Difference between revisions of "Tensor product of vector spaces"
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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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: {{Note|Any first-time readers should look at the [[#Abstract definition|abstract definition]] first}} | : {{Note|Any first-time readers should look at the [[#Abstract definition|abstract definition]] first}} | ||
==Definition== | ==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]] on {{M|\prod_{i\eq 1}^kV_k}}. We define the (abstract) ''tensor product'' of {{M|V_1,\ldots,V_k}} as{{rITSMJML}}: | + | 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} } }}<!-- | * {{M|V_1\otimes\cdots\otimes V_k:\eq\dfrac{\mathcal{F}(V_1\times\cdots\times V_k)}{\mathcal{R} } }}<!-- | ||
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**# {{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},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\} }} | **# {{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== | ==Abstract definition== | ||
==[[Basis for the tensor product|Basis]]== | ==[[Basis for the tensor product|Basis]]== |
Latest revision as of 21:33, 22 December 2016
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- Currently in the notes stage, see Notes:Tensor product
Contents
[hide]- Any first-time readers should look at the abstract definition first
Definition
Let F be a field and let ((Vi,F))ki=1 be a family of vector spaces over F. Let F(V1×⋯×Vk) denote the free vector space on ∏ki=1Vk. We define the (abstract) tensor product of V1,…,Vk as[1]:
- V1⊗⋯⊗Vk:=F(V1×⋯×Vk)R[Note 1] where R is defined as follows:
Abstract definition
Basis
Let F be a field and let ((Vi,F))ki=1 be a family of finite dimensional vector spaces. Let ni:=Dim(Vi) and e(i)1,…,e(i)ni denote a basis for Vi, then we claim[1]:
- B:={e(1)i1⊗⋯⊗e(k)ik | ∀j∈{1,…,k}⊂N[1≤ij≤nj]}
Is a basis for the tensor product of the family of vector spaces, V1⊗⋯⊗Vk
Note that the number of elements of B, denoted |B|, is ∏ki=1ni or ∏ki=1Dim(Vi), thus:
- Dim(V1⊗⋯⊗Vk)=∏ki=1ni[1]
Characteristic property
Let F be a field and let ((Vi,F))ki=1 be a family of finite dimensional vector spaces over F. Let (W,F) be another vector space over F. Then[1]:- If A:V1×⋯×Vk→W is any multilinear map
- there exists a unique linear map, ¯A:V1⊗⋯⊗Vk→X such that:
- ¯A∘p=A (that is: the diagram on the right commutes)
- there exists a unique linear map, ¯A:V1⊗⋯⊗Vk→X such that:
Where p:V1×⋯×Vk→V1⊗⋯⊗Vk by p:(v1,…,vk)↦v1⊗⋯⊗vk (and is p is multilinear)
Notes
- Jump up ↑ Take a moment to respect just how vast the space F(V1×⋯×Vk) is (especially if F:=R for example). Remember that this is not the space V1×⋯×Vk even though we write them as tuples. It is a huge space.
- TODO: Flesh out this note
-