MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Quillen gives the following projection formula his foundational paper on higher algebraic k-theory. (For simplicity, I assume all schemes are noetherian.)

Let $f: X \rightarrow Y$ be a proper map of finite tor dimension. Assume $X, Y$ have ample line bundles. Then, for $x \in K_0 X, y \in K_q' Y$ we have $$ f_*(x \cdot f^* y) = f_* (x) \cdot y$$

Where he says $f_*: K_0 X \rightarrow K_0 Y$ is the map of SGA 6 2.12.2 (no expose given).


Notion of finite tor dimension

Quillen defines a morphism $f: X \rightarrow Y$ to be "of finite tor dimension" if $\mathcal{O}_X$ is of finite tor dimension as a module over $f^{-1}(\mathcal{O}_Y)$.

Suppose $X, Y$ are quasiprojective. Then, (ref: Fulton/MacPherson Categorical framework for the study of singular spaces) being a morphism of finite tor dimension is the same as being a perfect morphism in the sense of SGA 6.

$\textbf{Question}$: I wasn't able to deduce if Quillen's notion of finite tor dimension is the same as being perfect in the sense of the stacks project (say if $X,Y$ have ample line bundles). Is this true?

share|cite|improve this question
Thanks for the comments so far. The question is still completely open. – abcd Jul 7 '13 at 5:15

I don't have enough rep to comment: usually the pushforward on grothendieck groups is defined using higher direct images (or derived categories is you prefer), so that $f_*[E] = \sum (-1)^i [R^if_* E]$.

For this to be well defined you need properness to still land in coherent sheaves and only finitely many $R^i$ (which is true for essentially any morphism of everything).

On the other hand you can define $f^*$ on the level of Grothendieck groups again by using the derived inverse image (which is essentially computing Tors for $f^{-1}O_Y$), here you use finite tor dimension.

share|cite|improve this answer
oh, and the projection formula then follows from the projection formula on the level of derived functors. – John Salvatierrez Jul 4 '13 at 23:15
by the way, what do you mean by $K'_q$? – John Salvatierrez Jul 4 '13 at 23:15
Thanks for the perspective. By $K'_q$ I mean Quillen's $K'$ theory, which agrees with $K-$theory on regular noetherian ... schemes. – abcd Jul 4 '13 at 23:20
where exactly in SGA are perfect morphisms defined? – John Salvatierrez Jul 7 '13 at 19:29
It's pretty hard to understand for me, but it is: Expose III, definition 4.1. It's on the top of page 256 of the scan available here: – abcd Jul 7 '13 at 20:53

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.