What does the Tensor Direct Product do?

[NotPrinceHamlet (Graham)]

Has anyone got a good reference for what the mathematical operation of the tensor direct product, i.e.http://upload.wikimedia.org/math/e/9...dfc2c9a876.png does?

I've got an ok book

http://www.amazon.com/Brief-Tensor-A.../dp/038794088X

and its all going along well, with me hoping to get enough maths under my belt to tackle GR, and then suddenly the author slips in that circled cross symbol near the end of a chapter with almost no preamble or definition of what operation it represents. Almost as if he thought I wouldnt notice!

As a result, suddenly after racing through the book I've hit a brick wall at that point - all was going swimmingly until then!

I'm particularly interested in developing a feel for *what* it does, not just its strict definition of what it is.

[sjastro]

Originally Posted by NotPrinceHamlet

Has anyone got a good reference for what the mathematical operation of the tensor direct product, i.e.http://upload.wikimedia.org/math/e/9...dfc2c9a876.png does?

I've got an ok book

http://www.amazon.com/Brief-Tensor-A.../dp/038794088X

and its all going along well, with me hoping to get enough maths under my belt to tackle GR, and then suddenly the author slips in that circled cross symbol near the end of a chapter with almost no preamble or definition of what operation it represents. Almost as if he thought I wouldnt notice!

As a result, suddenly after racing through the book I've hit a brick wall at that point - all was going swimmingly until then!

I'm particularly interested in developing a feel for *what* it does, not just its strict definition of what it is.

It's the operator for the tensor outer product as opposed to the inner product. The inner product is simply the dot product. (For vectors a.b= ab cos(A).)

Consider a tensor of rank 1 which is a vector. A vector is defined as having both magnitude and direction. The number of directions defines the rank.

The outer product of two vectors is a tensor of rank 2.
For example in 2-dimensional space consider vectors U and V.

U = u1a + u2b and V= v1a + v2b a and b are basis vectors.

The outer product UV = u1v1aa + u1v2ab + u2v1ba + u2v2bb

Each component of magnitude uivj now defined in 2 directions (aa, ab, ba, bb).

UV is therefore a tensor of rank 2.

The outer product of 2 tensors results in a tensor which is the sum of the ranks of the 2 tensors.

Hope this simplifies matters.

Steven

[NotPrinceHamlet (Graham)]

Got it. Thank you kindly - I appreciate your efforts.

I guess that there is a relationshio between a and aa, ab that will become apparent soon enough.

[NotPrinceHamlet (Graham)]

Hi sjastro,

Assuming that direct product is the same as outer product?

I tried to apply what you've said to the text in the book where I got stuck.

The book gives the simple example of a 2nd order tensor Proju that operates on a vector v to give the projection of v onto u:

i.e. Projuv = (v.u)u

Easy enough to prove and understand.

The author then goes on to make the breathtaking leap of stating

"

a generalisation of this is that the direct product uv of two vectors u and v is a tensor that sends any vector w into a new vector according to the rule

uv(w) = u(v.w)

"



The text is not using the (x) symbol - i.e. outer product - , so I'm lost as to how to interpret this, since the result is supposed to be a vector, not a second order tensor...

Does it mean

[ u (x) v ] (x) w or u (x) v . [w] or some thing else?

[Robh (Rob)]

Don't know a whole lot about tensors but this seems analogous to the dot product and cross product used in straight vector analysis.
In vector analysis the expression would read
(u x v).w = u.(v x w) where u, v and w are vectors.
Regards, Rob

[sjastro]

Originally Posted by NotPrinceHamlet

Hi sjastro,

Assuming that direct product is the same as outer product?

I tried to apply what you've said to the text in the book where I got stuck.

The book gives the simple example of a 2nd order tensor Proju that operates on a vector v to give the projection of v onto u:

i.e. Projuv = (v.u)u

Easy enough to prove and understand.

