How to use the assumption that a vector field is curl-free in a "convex" region,

Question

I don't seem to need this assumption in one of my proofs, but the problem statement gives it, so I think I had better try to use it.

Does a convex region imply that it is simply connected (but that the converse need not be true)?

And then a curl-free vector field in a convex, simply connected region, must be conservative -- and so it is a gradient vector field.

I.e., Perhaps I needed to know that the region was simply connected before claiming the vector field is conservative -- the curl-free assumption alone does not give enough information. What do you think?

Thanks,

Answer 1

• Yes, convex implies simply-connected. In fact, convex implies contractible: choose a point $x_0$ in the set. For any other point $x$, define $h(t,x) = (1 - t)x + tx_0$. $h$ is a contraction of the set to a point. Any closed curve in the set also contracts to a point under $h$, so the set is simply-connected.
• No, the converse is not necessarily true: consider a half-circle, for example.
• And yes, in general you need to know that the domain is simply-connected before you can prove that the integral around closed loops is $0$ when a derivative is $0$. Otherwise you can pick up a constant factor when you circle around a singularity. This holds no matter what type of derivative it is.