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](https://mathoverflow.net/q/443274)
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.



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


