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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.

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This is a question about linear algebra, not topology. Every skew-symmetric map $\mathbb{Z}^k \times \mathbb{Z}^k \to \mathbb{Z}^n$ occurs as cup product for some topological space: Just take a wedge of $k$ circles and glue on $n$ $2$-cells with appropriate boundaries. I don't know the answer, though. – David Speyer Mar 8 '11 at 13:58
You are right, in full generality this is a question of linear algebra but maybe one can say something when the question is restricted to some class of manifolds. For instance this seems to be true for Kähler manifolds and I want to see if this has something to do with Hodge decomposition. – mister_jones Mar 8 '11 at 14:25
You might be interested in the notion of a $g$-element -- this axiomatizes some of the properties of a Kahler class. I haven't thought through exactly what it would do for you in this case, though. – David Speyer Mar 8 '11 at 17:04
up vote 3 down vote accepted

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.

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This is a nice example and seems to show that in high dimensions such examples are not at all pathological. Do you think than we can obtain in this way a result like : "a generic finitely presentable group (with sufficiently generators and relations) does not verify $g_\mathcal{R} = g_\mathcal{C}$" ? – mister_jones Mar 9 '11 at 19:20

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