## A Weird Condition on Tableaux

**Point of Post: **In this post we discuss an interesting property between two tableaux which will ultimately help us construct the irreps of associated to each -frame.

*Motivation*

So, enough being cryptic. I promised that we will create a bijection in such a way that –it’s about time I explained roughly how. So, in our last post we created this interesting function . Our main goal to the construction is to show that up to normalization is a minimal projection from where we shall get our corresponded irrep. In the journey to prove this we will need a strange, un-motivated concept which has to do with the relationship between the rows of one tableau and another tableau .Luckily, the motivation and usefulness will become apparent shortly. That said, we can at least give a glance of why anyone would even care about this condition. In particular, we shall use this condition to prove that the irreps associated to two different -frames are different.

*The Condition*

We now define the condition we stated above. In particular, let be two – tableaux. We say that if there exists such that are in the same row of and the same column of . For example:

since are both in the same row of the right tableau and the same column of the left tableau. Note that this is really saying that the transposition is in . If we obviously write . To get the first reason why this condition would ever come up in relation to our last post we first note that:

**Theorem: ***Let be two -tableaux such that . Then, (where the product of these two elements of is meant to be interpreted as convolution as per the identification of with as previously discussed).*

**Proof: **We begin by noticing that

where we’ve used the fact that , , and is a bijection . Similarly one can show that

where (the only [really quite obvious] extra trick for the case is the fact that is a homomorphism). With this observation the proof becomes easy. In particular, since (where are the guaranteed elements of in the definition of ) we see that

from where the conclusion follows.

**Corollary: ***Let be two -tableaux with , then .*

*A Characterization of when *

We actually now find a characterization of when two -tableaux satisfy and they have the same frame. Indeed:

**Theorem: ***Let be two -tableaux with , then if and only if there exists and such that *

*recalling the -action on tableaux.*

**Proof: **Suppose first that as described above. Next, suppose that were in the same row in , then by definition are in the same row of . That said, we also have by assumption then that are in the same row of and in particular not in the same column. But, this implies by definition that (since ) that are in different rows of . Since were arbitrary the conclusion follows.

Conversely, suppose that . Let be two elements in the same column of , since we have that there exists a permutation such that are in the same column of . And, so there exists a permutation such that lie in the same position in as they do in . Noting that we can repeat this process and not screw any of the previous permutations up (since the permutations that shift rows and columns can move the things that are of interest now say some different than and leave fixed) we may conclude that there exists and such that

The conclusion follows.

**Corollary: ***Let are -frames with and , then there exists and such that *

**Proof: **This doesn’t follow immediately from the previous theorem since the order of the elements is different. But, we note that if are the elements of and guaranteed by the previous theorem such that then we see that where but we have by previous theorem then that

**References:**

1. Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Mathematical Society, 1996. Print

[…] Note that since cannot be written as by assumption we have by a previous theorem that . Thus, there exists which are in the same row of and the same column of . We have by […]

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