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 


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*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.
 A: 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.
Observe that the reverse implication is also inmediate from the above argument.
