Properties of Tensor Products - Cooperstein, Theorem 10.3

[Math Amateur]

I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

I am focused on Section 10.2 Properties of Tensor Products ... ...

I need help with an aspect of the proof of Theorem 10.3 regarding a property of tensor products ... ...The relevant part of Theorem 10.3 reads as follows:

In the above text from Cooperstein (Second Edition, page 355) we read the following:" ... ... The map $f$ is multilinear and therefore by the universality of $Y$ there is a linear map $T \ : \ Y \longrightarrow X$ such that$T(v_1 \otimes \ ... \ v_s \otimes w_1 \otimes \ ... \ w_t )$

$= (v_1 \otimes \ ... \ v_s ) \otimes (w_1 \otimes \ ... \ w_t )$

... ... ... "
My question is as follows:

What does Cooperstein mean by "the universality of $Y$" and how does the universality of $Y$ justify the existence of the linear map $T \ : \ Y \longrightarrow X$ ... and further, if $T$ does exist, then how do we know it has the form shown ...

Hope someone can help ...

Peter
*** Note ***

Presumably, Cooperstein is referring to some "universal mapping property" or "universal mapping problem" such as he describes in his Section 10.1 Introduction to Tensor Products as follows:

... ... BUT ... ... there is no equivalent of the logic surrounding the mapping $j$ ... unless we are supposed to assume the existence of $j$ and its relation to the existence of $T$ ... ?Indeed reading Cooperstein's definition of a tensor product ... it reads like the tensor product is the solution to the UMP ... but I am having some trouble fitting the definition and the UMP to the situation in Theorem 10.3 ...

Again, hope someone can help ...

Peter

 

[Samy_A]

I will state it with 2 vector spaces for simplicity.

The tensor product $V \otimes W$, where $V,W$ are vector spaces, has the following universal property:
If $f: V \times W \to U$ is a bilinear map to a vector space $U$, then there is a unique linear map $T : V \otimes W \to U$ such that $f(v,w)=T(v \otimes w)$.

(Prop 5.1 in https://people.maths.ox.ac.uk/hitchin/hitchinnotes/Differentiable_manifolds/Chapter_2.pdf)
(See also https://en.wikipedia.org/wiki/Tensor_product#Universal_property )

 

[Math Amateur]

Thanks Samy ... that looks pretty directly applicable to Theorem 10.3 ...

Appreciate your help ...

Peter

 

[Math Amateur]

I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

I am focused on Section 10.2 Properties of Tensor Products ... ...

I need help with an aspect of the proof of Theorem 10.3 regarding a property of tensor products ... ...The relevant part of Theorem 10.3 reads as follows:

In the above text from Cooperstein (Second Edition, page 355) we read the following:" ... ... The map $f$ is multilinear and therefore by the universality of $Y$ there is a linear map $T \ : \ Y \longrightarrow X$ such that$T(v_1 \otimes \ ... \ v_s \otimes w_1 \otimes \ ... \ w_t )$

$= (v_1 \otimes \ ... \ v_s ) \otimes (w_1 \otimes \ ... \ w_t )$

... ... ... "
My question is as follows:

What does Cooperstein mean by "the universality of $Y$" and how does the universality of $Y$ justify the existence of the linear map $T \ : \ Y \longrightarrow X$ ... and further, if $T$ does exist, then how do we know it has the form shown ...

Hope someone can help ...

Peter
*** Note ***

Presumably, Cooperstein is referring to some "universal mapping property" or "universal mapping problem" such as he describes in his Section 10.1 Introduction to Tensor Products as follows:

... ... BUT ... ... there is no equivalent of the logic surrounding the mapping $j$ ... unless we are supposed to assume the existence of $j$ and its relation to the existence of $T$ ... ?Indeed reading Cooperstein's definition of a tensor product ... it reads like the tensor product is the solution to the UMP ... but I am having some trouble fitting the definition and the UMP to the situation in Theorem 10.3 ...

Again, hope someone can help ...

Peter

I will state it with 2 vector spaces for simplicity.

The tensor product $V \otimes W$, where $V,W$ are vector spaces, has the following universal property:
If $f: V \times W \to U$ is a bilinear map to a vector space $U$, then there is a unique linear map $T : V \otimes W \to U$ such that $f(v,w)=T(v \otimes w)$.

(Prop 5.1 in https://people.maths.ox.ac.uk/hitchin/hitchinnotes/Differentiable_manifolds/Chapter_2.pdf)
(See also https://en.wikipedia.org/wiki/Tensor_product#Universal_property )

Hi Samy,

I was too quick to indicate that I understood what you said and the implications of what you said ... ...

... indeed ... I need some further help ...

I cannot see how what you have said leads to T having the property that

$T(v_1 \otimes \ ... \ v_s \otimes w_1 \otimes \ ... \ w_t )$

$= (v_1 \otimes \ ... \ v_s ) \otimes (w_1 \otimes \ ... \ w_t )$Can you explain why T has the above property?Hope you can help ...

Peter

 

[Samy_A]

It is a direct application of the universality of the tensor product. In sloppy language, that principle means that a multilinear map on a cartesian product of vector spaces defines ("can be extended to") a linear map on the tensor product of these vector spaces.

In the proof, they define $f: V_1 \times \dots \times V_s \times W_1 \times \dots \times W_t \to X=V \otimes W$ by $f(v_1,\dots ,v_s,w_1,\dots ,w_t)=(v_1 \otimes \dots \otimes v_s ) \otimes (w_1 \otimes \dots \otimes w_t)$.

This $f$ is clearly multilinear. Therefore, there exists a linear $T: V_1 \otimes \dots \otimes V_s \otimes W_1 \otimes \dots \otimes W_t \to X$ such that $T(v_1 \otimes \dots \otimes v_s \otimes w_1 \otimes \dots \otimes w_t) =f(v_1,\dots ,v_s,w_1,\dots ,w_t)=(v_1 \otimes \dots \otimes v_s ) \otimes (w_1 \otimes \dots \otimes w_t)$

 

[Math Amateur]

Oh! , OK Samy, Thanks ...

Most important principle ... seems to be used again and again in proofs of properties of tensor products ...

Peter