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

As we know, Grothendieck Riemann Roch only involves $K_{0}$. Is there any work generalizing this formula to (Quillen's Higher K)? If there is, what is the meaning for such kind of formula?

Thanks in advance

share|cite|improve this question
Hopefully, someone will give you a detailed answer. Such theorems have certainly been proved by Gillet, Soulé and others. The article by Tamme in the Beilinson conjectures volume may a good place to start. – Donu Arapura Feb 21 '11 at 22:15
up vote 12 down vote accepted

MR0624666 (83m:14013) Gillet, Henri Riemann-Roch theorems for higher algebraic K-theory. Adv. in Math. 40 (1981), no. 3, 203–289.

share|cite|improve this answer
Thanks for the reference – Shizhuo Zhang Feb 22 '11 at 0:31
There is also the much shorter announcement, also by Gillet, available freely at… – David Roberts Feb 22 '11 at 2:48

There is the recent paper Algebraic K-theory, $A^1$-homotopy and Riemann-Roch theorems by Riou, which gives a different proof of the results by Gillet referenced in the answer by profilesdroxford54.


In this article, we show that the combination of the constructions done in SGA 6 and the $A^1$-homotopy theory naturally leads to results on higher algebraic K-theory. This applies to the operations on algebraic K-theory, Chern characters and Riemann-Roch theorems.

To give you a rough idea, algebraic K-theory is representable in the stable homotopy category arising from $A^1$-homotopy theory of schemes over a regular scheme. As far as I understand it, Grothendieck-Riemann-Roch is just a shadow of the maps from the spectrum representing algebraic K-theory to the Eilenberg-MacLane spectrum. But the paper above considers more than just stable homotopy.

share|cite|improve this answer

Your Answer


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.