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 R

n, so from now we can consider in fact V ≡R

n.

Anyone linear application f: R

n → R is in the form

f(x)=v

tx, con v, x∈R

n
Ie, Dual space of R

n is really itself (or they are

isomorphics by transpose v

t to v ).

Lets B={e

1, ...,e

n}
a Basis of V, then
B*={e

1, ...,e

n} is a basis of Dual V

*, characterized by

e

je

i = 0, j≠i

e

je

i = 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

### Tensor definition

It is called s-contravariant and n-covariant Tensor or s-times contravariant n-times covariant tensor, or tensor kind (s, n) to a

multilinear Map like

T: V

* x ...(s times)... x V

* x V x ...(n times)... x V → R

If a tensor is (0, n) type, we say simply that it is n times covariant

If a tensor is type (s, 0) say simply that it is s times contravariant

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

u

1, ...,u

r∈V

*, v

1, ...,v

s ∈V → F(u

1, ...,u

r,v

1, ...,v

s)∈ R

Then, if B={e

1, ...,e

n}
is an Basis of V and B*={e

1,...,e

n}
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(e

j1,...,e

jr,e

i1,...,e

1r)

Therefore, a (r,s) tensor on R

n has n

r+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(u

1,...,u

r, u

1, ...,u

m, v

1, ...,v

s,v

1,...,v

n) =
T(u

1,...,u

r,v

1,...,v

s) .S(u

1, ...,u

m,v

1,...,v

n)

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 R

4 (z

1,z

2,z

3,z

4) and 3 vectors
(u

1,u

2,u

3,u

4)
(v

1,v

2,v

3,v

4)
(w

1,w

2,w

3,w

4)
to denote the application of the tensor in the vectors would have to write

∑

4i=1∑

4j=1∑

4k=1∑

4l=1R

lijk z

lu

iv

jw

k
with

Einstein summation convention this equation is reduced to

R

lijk z

lu

iv

jw

k
ie,

Einstein summation convention
consists in a duplicate index and dubindex indicates addition, for example
a tensor given by v

ijw

kl is a (2,2) tensor,
showever v

ikw

kl 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=v

iw

i
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.

Given two charts C = (φ=(x

1,...,x

m), U)
K = (ψ=(y

1,...,y

m), V)

There exists an application
ψoφ

-1 wich transforms coodinates (x

1,...,x

m) into coordinates (y

1,...,y

m)
and its differential matrix is written as

and its inverse

Each chart have the vector fields

and

and

Respectively and the relationship between them is

1)

2)

A very important tensor in differential geometry is called

Metric Tensor. We denote it by G=g

ij
Given an element v=v

i of V, then v is a (1,0)tensor. It is possible to
apply the

Metric Tensor as follows

w

j=g

ijv

i
Obtaining a (0,1) tensor w, ie, an dual space element.
We will call v

i the contravariant coordiantes of v
and w

j will call covariants ones.

Finally, this process of matching indexes with superindexes and sum them (again the

Einstein summation convention) is called contraction.

### Riemannian Manifolds

Is called a Riemannian a manifold to a differentiable manifold join a metric tensor acting on it, whose coefficients matrix is definite-positive,

When matrix coefficient of metric tensor is positive-semidefinite, is said that manifold is Semirimannian