Suppose $V$ is a vector bundle with structure group $SO(3)$, and suppose that it can be lifted to a $\text{Spin}(3) = SU(2)$ bundle (i.e. $w_2(V) = 0$). Let us call the lifted bundle $E$. Then it is stated on page 42 in The Geometry of Four-Manifolds by Donaldson and Kronheimer that we have the relation $p_1(V) = -4c_2(E)$. My question is:

How does one show this?

More generally, how does one compute the effect of going over to a lifted bundle on the characteristic classes? Generally one has $p_1(E) = c_1^2(E) - 2c_2(E)$. In our case the first term drops out, so that the claim in the book can also be written as $p_1(V) = 2p_1(E)$. This factor $2$ undoubtedly somehow comes from the covering homomorphism $SU(2) \rightarrow SO(3)$ which is 2 : 1, but how?

Probably related: on the same page he defines at the bottom for the so-called instanton number $\kappa = \frac1{8\pi^2}\int_M\text{Tr}(F^2)$, and then claims that this is $\kappa = c_2$ for $SU(r)$ bundles $E$ and $\kappa = -\frac14p_1$ for $SO(r)$ bundles $V$. There again is that factor 4; again, from the formula $p_1(E) = c_1^2(E) - 2c_2(E) = \left[\frac{-1}{4\pi^2}\text{Tr}(F^2)\right]$ one would expect this to be 2. I can see that this factor is chosen so that if one lifts the bundle that $\kappa$ does not change, but on the other hand, not every bundle is liftable, and for bundles which have both Chern classes and Pontryagin classes (such as complex $SU(r)$ bundles), I would expect that one would want the two formulae to give the same answer. As it is, they don't.

(I guess the real problem is I haven't managed to find any readable sources on $SO(r)$-bundles in the context of gauge theories. For special unitary groups there are sources in abundance, but for special orthogonal they are a lot harder to find.)

  • 3
    $\begingroup$ Johannes answered the question, but one comment that might help if you want a Chern-Weil answer is that $Tr$ means 2 different things. More explicitly, $Tr(F^2)$ is taken in 2x2 matrices in the $su(2)$ case and 3x3 matrices in the $so(3)$ case and the map on lie algebras $c:su(2)\to so(3)$ obtained by differentiating the covering map doesn't preserve this trace. If I recall correctly, for $A\in su(2)$, $Tr_3(c(A)^2)= -2Tr_2(A^2)$, giving your missing factor of 2. $\endgroup$ – Paul Mar 24 '11 at 13:40

Let me rephrase your question so that I understand it: let $P \to X$ be an $SU(2)$-principal bundle. Then we get a $2$-dimensional complex vector bundle $E \to X$ by $P \times_{SU(2)} C^2$ (with the defining representation). And we get a $3$-dimensional real vector bundle $V:= P \times_{SU(2)} R^3$ (with the adjoint representation). You (or Donaldson-Kronheimer) claim that $p_1(V)=-4c_2(E)$.

It is enough to consider the universal case $X=BSU(2)$. The map $BU(1) \to BSU(2)$ induced by the inclusion of the maximal torus is injective in cohomology, so it suffices to check your identity on $BU(1)$ (splitting principle). The pullback of $E$ to $BU(1)$ is $L \oplus L^{\ast}$, so its $c_2$ is a generator $\pm u$ of $H^4 (BU(1))$. The pullback of $V$ to $BU(1)$ is $L^2 \oplus \mathbb{R}$ (here the fact that $SU(2) \to SO(3)$ has degree comes in), hence the total Chern class of $V \otimes \mathbb{C}$ is $(1+2u)(1-2u)$, from which you can read off the Pontrjagin class. For the correct sign, you need a bit more care, though.

EDIT: The inclusion of the maximal torus is the group homomorphism $U(1) \to SU(2)$; $z \mapsto diag (z,z^{-1})$. Thus the defining rep. of $SU(2)$ restricts to a sum of the defining rep. of $U(1)$ and its dual, giving the pullback of $E$.

Consider the standard basis the Lie algebra $\mathfrak{su}(2)$ (http://en.wikipedia.org/wiki/Special_unitary_group#SU.282.29) and compute the action of $diag (z,z^{-1})$ via the adjoint rep in this basis. The result is that the adjoint rep. of $SU(2)$ restricts to a sum of the tensor square of the defining rep of $U(1)$ with the trivial real one-dim. rep. (this computation is basic in representation theory, it is the computation of the roots of $SU(2)$).

  • 1
    $\begingroup$ Although I'm not really familiar with universal bundles, this seems very helpful. One question (or two, really): why are those pullbacks $L\oplus L^*$ respectively $L^2\oplus\mathbb{R}$? In the first case, applying the splitting principle gives something like $L\oplus K$, so why is $K = L^*$? $\endgroup$ – miramo Mar 24 '11 at 16:10

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.