My question comes from
https://encyclopediaofmath.org/index.php?title=Affine_tensor
An affine tensor as described above is commonly called simply a tensor.
Is there any uncommon scenario?
I tried to compare to the wikipedia definitions of tensor. The most non-trivial condition seems to be (I think the Einstein summation rule is implied)
$$
A^s_j A′^i_s={\delta}^i_j, \tag{1}
$$
Eq. (1) can be derived if one imposes $(e^{i'},e_{j'} )= \delta_{j'}^{i'}$, $(e^{i},e_{j} ) =\delta_j^i$ in any basis. My derivation is
$$
(e^{i}, e_{j} ) = (e^{i'}, e_{j'} ) A^{'i}_{i'} A^{j'}_j
$$
$$
\delta^j_i = \delta^{i'}_{j'} A_{j}^{j'} A_{i'}^{'i} = A_{j}^{j'} A_{j'}^{'i} = A_{j}^{s} A_{s}^{'i} = A_{s}^{'i} A_{j}^{s}
$$
(renaming dummy indices and adjust order of matrix elements A)
So my questions are:
(a) Is the orthogonal relation (1) holds in any basis?
(b) Does the orthogonal relation (1) defines the affine type tensor?
(c) What are other definitions of tensors besides the affine in the encyclopediaofmath link ?
Editted, checked https://mathworld.wolfram.com/AffineTensor.html
where the determinant of a^i_j is nonzero.
seems the crucial part of affine tensor is non singluar transformation. If so, only question (a) remains.
Best Answer
Usually, affine in this context means that the transformation matrix is orthogonal which in turn means the transformation is linear by definition.
If you have the transformation
$$\bar{x}^j=A_{jh}x_h$$
The orthogonality condition is
$$AA^T=A^TA=I$$
or in other words
$$A_{jh}A_{kh}=\delta_{jk}$$
In the case of linear orthogonal transformations we have
$$\frac{\partial \bar{x}^j}{\partial x^h}=A_{jh}=\frac{\partial x^h}{\partial \bar{x}^j}$$
This simplifies matters greatly. In this setting "the inner product" or the "contraction" $A_jB_j$ is a scalar, while a contraction under more general circumstances requires an expression like $A^jB_j$ to render a true scalar.
Further, in a more general setting the relation
$$\frac{\partial \bar{x}^j}{\partial x^h}\frac{\partial x^h}{\partial \bar{x}^k}=\delta^j_k$$
still holds, however, you can no longer find the inverse by transposing the matrix.
Notice how I deliberately use a notation with upper- and lower indices for a more general setting and lower indices only for affine tensors (orthogonal linear transformations).