Isotropic subspaces in cohomology Hello,
Here is a problem I encountered in the study of Kähler manifolds but there is a natural generalisation of this for topological spaces.
If $X$ is a topological space, denote by $g_\mathbb{R}$ the real genus of $X$, that is the maximal dimension of an isotropic subspace in $H^1(X,\mathbb{R})$ (isotropic means that the cup-product restricted to this space is $0$). We can in the same way define $g_\mathbb{C}$ (here we take the complex dimension).
Now the question is : $g_\mathbb{C} = g_\mathbb{R}$ ?
This seems totally obvious : if one has a real isotropic space, then its complexification is a complex isotropic subspace but conversely I don't see how to construct a real isotropic subspace from a complex one.
This is true if $H^2$ has dimension $0$ or $1$ : $0$ is clear and for $1$ one can see the cup-product as a standard symplectic form (maybe degenerate), but in general ?
Thank you for your answers and sorry if I just missed something obvious.
 A: In the full linear algebra generality, the answer is no. 
Take four generic $2$-planes, $V_1$, $V_2$, $V_3$ and $V_4$ in $\mathbb{R}^4$. Over $\mathbb{C}$, there are always two $2$-planes $W$ such that $W \cap V_i \neq (0)$ for $1 \leq i \leq 4$. (This is the first nontrivial Schubert calculus example.) Choose the $V_i$ such that the $W$ are NOT defined over $\mathbb{R}$. If you want a concrete example, take
$\def\span{\mathrm{Span}_{\mathbb{R}}}$ $\span(e_1, e_2)$, $\span(e_3, e_4)$, $\span(e_1+e_3, e_2+e_4)$ and $\span(e_1+e_4, e_2-e_3)$; the two $W$'s are $\mathrm{Span}_{\mathbb{C}}(e_1 + ie_2, e_3 + i e_4)$ and its complex conjugate.
Let $\omega_i$ be a degenerate skew-symmetric bilinear form of rank $2$, for which $V_i$ is the kernel. For example, if $V_i$ is $\span(e_1, e_2)$, you could take $\omega_i = e_3^* \wedge e_4^*$. Lemmma: $W$ is $\omega_i$-isotropic if and only if $W$ has nontrivial intersection with $V_i$. 
So, with the $V_i$ as above, there are two complex isotropic $2$-planes, but no real ones.
