Equivalence between statements of Hodge conjecture - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:21:31Z http://mathoverflow.net/feeds/question/103365 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103365/equivalence-between-statements-of-hodge-conjecture Equivalence between statements of Hodge conjecture Shanmukha_Srinivasan 2012-07-28T07:07:55Z 2012-07-28T14:05:47Z <p>Dear everyone, </p> <p>I was unable to obtain the equivalence between the two statements of the Hodge conjecture. I searched for some previous questions that others asked here, to check whether someone has asked the same thing previously. But I didn't find any such instance, that is why I am asking. We know that Hodge conjecture gives some relation between the topological cycles and algebraic cycles. But I have read two different variations of the same conjecuture. I number my pointers. </p> <ol> <li>A fantastic description given by Prof.Dan Freed (<a href="http://www.ma.utexas.edu/users/dafr/HodgeConjecture/netscape_noframes.html" rel="nofollow">here</a>), which an undergraduate student can also understand. </li> <li>A bit tough description given by Prof.Pierre Deligne (<a href="http://www.claymath.org/millennium/Hodge_Conjecture/Official_Problem_Description.pdf" rel="nofollow">here</a>), with lot of technical terms and constructions. </li> </ol> <p>So I was befuddled in asking myself that how can one obtain equivalence between the both statements. </p> <p><strong>Dan Freed's Version :</strong> </p> <blockquote> <p>He considers a Topological cycle ( boundary less chains that are free to deform ) on a projective manifold. Then he says that the topological cycle is homologous to a rational combination of algebraic cycles, if and only if the topological cycle has a rotation number Zero. </p> </blockquote> <p><strong>P.Deligne's Version :</strong> </p> <blockquote> <p>On a projective non-singular algebraic variety over $\mathbb{C}$ , and Hodge Class is a rational combination of classes $\rm{Cl(Z)}$ of algebraic cycles. </p> </blockquote> <p>So now I have the following queries for my learned friends.</p> <ul> <li>How can one explain that both the statements are equivalent to each other ? One speaks about the rotation number and another doesn't even speak about it. How can one say that both the statements are valid ? I infact know that both the statements are valid ( as both the speakers are seminal mathematicians ) But how ? </li> <li>So can anyone explain me what the <em><a href="http://en.wikipedia.org/wiki/Rotation_number" rel="nofollow">Rotation number</a></em> has to do with the Hodge Conjecture ? I obtained some information about the rotation number from Wiki. But I am afraid , to decide whether Freed is speaking about the same rotation number ( given in wiki ) in his talk ? or something different ? </li> </ul> <p>I would be really honored to hear answers for both of them . Thank you one and all for sparing your time reading my question. </p> http://mathoverflow.net/questions/103365/equivalence-between-statements-of-hodge-conjecture/103385#103385 Answer by BS for Equivalence between statements of Hodge conjecture BS 2012-07-28T14:05:47Z 2012-07-28T14:05:47Z <p>The rotation number in question has to do with the behaviour of a differential form $\omega$ on the manifold $X$ under rotation $e^{i\theta}$ of tangent vectors : a complex $k$-form $\omega$ has rotation number $p-q$ iff $$\omega(e^{i\theta}v_1,\dots,e^{i\theta}v_k)=e^{i(p-q)\theta}\omega(v_1,\dots,v_k)$$ (with $p+q=k$, of course). </p> <p>This should explain the terminology, although it is not much in use today. Such a form is called a $(p,q)$-form, their space is denoted $\Omega^{p,q}(X)$, and the subspace of $H^{p+q}(X,\mathbb{C})$ (de Rham cohomology) of classes having a (closed) $(p,q)$-form representative is denoted $H^{p,q}(X)$. </p> <p>Hodge theory then gives a decomposition $H^k(X,\mathbb{C})=\bigoplus_{p+q=k} H^{p,q}(X)$. </p> <p>Now a topological cycle $C$ (or its class) of (real) codimension $2p$ has a <em>Poincaré dual</em> de Rham cohomology class $[\omega]$, for a closed $2p$ form $\omega$, such that intersection of cycles of <em>dimension</em> $2p$ with $C$ coincides with integration of $\omega$.</p> <p>"The rotation number of $C$ is zero" means that one can choose $\omega$ of type $(p,p)$.</p> <p>On the other hand, a cohomology class in $H^{2p}(X,\mathbb{Q})$ is said to be Hodge if under the Hodge decomposition $$H^{2p}(X,\mathbb{Q})\otimes\mathbb{C}\simeq \bigoplus_{i=-p}^{p} H^{p-i,p+i}(X)$$ it belongs to $H^{p,p}(X)$.</p> <p>I hope this clarifies the equivalence of the two statements you found.</p>