What does the Hodge conjecture mean? 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.
 A: Consider a Kähler  manifold  $X=X_n$ of dimension $n$, for example a projective algebraic manifold .
It is in particular a topological space and as such has cohomology groups $H^r(X,\mathbb C)$.   
A very down-to-earth  way of studying it is through De Rham cohomology.
Moreover the Kähler structure (= a special type of Riemannian structure) on $X$ permits us to define the vector space $H^{p,q}$ of harmonic forms of type $p,q$ and the  pleasant result is that they are canonical representatives of all cohomology classes: $$H^r_{DR}(X,\mathbb C)=\oplus_{p+q=r}H^{p,q}$$
Let me insist that the elements of $H^{p,q}$ are genuine differential forms, not equivalence classes modulo exact forms. Very satisfying!
This is due to Hodge and is a well-established theorem, not conjectural at all.
It has very interesting topological consequences, for example that the odd Betti numbers $b_{2k+1}=dim_{\mathbb C}H^{2k+1}_{DR}(X,\mathbb C)$ are even.
This allows one to prove, for example,  that the product $S^1\times S^3$ has no algebraic structure (although it has a holomorphic structure discovered by Hopf), because its first Betti number $b_1=1$ is odd.   
The Hodge conjecture claims to predict for a projective algebraic manifold $X$  which cohomology classes (with  coefficients in $\mathbb Q$ actually) come from algebraic subvarieties of $X$.
There are some obvious necessary conditions on these classes, since for example  algebraic subvarieties of $X$ of codimension $p$ will give cohomology classes necessarily in $H^{p,p}$.
Hodge's conjecture is that these conditions are sufficient.     
Since I certainly didn't want to subject myself to the ridicule of trying to improve or equal Deligne's presentation (link in Parsa's comment) I have just tried to write a few introductory words to his text.
You are now in his hands: good luck!
A: 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:

(Not the Hodge conjecture) A cohomology class $\omega$ in $H^{2n-2k}_{DR}(X)$ is of the form $\omega_{Z \to X}$ for a complex submanifold $Z$ if and only if $\omega$  can be represented by an $(n-k,n-k)$-form and lies in the image of $H^{2n-2k}(X, \mathbb{Z}) \to H_{DR}^{2n-2k}(X)$.

The actual Hodge conjecture The actual Hodge conjecture is

(The Hodge conjecture) Let $X$ be projective. A cohomology class $\omega$ in $H_{DR}^{2n-2k}(X)$ is of the form $\sum a_i \omega_{Z_i \to X}$, for some complex subvarieties $Z_i$ of $X$, and for some $a_i \in \mathbb{Q}$, if and only if $\omega$ can be represented by an $(n-k,n-k)$-form and lies in the image of $H^{2n-2k}(X, \mathbb{Q}) \to H^{2n-2k}_{DR}(X)$.

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

Atiyah and Hirzebruch, Algebraic Cycles on Complex Manifolds,
  Topology 1 (1962) p. 25-45

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.
