Is removing a grain of sand from an abelian sandpile also abelian? The abelian sandpile is a dynamical system that works as follows. We maintain a grid of cells, all initially empty. Each cell can hold zero, one, two, or three grains of sand. If a cell has four or more grains of sand, it topples, removing four grains of sand and adding one grain of sand to each of its four neighbors. This can in turn lead to other cells toppling. If the cell is on the border of the world, some of these sand grains may “fall off” the world and disappear.
This model is called the “abelian” sandpile because the order in which overfull cells topple makes no difference to the final configuration that the sand grains reach.
My question concerns an operation on sandpiles that is a sort of “inverse” of dropping a grain of sand onto a cell. We can imagine picking up a grain of sand from a location as follows. Remove one grain of sand from the location. If that cell now has a negative number of sand grains on it, we untopple the cell by picking up one grain of sand from each neighboring location putting those four sand grains back onto the cell. This might cause those other cells to untopple. If the cell has fewer than four neighbors because it’s on a border, we allow the cell to magically conjure new sand grains into existence for each neighbor.
My question is to what extent, if any, this inverse operation is also abelian. Is it necessarily the case that after removing a grain of sand from a cell that

*

*the process of performing untopples eventually terminates regardless of the order in which they’re performed, and

*the final state that is reached does not depend on the order of untopples performed?

I have no idea how to even think about this problem. The idea came to me when working through possible coding assignments for my students based on the regular abelian sandpile model when I was wondering if you could “clean up” a sandpile by repeatedly picking up sand grains that had already dropped onto the grid.
 A: We can have the "Positive" Abelian Sand-Pile = $X$ on top of the Corresponding [[ Mirror Universe ]] "Negative" Abelian(?) Sand-Pile = $Y$. We have to show that when $X$ is Abelian , $Y$ must be Abelian too.
Initially , let us have Cells $X(n,m) + Y(n,m) = 3$ counting the grains , having Constant total.
Now , what-ever Operation we do to X , let us do the Opposite to Y. Equivalently , what-ever Operation we do to Y , let us do the Opposite to X. That is :

When $X(n,m)$ increase by $1$ , then $Y(n,m)$ reduces by 1.
When $X(n,m)$ topples , then $Y(n,m)$ untopples.
When $X(n,m)$ unchanged , then $Y(n,m)$ unchanged.
When $X(n,m)$ reduces by $1$ , then $Y(n,m)$ increases by 1.

We can check that $X(n,m) + Y(n,m) = 3$ is Constant Invariant.
When $X$ is at final State , $Y must be at final State.
Hence when $X$ is Abelian , $Y$ must be Abelian too.
In other words :

When $X$ does not depend on order , $Y$ does not depend on order.
When $Y$ does not depend on order , $X$ does not depend on order.
When $X$ is at Steady State , $Y$ is at Steady State.
When $Y$ is at Steady State , $X$ is at Steady State.

We have Positive Abelian Sand-Pile Equivalent to Negative Abelian Sand-Pile !
I hope this is what you want !
