Tensor product of vector spaces
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- Currently in the notes stage, see Notes:Tensor product
Contents
[hide]- Any first-time readers should look at the abstract definition first
Definition
Let F be a field and let ((Vi,F))ki=1 be a family of vector spaces over F. Let F(V1×⋯×Vk) denote the free vector space on ∏ki=1Vk. We define the (abstract) tensor product of V1,…,Vk as[1]:
- V1⊗⋯⊗Vk:=F(V1×⋯×Vk)R[Note 1] where R is defined as follows:
Abstract definition
Basis
Let F be a field and let ((Vi,F))ki=1 be a family of finite dimensional vector spaces. Let ni:=Dim(Vi) and e(i)1,…,e(i)ni denote a basis for Vi, then we claim[1]:
- B:={e(1)i1⊗⋯⊗e(k)ik | ∀j∈{1,…,k}⊂N[1≤ij≤nj]}
Is a basis for the tensor product of the family of vector spaces, V1⊗⋯⊗Vk
Note that the number of elements of B, denoted |B|, is ∏ki=1ni or ∏ki=1Dim(Vi), thus:
- Dim(V1⊗⋯⊗Vk)=∏ki=1ni[1]
Characteristic property
Let F be a field and let ((Vi,F))ki=1 be a family of finite dimensional vector spaces over F. Let (W,F) be another vector space over F. Then[1]:- If A:V1×⋯×Vk→W is any multilinear map
- there exists a unique linear map, ¯A:V1⊗⋯⊗Vk→X such that:
- ¯A∘p=A (that is: the diagram on the right commutes)
- there exists a unique linear map, ¯A:V1⊗⋯⊗Vk→X such that:
Where p:V1×⋯×Vk→V1⊗⋯⊗Vk by p:(v1,…,vk)↦v1⊗⋯⊗vk (and is p is multilinear)
Notes
- Jump up ↑ Take a moment to respect just how vast the space F(V1×⋯×Vk) is (especially if F:=R for example). Remember that this is not the space V1×⋯×Vk even though we write them as tuples. It is a huge space.
- TODO: Flesh out this note
-