Last update: 27 August 2013
Theorem 2.45 where the sum is over partitions such that is a border strip of length and is the number of rows in
Since the function is symmetric, we have where has a 1 in the slot and 0 everywhere else.
The function is alternating, so if there is a repeated entry in the exponents. That is, it is zero if for some If this is not the case, then there exists some with so that satisfies for some partition In other words, rearranges the exponents so that they are in the form Then we have To see the relation to border strips, consider the case where and Then Since is alternating, we can apply as long as we multiply by Multiplying by switches row three and row four. If we do this and then write the result with left of the wall, we get Continuing in this way, i.e., multiplying by and then gives the diagram which is a border strip of length 6 that has 4 rows. For each we add a border strip of length at row and then sum over If adding such a strip does not produce a partition, then at some point in the process, we have violated the condition that all of the exponents (thus the length of each row) must be distinct, and we get 0. Thus, we have shown Dividing both sides by proves the theorem.
Corollary 2.47 for some positive integer
Use induction and the previous corollary.
Since we have finished the proof of the Frobenius formula 2.29.
Theorem 2.48 [Murnaghan-Nakayama Rule] Let and Then where the sum is over such that is a border strip of length and is the number of rows in this border strip.
From the Frobenius formula, we have where and is a border strip of length with rows. Comparing coefficients of gives the theorem.
Definition 2.49 The weight wt of a border strip is given by where is the number of rows in
Definition 2.50 Let be a partition. A strip tableau of shape is a sequence of partitions such that is a border strip of length The weight of is given by
With these definitions, the Murnaghan-Nakayama rule can be written in the following form
Theorem 2.51 [Murnaghan-Nakayama Rule II] where the sum is over all μ-border strip tableaux
Now let We can compute all of the strip tableaux as follows. This graph is called the Young lattice. It is characterized by the following properties:
|(1)||On the level, the nodes are labeled by partitions of|
|(2)||A node on the level is connected to a node on the level if the partitions differ by a single box.|
Fix a partition The paths from to that are always increasing in level are the strip tableaux of shape Moreover, from the Murnaghan-Nakayama rule, we see that the number of these paths is equal to which is the dimension of the irreducible labeled by
Definition 2.52 A standard tableau of shape is a filling in of the boxes in the Ferrers diagram of with the integers such that numbers increase in both the rows (moving left to right) and the columns (moving top to bottom).
The increasing paths in the Young diagram correspond exactly to standard tableaux. For example, the path is completely determined by the standard tableau and vice-versa.
Recall that from the semisimplicity of we have where denotes the number of standard tableaux of shape For example, in the 4th row of the Young diagram, we have
This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.