In "Floer Homology groups in Yang-Mills theory", Donaldson says that if we take an $U(2)$-vector bundle $E$ and we construct the bundle $\mathfrak{g}_E$ of trace-free, skew adjoint automorphisms of $E$, ($\mathfrak{g}_E \subset E^*\otimes E$) then we have the following relation $$w_2(\mathfrak{g}_E) = c_1(E) \ \text{mod}\ 2$$ $$p_1(\mathfrak{g}_E) = c_1(E)^2 - 4 c_2$$

Is there some neat way to see this? More generally, can we express the characteristic classes of a bundle associated to a principal bundle $P$ in terms of the characteristic classes of $P$?

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    $\begingroup$ Here is how these things usually go. One has to find a functorial relationship between a vector space $V$ and the vector space of trace-free, skew adjoint automorphisms of $V$. Then there is the same relationship between the corresponding vector bundles (see section 3 in Milnor-Stasheff). Finally there are standard relations between Chern, Stiefel-Whitney and Ponryagin classes (again, Milnor-Stasheff, exercise 14B and Corollary 15.5) which will probably imply the formulas in Donaldson's book. $\endgroup$ – Igor Belegradek Dec 3 '18 at 15:33
  • $\begingroup$ Thank you Igor. In my edition (2002) it is $-4c_2$ (pg. 146), but I agree with you that should be a $-2$. Can you recommend also some other references to undergo Donaldson book (apart from Milnor-Stasheff)? $\endgroup$ – Warlock of Firetop Mountain Dec 3 '18 at 17:35

As a real vector bundle, $E^{\ast} \otimes E$ decomposes as the direct sum of two copies of the bundle $\mathfrak{su}(E)$ of trace-free skew-adjoint endomorphisms and two copies of the trivial bundle. This follows from the corresponding decomposition of $V^{\ast} \otimes V$ where $V$ is the defining $2$-dimensional representation of $U(2)$, as follows. First, the trace map $\text{tr} : V^{\ast} \otimes V \to \mathbb{C}$ has kernel $\mathfrak{sl}(V)$. Second, $\mathfrak{sl}(V)$ is the complexification of $\mathfrak{su}(V)$, so decomposes (as a real representation of $U(2)$) as a direct sum $\mathfrak{sl}(V) = \mathfrak{su}(V) \oplus i \mathfrak{su}(V)$. So, overall we have

$$E^{\ast} \otimes E \cong \mathfrak{su}(E) \oplus \mathfrak{su}(E) \oplus \mathbb{C}$$

(as a real vector bundle). The Whitney sum formula gives

$$w(E^{\ast} \otimes E) = w(\mathfrak{su}(E))^2$$

from which it follows that $w_4(E^{\ast} \otimes E) = w_2(\mathfrak{su}(E))^2$. We know that $w_4 \equiv c_2 \bmod 2$, so it remains to compute $c_2(E^{\ast} \otimes E)$, which we can do by the splitting principle. If $E$ has Chern roots $\alpha, \beta$ (so $c_1(E) = \alpha + \beta, c_2(E) = \alpha \beta$), then $E^{\ast}$ has Chern roots $-\alpha, -\beta$, so $E^{\ast} \otimes E$ has Chern roots $0, 0, \alpha - \beta, \beta - \alpha$. This gives

$$c(E^{\ast} \otimes E) = (1 + \alpha - \beta)(1 + \beta - \alpha) = 1 - (\alpha - \beta)^2 = 1 + 2 \alpha \beta - \alpha^2 - \beta^2$$


$$c_2(E^{\ast} \otimes E) = - c_1(E)^2 + 4 c_2(E)$$


$$w_4(E^{\ast} \otimes E) \equiv c_1(E)^2 \equiv w_2(\mathfrak{su}(E))^2 \bmod 2.$$

To establish that we in fact have $w_2(\mathfrak{su}(E)) \equiv c_1(E) \bmod 2$ on the nose we need to think a bit about the cohomology of $BU(2)$. Its $\bmod 2$ cohomology is a polynomial ring on the Chern classes $c_1, c_2$, from which we can compute that the squaring map $H^2(BU(2), \mathbb{Z}_2) \to H^4(BU(2), \mathbb{Z}_2)$ is injective, so $w_2(\mathfrak{su}(E))$, which universally lives in $H^2(BU(2), \mathbb{Z}_2)$, is uniquely identified by its square.

The computation of the Pontryagin class is similar. In this blog post you can find the proof that $V$ is a complex vector bundle then the first Pontryagin class of its underlying real vector bundle is $p_1(V) = c_1(V)^2 - 2 c_2(V)$. We computed $c_2$ above, and we have $c_1(E^{\ast} \otimes E) = 0$, so altogether

$$p_1(E^{\ast} \otimes E) = 2 c_1(E)^2 - 8 c_2(E).$$

By the Whitney sum formula for Pontryagin classes, this class is equal to $2 p_1(\mathfrak{su}(E))$ modulo $2$-torsion. But again thinking about the cohomology of $BU(2)$, its integral cohomology is again a polynomial ring on $c_1$ and $c_2$, and in particular is torsion-free, so we can ignore $2$-torsion and divide by $2$, which gives

$$p_1(\mathfrak{su}(E)) = c_1(E)^2 - 4 c_2(E)$$

as desired.

More generally, can we express the characteristic classes of a bundle associated to a principal bundle $P$ in terms of the characteristic classes of $P$?

Yes, in the following sense. If $f : G \to H$ is any morphism of Lie groups, it induces a map $Bf : BG \to BH$ on classifying spaces. Applying this map allows you to associate an $H$-bundle to a $G$-bundle. This map further induces a map $H^{\ast}(Bf) : H^{\ast}(BH) \to H^{\ast}(BG)$ on cohomology, which universally describes characteristic classes for associated $H$-bundles in terms of characteristic classes for $G$-bundles. Above we're considering the map $BU(2) \to BGL_3(\mathbb{R})$ coming from the adjoint action of $U(2)$ on $\mathfrak{su}(2)$, and taking advantage of the fact that the cohomology of $BU(2)$ is very well-behaved.

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    $\begingroup$ Explicitly, if $\eta_1 \oplus \eta_2$ is the $U(2)$-bundle in question, the associated $SO(3)$-bundle is $\Bbb R \oplus (\eta_1 \otimes \eta_2^{-1})$, and your description shows how to compute the characteristic classes of this. $\endgroup$ – Mike Miller Dec 3 '18 at 22:34
  • $\begingroup$ Thanks Qiaochu, how is the map $Bf$ defined? I think that we would like that $h_{\mathfrak{su}(2)}=Bf \circ h_{U(2)}$ ($h_{\mathfrak{su}(2)}$ and $h_{U(2)}$ are respectively the classifying maps of the frame bundles of $\mathfrak{su}(E)$ and $E$), am I right? P.S. Do you have any good book to suggest on this subject? $\endgroup$ – Warlock of Firetop Mountain Dec 3 '18 at 23:47
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    $\begingroup$ @Warlock Yes, that is the desired property. Given any homomorphism $f: G \to H$, you have a natural way to construct a principal $H$-bundle out of any principal $G$-bundle, sending $P \mapsto P \times_G H$. In particular applying this to the universal bundle gives you a canonical map $BG \to BH$ classifying the new bundle, and it is clear that this map has the property you ask for. Actually doing the computation of what it does in cohomology usually looks like this: using known formulas for sums and tensor products and relating your construction to those. Sometimes you can run spectral seqs. $\endgroup$ – Mike Miller Dec 4 '18 at 0:48

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