# Curvature of dual symmetric space

Let $$S=G/K$$ and $$S^*=G^*/K^*$$ by symmetric spaces, with Riemannian metrics given by the Killing forms $$B$$ and $$B^*$$. We call $$S$$ and $$S^*$$ dual if

• there exists a Lie algebra isomorphism $$\phi\colon \mathfrak k \to \mathfrak k^*$$ such that $$B^*(\phi V, \phi W) = B(V,W) \text{ for all }V,W\in \mathfrak k$$
• there is a linear isometry $$\psi\colon \mathfrak p \to \mathfrak p^*$$ such that $$[\psi X,\psi Y]=-\phi[X,Y]\text{ for all } X,Y\in \mathfrak p$$

where $$\mathfrak g=\mathfrak k \oplus \mathfrak p$$ is the Cartan decomposition.

I have to show that $$S$$ and $$S^*$$ have opposite curvature. I already know that $$R(X,Y)Z=[[X,Y],Z]$$, so if I have two orthonormal vectors $$X,Y\in \mathfrak p$$, I get $$\operatorname{sec}(X,Y) = \langle R(X,Y)X,Y\rangle=\langle [[X,Y],X],Y\rangle = \langle [X,Y],[X,Y]\rangle$$. Therefore it remains to show $$-\langle [X,Y],[X,Y]\rangle_S = \langle [\psi X,\psi Y],[\psi X,\psi Y]\rangle_{S^*}$$

My problem seems to be that I'm not able to show that $$S$$ and $$S^*$$ are of different type. If I assume that, for example, $$S$$ is of non-compact type and $$S^*$$ is of compact type, then

$$\langle [\psi X,\psi Y],[\psi X,\psi Y]\rangle_{S^*} = -B^*([\psi X,\psi Y],[\psi X,\psi Y]) = -B^*(-\phi[X,Y],-\phi[X,Y])=-B([X,Y],[X,Y])=-\langle [X,Y],[X,Y]\rangle_S$$

But I have no idea how to show that, if $$S$$ is of non-compact type, then $$S^*$$ is of compact type. This however seems to be the main reason for the change of sign in the calculation. Any help?

EDIT: I'm not so sure anymore whether this problem is answerable with a positive answer. A counterexample would also be nice.

I would like to know where you take that definition of duality from. If you use the following notion of duality, it is very easy to prove that: if $$S$$ is of non-compact type, then $$S^*$$ is of compact type.
Let $$\mathfrak{g}:=\mathfrak{p}\oplus[\mathfrak{p},\mathfrak{p}]$$. Then, you can consider the complexification of $$\mathfrak{g}$$ and $$G$$. Let us call it $$\mathfrak{g}^{\mathbb{C}}$$ and $$G^{\mathbb{C}}$$, respectively. Hence, we can define $$\mathfrak{g}^*=\mathfrak{k}\oplus i\mathfrak{p}$$. It is easy to check that $$\mathfrak{g}^*$$ is a Lie subalgebra of $$\mathfrak{g}^{\mathbb{C}}$$. On the other hand, let us define $$S^*:=G^*/K,$$ where $$G^*\subset G^{\mathbb{C}}$$ is the connected subgroup of $$G^{\mathbb{C}}$$ with Lie algebra $$\mathfrak{g}^*$$. This is the dual symmetric space to $$S$$.
Notice that the Killing form of $$\mathfrak{g}$$ or $$\mathfrak{g}^*$$ is just the restriction of the Killing form of $$\mathfrak{g}^\mathbb{C}$$ to $$\mathfrak{g}$$ or to $$\mathfrak{g}^*$$, respectively. Then, the Killing form of $$\mathfrak{g}$$ is negative definite. Now for every non-zero $$X\in\mathfrak{p}$$, $$B_{g^*}(i X, i X)=B_{\mathfrak{g}^{\mathbb{C}}}(i X, i X)=-B_{\mathfrak{g}^{\mathbb{C}}}(X,X)=-B_{\mathfrak{g}}(X,X).$$
Thus, if for instance $$S$$ is of non compact-type, then, $$B_{g^*}(i X, i X)<0$$ for every non-zero $$X\in\mathfrak{p}$$, which implies that $$S^*$$ is of compact type.