MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been trying to understand signs and conventions in the Hodge theory. Clearly, I am no expert in this area. I apologize if I ask stupid question(s) and make wrong comment(s).

In Deligne's paper "Theorie de Hodge II", he defined the polarization of a $\mathbb Q$-Hodge structure $(V, h)$ of weight $n$ to be a homomorphism $\psi: V \otimes V \to \mathbb Q(-n)$ such that $(2 \pi \mathbf i)^n \psi(x, h(\mathbf i)y)$ is symmetric and positive definite on $V_\mathbb R$. Here $\mathrm i$ is a chosen root of $x^2+1=0$. The condition above is equivalent to $\psi(x,y) = (-1)^n\psi(y,x)$ and $(2 \pi \mathbf i)^n \mathbf i^{p-q}\psi_\mathbb C(z, \bar z)>0$ for all $z \in V^{pq}$ ($n=p+q$).

Clearly his choice of sign is so that, for a smooth complex projective variety $X$ of dimension $d$, the cup-product pairing on the primitive part of $H^n(X,\mathbb Q)$ given by $(x,y) \mapsto x \wedge y \wedge c_1(\mathcal O(1))^{d-n}$ is a polarization of weight $n$.

  1. My first question is probably my ignorance: I went back to check the Hodge-Riemann bilinear relation in Griffith-Harris' AG book. There is a sign of the form $(-1)^{n(n-1)/2}$ which doesn't seem to die no matter how I normalize things. Could someone give me a hint on how everything is converted? Or a reference where this is explained?

  2. Deligne claimed that if we had chosen the other root $-\mathbf i$ instead, the theory works out equivalently. But I didn't quite get this part. At least, if $\psi$ were a polarization before, we will have to take $(-1)^n \psi$ to get a polarization. I have a bit of difficult time to rap my head around this. There's gotta be something else changed here.

  3. Even more confusing to me: in the paper I just mentioned, Deligne's convention is that $h(z)$ acts on $V^{pq}$ by $z^p \bar z^q$. Later, he decided (for good cause of course) that $h(z)$ should act on $V^{pq}$ by $z^{-p} \bar z^{-q}$. For example, this is the case of his paper in Covallis on Shimura varieties. My feeling is that, due to the change of convention, one needs to change the positivity of $(2 \pi \mathbf i)^n \psi(x, h(\mathbf i)y)$ into the positivity of $(2 \pi \mathbf i)^n \psi(x, h(\mathbf i)^{-1}y)$. (The latter is the same as $(2 \pi \mathbf i)^n \psi( h(\mathbf i) x,y)$.) Deligne didn't seem to have addressed this change. But note that this has the same effect as changing $\mathbf i$ to $-\bf i$. Maybe this is just a convention thing? Or have I missed out something?

Many thanks in advance!

share|cite|improve this question
I've also been confused by this issue, but as for your question 1: the answer is that the natural pairing on primitive cohomology is a polarization up to $(-1)^{n(n+1)/2}$. This of course seems to disagree with your sign by a factor of $(-1)^n$! – Keerthi Madapusi Pera Dec 31 '12 at 7:11
G-H is unreliable for signs, so beware. Definitions must be $i$-independent or else the theory can't be compatible with the algebro-geometric one in which there's no choice of $i$ (as Deligne knows very well). But you seem to be miscomputing in #2: if you negate $i$ then you acquire $(-1)^n$ on the outside and $h(-1) = (-1)^n$ on the 2nd variable inside, canceling out as desired. (A more elegant "solution" is to introduce an $i$-independent Weil operator $C:V_{\mathbf{R}} \simeq V_{\mathbf{R}}(n)$ and consider $\psi(x,Cy)$, but this comment box is too small to get into that.) – user29720 Dec 31 '12 at 7:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.