The Unapologetic Mathematician

Mathematics for the interested outsider

Inner and Outer Corners

The next thing we need to take note of it the idea of an “inner corner” and an “outer corner” of a Ferrers diagram, and thus of a partition.

An inner corner of a Ferrers diagram is a cell that, if it’s removed, the rest of the diagram is still the Ferrers diagram of a partition. It must be the rightmost cell in its row, and the bottommost cell in its column. Similarly, an outer corner is one that, if it’s added to the diagram, the result is still the Ferrers diagram of a partition. This is a little more subtle: it must be just to the right of the end of a row, and just below the bottom of a column.

As an example, consider the partition (5,4,4,2), with Ferrers diagram

\displaystyle\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&&&\end{array}

We highlight the inner corners by shrinking them, and mark the outer corners with circles:

\displaystyle\begin{array}{cccccc}\bullet&\bullet&\bullet&\bullet&\cdot&\circ\\\bullet&\bullet&\bullet&\bullet&\circ&\\\bullet&\bullet&\bullet&\cdot&&\\\bullet&\cdot&\circ&&&\\\circ&&&&&\end{array}

That is, there are three ways we could remove a cell and still have the Ferrers diagram of a partition:

\displaystyle\begin{array}{cccc}\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&&\\\bullet&\bullet&&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&\bullet&\\\bullet&&&&\end{array}

And there are four ways that we could add a cell and still have the Ferrers diagram of a partition:

\displaystyle\begin{array}{cccccc}\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&&\\\bullet&\bullet&\bullet&\bullet&&\\\bullet&\bullet&&&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&&&\\\bullet&&&&\end{array}

If the first partition is \lambda, we write a generic partition that comes from removing a single inner corner by \lambda^-. Similarly, we write a generic partition that comes from adding a single outer corner by \lambda^+. In our case, if \lambda=(5,4,4,2), then the three possible \lambda^- partitions are (4,4,4,2), (5,4,3,2), and (5,4,4,1), while the four possible \lambda^+ partitions are (6,4,4,2), (5,5,4,2), (5,4,4,3), and (5,4,4,2,1).

Now, as a quick use of this concept, think about how to fill a Ferrers diagram to make a standard Young tableau. It should be clear that since n is the largest entry in the tableau, it must be in the rightmost cell of its row and the bottommost cell of its column in order for the tableau to be standard. Thus n must occur in an inner corner. This means that we can describe any standard tableau by picking which inner corner contains n, removing that corner, and filling the rest with a standard tableau with n-1 entries. Thus, the number of standard \lambda-tableaux is the sum of all the standard \lambda^--tableaux:

\displaystyle f^\lambda=\sum\limits_{\lambda^-}f^{\lambda^-}

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January 26, 2011 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups

2 Comments »

  1. [...] We want to “categorify” the relation we came up with last time: [...]

    Pingback by The Branching Rule « The Unapologetic Mathematician | January 27, 2011 | Reply

  2. [...] the steps from to give us all the as we add up dimensions. Comparing to the formula we’re categorifying, we see that this accounts for all of . And so there are no dimensions left for the steps from to [...]

    Pingback by The Branching Rule, Part 2 « The Unapologetic Mathematician | January 28, 2011 | Reply


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