Hodge conjecture

Question

Hello I'm trying to understand the idea behind Hodge conjecture and I have naive approach but what does it mean Hodge classes statement: $H^{2k}(X,\mathbb{Q}) \cap H^{k,k}(X)$? these symbols? My intuition says If $X$ was not Kähler manifold (just some real smooth manifold), this notation would denote the number of holes in that manifold? bq torus by de Rham theorem has homology group isomorphic to its cohomology group: $H_{1} \cong H^{1}$. So this is about interesection of manifolds with $2k$ holes and I do not know what $k,k$ means in this context (maybe deal with some complex geometry). If that is true why rational numbers instead integers? Can you recommend some popular science books on this topic?

Answer 1

Since the question is about intuition and basic understanding, much of what I say below is wrong, or at least imprecise (and the subtleties are sources of difficulty) but I believe it is morally correct.

In the context of the Hodge conjecture, it is useful to think of $H^{n-k}(M)$, for a (closed oriented) $n$-manifold $M$ as a vector space spanned by $k$-dimensional submanifolds of $M$, where two such linear combinations are equivalent if their difference is a "hole that can be filled in", ie (some linear combination of) the boundary of $(k + 1)$-dimensional submanifolds. This already needs rational coefficients rather than integer coefficients, so we will use $H^{n-k}(M)$ to mean $H^{n-k}(M; \mathbb{Q})$.

Let us assume that $M$ is a complex Kähler manifold (in particular $n = 2d$ where $d$ is its complex dimension). Then we can add some refinements to this data. Specifically, we can look at the equations defining the submanifolds and ask how they interact with complex conjugation, kind of like the difference between setting $\Re(z) = 0$ and $\Im(z) = 0$ (note that $\Re(\bar{z}) = \Re(z)$ and $\Im(\bar{z}) = -\Im(z)$). For the lack of better terms, call these odd and even and let $H^{p, q}$ to be the subspace (of $H^{n-k}$ if $n - k = p + q$) spanned by those submanifolds cut out by $p$ even and $q$ odd functions. This further decomposition of $H^{n-k}$ into $H^{p, q}$ also only makes sense with rational coefficients (and of course only for $M$ complex). I am also ignoring the difference between $H^{p, q}(M)$ and the $\mathbb{Q}$-span of $H^{p + q}(M; \mathbb{Z}) \cap H^{p, q}(M)$.

If your submanifolds are actually cut out by complex analytic functions then you're enforcing an equal number of even and odd conditions ($z = 0$ iff $\Re(z) = 0$ and $\Im(z) = 0$) so these all lie in $H^{p, p}$ (with $n - k = 2p$). The Hodge conjecture is about the converse, when $M$ is projective (ie is itself cut out by algebraic equations in complex projective space $\mathbb{C}P^N$ of some dimension $N$). Specifically, it asks if $H^{p, p}$ is $\mathbb{Q}$-spanned by (classes of) complex submanifolds of $M$. Under the assumptions on $M$, complex submanifolds are also algebraic, so the statement is often in terms of "algebraic cycles".