Characteristic property of the tensor product

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Important linear algebra here. Totally


[ilmath]\xymatrix{ V_1\times\cdots\times V_k \ar@2{->}[rr]^-A \ar@2{->}[d]_p & & W \\ V_1\otimes\cdots\otimes V_k \ar@{.>}[urr]_-{\overline{A} } }[/ilmath]
Diagram of the situation, the double-arrows is multilinear, the other is linear
Let [ilmath]\mathbb{F} [/ilmath] be a field and let [ilmath]\big((V_i,\mathbb{F})\big)_{i\eq 1}^k[/ilmath] be a family of finite dimensional vector spaces over [ilmath]\mathbb{F} [/ilmath]. Let [ilmath](W,\mathbb{F})[/ilmath] be another vector space over [ilmath]\mathbb{F} [/ilmath]. Then[1]:
  • If [ilmath]A:V_1\times\cdots\times V_k\rightarrow W[/ilmath] is any multilinear map
    • there exists a unique linear map, [ilmath]\overline{A}:V_1\otimes\cdots\otimes V_k\rightarrow X[/ilmath] such that:
      • [ilmath]\overline{A}\circ p\eq A[/ilmath] (that is: the diagram on the right commutes)

Where [ilmath]p:V_1\times\cdots\times V_k\rightarrow V_1\otimes\cdots\otimes V_k[/ilmath] by [ilmath]p:(v_1,\ldots,v_k)\mapsto v_1\otimes\cdots\otimes v_k[/ilmath] (and is [ilmath]p[/ilmath] is multilinear) (see claim 1 for the proof of this)

Proof of claims

Claim 1: [ilmath]p[/ilmath] is multilinear

  • Let [ilmath]i\in\{1,\ldots,k\} [/ilmath] be given
    • [ilmath]p(v_1,\ldots,v_{i-1},v_i+\lambda u,v_{i+1},\ldots,v_k)\eq v_1\otimes\cdots\otimes v_{i-1}\otimes (v_i+\lambda u)\otimes v_{i+1}\otimes\cdots\otimes v_k[/ilmath]
    [ilmath]\eq (v_1\otimes\cdots\otimes v_k)+(v_1\otimes \cdots\otimes v_{i-1}\otimes \lambda u\otimes v_{i+1}\otimes\cdots\otimes v_k)[/ilmath]
    [ilmath]\eq (v_1\otimes\cdots\otimes v_k)+\lambda(v_1\otimes\cdots\otimes v_{i-1}\otimes u\otimes v_{i+1}\otimes\cdots\otimes v_k)[/ilmath]
    • Next notice: [ilmath]p(v_1,\ldots,v_k)+\lambda p(v_1,\ldots,v_{i-1},u,v_{i+1},\ldots,v_k) [/ilmath]
    [ilmath]\eq (v_1\otimes\cdots\otimes v_k)+\lambda(v_1\otimes\cdots\otimes v_{i-1}\otimes u\otimes v_{i+1}\otimes \cdots \otimes v_k)[/ilmath]
    • We see: [ilmath]p(v_1,\ldots,v_{i-1},v_i+\lambda u,v_{i+1},\ldots,v_k)[/ilmath][ilmath]\eq p(v_1,\ldots,v_k)+\lambda p(v_1,\ldots,v_{i-1},u,v_{i+1},\ldots,v_k)[/ilmath]
  • Since [ilmath]i\in \{1,\ldots,k\} [/ilmath] was arbitrary, we see it is linear in each [ilmath]i[/ilmath] - the definition of a multilinear map


[ilmath]\xymatrix{ \mathcal{F}(V_1\times\cdots\times V_k) \ar@{-->}@/^1.8em/[rrr]^{\tilde{A} } \ar[dr]_\pi & V_1\times\cdots\times V_k \ar@{_{(}->}[l]_-i \ar@2{->}[d]_p \ar@2{->}[rr]^-A & & W \\ & V_1\otimes\cdots\otimes V_k \ar@{.>}[urr]_-{\overline{A} } }[/ilmath]
The gist of the proof is as follows:
  • First note that by the characteristic property of the free vector space that [ilmath]\overline{A} [/ilmath] extends uniquely to a linear map [ilmath]\tilde{A}:\mathcal{F}(V_1\times\cdots\times V_2)\rightarrow W[/ilmath]
  • Now we want to factor [ilmath]\overline{A} [/ilmath] through [ilmath]\pi[/ilmath], this will yield us another linear map:
    [math]\overline{A}:\underbrace{\dfrac{\mathcal{F}(V_1\times\cdots\times V_K)}{\mathcal{R} } }_{\text{i.e. }V_1\otimes\cdots\otimes V_k} \rightarrow W[/math] - as required
    • To do this we use the group factorisation theorem:
      • If [ilmath]\mathcal{R}\eq\text{Ker}(\pi)\subseteq\text{Ker}(\tilde{A})[/ilmath] then [ilmath]\tilde{A} [/ilmath] factors through [ilmath]\pi[/ilmath] to yield (in this case) [ilmath]\overline{A} [/ilmath], and furthermore [ilmath]\overline{A} [/ilmath] is a linear map.
        • Let [ilmath](v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)\in\mathcal{R} [/ilmath] be given
          • Then [ilmath]\tilde{A}(v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)[/ilmath] is:
            • [ilmath]\eq A(v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)[/ilmath] as it extends [ilmath]A[/ilmath] and [ilmath](v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)\in V_1\times\cdots\times V_k[/ilmath]
            • [ilmath]\eq aA(v_1,\ldots,v_k)[/ilmath] as [ilmath]A[/ilmath] is multilinear
            • [ilmath]\eq a\tilde{A}(v_1,\ldots,v_k)[/ilmath] as [ilmath]\tilde{A} [/ilmath] is an extension of [ilmath]A[/ilmath]
            • [ilmath]\eq \tilde{A}(a(v_1,\ldots,v_k))[/ilmath] as [ilmath]\tilde{A} [/ilmath] 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



  1. Introduction to Smooth Manifolds - John M. Lee