# Regrettable definition/notation

hello, I would like to hear some comments about this situation. There seems to be two incompatible definitions concerning "Tensor Product" and "Dyadic product" of two vectors. The mathematical definition states tha $a\otimes b$ is a bilinear funtion defined on $V^*\times V^*$ with values in R while the dyadic product and also sometimes called tensor product written as ab and sometimes $a\otimes b$ is defined as a linear operator defined on V with values in V
I reached this conclution after looking up for the mathematical definition in "Tensor Analysis on Manifolds" by Bishop and for the defintion used in physics in "Mechanics" by Symon, "Continuum Mechanics" by Chadwick,"Continuum Mechanics" by Spencer

## Answers and Replies

Simon Bridge
Homework Helper
You'll probably find that the bilinear function can be written as an operator.

were the bilinear function defined on VxV or V*xV it would be possible but since it is denfined on V*xV* I don't know how to do it.

stevendaryl
Staff Emeritus
hello, I would like to hear some comments about this situation. There seems to be two incompatible definitions concerning "Tensor Product" and "Dyadic product" of two vectors. The mathematical definition states tha $a\otimes b$ is a bilinear funtion defined on $V^*\times V^*$ with values in R while the dyadic product and also sometimes called tensor product written as ab and sometimes $a\otimes b$ is defined as a linear operator defined on V with values in V
I reached this conclution after looking up for the mathematical definition in "Tensor Analysis on Manifolds" by Bishop and for the defintion used in physics in "Mechanics" by Symon, "Continuum Mechanics" by Chadwick,"Continuum Mechanics" by Spencer

I can understand the first definition, but not the second. What I would think would be correct is this:

The tensor product of vectors $a$ and $b$ is a bilinear function of type $(V^*\times V^*) \rightarrow R$, which is also a linear function of type $V^*\rightarrow V$

(I'm using $A \rightarrow B$ to mean the type of functions that take objects of type $A$ and return objects of type $B$)

I can understand the first definition, but not the second. What I would think would be correct is this:

The tensor product of vectors $a$ and $b$ is a bilinear function of type $(V^*\times V^*) \rightarrow R$, which is also a linear function of type $V^*\rightarrow V$

(I'm using $A \rightarrow B$ to mean the type of functions that take objects of type $A$ and return objects of type $B$)

The definition states the action of ab on a vector c through the equation:

$a\otimes b (c)= (b.c)a$ where b.c is the escalar product

stevendaryl
Staff Emeritus
The definition states the action of ab on a vector c through the equation:

$a\otimes b (c)= (b.c)a$ where b.c is the escalar product

Ah! Then I understand the two definitions. They aren't exactly the same, but they are both natural. If you have a scalar product, then there is NO difference between $V$ and $V^*$. So all the following are equivalent:

1. $(V^* \times V^*) \rightarrow R$
2. $V^* \rightarrow (V^* \rightarrow R)$
3. $V^* \rightarrow V^{**}$
4. $V^* \rightarrow V$
5. $V \rightarrow V$

1-3 are always equivalent. 4 is equivalent for finite-dimensional vector spaces, and 5 is equivalent for finite dimensional vectors spaces with a scalar product.

How is 3 equivalent to 1 or 2?

stevendaryl
Staff Emeritus
How is 3 equivalent to 1 or 2?

For any vector space $V$, the dual space $V^*$ is defined to be the linear functions of type $V \rightarrow R$.

Right!

You'll probably find that the bilinear function can be written as an operator.
Yes, that's right. The piece of information I was not using is that the diad definition presupposes the existance of an internal product while the mathematical definition of tensor product does not. So taking into consideration the internal product we have a natural isomorphism between V* and V and finally both definitions coincide

Simon Bridge