I'm reading [RS] and I was wondering what kind of connection there is between the Maslov index for a pair of paths $\lambda_0,\lambda_1 \colon [a,b] \to \mathcal{L}(2n)$ as defined in [RS] and the boundary Maslov index for loops as defined in [MS] Appendix C page 554.

Some notation: $\mathcal{L}(2n)$ is the Lagrangian Grassmannian $\mathcal{L}(\Bbb R^{2n},\omega_{std})$, by $\mu$ I will denote the Maslov index for the pair of paths and $\mu'$ the one of a loop.

Is it true that for, say $\lambda_0$ is a loop and $\lambda_1$ a constant path then $$\mu(\lambda_0,\lambda_1)=\mu'(\lambda_0)$$ holds?

I think this should be the case since according to [RS] with $\mu$ I'm counting the intersections of $\lambda_0$ with $\Sigma(\lambda_1(a))$, the universal Maslov cycle, and it's true (according to my knowledge/memory) that the intersection number of $\lambda_0$ with it is equal to $[\lambda_0]\in \pi_1(\mathcal{L}(2n))\cong \Bbb Z$.

More interestingly I'm trying to prove (or disprove) that if $\lambda_0(a)=\lambda_1(a)$ and $\lambda_0(b)=\lambda_1(b)$ then $$\mu(\lambda_0,\lambda_1)=\mu'(\lambda_0\ast \overline{\lambda_1})$$

At first glance it looks reasonable, and a possible way to prove it would be to consider the Maslov index for Symplectic paths, So let $\Psi_i\colon [a,b]\to Sp(2n)$ such that $\lambda_i(s)=\Psi_i(s)\lambda_i(a)$. By naturality \begin{align*}\mu(\lambda_0,\lambda_1) &=\mu(\Psi_0(s)\lambda_0(a),\Psi_1(s)\lambda_1(a)) \\ &=\mu(\Psi_1^{-1}\circ\Psi_0(s)\lambda_0(a),\lambda_1(a)) \\ &=\mu'(\Psi_1^{-1}\circ\Psi_0(s)\lambda_0(a)) \end{align*}

But I can't relate $\Psi_1^{-1}\circ\Psi_0(s)$ with $\overline{\Psi}_1\ast \Psi_0$ as paths in $Sp(2n)$, which would prove my claim. Any suggestion?

References

[RS] J. Robbin, D. Salamon, The Maslov index for paths, Topology Volume 32, Issue 4, October 1993, Pages 827-844

[MS] D. McDuff, D. Salamon, J-Holomorphic curves and Symplectic Topology, AMS Colloquim Publications, Volume 25