I read from the Internet that according to the Hodge conjecture, a certain harmonic differential form in a projective, non-singular algebraic variety is a rational linear combination of the cohomology classes of algebraic cycles. Could anyone explain me what is that particular differential form? And how do one defines the cohomology class of algebraic cycles? I have read Atiyah's and McDonald's book on commutative algebra, two first chapters on Liu's "Algebraic geometry and Arithmetic curves" and fundamental groups on algebraic topology.
[Math] What does the Hodge conjecture mean
algebraic-geometryalgebraic-topologyopen-problem
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A compact Riemann surface $X$ is in particular a compact real orientable surface. These surfaces are classified by their genus.
That genus is indeed the number of handles cited in popular literature; more technically it is
$$g(X)=\frac {1}{2}\operatorname {rank} H_1(X,\mathbb Z) = \frac {1}{2}\operatorname {dim} _\mathbb C H^1_{DR}(X,\mathbb C) $$ in terms of singular homology or De Rham cohomology.
Under the pressure of arithmetic, geometers have been spurred to consider the analogue of compact Riemann surfaces over fields $k$ different from $\mathbb C$: complete smooth algebraic curves.
These have a genus that must be calculated without topology.
The modern definition is (for algebraically closed fields) $$ g(X)=\operatorname {dim} _k H^1(X, \mathcal O_X)= \operatorname {dim} _kH^0(X, \Omega _X)$$
in terms of the sheaf cohomology of the structural sheaf or of the sheaf of differential forms of the curve $X$.
Of course this geometric genus is always $\geq 0$.
There is a more general notion of genus applicable to higher dimensional and/or non-irreducible varieties over non algebraically closed fields: the arithmetic genus defined by $$p_a(X)=(-1)^{dim X}(\chi(X,\mathcal O_X)-1)\quad {(ARITH)}$$ (where $\chi(X,\mathcal O_X)$ is the Euler-Poincaré characteristic of the structure sheaf).
[ Hirzebruch and Serre have, for very good reasons, advocated the modified definition $p'_a(X)=(-1)^{dim X}\chi(X,\mathcal O_X)$, which Hirzebruch used in his ground-breaking book and Serre in his foundational FAC]
For smooth projective curves over an algebraically closed field $g(X)=p_a(X)\geq 0$ : no problem.
It is only in more general situations that the arithmetic genus $p_a(X)$ may indeed be $\lt 0$
Edit
The simplest example of a reducible variety with negative arithmetic genus is the disjoint union $X=X_1\bigsqcup X_2$ of two copies $X_i$ of $\mathbb P^1$.
The formula $(ARITH)$ displayed above yields: $p_a(X)=1-\chi(X,\mathcal O_X)=1-(dim_\mathbb C H^0(X,\mathcal O_X)-dim_\mathbb C H^1(X,\mathcal O_X))=1-(2-0)$
so that $$p_a(X)=p_a(\mathbb P^1\bigsqcup \mathbb P^1)=-1\lt0$$
No differential geometry or algebraic topology is necessary, though the former might help with motivation for some things, depending upon your mathematical inclinations. You need to know basic graduate abstract algebra (he develops most of the commutative algebra he needs so you don't need to know that in advance) and you need to know some basic point set topology (definition of a topological space, continuous map, Hausdorff, not much more). That's it. For serious.
Would it help to already know some commutative algebra? Yeah, probably, but it's not essential by any means. Liu's is a remarkably self-contained book. I consider it to be an excellent first book on modern algebraic geometry. Whether or not you should first learn some classical algebraic geometry depends on you and your tastes. Liu does do some of that (algebraic sets) in the second chapter. Being familiar with the classical theory might help with motivation, but I don't think it's necessary.
Best Answer
First of all, the Hodge conjecture is not about one particular differential form. It says that any differential form which satisfies certain conditions will be a $\mathbb{Q}$-linear combination of algebraic forms. There are some particular forms which satisfy those conditions but have not been proven to be such a linear combination.1 However, it is my very-uninformed impression that the hardest case of the Hodge conjecture is not those cases, but rather the possibility that some differential form might just happen to satisfy the Hodge conditions for no good reason, and that such a form might prove to be a counterexample.
It's hard to answer this question without knowing your background. I'm going to assume that you've had a good course in differential geometry, and that you know what a complex manifold is.
