Tensor product of vector spaces

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Currently in the notes stage, see Notes:Tensor product
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]:

  • V1Vk:=F(V1××Vk)R[Note 1] where R is defined as follows:
    • R denotes the span of all the union of the following two sets:
      1. {(v1,,vi1,avi,vi+1,,vk)a(v1,,vk) | i{1,,k}aFj{1,,k}[vjVj]}
      2. {(v1,,vi1,vi+vi,vi+1,,vk)(v1,,vk)(v1,,vi1,vi,vi+1,,vk) | i{1,,k}viVij{1,,k}[vjVj]}

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)i1e(k)ik | j{1,,k}N[1ijnj]}

Is a basis for the tensor product of the family of vector spaces, V1Vk


Note that the number of elements of B, denoted |B|, is ki=1ni or ki=1Dim(Vi), thus:

  • Dim(V1Vk)=ki=1ni[1]

Characteristic property

Diagram of the situation, the double-arrows is multilinear, the other is linear
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××VkW is any multilinear map
    • there exists a unique linear map, ¯A:V1VkX such that:
      • ¯Ap=A (that is: the diagram on the right commutes)

Where p:V1××VkV1Vk by p:(v1,,vk)v1vk (and is p is multilinear)

Notes

  1. 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

References

  1. Jump up to: 1.0 1.1 1.2 1.3 Introduction to Smooth Manifolds - John M. Lee