Correspondences acting on cohomology groups $H^*(X)$ & splittings

Question

Let $X$ be a smooth connected proper scheme over field $k$. It is known that correspondences $\alpha \subset X \times X$ regarded as objects in Chow groups $\text{CH}^*(X \times X)$ act on cohomology $H^*(X, \mathbb{Z})$ via "push and pull", namely we have a cycle class map

$$ \text{cl}: \text{CH}^*(X) \to H^*(X, \mathbb{Z} ) $$

which is compatible with pull-backs & push-forwards. If $p_1, p_2: X \times X \to X $ are the projections and $\alpha $ a correspondence in $X \times X$, then it acts as

$$ \alpha^*: H^*(X \mathbb{Z}) \to H^*(X \mathbb{Z}), \ \ \ \alpha^*(s) := p_{2*}(p^*_1(s) \cap \text{cl}(\alpha)) $$

Clearly this action extends linearly to action by rational Chow groups $\text{CH}_r^*(-)= \text{CH}^* \otimes \mathbb{Q} $ on rational cohomology.

Let now specialize: Assume $X$ is a smooth proper curve and $D \subset X$ an effective divisor of degree $d$, ie a "multisection". By dividing the degree $\beta:= \frac{1}{d} \cdot D $ becomes a $\mathbb{Q}$-divisor of degree $1$, so a "virtual" section.

So far I understand it correctly in the discussion here Dan Petersen uses that the induced action by such $\mathbb{Q}$-divisor of degree $1$ in terms from above as associated Chow correspondence induces a splitting of the cohomology. (Dan Petersen used this more generally in relative setting & associated action on derived object $Rf_* \mathbb{Q}$ but for sake of simplicity I would like to understand it in absolute case for a single curve as baby version of this mechanism.)

Question: I not see why this action gives a splitting of the cohomology? In other words does such $\mathbb{Q}$-divisor "normed" to degree $1$ induce an idempotent endomorphism corresponding to the associated action described above? It seems that the degree $1$ assumption is crucial, but I not see the connection. Then it would mean that every effective divisor turned into degree $1$ $\mathbb{Q}$-divisor by dividing it's degree would acting idempotently on the cohomology, ie give a splitting of the cohomology? That seems strange. Maybe I misunderstood somewhere Dan Petersen's argument.

Answer 1

Composition of a correspondence in $X \times Y$ and a correspondence in $Y \times Z$ is given by pulling both back to $X\times Y\times Z$, intersecting them, and pushing forward to $X \times Z$.

To check that $ D\times X \subset X\times X$ is idempotent under composition for $D$ of degree $1$, we pull back along two projections $X\times X \times X \to X\times X$, obtaining $D \times X \times X$ and $X\times D \times X$, then intersect $D \times X \times X$ and $X\times D \times X$, obtaining $D \times D \times X$, then finally pushforward $D \times D \times X$ along the projection to $X \times X$, which gives $D\times X$.

Here we use that for a cycle on $X \times Y \times Z$ obtained as a product of $A$ on $X\times Z$ with $B$ on $Y$, the pushforward to $X\times Z$ is $A$ times the degree of $B$. So this is where the degree enters into the picture.

Answer 2

This is something you can find discussed in many references on Chow motives. For example you can look at "On the motive of an algebraic surface" by Murre, §1.5 "Splitting off the trivial parts".

In general, if $X$ is a smooth projective variety then any zero-cycle $\mathfrak z$ on $X$ will define a correspondence $\mathfrak z \times X$ from $X$ to itself. It is idempotent iff $\deg \mathfrak z=1$, and working with $\mathbf Q$-coefficients we may assume this. This idempotent acts on several things: the motive $h(X)$, but also more closely related to your question, the object $C^\bullet(X)$ in the derived category. The image of this idempotent splits off $H^0$. Its transpose splits off $H^{2\dim(X)}$. If $X$ only has cohomology in three degrees we get a decomposition in the derived category. Now splitting the object $C^\bullet(X)$ is typically not so interesting, as every chain complex is quasi-isomorphic to its cohomology. But in a relative setting it is more nontrivial.