feb 27 '18 at 15:09
-1
$\begingroup$

I thought about this problem for a while and there could be an explanation like this:

First of all, we need an interpretation of the first Pontryagin class as an obstruction class. For a rank $n$ bundle $E^n$ over $X$, we choose $n$ generic sections $\sigma_1,\cdots,\sigma_n$. Consider the points $x$ where they span a subspace of $E_x$ of rank $n-2$. Such points $x$ form a co-dimension 4 cycle of $X$. Its Poincare dual is $P_1(E)$. For details, see the link https://mathoverflow.net/questions/117036/what-is-geometrically-the-pontryagin-class.

In our case, $n=3$ or $4$ and $\dim X=4$. So the set of points $x$ is finite and discrete.

To prove $P_1(\Lambda^+X)=P_1(TX)+2e(X)$, we choose a generic section $v$ of $TX$. Then away from the zero locus of $v$, we have a bundle isomorphism:

$ \phi:\Lambda^+X|_{X-Z(v)}\to v^{\perp} \subset TX|_{X-Z(v)} $

This means whenever we have three generic sections $\sigma_1,\sigma_2,\sigma_3$ of $\Lambda^+X$, we can apply $\phi$ and get sections of TX over $X-Z(v)$. Then add $v$, we get four sections.

Suppose $Z(v)=\{x_1,\cdots,x_m\}$. For each of them, we remove a four-ball around it. Outside these balls, $\{\sigma_1,\sigma_2,\sigma_3\}$ and $\{\phi(\sigma_i),v\}$ define the same obstruction cycles for $P_1$. Therefore, in order to compute $P_1(\Lambda^+X)-P_1(TX)$, it suffices to work with these four balls and we can deal with them separately.

In the end, we reduce to the case when $X=D^4$ and we prescribe a framing of $\Lambda^+X$ over $\partial X=S^3$. We take the section $v$ to be $v(x)=x$ (or $(x_1,x_2,x_3,-x_4)$, if the origin is a negative zero) for $x\in D^4\subset \mathbb{R}^4$ and so it vanishes at the origin. I guess it is not hard to verify the relation in this case, if we take care of the sign.

Do you think this is valid?

$\endgroup$
  • $\begingroup$ I understand everything until the last paragraph. How can we reduce to $X=D^4$? I believe we are considering Poincare duals of the zero loci, but once you restrict to a submanifold, the Poincare dual would change. Also I would like to remark that for $X=D^4$, there is no cohomology so the statement to prove holds trivially. $\endgroup$ – Chi Cheuk Tsang May 7 '18 at 1:34
  • $\begingroup$ Suppose we have $X=D^4$ and $\partial X=S^3$. We take $v(x)=x$. Over $S^3$ a framing of $v^\perp\subset S^3\times \mathbb{R}^4$ is given by $x\cdot i$, $x\cdot j$, $x\cdot k$ where I identified $\mathbb{R}^4$ with quaternion numbers $\mathbb{H}$. We are computing relative obstruction classes. For instance, for the Euler number, $v$ has a positive zero at the origin and it gives a +1 contribution. Sections $\{x, x\cdot i, x\cdot j, x\cdot k\}$ and induced sections of $\Lambda^+$ are defined only on $S^3$ at this moment. Extend them over $X$ and look at the points where the rank drops by 2. $\endgroup$ – Donghao May 7 '18 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.