# Holonomy of a triangle in an affine symmetric space

Let $$G/H$$ be an affine symmetric space with involution $$\sigma$$, and $$\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{h}$$ the Cartan decomposition of its Lie algebra. We can identify $$G/H$$ and $$\exp(\mathfrak{m})$$. Given $$p,q\in \exp(\mathfrak{m})$$, I am wondering how to compute the holonomy of the curve formed by the geodesic segments, $$e\rightarrow p$$, $$p\rightarrow q$$, $$q\rightarrow e$$, $$e$$ being the identity. If I understand correctly, the holonomy should be given by a $$Ad_k$$ with $$k\in H$$. Can we compute $$k$$ in closed form from $$p$$ and $$q$$? I suspect it might be possible but I didn't manage.

For each $$p\in G/H$$, let $$\iota_p:G/H\to G/H$$ be the geodesic inversion through $$p$$, i.e., the map that reverses all geodesics through $$p$$. Of course, $${\iota_p}^2$$ is the identity. If $$AB\subset G/H$$ is a geodesic segment with midpoint $$M$$, then the map $$R_A^B:G/H\to G/H$$ defined by $$R_A^B = \iota_M\circ\iota_A$$ is a symmetry of $$G/H$$, i.e., it is left action by an element of $$G$$, that carries $$A$$ to $$B$$. Note that $$R_A^B$$ depends on the actual geodesic segment $$AB$$, not just on the endpoints $$A$$ and $$B$$.
Now, it turns out that the derivative of $$R_A^B$$ at $$A$$, i.e., $$(R_A^B)'(A):T_A(G/H)\to T_B(G/H)$$, is the result of parallel translating tangent vectors at $$A$$ along the geodesic segment $$AB$$ to tangent vectors at $$B$$. As a result, if $$A$$, $$B$$, and $$C$$ are points in $$G/H$$ such that $$AB$$ is a geodesic segment with midpoint $$M$$, $$BC$$ is a geodesic segment with midpoint $$N$$, and $$CA$$ is a geodesic segment with midpoint $$P$$, then the holonomy around the sides of the resulting geodesic triangle (in the order $$A\to B\to C\to A$$) is given by $$R_C^A\circ R_B^C\circ R_A^B = \iota_P\circ\iota_C\circ\iota_N\circ\iota_B\circ\iota_M\circ\iota_A\,.$$ (Note that this map fixes $$A$$, so, as expected, it represents left action by an element of $$H$$.)
In case one has a reasonably explicit formula for the map $$\exp:{\frak{m}}\to G$$ and its inverse, one can work out an explicit formula for the holonomy in that case. For example, when $$G = \mathrm{SO}(n{+}1)$$ and $$H=\mathrm{SO}(n)$$, so that $$G/H = S^n$$, the unit vectors in $$\mathbb{R}^{n+1}$$, and $$A,B\in S^n$$ are not antipodal (i.e., $$A+B\not=0$$), one finds $$R_A^B(Z) = Z +2\,(A{\cdot}Z)\,B - \frac{(A{+}B)\cdot Z}{1+A{\cdot}B}\,(A+B).$$ Using this and a little spherical trigonometry, one sees that, for a geodesic triangle with vertices $$A$$, $$B$$, and $$C$$ (no pair antipodal), then the holonomy of the geodesic path $$A\to B\to C\to A$$ is trivial unless $$A$$, $$B$$, and $$C$$ are linearly independent, spanning a $$3$$-plane $$E\subset\mathbb{R}^{n+1}$$, in which case, the holonomy is rotation about $$A$$ in the $$3$$-plane $$E$$ by an angle equal to the (signed) area of the spherical triangle with vertices $$A$$, $$B$$, and $$C$$ in the $$2$$-sphere $$S^n\cap E$$, as expected.