Difference between revisions of "Tensor product of vector spaces"

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Latest revision as of 21:33, 22 December 2016

Currently in the notes stage, see Notes:Tensor product

Any first-time readers should look at the abstract definition first

Definition

Let $\mathbb{F}$ be a field and let $\big((V_i,\mathbb{F})\big)_{i\eq 1}^k$ be a family of vector spaces over $\mathbb{F}$. Let $\mathcal{F}(V_1\times\cdots\times V_k)$ denote the free vector space on $\prod_{i\eq 1}^kV_k$. We define the (abstract) tensor product of $V_1,\ldots,V_k$ as^[1]:

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

Abstract definition

Basis

Let $\mathbb{F}$ be a field and let $\big((V_i,\mathbb{F})\big)_{i\eq 1}^k$ be a family of finite dimensional vector spaces. Let $n_i:\eq\text{Dim}(V_i)$ and $e^{(i)}_1,\ldots,e^{(i)}_{n_i}$ denote a basis for $V_i$, then we claim^[1]:

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

Is a basis for the tensor product of the family of vector spaces, $V_1\otimes\cdots\otimes V_k$

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

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

Characteristic property

Let $\mathbb{F}$ be a field and let $\big((V_i,\mathbb{F})\big)_{i\eq 1}^k$ be a family of finite dimensional vector spaces over $\mathbb{F}$. Let $(W,\mathbb{F})$ be another vector space over $\mathbb{F}$. Then^[1]:

• If $A:V_1\times\cdots\times V_k\rightarrow W$ is any multilinear map
  • there exists a unique linear map, $\overline{A}:V_1\otimes\cdots\otimes V_k\rightarrow X$ such that:
    • $\overline{A}\circ p\eq A$ (that is: the diagram on the right commutes)

Where $p:V_1\times\cdots\times V_k\rightarrow V_1\otimes\cdots\otimes V_k$ by $p:(v_1,\ldots,v_k)\mapsto v_1\otimes\cdots\otimes v_k$ (and is $p$ is multilinear)

Notes

1. Jump up ↑ Take a moment to respect just how vast the space $\mathcal{F}(V_1\times\cdots\times V_k)$ is (especially if $\mathbb{F}:\eq\mathbb{R}$ for example). Remember that this is not the space $V_1\times\cdots\times V_k$ 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