Dear everyone,

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.

  1. A fantastic description given by Prof.Dan Freed (here), which an undergraduate student can also understand.
  2. A bit tough description given by Prof.Pierre Deligne (here), with lot of technical terms and constructions.

So I was befuddled in asking myself that how can one obtain equivalence between the both statements.

Dan Freed's Version :

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.Deligne's Version :

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.

So now I have the following queries for my learned friends.

  • 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 ?
  • So can anyone explain me what the Rotation number 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 ?

I would be really honored to hear answers for both of them . Thank you one and all for sparing your time reading my question.

  • $\begingroup$ A silly question, Is Dan Freed a member of MO community ? I have found that many seminal mathematicians of this century are members in MO, hence asked about Dan too. $\endgroup$ Jul 28, 2012 at 7:09
  • $\begingroup$ You can also look at Hodge's original paper ada00.math.uni-bielefeld.de/ICM/ICM1950.1/Main/… the condition of zero "rotation number" is on page 184 (although he doesn't use this word) $\endgroup$
    – YangMills
    Jul 28, 2012 at 14:12
  • $\begingroup$ @YangMills : Your link was quite informative. Thank you. $\endgroup$ Jul 28, 2012 at 15:19

1 Answer 1


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).

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 $H^{p,q}(X)$.

Hodge theory then gives a decomposition $H^k(X,\mathbb{C})=\bigoplus_{p+q=k} H^{p,q}(X)$.

Now a topological cycle $C$ (or its class) of (real) codimension $2p$ has a Poincaré dual de Rham cohomology class $[\omega]$, for a closed $2p$ form $\omega$, such that intersection of cycles of dimension $2p$ with $C$ coincides with integration of $\omega$.

"The rotation number of $C$ is zero" means that one can choose $\omega$ of type $(p,p)$.

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 $H^{p,p}(X)$.

I hope this clarifies the equivalence of the two statements you found.

  • $\begingroup$ Clear Answer. Thank you, very gentle introduction to beginners and nice motivation. $\endgroup$ Jul 28, 2012 at 15:20

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