# The Hodge Conjecture

 Importance: Outstanding ✭✭✭✭
 Author(s): Hodge, W. V. D.
 Subject: Geometry » Algebraic Geometry
 Keywords: Hodge Theory Millenium Problems
 Prize: $1,000,000 from the Clay Mathematics Institute  Posted by: Charles on: July 13th, 2008 \begin{conjecture} Let$X$be a complex projective variety. Then every Hodge class is a rational linear combination of the cohomology classes of complex subvarieties of$X$. \end{conjecture} % You may use many features of TeX, such as % arbitrary math (between$...$and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{http://www.research.att.com/~njas/sequences/} A complex projective variety is the set of zeros of a finite collection of homogeneous polynomials on projective space, and we are concerned with the singular cohomology ring. There is a well known Hodge Decomposition of the cohomology into groups$H^{p,q}(X.\mathbb{C})$which hare holomorphic in$p$variables and antiholomorphic in$q$variables with the property that$\oplus_{p+q=k}H^{p,q}=H^k$. So we define the Hodge classes to be those in the intersection$H^{k,k}(X,\mathbb{C})\cap H^{2k}(X,\mathbb{Q})$. It is fairly easy to show that the cohomology class of a subvariety is Hodge. We say that a cycle is \emph{algebraic} if it is a rational linear combination of the classes of subvarieties. So every algebraic cycle is Hodge. In dimension one, we have the following result: \begin{theorem}[Lefshetz (1,1) Theorem] Any element of$H^2(X,\mathbb{Q})\cap H^{1,1}$is the cohomology class of a divisor, and so is algebraic. \end{theorem} It's also true that if the Hodge Conjecture holds for cycles of degree$p2n-p\$. So this and the (1,1) Theorem show that the Hodge Conjecture is true for complex curves, surfaces and threefolds.