Explicit construction of a (the?) dual symmetric space

Question

I am looking for a reference, proof or disproof of the fact that every Riemannian globally symmetric space of compact (non-compact) type has a "dual", which is of non-compact (compact) type.

More precisely, in [1], there is the construction of a "dual Lie algebra" of an orthogonally symmetric Lie algebra $(\mathfrak{g}, \Theta)$; the dual Lie algebra is given as $$\mathfrak{g}^* := \mathfrak{u} \oplus i\mathfrak{p} \subset \mathfrak{g}^{\mathbb{C}}, $$ where $$\mathfrak{g} = \mathfrak{u} \oplus \mathfrak{p} $$ is the Cartan decomposition of $\mathfrak{g}$, with $$\mathfrak{u} := \ker(\Theta - \mathrm{Id}), \quad \mathfrak{p} := \ker(\Theta + \mathrm{Id}), $$ $\mathfrak{g}^{\mathbb{C}}$ is the complexification of $\mathfrak{g}$, and the Cartan involution $\Theta^*$ on $\mathfrak{g}^*$ is given by the complex conjugation map.

Helgason then proves several properties of this duality, one of them being that $(\mathfrak{g}, \Theta)$ is of compact (non-compact type) if and only if $(\mathfrak{g}^*, \Theta^*)$ is of non-compact (compact type). Naturally, I would expect that this duality can be used to study the geometry of these spaces, not just the algebra.

More precisely, we can make the definition that two Riemannian globally symmetric spaces are dual to each other if the corresponding orthogonally symmetric Lie algebras are dual to each other. However, does a globally symmetric space $(M, g)$ (with corresponding Riemannian symmetric pair $(G, K)$) of, say, compact type, always admit a non-compact dual (defined as above)? Moreover, up to what extent is this dual, if it exists, unique? I expect it should be unique up to isometry.

I have not been able to find a reference where the above questions are answered. In Curvature of dual symmetric space on Math Stack Exchange, @J. Salieri gives an answer in which the dual is "somehow constructed": the dual symmetric space is supposedly $G^*/K$, where $G^*$ is the unique connected Lie subgroup of $G^{\mathbb{C}}$ (the complexification of $G$) with Lie algebra equal to $\mathfrak{g}^*$ (a similar construction can be seen at How does duality of symmetric spaces explain the hyperbolic cosine theorem?). The answer to Compact dual of a noncompact Lie group shows some left-out details more explicitly, but I still do not fully understand it.

More precisely, I do not understand what kind of complexifications these are, and how $K$ is identified as a "stabilizer subgroup" of $G^*$. I know about the universal complexification of a Lie group, but the "inclusion map" from $G$ into its universal complexification need not be injective; but even it if were, it is not clear to me how to see $K$ as a subgroup of $G^*$.

[1] S. Helgason, Differential Geometry, Lie Groups, and Symmetric Spaces.

Answer 1

The only problem that we may run into when converting the algebraic construction to a geometric one is that of coverings. To avoid this for example we could restrict ourselves to simply connected symmetric spaces where this duality will be clear: There is a unique simply connected Riemannian symmetric space whose corresponding symmetric Lie algebra is dual to our original one and is of compact/noncompact type as our original is of noncompact/compact type. Each symmetric space has a universal cover which is simply connected.

Alternatively, if we choose $G$ to be linear so that we don't have to worry about universal complexifications and can just consider the honest complexification, the construction you mention by @J. Salieri works clearly and there are no issues. You can consider $G,G^*,K \subset G^\mathbb{C}$ and so the dual is straightforward to construct (and since we are constructing it, it is inherently unique in the appropriate sense).

I suspect there may be symmetric spaces that are left without a dual if perhaps there are more Lie groups with Lie algebra $\mathfrak{g}$ than there are with Lie algebra $\mathfrak{g}^*$ but their universal cover will always have a dual even so.