The author then goes on to make the breathtaking leap of stating

"

a generalisation of this is that the direct product uv of two vectors u and v is a tensor that sends any vector w into a new vector according to the rule

uv(w) = u(v.w)

"



The text is not using the (x) symbol - i.e. outer product - , so I'm lost as to how to interpret this, since the result is supposed to be a vector, not a second order tensor...

Does it mean

[ u (x) v ] (x) w or u (x) v . [w] or some thing else?

The circle/cross terminology used for the outer product doesn't appear to be standard. Usually vectors (tensors) written next to each other are assumed to have the outer product operation performed.

The equation uv(w) = u(v.w) doesn't make any sense.

If the LHS of the equation is interpreted as the outer product of 3 vectors uvw, this results in a tensor of rank 3 (the sum of the rank of each vector is 1+1+1=3).

The RHS gives a vector of rank 1. (v.w) is a dot product giving a scalar of rank 0. The direct product of this scalar with u gives a rank of 1 (0+1=1) which is a vector.

What I suspect the equation should read is:

uv.w = u(v.w)
Note that u(v.w) doesn't equal (u.v)w.

This is a special case of an outer product of three vectors u, v, w followed by a contraction. The inner (dot) product provides the contraction. Contraction of a tensor of rank n results in a tensor of rank (n-2).

Contraction of tensors are used to simplify Einstein's equations.

Regards

Steven.

[sjastro]

Originally Posted by Robh

Don't know a whole lot about tensors but this seems analogous to the dot product and cross product used in straight vector analysis.
In vector analysis the expression would read
(u x v).w = u.(v x w) where u, v and w are vectors.
Regards, Rob

Hello Rob,

The outer product for tensors is a unique operation. It has no relationship to dot or cross products for vectors.

Regards

Steven

[Robh (Rob)]

Originally Posted by sjastro

What I suspect the equation should read is:

uv.w = u(v.w)

Steven,
You got me curious enough to research some of the maths here.
I've gotten as far as dyad products UV and their scalar matrix forms.
I've at least confirmed your expression.
This is a proof for vectors in two dimensions.
If U=(u1 u2), V=(v1 v2), W=(w1 w2) are three vectors in 2D then the dyad matrix UV is
(u1v1 u1v2)
(u2v1 u2v2)

UV.W is
(u1v1w1+u1v2w2 u2v1w1+u2v2w2)=(v1w1+v2w2)(u1 u2)=kU, a vector with magnitude k and sense direction U.

V.W=v1w1+v2w2=k
So U(V.W)=Uk
As kU=Uk then UV.W=U(V.W)

Fascinating stuff. Have no idea where it leads.
But thanks for the challenge, Rob.

[NotPrinceHamlet (Graham)]

With a bit of perseverance, perspiration and lots of train trips (thats my study time), we'll get to Einstein's field equations yet...

[sjastro]

Originally Posted by Robh

Fascinating stuff. Have no idea where it leads.
But thanks for the challenge, Rob.

It all leads to what the fabric of the Universe is made from.

Steven

[Peter Ward]

Originally Posted by sjastro

It all leads to what the fabric of the Universe is made from.

Steven

Steve,

I found linear algebra tough enough..a lifetime ago in Maths 100.

Multi-linear! Clearly you have a handle on it. I'm impressed!

Regards
Peter

[g__day (Matthew)]

Wait till you have to deal with N >> 3 dimensional maths - or integrals and curls around singularities (the -1 / 4 * pi * i stuff - ugh!).

Second year Uni maths or physics I take it you're doing?

[NotPrinceHamlet (Graham)]

Nope - did that around 20 years ago, and more. But its got very, very rusty over the last two decades.

[Nesti (Mark)]

Putting the GR math aside for just one second, have you ever wondered what gravitation really is?

From the moment you hit gauss's [negative] divergence, it really makes you stop and think, what's dragging the spacetime carpet into the cellar, furniture and all, and without effort? Is the continuum energy being consumed, like water down a drain, taking the soap bubbles with it?

