The Unapologetic Mathematician

Mathematics for the interested outsider

The Dominance Order on Tabloids

Sorry, this should have gone up last Friday.

If \{t\} is a Young tabloid with shape \lambda\vdash n, we can define tabloids \{t^i\} for each i from 1 to n by letting \{t^i\} be formed by the entries in \{t\} less than or equal to i. We define \lambda^i to be the shape of \{t^i\} as a composition. For example, if we have

\displaystyle\{t\}=\begin{array}{cc}\cline{1-2}2&3\\\cline{1-2}1&4\\\cline{1-2}\end{array}

then we define

\displaystyle\begin{aligned}\{t^1\}&=\begin{array}{c}\cline{1-1}\\\cline{1-1}1\\\cline{1-1}\end{array}\\\lambda^1&=(0,1)\\\{t^2\}&=\begin{array}{c}\cline{1-1}2\\\cline{1-1}1\\\cline{1-1}\end{array}\\\lambda^2&=(1,1)\\\{t^3\}&=\begin{array}{cc}\cline{1-2}2&3\\\cline{1-2}1&\\\cline{1-1}\end{array}\\\lambda^3&=(2,1)\\\{t^4\}&=\begin{array}{cc}\cline{1-2}2&3\\\cline{1-2}1&4\\\cline{1-2}\end{array}\\\lambda^4&=(2,2)\end{aligned}

Along the way we see why we might want to consider a composition like (0,1) with a zero part.

Anyway, now we define a dominance order on tabloids. If \{s\} and \{t\} are two tabloids with composition sequences \lambda^i and \mu^i, respectively, then we say \{s\} “dominates” \{t\} — and we write \{s\}\triangleright\{t\} — if \lambda^i dominates \mu^i for all i.

As a (big!) example, we can write down the dominance order on all tabloids of shape (2,2):

\displaystyle\begin{array}{ccccc}&&\begin{array}{cc}\cline{1-2}1&2\\\cline{1-2}3&4\\\cline{1-2}\end{array}&&\\&&\uparrow&&\\&&\begin{array}{cc}\cline{1-2}1&3\\\cline{1-2}2&4\\\cline{1-2}\end{array}&&\\&\nearrow&&\nwarrow&\\\begin{array}{cc}\cline{1-2}2&3\\\cline{1-2}1&4\\\cline{1-2}\end{array}&&&&\begin{array}{cc}\cline{1-2}1&4\\\cline{1-2}2&3\\\cline{1-2}\end{array}\\&\nwarrow&&\nearrow&\\&&\begin{array}{cc}\cline{1-2}2&4\\\cline{1-2}1&3\\\cline{1-2}\end{array}&&\\&&\uparrow&&\\&&\begin{array}{cc}\cline{1-2}3&4\\\cline{1-2}1&2\\\cline{1-2}\end{array}&&\end{array}

It’s an exercise to verify that these are indeed all the tabloids with this shape. For each arrow, we can verify the dominance. As an example, let’s show that

\displaystyle\begin{array}{cc}\cline{1-2}1&2\\\cline{1-2}3&4\\\cline{1-2}\end{array}\trianglerighteq\begin{array}{cc}\cline{1-2}2&3\\\cline{1-2}1&4\\\cline{1-2}\end{array}

First, let’s write down their composition sequences:

\displaystyle\begin{array}{c|cc}i&\lambda^i&\mu^i\\\cline{1-3}1&(1)&(0,1)\\2&(2)&(1,1)\\3&(2,1)&(2,1)\\4&(2,2)&(2,2)\end{array}

Now it should be easy to see on each row that \lambda^i\trianglerighteq\mu^i. As another example, let’s try to compare \begin{array}{cc}\cline{1-2}2&3\\\cline{1-2}1&4\\\cline{1-2}\end{array} and \begin{array}{cc}\cline{1-2}1&4\\\cline{1-2}2&3\\\cline{1-2}\end{array}. Again, we write down their composition sequences:

\displaystyle\begin{array}{c|cc}i&\lambda^i&\mu^i\\\cline{1-3}1&(0,1)&(1)\\2&(1,1)&(1,1)\\3&(2,1)&(1,2)\\4&(2,2)&(2,2)\end{array}

We see that \lambda^1\not\trianglerighteq\mu^1, but \mu^3\not\trianglerighteq\lambda^3. Thus neither tabloid dominates the other. The other examples to verify this diagram are all similarly straightforward.

January 10, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 4 Comments

   

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