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


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 vector spaces over [ilmath]\mathbb{F} [/ilmath]. Let [ilmath]\mathcal{F}(V_1\times\cdots\times V_k)[/ilmath] denote the free vector space on [ilmath]\prod_{i\eq 1}^kV_k[/ilmath]. We define the (abstract) tensor product of [ilmath]V_1,\ldots,V_k[/ilmath] as[1]:

  • [ilmath]V_1\otimes\cdots\otimes V_k:\eq\dfrac{\mathcal{F}(V_1\times\cdots\times V_k)}{\mathcal{R} } [/ilmath][Note 1] where [ilmath]\mathcal{R} [/ilmath] is defined as follows:
    • [ilmath]\mathcal{R} [/ilmath] denotes the span of all the union of the following two sets:
      1. [ilmath]\big\{ (v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)-a(v_1,\ldots,v_k)\ \big\vert\ i\in\{1,\ldots,k\}\wedge a\in\mathbb{F}\wedge\forall j\in\{1,\ldots,k\}[v_j\in V_j]\big\} [/ilmath]
      2. [ilmath]\big\{(v_1,\ldots,v_{i-1},v_i+v'_i,v_{i+1},\ldots,v_k)-(v_1,\ldots,v_k)-(v_1,\ldots,v_{i-1},v'_i,v_{i+1},\ldots,v_k)\ \big\vert\ i\in\{1,\ldots,k\}\wedge v'_i\in V_i\wedge\forall j\in\{1,\ldots,k\}[v_j\in V_j]\big\} [/ilmath]

Abstract definition


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. Let [ilmath]n_i:\eq\text{Dim}(V_i)[/ilmath] and [ilmath]e^{(i)}_1,\ldots,e^{(i)}_{n_i} [/ilmath] denote a basis for [ilmath]V_i[/ilmath], then we claim[1]:

  • [math]\mathcal{B}:\eq\left\{e^{(1)}_{i_1}\otimes\cdots\otimes e^{(k)}_{i_k}\ \big\vert\ \forall j\in\{1,\ldots,k\}\subset\mathbb{N}[1\le i_j\le n_j]\right\} [/math]

Is a basis for the tensor product of the family of vector spaces, [ilmath]V_1\otimes\cdots\otimes V_k[/ilmath]

Note that the number of elements of [ilmath]\mathcal{B} [/ilmath], denoted [ilmath]\vert\mathcal{B}\vert[/ilmath], is [ilmath]\prod_{i\eq 1}^kn_i[/ilmath] or [ilmath]\prod_{i\eq 1}^k\text{Dim}(V_i)[/ilmath], thus:

  • [ilmath]\text{Dim}(V_1\otimes\cdots\otimes V_k)\eq\prod_{i\eq 1}^k n_i[/ilmath][1]

Characteristic property

[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)


  1. Take a moment to respect just how vast the space [ilmath]\mathcal{F}(V_1\times\cdots\times V_k)[/ilmath] is (especially if [ilmath]\mathbb{F}:\eq\mathbb{R} [/ilmath] for example). Remember that this is not the space [ilmath]V_1\times\cdots\times V_k[/ilmath] even though we write them as tuples. It is a huge space.
    • TODO: Flesh out this note


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