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From: ganesh on 20 Sep 2009 08:55
hi,
How does one compute the dimension of a nth ranked symmetric
tensor?
From: Omega Cubed on 20 Sep 2009 11:41
["Followup-To:" header set to sci.math.]
On 2009-09-20, ganesh <ganeshs138(a)gmail.com> wrote:
> How does one compute the dimension of a nth ranked symmetric
> tensor?

The dimension? Do you mean the dimension of the space of symmetric
tensors of rank n? (Let me know if I am answering the wrong question.)

Let V be a m-dimensional vector space, with a basis {v1... vm}. The
space Sym[tensor^n V] has a basis given by

{ (v1)^a1 . (v2)^a2 . ... . (vm)^am }

where products are taken with respect to the symmetric tensor product,
and the exponents a1... am sum up to n.

Thus it is clear that Sym[tensor^n V] is canonically isomorphic to the
space of homogeneous polynomials of degree n over m variables. The
dimension of this is well known as [ m + n - 1 choose n ]. To see that
it is true, notice that the number of monomials of degree n over m
variables is equivalent to the number of distinct m-tuples whose sum
is n. The problem now reduces to a problem in combinatorics: See
Theorem 2 at
http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29

W


From: Schoenfeld on 21 Sep 2009 10:17
On Sep 20, 10:55 pm, ganesh <ganeshs...(a)gmail.com> wrote:
> hi,
>    How does one compute the dimension of a nth ranked symmetric
> tensor?

A rotation matrix in 2-space has 2 dimensions with 2 components along
each dimension.

A rotation matrix in 3-space has 2 dimensions with 3 components along
each dimension.

A rotation matrix in M-space has 2 dimensions with M components along
each dimension.

A rank-N tensor is an N-dimensional generalization of a square matrix
(holding well-defined relations to the Jacobian transformation).

So, a rank-N tensor in M-space has N dimensions with M components
along each dimension.

The relations to the Jacobian allow tensors to be manipulated
independently of a coordinate system. Tensors, as abstract objects,
have an existence outside of a geometry.

That they can describe geometry without being subject to it earns them
"fundamental" status (at least in the eyes of Riemann).
From: ganesh on 23 Sep 2009 00:28
hi,

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?

ganesh



On Sep 20, 8:41 pm, Omega Cubed <omega_cu...(a)netzero.net> wrote:
> ["Followup-To:" header set to sci.math.]
> On 2009-09-20, ganesh <ganeshs...(a)gmail.com> wrote:
>
> >    How does one compute the dimension of a nth ranked symmetric
> > tensor?
>
> The dimension? Do you mean the dimension of the space of symmetric
> tensors of rank n? (Let me know if I am answering the wrong question.)

>
> Let V be a m-dimensional vector space, with a basis {v1... vm}. The
> space Sym[tensor^n V] has a basis given by
>
> { (v1)^a1 . (v2)^a2 . ... . (vm)^am }
>
> where products are taken with respect to the symmetric tensor product,
> and the exponents a1... am sum up to n.
>
> Thus it is clear that Sym[tensor^n V] is canonically isomorphic to the
> space of homogeneous polynomials of degree n over m variables. The
> dimension of this is well known as [ m + n - 1 choose n ]. To see that
> it is true, notice that the number of monomials of degree n over m
> variables is equivalent to the number of distinct m-tuples whose sum
> is n. The problem now reduces to a problem in combinatorics: See
> Theorem 2 athttp://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29
>
> W

From: Omega Cubed on 23 Sep 2009 13:40
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
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