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I apologize in advance, since I am probably doing a very naive mistake in my computation. I am learning about pure (Chow / Grothendieck) motives. One of the first steps is to consider the category where objects are smooth projective varieties and morphisms from $X$ to $Y$ are defined to be correspondences of degree $0$ from $X$ to $Y$, i.e., algebraic cycles on $X \times Y$ of codimension $\dim X$ (up to some adequate equivalence relation). Composition works as follows: given a cycle on $X \times Y$ and one on $Y \times Z$, I pull them back to $X \times Y \times Z$ and then intersect them and push forward the result to $X \times Z$. The identity on $X$ is the graph $\Delta$ of the diagonal morphism. A projector is a correspondence whose square is equal to itself. Now, I can prove that given a curve $X$ and a point $P$ on $X$, both $p_0 := P \times X$ and $p_{2} := X \times P$ are projectors on $X$. But I am stuck proving that $p_1 := \Delta - p_0 - p_{2}$ is also a projector. Can someone please point out where I am making a mistake? I am computing

$$p_1^2 = \Delta^2 + p_0^2+p_2^2 -\Delta \circ p_0 - \Delta \circ p_2-p_0 \circ \Delta - p_2 \circ \Delta + p_0 \circ p_2 + p_2 \circ p_0$$ $$= \Delta +p_0 + p_2 -p_0-p_2-p_0-p_2+X \times X +P\times P $$ $$ = p_1 + X \times X+P\times P. $$

But I can really not see why the last two summands should be $0$. The mistake is probably in the computation of $p_0 \circ p_2$ and $p_2 \circ p_0$, since I do not even obtain cycles of the right codimension. But it seems to me that, when I compute $p_2 \circ p_0$, pulling back $p_0$ (first two factors) gives $P \times X \times X$, pulling back $p_2$ (last two factors) gives $X \times X \times P$, their intersection is $P \times X \times P$ and pushing forward to the first and third factor gives $P \times P$. And similarly for $p_0 \circ p_2$. What am I doing wrong?

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    $\begingroup$ Your equality is in $CH^1(X\times X)$, how could you get $X\times X$ or $P\times P$??? $\endgroup$
    – abx
    Commented Jan 10, 2020 at 13:23
  • $\begingroup$ @abx I remarked by myself in the question that this cannot possibly make sense. But I don’t see why. So the question is, how can I correct it? How does the correct argument go? $\endgroup$
    – 57Jimmy
    Commented Jan 10, 2020 at 13:53
  • $\begingroup$ Indeed, $p_0 \circ p_2$ and $p_2 \circ p_0$ are $0$. For $p_0 \circ p_2$, we have to cap $P \times X \times X$ with $X \times X \times P$ and pushforward onto the first and third factor. The cap product is just the intersection $P \times X \times P$. However, the projection of $P \times X \times P$ onto the first and third factor has positive dimensional fibers, so the pushforward of $[P \times X \times P]$ is $0$. (continued) $\endgroup$
    – David E Speyer
    Commented Jan 10, 2020 at 14:23
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    $\begingroup$ For $p_0 \circ p_2$, we have to cap $[X \times P \times X]$ with itself. But you can't compute a self cap-product by literally intersecting the subvariety with itself! You have to either look at the normal bundle, or else replace one of the two copies with an equivalent cycle. If your equivalence relation is algebraic equivalence (or anything coarser), then the second is easier. Let $Q$ be a different point than $P$, then $[X \times P \times X] = [X \times Q \times X]$ and $(X \times P \times X) \cap (X \times Q \times X)=0$, so the self intersection of $[X \times P \times X]$ is zero. $\endgroup$
    – David E Speyer
    Commented Jan 10, 2020 at 14:25
  • $\begingroup$ Thanks a lot, I was indeed using a very naive notion of intersection and push forward. Now everything is clear $\endgroup$
    – 57Jimmy
    Commented Jan 10, 2020 at 23:52

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