What I also find remarkable, is that it takes a force to alter the speed or direction of a massed object in uniform motion or at rest, right?! So, if gravitation offers force free acceleration by pulling spacetime in, towards mass - from a solitary electron to a black hole - then what's doing the pulling, what's doing the bending and warping as another massed object, in force free uniform motion, cruises along its Geodesic, and bends around the massed body? Also, the concept that laws are 'universal', seems to stop right there, since laws are carried alongside the Christoffels (representing the forces) and are not carried across spacetime. Remember this is where Einstein's 'get out of jail free' card came into play, he said that the notion of an absolute space or absolute time, for SR and GR, must be abolished, which was right after he used just that notion, a "Local Time" to formulate SR in the first place (Clock Synchronization, as per Lorentz and FitzGerald 's interpretation of Michelson and Morley's 1887 experiment). Seriously, it's the [Minkowski] dog chasing its own tail (did you like my little Polytechnikum joke there?).

We all know that spacetime is drawn toward the presence of mass, but how does spacetime know it is there, how much is there, and the density, which in turn affects the rate at which spacetime is pulled (metric value)?

Seriously, it's almost as though mass is a heatsink for energy. I can't bring myself to believe in either the Higgs Field or Particle, and would happily put a 100 pound bet that the LHC doesn't find it (Stephen Hawking style).

Sorry to interrupt your Gij's and Gji's (Riemann's equality was it?)

[Zaps]

Originally Posted by Nesti

...[W]hat's dragging the spacetime carpet into the cellar, furniture and all, and without effort?

...[W]hat's doing the pulling, what's doing the bending and warping as another massed object, in force free uniform motion, cruises along its Geodesic, and bends around the massed body?

...[H]ow does spacetime know it is there, how much is there, and the density, which in turn affects the rate at which spacetime is pulled (metric value)?

God's will, obviously.

[renormalised (Carl)]

So, if gravitation offers force free acceleration by pulling spacetime in, towards mass - from a solitary electron to a black hole - then what's doing the pulling, what's doing the bending and warping as another massed object, in force free uniform motion, cruises along its Geodesic, and bends around the massed body?

Quantum fairies playing on cosmic strings in a supersymmetrical orchestra....didn't you know that??!!

They're playing the "Higgs Concerto in A and/or C#"

[davidrh (David)]

Aha, I think I found the ideal place to try and get an answer to my question. Doing multilinear algebra with rank-2 tensors is all fine, but I am unable to get a straight (and simple example) for rank-3 (and hence a generalisation to rank-n).

I have a tensor A_{ijk} where i = 1..10, j=1..5 and k=1..15
and tensor B_{ijk}

Taking a direct step from rank-2 matrices, I want to compute the inner product of Z=AB'
how? and what is the size of Z is it 10x15x10? or...

Actually being even more precise of what I am trying to write out simply is I have A_{ijk} = x_i y_j h_k
and B = A;

So how do I write the dot product Z in terms of x,y and h?

Much appreciated!
David

[sjastro]

Originally Posted by davidrh

Aha, I think I found the ideal place to try and get an answer to my question. Doing multilinear algebra with rank-2 tensors is all fine, but I am unable to get a straight (and simple example) for rank-3 (and hence a generalisation to rank-n).

I have a tensor A_{ijk} where i = 1..10, j=1..5 and k=1..15
and tensor B_{ijk}

Taking a direct step from rank-2 matrices, I want to compute the inner product of Z=AB'
how? and what is the size of Z is it 10x15x10? or...

Actually being even more precise of what I am trying to write out simply is I have A_{ijk} = x_i y_j h_k
and B = A;

So how do I write the dot product Z in terms of x,y and h?

Much appreciated!
David

I'm afraid the formula you seek doesn't exist. For tensors greater than rank 2 involves messy calculations of each coefficient.

First of all matrix algebra can only be used on tensors up to rank 2.

