Pontryagin class of self-dual forms on a 4-manifold 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$$p_1(X)=-c_2(TX \otimes \mathbb{C})=-\sum_{i&lt;j} a_ia_j$$ $$p_1(\wedge^2_+ X)=-(a_1+a_2)(a_1+a_3)-(a_1+a_2)(a_1+a_4)-(a_1+a_2)(a_1+a_4)$$
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.
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?
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?