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.



Isotropic subspaces in cohomology
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.


