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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.

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up vote 5 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.

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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

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