2025-02-16

1007: Tensors Space w.r.t. Field and k Vectors Spaces and Vectors Space over Field

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definition of tensors space w.r.t. field and k vectors spaces and vectors space over field

Topics


About: vectors space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a definition of tensors space with respect to field and k vectors spaces and vectors space over field.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

Main Body

1: Structured Description


Here is the rules of Structured Description.

Entities:
F: ∈{ the fields }
{V1,...,Vk,W}: ⊆{ the F vectors spaces }
∗L(V1,...,Vk:W): ={f:V1,...,Vk→W:∀j∈{1,...,k},∀v1∈V1,...,∀vk∈Vk,∀vj′∈Vj,∀r,r′∈F(f(v1,...,rvj+r′vj′,...,vk)=rf(v1,...,vj,...,vk)+r′f(v1,...,vj′,...,vk))}, ∈{ the F vectors spaces } with the addition and the scalar multiplication specified below
//

Conditions:
∀f1,f2∈L(V1,...,Vk:W)((f1+f2)(v1,...,vk)=f1(v1,...,vk)+f2(v1,...,vk))
∧
∀f∈L(V1,...,Vk:W),∀r∈F((rf)(v1,...,vk)=r(f(v1,...,vk))
//

2: Note


In other words, f is a multi-linear map.

F needs to be the same for all the vectors spaces, because otherwise the linearity would not make sense: "rf(v1,...,vj,...,vk)" would not make sense if W was not an F vectors space, so, each Vj needs to be over the field of W. In fact, there can be the argument that the field of W does not need to be exactly the filed of Vj but needs to be an extension of the field of Vj, which may be indeed possible, but we do not see any immediate necessity to allow that case complicating the situation.

Let us see that L(V1,...,Vk:W) is indeed an F vectors space.

1) for any elements, f1,f2∈L(V1,...,Vk:W), f1+f2∈L(V1,...,Vk:W) (closed-ness under addition): f1+f2 is :V1,...,Vk→W; (f1+f2)(v1,...,rvj+r′vj′,...,vk)=f1(v1,...,rvj+r′vj′,...,vk)+f2(v1,...,rvj+r′vj′,...,vk)=rf1(v1,...,vj,...,vk)+r′f1(v1,...,vj′,...,vk)+rf2(v1,...,vj,...,vk)+r′f2(v1,...,vj′,...,vk)=rf1(v1,...,vj,...,vk)+rf2(v1,...,vj,...,vk)+r′f1(v1,...,vj′,...,vk)+r′f2(v1,...,vj′,...,vk)=r(f1(v1,...,vj,...,vk)+f2(v1,...,vj,...,vk))+r′(f1(v1,...,vj′,...,vk)+f2(v1,...,vj′,...,vk))=r(f1+f2)(v1,...,vj,...,vk)+r′(f1+f2)(v1,...,vj′,...,vk).

2) for any elements, f1,f2∈L(V1,...,Vk:W), f1+f2=f2+f1 (commutativity of addition): (f1+f2)(v1,...,vk)=f1(v1,...,vk)+f2(v1,...,vk)=f2(v1,...,vk)+f1(v1,...,vk)=(f2+f1)(v1,...,vk).

3) for any elements, f1,f2,f3∈L(V1,...,Vk:W), (f1+f2)+f3=f1+(f2+f3) (associativity of additions): ((f1+f2)+f3)(v1,...,vk)=(f1+f2)(v1,...,vk)+f3(v1,...,vk)=f1(v1,...,vk)+f2(v1,...,vk)+f3(v1,...,vk)=f1(v1,...,vk)+(f2(v1,...,vk)+f3(v1,...,vk))=f1(v1,...,vk)+(f2+f3)(v1,...,vk)=(f1+(f2+f3))(v1,...,vk).

4) there is a 0 element, 0∈L(V1,...,Vk:W), such that for any f∈L(V1,...,Vk:W), f+0=f (existence of 0 vector): the 0 map, f0∈L(V1,...,Vk:W), is 0, because (f+f0)(v1,...,vk)=f(v1,...,vk)+f0(v1,...,vk)=f(v1,...,vk)+0=f(v1,...,vk).

5) for any element, f∈L(V1,...,Vk:W), there is an inverse element, f′∈L(V1,...,Vk:W), such that f′+f=0 (existence of inverse vector): −f∈L(V1,...,Vk:W) is f′, because (−f+f)(v1,...,vk)=−f(v1,...,vk)+f(v1,...,vk)=0=f0(v1,...,vk).

6) for any element, f∈L(V1,...,Vk:W), and any scalar, r∈F, r.f∈L(V1,...,Vk:W) (closed-ness under scalar multiplication): r.f is :V1,...,Vk→W; (r.f)(v1,...,svj+s′vj′,...,vk)=rf(v1,...,svj+s′vj′,...,vk)=r(sf(v1,...,vj,...,vk)+s′f(v1,...,vj′,...,vk))=srf(v1,...,vj,...,vk)+s′rf(v1,...,vj′,...,vk)=s(rf)(v1,...,vj,...,vk)+s′(rf)(v1,...,vj′,...,vk).

7) for any element, f∈L(V1,...,Vk:W), and any scalars, r1,r2∈F, (r1+r2).f=r1.f+r2.f (scalar multiplication distributability for scalars addition): ((r1+r2).f)(v1,...vk)=(r1+r2)f(v1,...vk)=r1f(v1,...vk)+r2f(v1,...vk)=(r1f)(v1,...vk)+(r2f)(v1,...vk)=(r1.f+r2.f)(v1,...vk).

8) for any elements, f1,f2∈L(V1,...,Vk:W), and any scalar, r∈F, r.(f1+f2)=r.f1+r.f2 (scalar multiplication distributability for vectors addition): (r.(f1+f2))(v1,...vk)=r(f1+f2)(v1,...vk)=r(f1(v1,...vk)+f2(v1,...vk))=rf1(v1,...vk)+rf2(v1,...vk)=(rf1)(v1,...vk)+(rf2)(v1,...vk)=(r.f1+r.f2)(v1,...vk).

9) for any element, f∈L(V1,...,Vk:W), and any scalars, r1,r2∈F, (r1r2).f=r1.(r2.f) (associativity of scalar multiplications): ((r1r2).f)(v1,...vk)=(r1r2)f(v1,...vk)=r1(r2f(v1,...vk))=r1((r2f)(v1,...vk))=(r1.(r2.f))(v1,...vk).

10) for any element, f∈L(V1,...,Vk:W), 1.f=f (identity of 1 multiplication): (1.f)(v1,...vk)=1f(v1,...vk)=f(v1,...vk).

Each element of L(V1,...,Vk:W) is called "tensor".

A typical example is L(V1,...,Vk:F): F is an F vectors space.

A more typical example is L(V:F):=V∗, called "the covectors space of V" or "the dual space of V".

Each element of V∗ is called "covector".

Another typical example is Tqp(V):=L(V∗,...,V∗,V,...,V:F), where there are p V∗ s and q V s, called "the (p,q)-tensors space of V".

V∗=T10(V).

As L(V1,...,Vk:W) is an F vectors space, it admits a basis, and any tensor has the components with respect to the basis. Such a set of components is sometimes called "tensor", but the set of components is exactly the set of the components of the tensor with respect to the basis, not "tensor".

References
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