Tensor field

Definition....
...

Anyway, given $U$ an open set of a manifold $M$ , there are two defining properties for a map

S : X (U) \times X (U) \times X (U) \times \dots \to X (U) \times X (U) \times \dots

to be a tensor:

The output depends linearly on each input vector field.
The output only depends on the input vectors evaluated at a point, that is, we can create a map

S_{p} : T_{p} M \times T_{p} M \times \dots \to T_{p} M \times T_{p} M \times \dots

Change of coordinates

Suppose you have a 2 contravariant tensor

T = T^{i j} e_{i} \otimes e_{j}

in the canonical basis ${e_{i}}$ for $R^{3}$ . If we apply a basis change given by the matrix $M$ (that is to say, the observer moves his point of view), the vectors change its coordinates $(v^{1}, v^{2}, v^{3}) = v^{i} e_{i}$ to

N \cdot (\begin{matrix} v^{1} \\ v^{2} \\ v^{3} \end{matrix})

where $N = M^{- 1}$ .

But, what about our 2-tensor $T$ ?
Doing the maths you can conclude that if we express the coordinates $T^{i j}$ in a matrix, the new matrix is

N T N^{T}

In index notation this can be written

N_{j}^{a} N_{i}^{b} T^{i j}

This formula works in any rank other that 2.