Differential forms and submanifolds Let $X$ be a compact oriented manifold (nothing complex yet) and let $Z$ be a compact oriented closed submanifold. Let $n=\dim X$ and $k=\dim Z$. Write $\Omega^\ell(X)$ for the vector space of differential forms on $X$ of degree $\ell$. Write $Z^{\ell}(X)$ for the closed forms, $B^{\ell}(X)$ for $d \Omega^{\ell-1}(X)$ and $H_{DR}^{\ell}(X)$ for the deRham cohomology $Z^{\ell}(X)/B^{\ell}(X)$.
We have a linear map $Z^k(X) \to \mathbb{R}$ given by $\eta \mapsto \int_Z \eta|_Z$. This map descends to $H_{DR}^k(X)$, because $\int_Z (d \alpha)|_Z = \int_Z d (\alpha|_Z) = \int_{\partial Z} \alpha|_{\partial Z} =0$. (Remember that $\partial Z=\emptyset$.)
By Poincare duality, every linear map $H_{DR}^k(X) \to \mathbb{R}$ is of the form $\eta \mapsto \int_X \omega \wedge \eta$ for some $\omega \in Z^{n-k}(X)$. Moreover, changing $\omega$ by a class in $B^{n-k}(X)$ does not change this map, so one can think of $\omega$ as a class in $H_{DR}^{n-k}(X)$. (The Wikipedia article on Poincare duality is less clear than I would like on this point. Suggestions?)
Let $\omega_{Z \hookrightarrow X}$ be the class in $H_{DR}^{n-k}(X)$ such that $$\int_Z \eta|_Z = \int_X \omega_{Z \hookrightarrow X} \wedge \eta \quad (\dagger)$$ for any $\eta \in Z^k(X)$.
Explicit formulas for $\omega_{Z \hookrightarrow X}$ can be found in differential geometry texts. For our purposes, the important fact is $(\dagger)$. One thing which is good to know is that the differential form $\omega_{Z \hookrightarrow X}$ can be taken to be zero away from a neighborhood of $Z$.
We have defined $\omega_{Z \hookrightarrow X}$ to be in $H_{DR}^{n-k}(X)$. In fact, it will always lie in the image of $H^{n-k}(X, \mathbb{Z})$ within $H_{DR}^{n-k}(X)$. I will not try to explain this, since even defining this map requires a fair amount of algebraic topology and differential geometry. It turns out that, for $X$ compact and oriented, $\eta$ is in the image of $H^p(X, \mathbb{Z})$ if and only if $\int_{\alpha} \eta \in \mathbb{Z}$ when the cycle $\alpha$ runs over a basis of $H_p(X, \mathbb{Z})$. This is not the standard definition, but it might be the most elementary one. By the way, one of the highest voted unanswered questions on math.SE is about working with this condition.
Complex manifolds and algebraic classes Now suppose that $X$ is a complex manifold. Morally speaking, the Hodge conjecture describes which differential forms can appear as $\omega_{Z \hookrightarrow X}$ with $Z$ a complex submanifold of $X$.
Therefore, in this section of the answer, let $Z$ be a complex submanifold of $X$. Let the dimensions of $Z$ and $X$, as real manifolds, be $2k$ and $2n$. We will discuss conditions which $\omega_{Z \to X}$ should obey. In the following section, I will actually state the Hodge conjecture, which is a bit more technical than you would guess from this section.
First of all, as described above, $Z$ should be in the image of $H^{2n-2k}(X, \mathbb{Z}) \to H_{DR}^{2n-2k}(X)$.
Since $X$ is a complex manifold, it makes sense to multiple an element of the tangent bundle by a complex number, so the multiplicative group $\mathbb{C}^{\times}$ acts on $T_{\ast} X$. Thinking of differential forms as multilinear forms on $T_{\ast} X$, the group $\mathbb{C}^{\times}$ also acts on $\Omega^{\ell}(X)$. A differential form $\eta$ is called a $(p,p)$ form if $a(\eta) = |a|^{2p} \eta$ for $a \in \mathbb{C}^{\times}$.
I should probably explain the origin of this terminology. If we tensor the vector space of $\ell$-forms with $\mathbb{C}$, then the action of $\mathbb{C}^{\times}$ diagonalizes. We have $\Omega^{\ell}(X) \otimes \mathbb{C} = \bigoplus_{p=0}^{\ell} \Omega^{p, \ell-p}(X)$ where $\eta$ is in $\Omega^{p, q}(X)$ if $a(\eta) = a^p \overline{a}^q \eta$. This idea will surely be important in any proof of the Hodge conjecture, but to understand the statement you don't strictly need this paragraph.
