On Sat, 27 Aug 2005 12:03:31 -0700, Schoenfeld wrote:

Suppose A is a rank-n tensor.
Question 1: What is the name given to the embedded rank (n-1) tensor
Z found along the diagonal of A? (assume that dim(A) = dim(Z)^2 )

Question 2: What is the symbology/formula used to extract this
diagonal (n-1) tensor?

Example:
Suppose A was the rank-2 tensor:

A = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]

The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
proper name for this entity?

And for the other question:
If,
A = B x C
where
x denotes tensor product
B is an n/2 rank tensor with dimension SQRT(dim(A))
C is an n/2 rank tensor with dimension SQRT(dim(A))

How can I extract this "embedded diagonal rank (n-1) tensor" from A?

Any help appreciated.

A tensor is an element of a linear space that is invariant
under transformations of some kind. For example in 3 space
where we take the transformations to be rotations a vector
(three component thingy) is transformed into another vector
upon rotation. In QM with rotations we may look at a vector
as spin-1 under rotation. A matrix, like your A may be looked
at as the sum of products of spin-1 vectors. Now in general

1 x 1 = 2 + 1 + 0

which says the the invariant spaces obtained by the direct
product of two spin-1 object are spin-2, spin-1 and spin-0
which are of dimension (2S+1) 5, 3, and 1 which sums to
9 which is exactly the dimension of your matrix. This says
that one may split your matrix into 3 components, one spin-2
tensor which is a symmetric traceless matrix of vector dimension
5, a vector of dimension 3 which is the anti-symmetric component
of your matrix, and a trace part or scalar of dimension 1. Therefore,
the invariant tensor components of your matrix are

S(5) = [ -14, 3, 5 ]
[ 3, -10, 7 ]
[ 5, 7, -6]

A(3) = [ 0, -1, -2 ]
[ 1, 0, -1 ]
[ 2, 1, 0 ]

T(1) = [-15, 0, 0]
[ 0,-15, 0 ]
[ 0, 0, -15]

note that under rotations S(5) rotates into another
symmetric tensor, A(3) rotates into another antisymmetric
tensor and T(1) rotates into itself.

In article .com,
Schoenfeld wrote:

Robert Low wrote:
TMG wrote:
Weren't you just told "Thanks for you help, please avoid any more in
future."?
I understand that school's back in session, but what happened to manners?

Anybody whose response to being told they're wrong is to tell you
to be quiet needs all the help you can throw at them :-)

Robert, I understand these components can be used to form such n-1
tensor, this is why i called it "embedded". In the same way the
components of a vector can be used to form a scalar. This is clearly
what I implied. That you persist with your objection implies this
conversation is useless and thread unfortunately corrupted.



The components of a vector form a scalar (a rank-0 tensor) if you contract
them. E.g. |S|^2 = x^2 + y^2 + z^2, which is a length that remains
unchanged under rotations, translations, and changes of coordinate system.

But rotate the coordinate system and the first component of S will sure
change. The first component, by itself, is a scalar in the other sense--
it's just a number.


--
"It is the weak who are cruel. Gentleness can only be expected from the
strong." -- Leo Roskin

In article .com,
Schoenfeld wrote:

Stephen Montgomery-Smith wrote:


No, because trace(A) = Z . SUM{i=1}^(SQRT(dim(A))) Bi). I needed Z not
the trace(A). I found the solution anyway, it's related to the
hardamard product. Thanks anyway.

Can you explain a little more? What does it represent, what do you do
with it?


--
"Voice or no voice, the people can always be brought to the bidding of
the leaders. This is easy. All you have to do is to tell them they
are being attacked, and denounce the pacifists for lack of patriotism
and exposing the country to danger." -- Hermann Goering

Robert Low wrote:
Schoenfeld wrote:
Robert Low wrote:
It's exacerbated by the plain fact
that he hasn't a clue what a tensor is (but interprets
that as us being a bunch of recitation automata).

In any case, I'm tempted to agree with him that the
conversation is useless.
A review of your post reveals nothing quantative, just emotional
opinion. I am treating a tensor as a multidimensional array of scalars.

Ah, that's your problem then. You see, a tensor is not
an array of scalars. Or then again, perhaps you don't.
But hey, that's just me emoting.

No, it's just you being wrong.

http://en.wikipedia.org/wiki/Tensor
"The classical approach views tensors as multidimensional arrays that
are n-dimensional generalizations of scalars, 1-dimensional vectors and
2-dimensional matrices. The "components" of the tensor are the indices
of the array."


If ignorance were a disability, you'd be on the full pension.

Gee, I'm glad only one of us is rude and ignorant. Just
think how things would degenerate if you were too...

