Struggling to see relativity through newly aquired math (geometrical) glasses So, I've been following the Lectures on the Geometric Anatomy of Theoretical Physics by
Frederic Schuller on Youtube (I'm about half way in) and now I've been looking at some special relativity (something I haven't really touched in 1-2 years). Let us start with a short naration about what I think I have clarified on my own so far (I welcome any remarks and clarifications):

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*We live in some space $M$, which is a 4-dimensional smooth manifold. What we commonly call a position 4-vector $x^\mu$ at a point $p$ in space is an element of the tangent space $T_pM$ (taken as a vector space over $T_p \mathbb{R}\cong\mathbb{R}$), $p\in M$.

*The relativistic interval is a bilinear map $ds^2 \in T^0_2(T_pM)$, that we can write explicitly as $ds^2 = g_{^{\mu\nu}} dx^\mu dx^\nu$, where $g_{\mu\nu}$ is the metric and $\{dx^\alpha \}$ is the basis in $T^*_pM$ (the cotangent space) that is determined by our choice of chart; i.e. $ds^2$ eats two vectors and outputs a scalar (in particular, the norm of a 4-vector). The way $dx^\mu$ works is such that $dx^\mu (x)=x^\mu$, for $x\in T_pM$, which enables us to write stuff like $x^2 = g_{^{\mu\nu}} x^\mu x^\nu$.

*Lorentz transformations are a special subset of $End(T_pM)$ (the set of endomorphisms on the tanget space), namely those that are invertible and leave the norm of a 4-vector invariant. Since $End(T_pM) \cong T^1_1(T_pM)$ we can see our Lorentz transformation as a map $\Lambda$ that eats a covector and a vector and outputs a scalar: $\Lambda = \Lambda^\mu_{\;\nu}\left(\frac{\partial}{\partial x^\mu} \right)_p \otimes dx^\nu$ (here $\left(\frac{\partial}{\partial x^\mu} \right)_p$ is the basis in $T_pM$ determined by our choice of chart), with $\Lambda^\mu_{\;\nu} = \Lambda\left(dx^\mu ,\left(\frac{\partial}{\partial x^\nu} \right)_p\right)$. This convention is what enables to let us treat Lorentz transformations as matrices.

Now, here are the points that I'm struggling with:

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*What is the transpose, really? In special relativity there is a common notation for the inverse of the Lorentz transforms, namely $\left(\Lambda^{-1} \right)^\mu_{\;\nu} = \Lambda_\nu^{\;\mu}$ (I'm not a big fan of this, since it can become a mess in handwriting, especially when the one who's writting has such a nice handwritting that it makes you wonder how he missed his carrier as a physician; but convention is convention). Here my lack of recent experience with special relativity might come into play. This relation appears a lot in special relativity textbooks: $\Lambda^T g \Lambda = g$, that is often refered to as a matrix relation (which is already confusing me, since it assumes that $g$ is a matrix made from the values $g_{\mu\nu}$, which form a bilinear map, but we've already defined matrices to be representations of endomorfisms). In the proof for this relation I always see $\Lambda^\mu_{\;\nu} = \left(\Lambda^T\right)_\nu^{\;\mu}$. This completely confuses me, as it seems to imply that $\Lambda = (\Lambda^T)^{-1}$? My best guess is that I completely fail to understand how the transpose of an object like $\Lambda^\mu_{\;\nu}$ is properly defined.

*The second question plays on a point I already touched when explaining the first question: If $g_{\mu\nu}$ are the coeffietients of a bilinear map, and the usual determinant is defined for endomorphisms, how is the determinant of the metric defined? (I see this quantity a lot, especially when people talk about the action in special relativity)

So, are there any phisicists that know these mathematical intricacies and can make this stuff clear for me? (Or at least maybe someone knows a book/some notes where this stuff is layed out?)
 A: The position $x^\mu$ is not a four-vector - it is a co-ordinate. $x^\mu=\phi(p)$ where $\phi$ is a chart and $p\in M$ is an event in Minkowski space. Since Minkowski space is globally like $\mathbf{R}^n$ we only need the one chart.
Likewise a Lorentz transformation $\Lambda$ corresponds to a change of chart.
$$\Lambda=\phi_{\rm New}\circ \phi_{\rm Old}^{-1}:\mathbf{R}^n\to\mathbf{R}^n . $$
Here $\phi_{\rm Old}$ is the old chart and $\phi_{\rm New}$ is the new chart. This corresponds to a function on coordinates (not tangent spaces)
$$\Lambda : x^\mu \mapsto (x')^\mu$$
specifically to be a Lorentz transformation we require that the Jacobian of this transformation (which is what acts on the tangent space) leave the norms of four-vectors unchanged. In Cartesian coordinates Lorentz transformations are linear and equal to their Jacobians and thus the notation $\Lambda$ gets used for both the coordinate transformation and also the matrix that mixes the components of vectors.
A: In special relativity, the metric in standard $(t, x, y, z)$ coordinates is just the identity matrix, with either the tt component equal to -1, or all of the spatial indices equal to -1.
In this basis, when you compute the lorentz transformation matrices, you will find that the inverse and the transpose are the same thing, because lorentz boosts are just hyperbolic rotations that preserve the metric with the minus sign in it.
