Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $E$ is a vector bundle with structure group $G$ and let $P = F(E)$ be the frame bundle. Let $\mathfrak{g}_P$ denote the associated bundle to the adjoint representation of $G$ on its Lie algebra (i.e. $\mathfrak{g}_P = P \times_G \mathfrak{g}$ where $\mathfrak{g}$ is the Lie algebra of $G$). Given a section of $\mathfrak{g}_P$, I should be able to "exponentiate" it pointwise to get a gauge transformation. How is this defined? I couldn't come up with anything well-defined.

For some context, I was reading Chapter 2 of Donaldson and Kronheimer's "The Geometry of 4-Manifolds," and they mention this pointwise exponential in passing on p. 33. I'm guessing they assume the reader is familiar with it from a more elementary text, but I looked in a few other books and couldn't find it.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

I'm not sure what it is that you tried, but the exponential map should work. First of all, let $U_i$ be a trivialising cover for $P$ and its associated bundles. A section through $\mathfrak{g}_P$ is given by functions $\omega_i: U_i \to \mathfrak{g}$ which, on overlaps, transform according to

$$\omega_i(p) = \operatorname{Ad}_{g_{ij}(p)} \omega_j(p) \qquad \forall p \in U_i \cap U_j,$$

where I use $\operatorname{Ad}$ to mean the adjoint representation of $G$ on its Lie algebra $\mathfrak{g}$.

Now simply compose $\omega_i$ with the exponential map $\exp: \mathfrak{g} \to G$, resulting in functions $\exp\omega_i : U_i \to G$ which, on overlaps, transform according to

$$\exp\omega_i(p) = g_{ij}(p) \exp\omega_j(p) g_{ij}(p)^{-1} \qquad \forall p \in U_i \cap U_j.$$

But this is just a section through the associated fibre bundle usually denoted $\operatorname{Ad} P$, and that is the same thing as a gauge transformation.

share|improve this answer
Thanks- that's exactly what I tried, except that I was thinking in terms of matrices for simplicity and I forgot the relation $Y(e^X)Y^{-1} = e^{YXY^{-1}}$. Actually, this relation seems a bit mysterious to me in its matrix form; it seems like it's easier to see when exp is defined abstractly using one-parameter subgroups. Thanks for clearing things up! –  Andy Manion Jul 30 '10 at 5:19
Isn't $Y(e^X)Y^{-1} = e^{YXY^{-1}}$ obvious from the Taylor series expansion of $e^X$? –  Deane Yang Aug 1 '10 at 4:54
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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