# Explicit construction of a (the?) dual symmetric space

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.

• You can, and when referring to a specific answer ideally should, link to answers, not just to questions. I have edited accordingly. Aug 22 at 17:17
• @LSpice Thank you for the feedback and edit! This indeed makes the text easier to "navigate".
– S.T.
Aug 22 at 20:01

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.

• Why does constructing something make it unique? Sep 16 at 12:04
• If the definition uses the explicit construction of the object then it is unique up to the choices that the definition requires in that construction. In this case, that is the choice of orthogonal symmetric Lie algebra $(\mathfrak{g},\Theta)$. So this construction is unique up to a choice of Lie algebra: $\mathfrak{g}$ (all of which are isomorphic) and a matching involution on $\mathfrak{g}$: $\Theta$ (which are all conjugate under the action of the group). We also need a choice of $G^\mathbb{C}$ and so on but e.g. $K, G^*$ are defined by our other choices by the subgroups-subalgebras theorem Sep 16 at 12:32
• Of course for an orthogonal symmetric Lie algebra $(\mathfrak{g}, \Theta)$, we can take its dual $(\mathfrak{g}^*, \Theta^*)$, take the unique simply-connected (real) Lie group $G^*$ with this Lie algebra, and then take $K^*$ to be the unique connected Lie subgroup with Lie algebra equal to $\ker(\Theta^* - 1)$. But why can $K$ be identified with $K^*$ (as isomorphic maybe?). They have the same Lie algebra, but why are they "basically the same stabilizer group"?
– S.T.
Sep 18 at 11:45
• Moreover, in the linear case, we can indeed just consider the usual complexification of linear Lie groups as just viewing them as complex matrices. However, I would like to see something that works in the general case. Why do you think there may be symmetric spaces without duals?
– S.T.
Sep 18 at 11:46
• My only concern is that if we fix a simply connected symmetric space and its dual that the (not simply connected) symmetric spaces which have those as covers might not pair up neatly. This may be an unfounded concern but I don't see a way to prove that easily. Most sources I see simplify either to the linear case or the simply connected case so I haven't found a clear result in the more general picture Sep 18 at 13:38