I'm aware of the great body of work on Legendrian knot theory in contact geometry, but suppose I'm curious just about homotopy and not isotopy. How does one understand the space of Legendrian loops based at a point in a contact manifold? Can that be made into a "Legendrian fundamental group" somehow?
I've heard that h-principles are somehow involved, but I'm not sure what the punchline is.
In general, the (parametric) h-principle for Legendrian immersions implies that Legendrian immersions f:L->(M,\xi) are classified up to homotopy (through Legendrian immersions) by the following bundle-theoretic invariant: Choosing a compatible almost complex structure on \xi allows one to complexify the differential of f to an isomorphism d_C f: TL\otimes C -> f*\xi, and the relevant invariant is the homotopy class of this isomorphism of complex vector bundles (of course this is independent of the almost complex structure since the space of compatible almost complex structures is contractible).
The above holds in any contact manifold (M,\xi) of arbitrary dimension. Of course when M is 3-dimensional and L is S^1, f^*\xi is the unique complex line bundle over S^1, automorphisms of which are parametrized up to homotopy by pi_1(U(1))=Z. So (given that the h-principle also implies that any loop in a 3-manifold is homotopic to a Legendrian loop) it appears to always be the case that the "Legendrian fundamental group" surjects onto the standard fundamental group, with kernel Z.
When M=R^3 this invariant is equivalent to the rotation number that Steven mentioned. There's a proof of the relevant h-principle in the book by Eliashberg and Mishachev. The above discussion is partly based on Section 3.3 of arXiv:0210124 by Ekholm-Etnyre-Sullivan.
I don't have a general answer, but for the standard tight contact structure \xi on R^3, see "A contact geometric proof of the Whitney-Graustein Theorem" by Geiges (arXiv:0801.0046). Proposition 4 says that regular Legendrian curves in (R^3, \xi) are classified up to homotopy through Legendrian curves by their rotation number, so I guess the "Legendrian fundamental group" should be Z in this case. The proof is less than a page long and very straightforward.
