# Multidimensional Young diagrams

Consider a Young diagram defined as follows:

A Young diagram (also called a Ferrers diagram, particularly when represented using dots) is a finite collection of boxes, or cells, arranged in left-justified rows, with the row lengths in non-increasing order. Listing the number of boxes in each row gives a partition $$\lambda$$ of a non-negative integer $$n$$, the total number of boxes of the diagram.

For example we may write 1+4+5=10: Question: Are there higher-dimensional versions, using cubes, such that the "faces" of the diagram are each themselves Young diagrams?

Here is an example, with three distinct faces, each representing diagrams: 1+2+3+3, 2+2+3+3, and 0+0+4+4. The faces are the Young diagrams on the faces of the cube in this case. It has 6 faces, and three pairs of (up,right,in), each a Young diagram. In 2d, there is only 1 face (1 diagram). In 3d, one has a cube with six faces, but only three are unique diagrams. One diagram in the 3d case is forced from the other two (up,right,up,right....up and up,up,in,in,up lead to the other necessarily being right,right,in,in,right). If so, is there a way of writing an integer in terms of the diagram, in the same way as an integer can be represented via one of many Young diagrams (i.e. integer partitions)? This would represent a restricted integer partition, but in a relatively unusual way.

For example the image below would represent the integer partition ((2+2) + (3+3)) + ((2+2) + (3+3)) = 20.

• I don't see the relation between the 3D illustration and the numbers in the question. I don't see what you mean by faces. What would (in dimension 2) you consider to be a face of a Young diagram? – Marc van Leeuwen Jun 29 '20 at 13:06
• In 2d, there is only 1 face (1 diagram). In 3d, one has a cube with six faces, but only three are unique diagrams. One diagram in the 3d case is forced from the other two (up,right,up,right....up and up,up,in,in,up lead to the other necessarily being right,right,in,in,right). – apkg Jun 29 '20 at 13:09

## 1 Answer

The three dimensional form is called a "plane partition". That is because it can be represented as a histogram, filling the squares of a Young diagram (the base) with numbers representing the height of the column on that square. These numbers are weakly decreasing along rows and columns, just like numbers in an ordinary partition are weakly decreasing. I do not know any name for higher dimensional variants.

• So a plane partition in 3d is necessarily weakly decreasing along the row/column pairs? – apkg Jun 29 '20 at 13:14
• @apkg Well, that's the definition. You can have all kinds of variations with modified names like reverse plane partitions (increasing conditions) – Marc van Leeuwen Jun 29 '20 at 14:22
• Not all histograms like this are weakly decreasing along rows and columns. For example, in the lattice path graphic in the OP, if I turn up,right,up,right,in,in,up,in,right,in,right,up (swapping, therefore the fourth-last and thrid-last turns), one of the final columns is removed, and I have (2+2+3)+(2+2+3+3)=17. Now, though, it is not weakly decreasing. Basically it is difficult to define the area directly based on a "product" of Young diagrams, since it does not necessarily produce a plane partition. Is there perhaps a name when it is not weakly decreasing? – apkg Jun 30 '20 at 9:24
• @apkg My problem is that I don't understand your construction, and you don't explain it very clearly. Apparently you start with a lattice path across a 3D block (I understand what that is), and then have some rule (which I don't follow) saying which small cubes are somehow "below" the path. I think that even if I did understand what your rule is, it would not define a plane partition, but something else. I understand that if you take the shadow of your structure in 3 directions,you get a Young diagram each time, but that information is not sufficient to deduce your rule from. – Marc van Leeuwen Jun 30 '20 at 15:48
• The rule is not really stated in the question, which is a bit unhelpful, sorry. The idea is to project the faces of the cubes in the large box onto the faces of the box. Then, if those faces are below the respective projections of the path onto the respective face, the cube is included. Sorry I will edit the question to make it more clear. – apkg Jun 30 '20 at 15:51