Regrettable definition/notation

  • Thread starter facenian
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In summary, the two definitions of the tensor product are not exactly the same, but they are both natural.
  • #1
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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 [itex]a\otimes b[/itex] is a bilinear funtion defined on [itex]V^*\times V^*[/itex] with values in R while the dyadic product and also sometimes called tensor product written as ab and sometimes [itex]a\otimes b[/itex] 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
 
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  • #2
You'll probably find that the bilinear function can be written as an operator.
 
  • #3
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.
 
  • #4
facenian said:
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 [itex]a\otimes b[/itex] is a bilinear funtion defined on [itex]V^*\times V^*[/itex] with values in R while the dyadic product and also sometimes called tensor product written as ab and sometimes [itex]a\otimes b[/itex] 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 [itex]a[/itex] and [itex]b[/itex] is a bilinear function of type [itex](V^*\times V^*) \rightarrow R[/itex], which is also a linear function of type [itex]V^*\rightarrow V[/itex]

(I'm using [itex]A \rightarrow B[/itex] to mean the type of functions that take objects of type [itex]A[/itex] and return objects of type [itex]B[/itex])
 
  • #5
stevendaryl said:
I can understand the first definition, but not the second. What I would think would be correct is this:

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

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

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

[itex]a\otimes b (c)= (b.c)a[/itex] where b.c is the escalar product
 
  • #6
facenian said:
The definition states the action of ab on a vector c through the equation:

[itex]a\otimes b (c)= (b.c)a[/itex] 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 [itex]V[/itex] and [itex]V^*[/itex]. So all the following are equivalent:

  1. [itex](V^* \times V^*) \rightarrow R[/itex]
  2. [itex]V^* \rightarrow (V^* \rightarrow R)[/itex]
  3. [itex]V^* \rightarrow V^{**}[/itex]
  4. [itex]V^* \rightarrow V[/itex]
  5. [itex]V \rightarrow V[/itex]

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.
 
  • #7
How is 3 equivalent to 1 or 2?
 
  • #8
dauto said:
How is 3 equivalent to 1 or 2?

For any vector space [itex]V[/itex], the dual space [itex]V^*[/itex] is defined to be the linear functions of type [itex]V \rightarrow R[/itex].
 
  • #9
Right!
 
  • #10
Simon Bridge said:
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 existence 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
 
  • #11
Well done: I knew you'd get there :)
 
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1. What is the definition of "Regrettable"?

The definition of "regrettable" is something that causes feelings of regret or disappointment.

2. Can "regrettable" be used in a positive context?

No, "regrettable" is generally used to describe something negative or unfortunate.

3. How is "regrettable" different from "regretful"?

"Regrettable" is an adjective that describes something that is causing regret. "Regretful" is an adjective that describes a person who is feeling regret.

4. Is there a difference between "regrettable" and "unfortunate"?

While both words can describe something negative, "regrettable" specifically implies that there is a sense of disappointment or regret attached to the situation.

5. Can "regrettable" be used to describe a person?

Yes, "regrettable" can be used to describe a person's actions, choices, or behavior that causes regret or disappointment.

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