2025-02-23

1021: Tensor Product of k Vectors Spaces over Field

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definition of tensor product of k vectors spaces over field

Topics


About: vectors space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a definition of tensor product of k vectors spaces over field.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

Main Body

1: Structured Description


Here is the rules of Structured Description.

Entities:
F: ∈{ the fields }
{V1,...,Vk}: ⊆{ the F vectors spaces }
V1×...×Vk: = the product set 
F(V1×...×Vk,F): = the free vectors space 
S: ={((v1,...,rvj,...,vk))−r((v1,...,vk))∈F(V1×...×Vk)|r∈F,v1∈V1,...,vk∈Vk}∪{((v1,...,vj+vj′,...,vk))−((v1,...,vj,...,vk))−((v1,...,vj′,...,vk))∈F(V1×...×Vk)|v1∈V1,...,vk∈Vk,vj′∈Vj}
(S): = the sub-'vectors space' generated by subset of vectors space 
∗V1⊗...⊗Vk: =F(V1×...×Vk)/(S), the quotient vectors space
//

Conditions:
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For each (v1,...,vk)∈V1×...×Vk, [f(v1,...,vk)]=[((v1,...,vk))] (where f(v1,...,vk):V1×...×Vk→F∈F(V1×...×Vk,F) is the function that maps (v1,...,vk) to 1 and maps the other elements to 0) is often denoted as v1⊗...⊗vk and called "tensor product of v1,...,vk", but one finds them misleading to be confused with tensor product of tensors and prefers denoting just [((v1,...,vk))]: for the reason why we use the notation like ((v1,...,vk)), see Note for the definition of free vectors space on set: (v1,...,vk)∈V1×...×Vk but ((v1,...,vk))∈F(V1×...×Vk,F).

2: Note


We need to be careful not to do like "[r((v1,...,vk))+r′((v1′,...,vk′))]=[(r(v1,...,vk))+(r′(v1′,...,vk′))]=[((rv1,...,rvk))+((r′v1′,...,r′vk′))]=[((rv1,...,rvk)+(r′v1′,...,r′vk′))]=[((rv1+r′v1′,...,rvk+r′vk′))]", which is wrong: the reason is described in Note for the definition of free vectors space on set: the 1st equal is wrong because r((v1,...,vk))+r′((v1′,...,vk′)) is the function that maps (v1,...,vk) to r and maps (v1′,...,vk′) to r′, while (r(v1,...,vk))+(r′(v1′,...,vk′)) is the function that maps r(v1,...,vk) to 1 and maps r′(v1′,...,vk′) to 1 but maps (v1,...,vk) to 0 and maps (v1′,...,vk′) to 0; the 3rd equal is wrong because ((rv1,...,rvk))+((r′v1′,...,r′vk′)) is the function that maps (rv1,...,rvk) to 1 and maps (r′v1′,...,r′vk′) to 1, while ((rv1,...,rvk)+(r′v1′,...,r′vk′)) is the function that maps (rv1,...,rvk)+(r′v1′,...,r′vk′) to 1 but maps (rv1,...,rvk) to 0 and maps (r′v1′,...,r′vk′) to 0.

These are 2 legitimate rules: 1) [((v1,...,rvj,...,vk))]=r[((v1,...,vk))]; 2) [((v1,...,vj+vj′,...,vk))]=[((v1,...,vj,...,vk))]+[((v1,...,vj′,...,vk))].

1) is because [((v1,...,rvj,...,vk))]=[((v1,...,rvj,...,vk))−(((v1,...,rvj,...,vk))−r((v1,...,vk)))]=[r((v1,...,vk))]=r[((v1,...,vk))].

2) is because [((v1,...,vj+vj′,...,vk))]=[((v1,...,vj+vj′,...,vk))−(((v1,...,vj+vj′,...,vk))−((v1,...,vj,...,vk))−((v1,...,vj′,...,vk)))]=[((v1,...,vj,...,vk))+((v1,...,vj′,...,vk))]=[((v1,...,vj,...,vk))]+[((v1,...,vj′,...,vk))].

References
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