Tensors and Tensorial Algebra

When we handle multiple variable in Linear algebras we need to use a generalized concept of Vector, matrix and constants, this concep is called Tensor.
To define it we must begin to remaind that all finite-dimension vector space V is isomorphic to Rn, so from now we can consider in fact V ≡Rn.

Anyone linear application f: Rn → R is in the form

f(x)=vtx, con v, x∈Rn

Ie, Dual space of Rn is really itself (or they are isomorphics by transpose vt to v ).

Lets B={e1, ...,en} a Basis of V, then B*={e1, ...,en} is a basis of Dual V*, characterized by

ejei = 0, j≠i
ejei = 1, j=i

This isomorphism, we call φ between V and its dual V*, allows us to consider any linear application in the form.

f: V x ...(n times)... x V → V

as

F: V* x V x ...(n times)... x V → R

Being F=φ◊f (composition of f and φ) , ie, after applying f, we apply the dual element to obtain a scalar.

So, we can consider Scalar Multilineal Functions simply, by multiplying by elements of the dual

So, the endomorphism that we considered at diagonalizations or Jordan Cannonical form sections were
F : V → V
x∈V → F(x) = Ax∈V
by our definition F is an (1,1) tensor.

Also a vector is a tensor kind (1,0) because it applies any row vector (dual element) to a constant.

The scalar product in V is a tensor of type (0,2), because it applies two vectors to a constant.

Finally we consider that a constant is a (0,0) tensor.

Let T be a (r, s) tensor, therefore we will consider ∏ = V* x ...(r times)... x V* x V x ...(s times)... x V
T: ∏ → R
u1, ...,ur∈V*, v1, ...,vs ∈V → F(u1, ...,ur,v1, ...,vs)∈ R

Then, if B={e1, ...,en} is an Basis of V and B*={e1,...,en} is a basis of dual V*, we can consider the basis of ∏ as β=B*∪B, then we define the coordinates of T at basis β as

= T(ej1,...,ejr,ei1,...,e1r)

Therefore, a (r,s) tensor on Rn has nr+s components

Given two bases in ∏, β y β there is a 'change of basis matrix' for covariant and contravariant coordinates given respectively by , Then given a tensor T to change the Coordinate of β a β'



Note the Einstein summation convention

There is a tensorial product denoted por ⊗, defined as follows, if T is a tensor (r, s) and S is a tensor (m, n), tensorial product is a (r + m, s + n) tensor consisting of

T⊗S(u1,...,ur, u1, ...,um, v1, ...,vs,v1,...,vn) = T(u1,...,ur,v1,...,vs) .S(u1, ...,um,v1,...,vn)

The notation tensotrial can get very complicated because we are dealing with covariant and contravariant coordinates of different dimensions, such as the Riemann tensor is a tensor (1.3) as it applies a contravector of R4 (z1,z2,z3,z4) and 3 vectors (u1,u2,u3,u4) (v1,v2,v3,v4) (w1,w2,w3,w4) to denote the application of the tensor in the vectors would have to write

    ∑4i=1∑4j=1∑4k=1∑4l=1Rlijk zluivjwk

withEinstein summation convention this equation is reduced to

    Rlijk zluivjwk

ie, Einstein summation convention consists in a duplicate index and dubindex indicates addition, for example a tensor given by vijwkl is a (2,2) tensor, showever vikwkl is a (1,1) tensor.

Therefore we can consider a constant k is a (0,0) tensor, because it is the contraction of a (1,0) tensor with a (0,1) tensor

k=viwi

In a differentiable manifold M, one can define a tensor field or just a tensor of kind (r,s) as an application which assigns at each point p of a manifold an (r,s) tensor such us

T: Tp(M)* x ...(r veces)... x Tp(M)* x Tp(M) x ...(s veces)... x Tp(M) → R

That is, taking the tangent vector space as a vector space where tensor operates.