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

I'm a little confused and in need of some clarification about the relationship between algebraic and holomorphic differential forms:

(1) What is the exact definition of the module of differential forms of a complex projective variety?

(2) What is the definition of its differential?

(3) Am I right in assuming that the algebraic forms are a submodule of the holomorphic forms with the two differentials coinciding in some obvious sense?

share|improve this question
add comment

1 Answer 1

up vote 5 down vote accepted

Yes, every algebraic differential form is holomorphic and yes, the differential preserves the algebraic differential forms. If you are interested in projective smooth varieties then every holomorphic differential form is automatically algebraic thanks to Serre's GAGA. This answers (3).

Concerning (1) and (2) I suggest that you consult some standard reference as Hartshorne's Algebraic Geometry.

Edit: As pointed out by Mariano in the comments below there are subtle points when one compares Kahler differentials and holomorphic differentials. I have to confess that I have not thought about them when I first posted my answer above.

Algebraic differential $1$-forms over a Zariski open set $U$ are elements of the module generated by $adb$ with $a$ and $b$ regular functions over $U$ (hence algebraic) by the relations $d(ab) = adb + b da$, $d (a + b) = da + db$ and $d \lambda =0$ for any complex number $\lambda$. Since these are the rules of calculus there is a natural map to the module of holomorphic $1$-forms over $U$. This map is injective, since the regular functions on $U$ are not very different from quotients of polynomials.

If instead of considering the ring $B$ of regular functions over $U$ one considers the ring $A$ of holomorphic functions over $U$ then one can still consider its $A$-module of Kahler differentials. If $U$ has sufficiently many holomorphic functions, for instance if $U$ is Stein, then one now has a surjective map to the holomorphic $1$-forms on $U$ which is no longer injective as pointed out by Georges Elencwajg in this other MO question.

share|improve this answer
2  
But holomorphic forms and differential forms are different beasts. Searching MO will find a discussion regarding $d\mathrm{sin}$... –  Mariano Suárez-Alvarez Jan 4 '10 at 18:13
1  
The discussion that Mariano is referring to can be found here: mathoverflow.net/questions/6074/… –  Ben Linowitz Jan 4 '10 at 18:19
    
@jvp So, just to be sure, you're saying that for projective smooth complex varieties: algebraic and holomorphic differential forms coincide? –  Aston Smythe Jan 4 '10 at 18:33
2  
@Aston Smythe: For projective smooth complex varieties, global algebraic forms and global holomorphic forms coincide. This is part of GAGA, as jvp says. –  David Speyer Jan 4 '10 at 18:49
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.