Im trying to understand the significance of the manifold being simply connected for the following (or any really) case to do with basic yang mills theory.
We are considering a U(1) line bundle, L, over a simply connected manifold, M, with a positive definite metric. Then given U(1) is abelian, we know that ad(L) is given by the trivial bundle $M \times i \mathbb{R}$. So the curvature f of a connection d' is an ordinary 2-form. For this case, the Yang mills equations reduce to:
$df=0\\ d*f=0$
Then given M is simply connected, every gauge transformation on L can be written as $s=e^{iu}$ for some function u.
This last line, I dont understand how they came to that conclusion by using the fact that M is simply connected. Why is this result not true if M is not simply connected?
(For reference, using bottom of pg 37 from https://books.google.com.au/books?id=X5HTBwAAQBAJ&pg=PA37&lpg=PA37&dq=we+study+positive+definite+metrics+hence+elliptic+versions+of+the&source=bl&ots=1r7OsypGfg&sig=Kf4giX6rAJFJs-9MuHjFUTg4MbE&hl=en&sa=X&ved=0CB4Q6AEwAGoVChMI-9Cap7qvyAIVRJ6mCh2uwg2H#v=onepage&q=we%20study%20positive%20definite%20metrics%20hence%20elliptic%20versions%20of%20the&f=false)
A gauge transformation is equivalent to a section of the bundle $P\times_G G$, where $G$ acts by conjugation on the second factor (see page 32 of the reference you cited above). Since $G=U(1)$, conjugation is trivial and thus $P\times_G G=M\times G$. A section of this bundle is equivalent to a function $s: M\rightarrow G=U(1)$. The group $U(1)$ is $S^1$, so we have a function from the simply connected manifold $M$ to the circle. The key point now comes from covering space theory, which (in this special case) says that we can factor our map $s$ through the universal cover (as the induced map $s_*$ on $\pi_1$ is trivial given that $\pi_1(M)=0$) of $S^1$, which is $\mathbb{R}$. Our map $s$ then looks like
$M\xrightarrow{u} \mathbb{R} \xrightarrow{\exp(i\cdot)} S^1 $
i.e. $s=e^{iu}$.