let V be a finite dimensional C-vector space and denote by Sym(d,V) the d-fold symmetric tensors, ie Sym(d,V) is the dth-symmetric power of V.
This is in a natural way a GL(V) representation.
How do I show that it's irreducible?
>>8185669
if the matrix A has distinct EigValues x1,...,xn and v1,...,vn is a basis of V consisting of EigVectors of A, the the induced endomorphism on Sym(d,V) has EigVectors and values
[math]x_{i_1}, \hdots ,x_{i_d}[/math]
[math]v_{i_1}, \hdots,v_{i_d}[/math]
for all sequences
[math]i_1 \leq \hdots \leq i_d[/math]
>>8185731
and if W is a subrepresentation of Sym(d,V), then it must be spanned by a subset of these eigenvectors.
>>8185731
of course eigenvals are
[math]x_{i_1} \cdot ... \cdot x_{i_d}[/math]
eigenvectors
[math]v_{i_1} \cdot ... \cdot v_{i_d}[/math]
for all sequences
[math]i_1 \leq ... \leq i_d[/math]
>>8185669
assume its not irreducible
get a contradiction
>>8185669
I don't think Young's Lattice represents partitioning integers well.
>>8185765
I think it does, and it's very relevant to the irreducible representations of GL(V).
back to the proof
>>8185737
cont.
now look at the elementary matrices [math]E_{ij}[/math] defined such that
[math]E_{ij} v_k = \delta_{jk] v_i[/math]
since W is invariant under the endomorphisms [math]1 + E_{ij}[/math] it follows.. hm.. something is supposed to follow here, not sure how to finish
>>8185807
[math]E_{ij} v_k = \delta_{jk} v_i[/math]
/sci/ really needs a preview button.
ANYWHO, not sure how W's invariance under the endomorphisms [math]1+E_{ij}[/math] implies that all eigenvectors of the _induced_ endomorphism must already lie in W, ie that W=Sym(d,V).
Say W was spanned by [math]v_1^d[/math] then with the above construction I get that all [math]v_j^d[/math] would also have to be in W, but those arent nearly all the eigenvectors
>>8185839
>/sci/ really needs a preview button.
>what is the TeX button in the reply window
Learn how to use a website newfag.
Is it really irreducible unless dim V = 1?
Because for any basis vector v \in V, v \otimes v \otimes v ... \otimes v should span a proper submodule.
>>8185968
it is always irreducible, yes.
also a single vector [math]v^d=v \otimes ... \otimes v[/math] can never span an invariant subspace, because it would have to be an eigenvector of every single element of GL(V)
