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 ?
2 Answers
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.
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$\begingroup$ Dear @Sanath Devalapurkar: To write bold text, you can use markdown as follows:
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code is not meant to format text. $\endgroup$ Mar 27, 2014 at 20:51 -
$\begingroup$ @RicardoAndrade Thanks - I will edit my answer. $\endgroup$– user62675Mar 27, 2014 at 22:22
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$\begingroup$ 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" ! $\endgroup$ Apr 7, 2014 at 7:15
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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-tzuMay 25, 2017 at 8:53
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.