Prev: N[(1/2)-k*(k-1/2)*(-3)/Factorial[k], 10000]
Next: product of tangents problem
From: Axel Vogt on 23 Sep 2009 14:19
Omega Cubed wrote:
> On 2009-09-23, ganesh <ganeshs138(a)gmail.com> wrote:
>> you answered the right question, and I think I understood.
>> How does this change for an asymetric tensor?
>> v1^a1.v2^a2 and v2^a2.v1^a1 would be two different axis for an
>> asymetric tensor, but for a symetric tensor these are the same. Is
>> this right?
>
> The space of fully antisymetric tensor of rank n over a vector space of
> dimension m has [ m choose n ] dimensions.
>
> I am too lazy to do the explanation here, so go look up "alternating
> forms" somewhere. The reason is really simple.
>
> In general, v1 tensor v2 and v2 tensor v1 are two different tensors.
> In the symmetric case, the reason we only count them once is that by
> the symmetry, v1 tensor v2 by itself is not a symmetric tensor. Or
> that if a symmetric tensor has a v1 tensor v2 component, it must also
> have a v2 tensor v1 component with the same coefficient. One cannot
> appear without the other.
>
> In the most generality, the space of rank n tensors over a vector
> space of m dimensions has dimension m^n. There are some interesting
> subspaces in there if you apply various types of symmetry assumptions.
> For example, you can ask: what is the subspace of tensors such that
> the first two spots is symmetric, the third spot and the fourth spot
> is antisymmetric, the fifth and sixth antisymmetric, but if you
> simultaneously swap the third and fifth AND the fourth and sixth it is
> symmetric. These are some seriously hard questions. If you are
> interested in it, the best (and I think only) source is the book by
> Hermann Weyl called "Classical Groups: their invariants and
> representations".
>
> W

Bourbaki, Algebre III should have, I think it is the one which
treats the tensor algebras
First  |  Prev  | 
Pages: 1 2
Prev: N[(1/2)-k*(k-1/2)*(-3)/Factorial[k], 10000]
Next: product of tangents problem