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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.

Thank you for your help.

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1 Answer 1

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There is a general formula, essentially due to Cartan:

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.

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