Example 2.1.9 Let us compute the full set of characters of these modules when n 3. In this case the only partitions are A = (3), (2, 1), and (1³). From the previous examples, the modules M correspond to the trivial, defining, and regular representations of S3, respectively. These character values have all been computed in Chapter 1. Denote the character of M by and represent the conjugacy class of S, corresponding to by K₁. Then we have the following table of characters: (3) (2,1) Ø(1³) K(13) K(21) K(3) 1 0 0- 1 3 6 1 1 0 Tro

Classical Dynamics of Particles and Systems
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Author:Stephen T. Thornton, Jerry B. Marion
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Chapter7: Hamilton's Principle-lagrangian And Hamiltonian Dynamics
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Could you please explain example 2.1.9 with the table.

 

 

w order and a
56 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
Definition 2.1.5 Suppose An. Let
M = C{{₁},... {tk}}.
where (t₁)....() is a complete list of A-tabloids. Then MA is called the
permutation module corresponding to A. ■
Since we are considering only row equivalence classes, we could list the rows of
MA in any order and produce an isomorphic module. Thus M" is defined for
any composition (ordered partition) . As will be seen in the next examples.
we have already met three of these modules.
Example 2.1.6 If A = (n), then
with the trivial action. ■
M() = C{12}
Example 2.1.7 Now consider A = (122). Each equivalence class {t} consists
of a single tableau, and this tableau can be identified with a permutation in
one-line notation (by taking the transpose, if you wish). Since the action of
Sn is preserved,
M(1) CS
and the regular representation presents itself. ■
Example 2.1.8 Finally, if λ = (n-1, 1), then each A-tabloid is uniquely
determined by the element in its second row, which is a number from 1 to n.
As in Example 1.6.3, this sets up a module isomorphism
M(-1,1)
C{1,2,...,n},
so we have recovered the defining representation. ■
Example 2.1.9 Let us compute the full set of characters of these modules
when n = 3. In this case the only partitions are A = (3), (2, 1), and (1³). From
the previous examples, the modules M correspond to the trivial, defining,
and regular representations of S3, respectively. These character values have
all been computed in Chapter 1. Denote the character of M by and
represent the conjugacy class of S, corresponding to u by K. Then we have
the following table of characters:
(3)
(2,1)
K(13)
1
3
6
K(2,1)
1
1
0
K(3)
1
0
0
Transcribed Image Text:w order and a 56 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP Definition 2.1.5 Suppose An. Let M = C{{₁},... {tk}}. where (t₁)....() is a complete list of A-tabloids. Then MA is called the permutation module corresponding to A. ■ Since we are considering only row equivalence classes, we could list the rows of MA in any order and produce an isomorphic module. Thus M" is defined for any composition (ordered partition) . As will be seen in the next examples. we have already met three of these modules. Example 2.1.6 If A = (n), then with the trivial action. ■ M() = C{12} Example 2.1.7 Now consider A = (122). Each equivalence class {t} consists of a single tableau, and this tableau can be identified with a permutation in one-line notation (by taking the transpose, if you wish). Since the action of Sn is preserved, M(1) CS and the regular representation presents itself. ■ Example 2.1.8 Finally, if λ = (n-1, 1), then each A-tabloid is uniquely determined by the element in its second row, which is a number from 1 to n. As in Example 1.6.3, this sets up a module isomorphism M(-1,1) C{1,2,...,n}, so we have recovered the defining representation. ■ Example 2.1.9 Let us compute the full set of characters of these modules when n = 3. In this case the only partitions are A = (3), (2, 1), and (1³). From the previous examples, the modules M correspond to the trivial, defining, and regular representations of S3, respectively. These character values have all been computed in Chapter 1. Denote the character of M by and represent the conjugacy class of S, corresponding to u by K. Then we have the following table of characters: (3) (2,1) K(13) 1 3 6 K(2,1) 1 1 0 K(3) 1 0 0
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