7
$\begingroup$

The Riemann-Roch Theorem, the Grothendieck-Riemann-Roch Theorem , the Grothendieck-Hirzebruch-Riemann-Roch Theorem , all of them are well explained at Wikipedia .I would like to understand the meaning of the "Riemann-Roch without denominators". Who first established such a theorem ? What was the motive ?

$\endgroup$
1
  • $\begingroup$ This doesn't look like a research question... $\endgroup$ Mar 26, 2014 at 21:00

2 Answers 2

9
$\begingroup$

Some History

I believe the theorem was formulated by Grothendeick, Theorie des intersections et theoreme de Riemann-Roch. Seminaire de geometrie algebrique du Bois-Marie 1966/67 (SGA 6), Lect. Notes Math. 225, Springer-Verlag, Berlin-Heidelberg-New York (1971), Expose XIV, (3.1), p. 670. It was proved by Jouanolou in Jouanolou, J.-P., Riemann-Roch sans denominateurs. (French) Invent. Math. 11, (1970) 15-26. In Gillet, H., Riemann-Roch theorems for higher algebraic K-theory. Adv. Math. 40, (1981) 203-289., Gillet extended it , but kept the name, to the Chern class maps for the higher algebraic K-groups with values in cohomology theories satisfying certain axioms.

Meaning of the ''Riemann-Roch without denominators''

Roughly, the following is what the Riemann-Roch theorem without denominators means (from O. B. Podkopaev, E. K. Shinder, On the Riemann–Roch Theorem Without Denominators, St. Petersburg Math. J., 6 Vol. 18 (2007), No. 6, Pages 1021–1027):

The Riemann–Roch formula without denominators for a closed embedding $i : Y \hookrightarrow X$ of codimension $d$ expresses the Chern class $c_d(i_∗\mathcal{O}_Y)$ in terms of the class $[Y ] ∈ CH^d(X)$.

The following is the Riemann-Roch theorem without denominators:

Theorem: Let $\frak X$ be a nonsingular variety over a field $\mathbb{F}$, and let $i:Y\hookrightarrow\frak X$ be a closed embedding of an irreducible subvariety $Y$, which has codimension $d$. Then, in $CH^d(\frak X)$, $$c_d(i_*\mathcal{O}_Y)=(-1)^{d-1}(d-1)![Y]$$

The proof is very long and hence I refer you to the paper cited above, http://www.maths.ed.ac.uk/~aar/papers/gillet.pdf, and W. Fulton, Intersection Theory, Second Edition, Springer-Verlag, 1998.

$\endgroup$
6
  • $\begingroup$ A nice and complete answer. $\endgroup$
    – Al-Amrani
    Mar 27, 2014 at 7:29
  • $\begingroup$ Dear @Sanath Devalapurkar: To write bold text, you can use markdown as follows: **bold text** produces bold text. To produce headers, you can use # header or ## header or ### header. Mathjax $...$ code is not meant to format text. $\endgroup$ Mar 27, 2014 at 20:51
  • $\begingroup$ @RicardoAndrade Thanks - I will edit my answer. $\endgroup$
    – user62675
    Mar 27, 2014 at 22:22
  • $\begingroup$ The mentioned "Riemann-Roch theorem without denominators" by O. B. Podkopaev, E. K. Shinder (St. Petersburg Math. J., 6 Vol. 18 (2007), No. 6, Pages 1021–1027),could be renamed "Riemann-Roch without Jouanolou" ! $\endgroup$
    – Al-Amrani
    Apr 7, 2014 at 7:15
  • 1
    $\begingroup$ @user62675 Does this theorem assume the subvariety $Y$ to be nonsingular? In all the versions of Riemann–Roch without denominators I saw, there is the assumption that the closed embedding is regular. What I really want to know is that is it always true that $c_j(i_*E)=0$ for $0<j<d$ when $Y$ is only a closed subvariety but may be singular. I don't find an affirmative answer to this, but the theorem you listed here seem to answer it. Could you tell me the source of your theorem? $\endgroup$
    – Lao-tzu
    May 25, 2017 at 8:53
6
$\begingroup$

The theorem was conjectured by Grothendieck and proved by J.P. Jouanolou, Riemann-Roch sans dénominateurs (1970). By clearing the denominators you can compute Chern classes, not just the Chern character.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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