Gregory L. Hansen wrote:
In article .com,
Schoenfeld wrote:

Stephen Montgomery-Smith wrote:


No, because trace(A) = Z . SUM{i=1}^(SQRT(dim(A))) Bi). I needed Z not
the trace(A). I found the solution anyway, it's related to the
hardamard product. Thanks anyway.

Can you explain a little more? What does it represent, what do you do
with it?

I use it extensively in my formula to predict the momemtum gain of my
perpetual motion machine (it also arises in my gold cold-fusion
equations).




--
"Voice or no voice, the people can always be brought to the bidding of
the leaders. This is easy. All you have to do is to tell them they
are being attacked, and denounce the pacifists for lack of patriotism
and exposing the country to danger." -- Hermann Goering

"Martin Hogbin" wrote in message ...

"Schoenfeld" wrote in message oups.com...

Robert Low wrote:
Schoenfeld wrote:
Suppose A was the rank-2 tensor:

A = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]

The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
proper name for this entity?

'Not a tensor', as another poster already told you. A rank 1 tensor
has to be either a vector or a covector, and this is neither.

It is something relevant, but obviously there is no point continuing
discussion. Thanks for you help, please avoid any more in future.

That is one of the most ungrateful replies I have ever
seen and a perfect example of how not to get help
from a newsgroup.


Indeed.
This (and like you say everything he says in the thread) is a
school example of arrogance and ignorance.
It deserves a nice:
http://users.pandora.be/vdmoortel/di...es/Thanks.html

I also have re-entered his previous entries:
http://users.pandora.be/vdmoortel/di...Confusion.html
http://users.pandora.be/vdmoortel/di...s/DisCont.html
Same style - Same Schoensmeer.

Dirk Vdm


I note that you continue in the same manner below.

Martin Hogbin

"Robert Low" wrote in message ...
Schoenfeld wrote:

[snip]

If ignorance were a disability, you'd be on the full pension.

Gee, I'm glad only one of us is rude and ignorant. Just
think how things would degenerate if you were too...

:-)

Dirk Vdm

Robert Low wrote:

It's exacerbated by the plain fact
that he hasn't a clue what a tensor is (but interprets
that as us being a bunch of recitation automata).

I've been waiting for him to use that standard word "regurgitate"(TM)
that invariably pops up in discussions like these :-)

In any case, I'm tempted to agree with him that the
conversation is useless.

:-)

--
Jan Bielawski

Schoenfeld wrote:
Robert Low wrote:
Schoenfeld wrote:
Robert Low wrote:
It's exacerbated by the plain fact
that he hasn't a clue what a tensor is (but interprets
that as us being a bunch of recitation automata).

In any case, I'm tempted to agree with him that the
conversation is useless.
A review of your post reveals nothing quantative, just emotional
opinion. I am treating a tensor as a multidimensional array of scalars.

Ah, that's your problem then. You see, a tensor is not
an array of scalars. Or then again, perhaps you don't.
But hey, that's just me emoting.

No, it's just you being wrong.

http://en.wikipedia.org/wiki/Tensor
"The classical approach views tensors as multidimensional arrays that
are n-dimensional generalizations of scalars, 1-dimensional vectors and
2-dimensional matrices. The "components" of the tensor are the indices
of the array."

This definition is incorrect. Whoever wrote it was either being
careless or a moron. Tensors are not arrays of scalars. And
"components" are definitely not "indices of the array". I checked the
rest of that web page and its quality of presentation is abysmal. Don't
use math dictionaries or lexicons to learn math, it's a waste of time.
Use real textbooks.

--
Jan Bielawski

Schoenfeld wrote:
Robert Low wrote:
Ah, that's your problem then. You see, a tensor is not
an array of scalars.
No, it's just you being wrong.

Bwahahahahaha!

OK, so let V be R^2, with the usual basis e_1, e_2,
which we'll represent by [1 0]^T and [0 1]^T. A
is in V tensor V, and is given in this basis by

A=[1 2;3 4]

You claim that the component A_11, which is 1
in this basis, is a scalar. I suggest you
change basis to f_1=0.5 e_1 and f_2=0.5 e_2,
and compute A in this new basis. If A_11 is
a scalar, it will still be equal to 1 in
this basis.

I look forward to your explicit computation.

http://en.wikipedia.org/wiki/Tensor
"The classical approach views tensors as multidimensional arrays that
are n-dimensional generalizations of scalars, 1-dimensional vectors and
2-dimensional matrices. The "components" of the tensor are the indices
of the array."

And the relevance of that paragraph to your claim that
a tensor is an array of scalars is...what, exactly?

In any case, the irony of a person who claimed that people who
disagree with him simply learned definitions by rote out
of a book subsequently quoting a wiki article in an
attempt to substantiate his position has not escaped me.