It turns out that we can always take $\omega_{Z \to X}$ to be a $(n-k,n-k)$-form. (Recall that $\omega_{Z \to X}$ is only defined modulo exact forms, so I shouldn't say that it is a particular differential form.) Sketch of proof: Because $Z$ is a complex submanifold, the action of $\mathbb{C}^{\times}$ preserves $T_{\ast} Z$ within $T_{\ast} X$. Also, one can show that a top-dimensional form on an $m$-dimensional complex manifold is always an $(m,m)$-form. If you trace through the effects of this, you get $$\int_X a(\omega) \wedge a(\eta) = \int_X a(\omega \wedge \eta) = |a|^{2n} \int_X \omega \wedge \eta$$ and also $$\int_X \omega \wedge a(\eta) = \int_Z a(\eta)|_Z = \int_Z a(\eta|_Z) = |a|^{2k} \int_Z \eta = |a|^{2k} \int_X \omega \wedge \eta.$$ The second equality in the latter chain of equalities is where we use that $Z$ is a complex submanifold.
So $\int_X a(\omega) \wedge a(\eta) = |a|^{2n-2k} \int_X \omega \wedge a(\eta)$. From this, it follow that we can arrange that $a(\omega) = |a|^{2n-2k} \omega$. $\square$
So a reasonable guess as to what the Hodge conjecture might say is:
The actual Hodge conjecture The actual Hodge conjecture is
So we have the following changes. In most cases, I don't have a strong intuition for why these changes were needed, but this is the version which has been open for decades and is worth $10^6$ dollars. I believe that there are known counterexamples indicating the need for each of these changes, but I don't know them.
$X$ must be projective, meaning that it is a closed submanifold of $\mathbb{CP}^N$ for some $N$. There are many theorems which hold for projective manifolds and not for more general compact complex manifolds.
We are allowed a linear combination of several $Z_i$'s. This is because the conditions of being a $(n-k,n-k)$-form and of lying in the image of $H(X, \mathbb{Z})$ are both closed under addition, and the condition of representing a complex submanifold is not. (There are various positivity conditions for the latter.)
We allow the $Z_i$ to be subvarieties rather than complex submanifolds. This means that the $Z_i$ are locally defined by the vanishing of holomorphic functions, but are allowed to have singularities.
We use $\mathbb{Q}$ instead of $\mathbb{Z}$. The need for this modification was shown in
Remark on Explicitness If you are given a complex manifold $X$ and a differential form $\omega$ in degree $2p$, testing whether $\omega$ is a $(p,p)$-form is straightforward. Testing whether $\omega$ is in $H(X, \mathbb{Q})$ is equivalent to checking $\int_{\alpha} \omega \in \mathbb{Q}$ for cycles $\alpha$ running over a basis of $H_{2p}(X, \mathbb{Z})$. Since, in general, integrals can only be computed numerically, and since there is no test for whether a floating point number is rational, this can be very hard. The same situation comes up in reverse if you triangulate $X$ and give a class in $H^{\ast}(X, \mathbb{Q})$ as a cocycle for the triangulation: In this case, it is easy to see that the cycle is in $H^{\ast}(X, \mathbb{Q})$, but the condition that it be representable by a $(p,p)$ form is equivalent to certain integrals vanishing. When I said that some form might satisfy the Hodge conditions for no good reason, I was thinking of the possibility that some complicated integral might be a rational number/might be zero for no good reason. I think it should be clear that this is hard to rule out!
Footnote 1: Here I am thinking of Grothendieck's Standard Conjectures. For example, the following is Grothendieck's Conjecture C: Let $X$ be projective. By Kunneth, we have $H_{DR}^{m}(X \times X) = \bigoplus_i H_{DR}^{i}(X) \otimes H_{DR}^{m-i}(X)$. Let $\pi_{i,j}$ be the projection $H_{DR}^{i+j}(X \times X) \to H_{DR}^i(X) \otimes H_{DR}^j(X)$. Let $\Delta \subset X \times X$ be the diagonal. Conjecture C states that $\pi_{i, 2n-i}(\omega_{\Delta \to X \times X})$ is of the form $\sum a_i \omega_{Z_i \to X}$. It is relatively straightforward to show that the $\pi_{i, 2n-i}$ operators preserve the conditions of the Hodge conjecture. But, while proving Conjecture C would definitely get you tenure, I do not believe that this sort of case is considered the hardest part of the Hodge conjecture.