Vectors or tensors of rank 1 can be defined as a row or column matrix.

Tensors of rank 2 are defined by a MxN matrix. If M=N this can define the dimension of space, while each component in the matrix is defined by a pair of indices (ij for example) which defines the rank.

Secondly the dot product cannot be used on tensors which are not of rank 1.
The generic operation for tensors is the inner product. This involves the outer product of each tensor followed by a dot product with a vector.

For example suppose C=ABC and D=EFG. C and D are tensors of rank 3, ABC and EFG are the respective vector (tensor 1) factors.

The outer product CD=ABCEFG is a tensor of rank 6.

The inner product with a vector U is applied to either A, B, C, D, E, F or G. Since the dot product will be different in each case there is obviously no hard and fast rules for calculations.

The only consistant point is that since the dot product of 2 vectors is a scalar the rank of the tensor is reduced from 6 to 4.

In general if the rank of the tensor is n, the dot product reduces the rank to n-2. This is known as contraction.

Hope this is of help.

Steven

[davidrh (David)]

Originally Posted by sjastro

I'm afraid the formula you seek doesn't exist. For tensors greater than rank 2 involves messy calculations of each coefficient.

First of all matrix algebra can only be used on tensors up to rank 2.

Vectors or tensors of rank 1 can be defined as a row or column matrix.

Tensors of rank 2 are defined by a MxN matrix. If M=N this can define the dimension of space, while each component in the matrix is defined by a pair of indices (ij for example) which defines the rank.

Secondly the dot product cannot be used on tensors which are not of rank 1.
The generic operation for tensors is the inner product. This involves the outer product of each tensor followed by a dot product with a vector.

For example suppose C=ABC and D=EFG. C and D are tensors of rank 3, ABC and EFG are the respective vector (tensor 1) factors.

The outer product CD=ABCEFG is a tensor of rank 6.

The inner product with a vector U is applied to either A, B, C, D, E, F or G. Since the dot product will be different in each case there is obviously no hard and fast rules for calculations.

The only consistant point is that since the dot product of 2 vectors is a scalar the rank of the tensor is reduced from 6 to 4.

In general if the rank of the tensor is n, the dot product reduces the rank to n-2. This is known as contraction.

Hope this is of help.

Steven

Thanks for the help Steven. So, following this line - where did vector U come from? and if I am understanding correctly, the procedure is to compute the outer product of the two tensor followed by an inner product with a vector.

Or were you just giving an example of a consistent computation which is the inner product of vector time tensor?

I have found code to do multi-dimensional matrix inner product (although don't know how correct this is) called multiprob http://www.mathworks.com/matlabcentr...eexchange/8773

Thanks,
David

[sjastro]

Originally Posted by davidrh

Thanks for the help Steven. So, following this line - where did vector U come from? and if I am understanding correctly, the procedure is to compute the outer product of the two tensor followed by an inner product with a vector.

Or were you just giving an example of a consistent computation which is the inner product of vector time tensor?

I shoudn't have used this terminology. Forget about the U.
The outer product is the tensor is CD=ABCEFG which is a tensor of rank 6.

The tensor can then be contracted by having a dot product between any pair of vectors involving A B C E F G (A.B, B.C etc).

The inner product between tensors is the outer product between the vectors followed by a contraction by a dot product between any pair of vectors.

Since the dot product of 2 vectors is a scalar (tensor of rank 0), the general format of the contracted tensor becomes CD=WXYZ which is a tensor of rank 4.
A further contraction results in a tensor of rank 2.

So if you start with a tensor of rank n each successive contraction reduces the rank by a value of 2.

Sorry for the confusion.

I have found code to do multi-dimensional matrix inner product (although don't know how correct this is) called multiprob http://www.mathworks.com/matlabcentr...eexchange/8773

I'm not familiar with handling multi dimensional arrays, or the meaning of matrix inner product in this context.

Regards

Steven