I'm looking at a paper by S.K. Donaldson http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCgQFjAA&url=http%3A%2F%2Fwww2.imperial.ac.uk%2F~skdona%2Fdonaldson-ams.ps&ei=AzoyU9zXKKWC2wWuoIDQDQ&usg=AFQjCNFimn32IRVXGI8YY7IKdR6sd3rwVg&bvm=bv.63738703,d.b2I and here is my second question.
A Lie group acts on a symplectic manifold $(X,\omega)$. Then the derivative of the action is a map $$ r:\mathfrak{g}\rightarrow Vect(X), $$ and a moment map for the action is a map $$ \mu:X\rightarrow \mathfrak{g}^*, $$ whose derivative $$ d\mu:TX\rightarrow \mathfrak{g}^* $$ is the transpose of $r$. What does Donaldson mean by transpose and why transpose?
If $V$ and $W$ are vector spaces and $T: V \longrightarrow W$ is a linear map, then the transpose of $T$ is the map $$T^\ast: W^\ast \longrightarrow V^\ast,$$ $$T^\ast(\phi) = \phi \circ T.$$
In our case, the transpose of $r$ is a map $$r^\ast: \mathrm{Vect}(X)^\ast \longrightarrow \mathfrak{g}^\ast.$$ We have that $\mathrm{Vect}(X)^\ast = \Omega^1(X)$, the space of $1$-forms on $X$. Note that $\mathrm{Vect}(X)$ is the space of sections of $TX$ and $\Omega^1(X)$ is the space of sections of $T^\ast X$. Using the nondegeneracy of the symplectic form, we have a distinguished bundle isomorphism $\varphi: TX \xrightarrow{~\cong~} T^\ast X$. This in turn induces an isomorphism $\tilde{\varphi}: \mathrm{Vect}(X) \xrightarrow{~\cong~} \Omega^1(X)$. Hence we have a map $$r^\ast \circ \tilde{\varphi}^{-1}: \mathrm{Vect}(X) \longrightarrow \mathfrak{g}^\ast.$$ We may think of this as a map $$r^\ast \circ \tilde{\varphi}^{-1}: TX \longrightarrow \mathfrak{g}^\ast.$$ Then what Donaldson is saying is that this $r^\ast \circ \tilde{\varphi}^{-1}$ is the same as the differential of the moment map, $d\mu$.
If you haven't already, I would recommend reading the part of McDuff and Salamon's Introduction to Symplectic Topology on moment maps and Hamiltonian reduction before getting into Donaldson's paper. They set up the relevant background in more detail and even sketch some of Donaldson's results.