Basis for the tensor product

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Important for manifolds or something


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]


  1. The proposed "basis" actually spans [ilmath]V_1\otimes\cdots\otimes V_k[/ilmath], i.e [ilmath]V_1\otimes\cdots\otimes V_k\subseteq\text{Span}(\mathcal{B}) [/ilmath]
    • Let [ilmath]A\in V_1\otimes\cdots\otimes V_k[/ilmath] be given. Then:
      • there is an [ilmath]m\in\mathbb{N} [/ilmath] such that [math]A\eq\sum_{\alpha\eq 1}^m \lambda_\alpha (v_{\alpha,1}\otimes\cdots\otimes v_{\alpha,k})[/math] for some [ilmath]\lambda_\alpha\in\mathbb{F} [/ilmath] and [ilmath]v_{\alpha,i}\in V_i[/ilmath]
      • But each [math]v_{\alpha,j}\eq\sum^{n_j}_{i_j\eq 1}v_{\alpha,j,i_j}e^{(j)}_{i_j} [/math] (where [ilmath]e^{(j)}_{i_j} [/ilmath] is the [ilmath]i_j[/ilmath]th basis vector of [ilmath]V_j[/ilmath] and [ilmath]v_{\alpha,j,i_j} [/ilmath] the [ilmath]i_j[/ilmath]th coefficient of [ilmath]v_{\alpha,j} [/ilmath])
      • Thus: [math]A\eq\sum^m_{\alpha\eq 1}\lambda_\alpha\left(\left(\sum_{i_1\eq 1}^{n_1}v_{\alpha,1,i_1}e^{(1)}_{i_1}\right)\otimes\cdots\otimes\left(\sum_{i_k\eq 1}^{n_k}v_{\alpha,k,i_k}e^{(k)}_{i_k}\right)\right)[/math]
      • [math]\eq\sum^m_{\alpha\eq 1}\lambda_\alpha\sum^{n_1}_{i_1\eq 1}v_{\alpha,1,i_1}\left(e^{(1)}_{i_1}\otimes\left(\sum_{i_2\eq 1}^{n_2}v_{\alpha,2,i_2}e^{(2)}_{i_2}\right)\otimes\cdots\otimes\left(\sum_{i_k\eq 1}^{n_k}v_{\alpha,k,i_k}e^{(k)}_{i_k}\right)\right)[/math]
      • [math]\eq\sum^m_{\alpha\eq 1}\sum^{n_1}_{i_1\eq 1}\lambda_\alpha v_{\alpha,1,i_1}\left(e^{(1)}_{i_1}\otimes\left(\sum_{i_2\eq 1}^{n_2}v_{\alpha,2,i_2}e^{(2)}_{i_2}\right)\otimes\cdots\otimes\left(\sum_{i_k\eq 1}^{n_k}v_{\alpha,k,i_k}e^{(k)}_{i_k}\right)\right) [/math]
      • [math]\eq\sum^m_{\alpha\eq 1}\ \underbrace{\sum^{n_1}_{i_1\eq 1}\cdots\sum^{n_k}_{i_k\eq 1} }\ \lambda_\alpha\ \underbrace{ v_{\alpha,1,i_1}\cdots v_{\alpha,k,i_k} }\ \left(e^{(1)}_{i_1}\otimes\cdots\otimes e^{(k)}_{i_k}\right)[/math] where the elements with the underbrace range over [ilmath]\{1,\ldots,k\} [/ilmath]
      • [math]\eq\sum^m_{\alpha\eq 1}\ \underbrace{\sum_{\begin{array}{c}i_1,\ldots,i_k \\ \forall j\in\{1,\ldots,k\}[1\le i_j\le \text{Dim}(V_j)]\end{array} } }_{\text{Finitely many - }\left(\prod^k_{\gamma\eq 1}\text{Dim}(V_\gamma)\right)\text{ - terms} }\ \overbrace{\lambda_\alpha\ \prod_{\beta\eq 1}^k v_{\alpha,\beta,i_\beta} }^{\in\mathbb{F} }\ \underbrace{\left(e^{(1)}_{i_1}\otimes\cdots\otimes e^{(k)}_{i_k}\right)}_{\in\mathcal{B} } [/math]
    • So it is clear that [ilmath]V_1\otimes\cdots\otimes V_k\subseteq \text{Span}(\mathcal{B})[/ilmath]
  2. Linear independence
Grade: A*
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At least do outline, I've done it once


  1. 1.0 1.1 Introduction to Smooth Manifolds - John M. Lee
Grade: A
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Smooth Manifolds, and Linear Algebra via Exterior Products