The Sign Lemma
As we move towards proving the useful properties of Specht modules, we will find the following collection of results helpful. Through them all, let 
be a subgroup, and also consider the 
-invariant inner product on 
for which the distinct Young tabloids form an orthonormal basis.
First, if 
, then
where 
is the alternating sum of the elements of 
. The proof basically runs the same as when we showed that 
where 
has shape 
.
Next, for any vectors 
we have
Indeed, we can calculate
where we have used the facts that 
, and that as 
runs over a group, so does 
.
Next, if the swap 
, then we have the factorization
for some ![k\in\mathbb{C}[S_n]](https://s0.wp.com/latex.php?latex=k%5Cin%5Cmathbb%7BC%7D%5BS_n%5D&bg=e6e6e6&fg=333333&s=0&c=20201002)
. To see this, consider the subgroup 
, and pick a transversal. That is, write as a disjoint union:
but then we can write the alternating sum
as we stated.
Finally, if is some tableau with 
and 
in the same row, and if the swap , then
Our hypothesis tells us that 
. We can thus use the above factorization to write
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December 29, 2010 -
Posted by John Armstrong |
Algebra, Representation Theory, Representations of Symmetric Groups
The Unapologetic Mathematician 
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Hi, I have a question, if Specht modules are cyclic, why do we need to find a basis? I know you have just explain that and maybe this is a basic question, but I have troubles with that idea. If one polytabloid generate all Specht module, then what is the point to find a basis?
The module may be generated as an
-module by a single element, but not as a vector space.
In fact, that’s exactly what makes the representation theory of (the group algebra of)
interesting while the representation theory of a field is boring: the only interesting thing about a field representation (vector space) is its dimension — the number of generators — and each generator behaves the same as every other. 
-modules have a lot more structure than that.