Characteristic property of the tensor product

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Stub grade: A*
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Important linear algebra here. Totally

Statement

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) (see claim 1 for the proof of this)

Proof of claims

Claim 1: p is multilinear

  • Let i{1,,k} be given
    • p(v1,,vi1,vi+λu,vi+1,,vk)=v1vi1(vi+λu)vi+1vk
    =(v1vk)+(v1vi1λuvi+1vk)
    =(v1vk)+λ(v1vi1uvi+1vk)
    • Next notice: p(v1,,vk)+λp(v1,,vi1,u,vi+1,,vk)
    =(v1vk)+λ(v1vi1uvi+1vk)
    • We see: p(v1,,vi1,vi+λu,vi+1,,vk)=p(v1,,vk)+λp(v1,,vi1,u,vi+1,,vk)
  • Since i{1,,k} was arbitrary, we see it is linear in each i - the definition of a multilinear map

Proof

Diagram
The gist of the proof is as follows:
  • First note that by the characteristic property of the free vector space that ¯A extends uniquely to a linear map ˜A:F(V1××V2)W
  • Now we want to factor ¯A through π, this will yield us another linear map:
    ¯A:F(V1××VK)Ri.e. V1VkW - as required
    • To do this we use the group factorisation theorem:
      • If R=Ker(π)Ker(˜A) then ˜A factors through π to yield (in this case) ¯A, and furthermore ¯A is a linear map.
        • Let (v1,,vi1,avi,vi+1,,vk)R be given
          • Then ˜A(v1,,vi1,avi,vi+1,,vk) is:
            • =A(v1,,vi1,avi,vi+1,,vk) as it extends A and (v1,,vi1,avi,vi+1,,vk)V1××Vk
            • =aA(v1,,vk) as A is multilinear
            • =a˜A(v1,,vk) as ˜A is an extension of A
            • =˜A(a(v1,,vk)) as ˜A is linear
Grade: A
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The gist is above. Tidy up a bit, basically done, not quite sure where to go, demote to E once the outline is complete

Notes

References

  1. Jump up Introduction to Smooth Manifolds - John M. Lee