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Below, I have added an answer to your question about the case of a complex semi-simple Lie group. It turns out that the answer is 'no' even in this case. The answer to your general question is 'no'. …

A particularly simple non-homogeneous example in which one can explicitly integrate the Jacobi equations is the complete metric on $\mathbb{R}^2$ given by $$ g = (x^2{+}y^2{+}2)\bigl(\mathrm{d}x^2+\ma …

Actually, I just remembered that your question is addressed in Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces. Look at Theorem 11.1 of Chapter VII, which implies what you want, sin …

I think that this is treated in Helgason's Differential Geometry, Lie Groups and Symmetric Spaces. The point is that, for any Riemannian symmetric space $G/K$, one has the notion of the rank $r$ of t …

You might want to look at the book "Mumford-Tate Groups and Domains: Their Geometry and Arithmetic", by Mark Green, Phillip Griffiths, and Matt Kerr. You should also look at the recent work of Collee …

The affine group of $(M,\nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Dif …

In this particular case, the invariant theory is pretty simple, so there is an explicit description. Recall that, in general, for a homogeneous space $M=G/H$, a $G$-invariant, $m$-th order linear d …

I'm not sure what you'll count as 'elegant' or as a 'description', but here's something you may find interesting or useful: Recall that the space of curvature tensors $\mathcal{K}(V)$ on an inner pro …

I suppose that $S^2_B(\Lambda^2(\mathbb{R}^n))$ means the subspace of $S^2(\Lambda^2(\mathbb{R}^n))$ that satisfies the Bianchi identity, i.e., the kernel of the natural map $S^2(\Lambda^2(\mathbb{R}^ …

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 c …

I'm sorry if this is obvious to everyone, but I thought that it was worth mentioning: I don't know the answer to the question asked, but the answer to the easier question, "Does the space $\mathrm{E} …

One reference is in Helgason's 1984 book Groups and Geometric Analysis. The result you want appears there as Corollary 5.12. The notation he uses is $X=G/K$ is a symmetric space where $G$ is connecte …

As I mentioned in my comment, a pair of orthogonal $2$-discs in $\mathbb{R}^4$ is known to be minimizing among all orientable $2$-currents with the same boundary by the technique of calibrations: One …

This is true. One can use a few facts from Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces to show that, indeed, $K = \mathrm{Stab}_G(Z)$. Since $M=G/K$ is an irreducible Hermitian …