Difference between revisions of "Tensor product of vector spaces"

Revision as of 16:30, 23 August 2015 (view source) Alec (Talk | contribs) m (Reverted edits by JessicaBelinda133 (talk) to last revision by Alec) ← Older edit		Latest revision as of 21:33, 22 December 2016 (view source) Alec (Talk | contribs) m (Alec moved page Tensor product to Tensor product of vector spaces without leaving a redirect: There's another tensor product - at 360 views ish)
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−	Currently in the notes stage, see [[Notes:Tensor product]]	+	{{Stub page|grade=A*|msg=Demote once more of it is finished!}}
		+	: {{strike|Currently in the notes stage, see [[Notes:Tensor product]]}}
		+	__TOC__
		+	: {{Note|Any first-time readers should look at the [[#Abstract definition|abstract definition]] first}}
		+	==Definition==
		+	Let {{M|\mathbb{F} }} be a [[field]] and let {{M|\big((V_i,\mathbb{F})\big)_{i\eq 1}^k}} be a family of [[vector spaces]] over {{M|\mathbb{F} }}. Let {{M|\mathcal{F}(V_1\times\cdots\times V_k)}} denote the [[free vector space generated by|free vector space]] on {{M|\prod_{i\eq 1}^kV_k}}. We define the (abstract) ''tensor product'' of {{M|V_1,\ldots,V_k}} as{{rITSMJML}}:
		+	* {{M|V_1\otimes\cdots\otimes V_k:\eq\dfrac{\mathcal{F}(V_1\times\cdots\times V_k)}{\mathcal{R} } }}<!--
		+
		+	NOTE ABOUT THE SIZE OF F(V_1X...XV_k)
		+
		+	--><ref group="Note">Take a moment to respect just how vast the space {{M|\mathcal{F}(V_1\times\cdots\times V_k)}} is (especially if {{M|\mathbb{F}:\eq\mathbb{R} }} for example). Remember that this is ''not'' the space {{M|V_1\times\cdots\times V_k}} even though we write them as [[tuples]]. It is a ''huge'' space.
		+	* {{XXX|Flesh out this note}}</ref><!--
		+
		+	END OF SIZEOF F(V_1X...XV_k) NOTE
		+
		+	--> where {{M|\mathcal{R} }} is defined as follows:
		+	** {{M|\mathcal{R} }} denotes the {{link|span|vector space}} of all the union of the following two [[sets]]:
		+	**# {{M|\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\} }}
		+	**# {{M|\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 for the tensor product|Basis]]==
		+	{{:Basis for the tensor product/Statement}}
		+	==[[Characteristic property of the tensor product|Characteristic property]]==
		+	{{:Characteristic property of the tensor product/Statement}}
		+	==Notes==
		+	<references group="Note"/>
		+	==References==
		+	<references/>
		+	{{Definition|Linear Algebra|Abstract Algebra}}

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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