Let $(M, \omega)$ be a symplectic manifold and $L \subseteq M$ - a Lagrangian submanifold. I am trying to understand under what circumstances the Maslov homomorphism $I\_{\mu, L} \colon \pi\_2(M, L) \to \mathbb{Z}$ is in fact induced by an element $\mu\_L \in H^2(M, L;\mathbb{Z})$. I am encountering a few related issues:


1. If I have a map $f \colon (S, \partial S) \to (M, L)$ from an oriented surface $S$ with non-empty boundary, I can symplectically trivialise $(f^\*TM, \omega)$ and calculate the Maslov index of $f\vert\_{\partial S}^\*TL$ using this trivialisation and the natural orientations of the different boundary components of $S$. But since $Sp(2n)$ is not simply connected, this might depend on the choice of trivialisation if $S$ has non-zero genus.
2. A related question is when is a class $a \in H\_2(M, L; \mathbb{Z})$ representable by a surface as above?
3. Even if the Hurewicz homomorphism $h\colon \pi\_2(M, L) \to H\_2(M, L;\mathbb{Z})$ is surjective, so I can represent every relative homology class by a disc, do I have that the Maslov index of a disc $a \in \pi\_2(M, L)$ depends in fact only on $h(a)$?



Maslov class as a relative cohomology class in $H^2(M, L)$
I think it is always true, and one can explicitly define the class as follows, at least in the compact case (this construction is surely well-known to some people, but I'm not sure where to find it written down).


Let $M^{2n}$ be an almost complex manifold and $L^n \subseteq M$ a totally real submanifold (of course in the symplectic/Lagrangian case you just pick a compatible almost complex structure before starting). Over $M$ we have the complex line bundle $$E := \Lambda^n\_{\mathbb{C}} TM,$$ and over $L$ we have the real line bundle $$F := \Lambda^n\_{\mathbb{R}} TL.$$ $F$ is a subbundle of $E|\_L$ in the obvious sense.


If $L$ is oriented then we can define $\mu\_L$ to be twice the Poincaré dual of the vanishing locus of a generic smooth section $s$ of $E$ such that $s|\_L$ lies in $F$ and is nowhere-zero. In the non-orientable case, $F$ doesn't have a nowhere-zero section, but we can replace $E$ and $F$ by their tensor squares and then omit the factor of two after taking Poincaré duals. It is clear from these definitions that $\mu\_L$ maps to $2c\_1(X)$ in $H^2(X)$ and that when $L$ is orientable $\mu\_L$ is divisible by two. One can also check by hand that it does indeed induce the Maslov homomorphism.


Without assuming compactness of $M$ and $L$ one probably has to work a bit more; morally the definition has to be at least as hard as that of $c\_1(X)$. You may also needs to make some reasonable assumption on the niceness of the embedding of $L$ in $M$. If the pair $(M, L)$ is a relative CW complex then $E$ (or its tensor square when $L$ is non-orientable) can be represented by a map $$M \rightarrow BU(1)$$ extended cell-by-cell from a constant map $L \rightarrow EU(1)$, by using $F$ (or its square) to give a canonical homotopy class of trivialisation over $L$. You could then define $\mu\_L$ to be twice (or not) the class of the pullback of a cocycle representing $c\_1$ of the tautological bundle. Here it's less obvious that this class is independent of choices and induces the Maslov homomorphism, but it can probably be checked.


