Depending on which version of the Traveling Salesman Problem (TSP) you're looking at, it's either NP-hard (finding the shortest possible route), or NP-complete (called the decision version of TSP; given a length $L$, determine if any tour of a given graph has length less than or equal to $L$).
The difference between the two is subtle, but important: NP-hard problems are at least as hard as any problem in NP. The way you show a problem is at least as hard as another problem is to use what's called a reduction. A reduction is a way of transforming your given problem into the problem you'd like to compare it with. In this case, if you want to compare TSP to, say, a known NP problem like 3-SAT (satisfiability of a Boolean statement consisting of clauses containing 3 literals), you would try to reduce TSP to 3-SAT. The idea is to convert an instance of a problem $a$ that we don't yet know how to solve into a problem $b$ that we do know how to solve, such that the solution to $b$ can be used to determine a solution to $a$. Then, by converting $a$ to $b$ and solving $b$, we've constructed an algorithm to solve $a$. If we know how long the conversion process -- that is, the reduction -- takes, and we know how long it takes to solve problem $b$, then we can figure out an upper bound on how long it takes to solve the problem $a$.
The most common and useful kind of reduction I saw in my introductory graduate-level theory of computation course was called a polynomial-time reduction, which is a reduction process whose algorithm takes polynomial time, as a function of some metric of problem size, to produce its output. This concept is then used to define what NP-complete means. A problem $p$ is NP-complete if it is in NP, and for every problem in NP, there exists a polynomial-time reduction from that problem to problem $p$. NP-hard can be defined a similar way: a problem $p$ is NP-hard if there exists a polynomial time reduction from an NP-complete problem to $p$.
Now that all of that terminology is out of the way, to answer your question:
- If you're talking about the NP-complete decision version of TSP, then any problem in NP can be reduced in polynomial time to TSP. Global optimization problems tend to be NP-hard (though I don't know for sure that all of them are, I do know that nonconvex optimization problem are NP-hard). Since NP-hard problems by definition are polynomial time reductions of NP-complete problems, TSP can be polynomial time reduced to NP-hard global optimization problems.
- If you're talking about the NP-hard version of TSP, then no statements can be made about reductions of NP-hard global optimization problems.
- There is nothing to suggest that NP-hard optimization problems can be reduced to either version of TSP. However, it is known that 0-1 integer programming is an NP-complete problem, so it can be reduced to TSP.
A good, but dense, reference is Michael Sipser's Introduction to the Theory of Computation, 2nd edition. (Disclaimer: Professor Sipser taught my theory of computation class.)