Defining tensor transpose without representing them as matrices In the comments of this post there was a discussion about why I hesitate to use the conventional tensor notation. There I briefly mentioned that I find it illogical and inconsistent. One of my main issues is the transpose of a tensor and what it entails.
As far as I have understood tensors, in the context of differential geometry, are abstract entities forming vector spaces. Transpose operator as far as I know applies to a 2D table or as we know it a matrix. I can't find a universal definition for transpose of a tensor. I would appreciate if you could help me know if there is any without considering a coordinate system and representing them as matrices or in Einstein notation?
 A: Yes.
There is something very important (I would say it is the most important thing) to consider: if $V$ is a vector space, then there are two distinct types of objects that can be represented with a matrix.


*

*Bilinear forms. I.e, elements of $T^{0,2}V$ or $T^{2,0}V$. The first ones are maps from ${V}^{\times 2} \to \mathbb{R}$, and the second ones are maps from ${V^{*}}^{\times 2}\to\mathbb{R}$

*Endomorphisms. I.e, elements of $T^{1,1}V$. These objects are maps from $V\times V^{*}\to \mathbb{R}$. However, the important thing here is that this space is isomorphic to $\text{Hom}(V,V)$.


Well, having said this, it would be very reasonable to think that the concept of the "transpose of a matrix" will do different things on each of these objecs. And that is indeed the case.

First, as someone mentioned in the comments, if $\phi\in T^{0,2}V$ (say) then we define its transpose as the operation of braiding its slots. This means that for all vectors $v,w\in V$:
$$\phi^{T}(v,w) = \phi(w,v)$$
Using abstract index notation, this can be written as:
$$(\phi^T)_{ab} := \phi_{ba}$$
The same can be done anologously with an element of $T^{2,0}V$. Note that this definition does not need additional structure. It is a canonical operation in any vector space.

Okay, that was easy. Now for $\phi\in T^{1,1}V$. In this case, there is no canonical identification available. However, if we introduce a metric $g$ in our vector space, we can define the adjoint of $\phi$ with respect to $g$ (denoted $\phi^{\text{Ad}_g}$) as the unique map such that for all vectors $v,w \in V$
$$g(v,\phi(w)) = g(\phi^{\text{Ad}_g}(v),w)$$
If you make the calculation in abstract index notation, you can see this reduces to:
$${(\phi^{\text{Ad}_g})^{a}}_{b} = {\phi^{c}}_{d}g^{ad}g_{cb}$$
Now, it is noteworthy that a lot of confusion arises from the "raising and lowering of indices" notation (${\phi^{c}}_{d}g^{ad}g_{cb} = {\phi_{b}}^{a}$)
since the previous definition reduces to
$${(\phi^{\text{Ad}_g})^{a}}_{b} = {\phi_{b}}^{a}$$
This, obviously, is the justification for keeping the horizontal spacing of the indices: the order of the indices does matter.
However, this notation kind of hides the fact that there is a metric involved, and it is by this reason I don't like it a lot.

Well, those are the abstract definitions.
If you choose an arbitrary basis for your vector space $V$ and write the components of a bilinear form, you can see that the abstract operation of braiding its entries corresponds to interchanging its rows and columns.
If you choose an orthonormal basis for your inner product space $V$ and write the components of an endomorphism, you can see that the abstract operation of the taking the adjoint with respect to the metric corresponds to interchanging its rows and columns.
By using matrices, one does not see the differences between these two objects, and normally one cannot see the difference between the two concepts that give rise to the "transpose of a matrix" (namely, braiding vs adjointness).
This is a similar issue to that of the determinant: $\det(\phi)$ is only an invariant with respect to changes of basis if $\phi$ is an endomorphism. One can take the (quote) "determinant" (unquote) of a bilinear form by means of performing that very well known recursive algorithm on the entries of the representing matrix, but the resulting scalar depends on the choice of basis for the vector space.
