Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm currently taking a course in Hodge theory ... and I wonder if all the splittings in $\{i,-i\}$ Eigenvalue pairs come from the Galois group action (of the extension $\mathbb{R}\rightarrow\mathbb{C}$) - it seems to me like that (and I couldn't find such a statement in my textbook).

Is this true? If yes, is this a good way to think of Hodge decomposition or does one need more data than just the Galois group? If not, what is my misconception?

I thought (if my assumption is true), this would be a way to generalize to other algebraic field extensions.. are there analogues of Hodge theory for any algebraic field extension? Does it involve the Galois group?

If this question isn't "researchy" enough, just close it ... I will come back asking questions in a year then :-)

share|improve this question
1  
Konrad- Questions don't need to be "researchy." They need to be interesting; the main thing you should ask yourself before posting a question is "wait, is this probably really easy to Google/find on Wikipedia?" –  Ben Webster Nov 4 '09 at 15:30
6  
By the way, in case it wasn't clear, I think this is a perfectly reasonable question. –  Ben Webster Nov 4 '09 at 16:03
add comment

2 Answers 2

up vote 15 down vote accepted

You are correct: there is a connection to the Galois theory of $\mathbb{C}/\mathbb{R}$ here.

To give a Hodge structure on a real vector space $V$ -- i.e., a direct sum decomposition of its complexification into $(p,q)$ subspaces such that $H^{q,p}$ is the complex conjugate of $H^{p,q}$ -- is equivalent to giving an action of $G = \operatorname{Res}_{\mathbb{C}/\mathbb{R}} \mathbb{C}^{\times}$ on $V$. Here $\operatorname{Res}_{\mathbb{C}/\mathbb{R}} \mathbb{C}^{\times}$ means the "restriction of scalars" from $\mathbb{C}/\mathbb{R}$ of the complex multiplicative group $\mathbb{C}^{\times}$. In plainer terms, it means that we view $\mathbb{C}^{\times}$ not as a one-dimensional complex algebraic group, but as a 2-dimensional real algebraic group, a "nonsplit torus". Then the fact that we have a homomorphism of real groups

$$G \to \operatorname{GL}(V)$$

implies an extra condition on the complexified representation $\mathbb{C}^{\times} \to \operatorname{GL}(V \otimes \mathbb{C})$: namely that the space $V^{p,q}$ on which $z$ in $\mathbb{C}^{\times}$ acts as $z^{p} \overline{z}^q$ is the complex conjugate of the space $V^{q,p}$.

A brief (but accurate!) discussion of this can be found at

http://en.wikipedia.org/wiki/Hodge_structure#Hodge_structures

share|improve this answer
    
Now that I have read a little bit of Serres Linear Representation Theory, I finally understood what you wrote. Thanks, that answers parts of my question! Maybe that should be it's own question, but what about the analogues of Hodge decomposition for other algebraic field extensions? The Wikipedia article remains unclear to me (at least the part "Applications") and I didn't find anything related in Claire Voisins Hodge Theory I... but maybe I should just continue learning along the standard path and then everything will become clear. If so, please tell me :-) –  Konrad Voelkel Nov 6 '09 at 15:43
add comment

As far as I understand acc. to Gelfand/Manin "Homological Algebra" p. 140, the idea of Hodge structures comes from Galois representations in arithmetics and was then by Deligne's "yoga des poids" transfered to complex varieties. But there it is (or was? The book was written 20 years ago) unclear which symmetry group (called "Hodge symmetries" in the book) lurks behind it.

share|improve this answer
    
This sounds very interesting. Do Gelfand/Manin explain this a little but furhter (and maybe accessible without knowing much arithmetics)? Sadly, at the moment I have no access to this book. Next week I'll take a look at it. –  Konrad Voelkel Nov 6 '09 at 15:46
    
This is a fairly standard and interesting topic, and instead of continuing in the comments I encourage you to post a new question. Others will answer/read it that way. But you should first look at the book indeed. –  Ilya Nikokoshev Nov 8 '09 at 22:54
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.