Lowering indices of the coordinate function $x^\mu$ In QFT, we frequently encounter expressions with upper or lower indices. I wonder how can one lower the index of the coordinate function $x^\mu$ in terms of differential geometric language.
Let $M$ be a (pseudo-)Riemannian manifold. Thanks to the metric tensor $g: T_pM\times T_pM \to \mathbb R$, we can identify $T_pM$ with $T^*_pM$. This is why we can perform operations like
$$\partial^\mu = g^{\mu\nu} \partial_\nu,$$
where the vector field $\partial_\mu \in TM$ naturally has lower indices but can be changed to have upper indices.
However, what about the coordinate function $x^\mu$? In many texts, one does the manipulation
$$x_\mu = g_{\mu\nu} x^\nu$$
by "extrapolating" the musical isomorphism. How can I understand this operation in terms of differential geometric language? Maybe the vector space structure of $M$ (say, $M=\mathbb R^{3,1}$) is relevant.
 A: There are several stages of abuse of notation going on here. But let me preface this by saying: one should NEVER lower/raise the indices of the coordinate functions of a general manifold. Even in $\Bbb{R}^n$, one should not raise/lower indices of the coordinate functions. The only exception is if one is in $\Bbb{R}^n$, and using Cartesian coordinates, and one knows what they're actually doing (unless one wants to blindly apply formulae).

Suppose we have a smooth manifold $M$, and a coordinate chart $(U,\alpha = (x^1(\cdot),\dots, x^n(\cdot)))$ on $M$. Then, on the tangent bundle $\pi_{TM}:TM\to M$, one has an adapted coordinate chart $(TU, T\alpha)$, where we label the coordinate functions as $T\alpha = (x^1\circ \pi_{TM},\dots, x^n\circ \pi_{TM}, \xi^1,\dots, \xi^n)$. Just so we're on the same page, the definition of $\xi^i$ is that for any $v\in TU$,
\begin{align}
\xi^i(v)&:=dx^i_{\pi_{TM}(v)}(v)\\
&= \text{$i^{th}$ coefficient in the expansion $v=\sum_{j=1}^nv^j \frac{\partial}{\partial x^j}\bigg|_{\pi_{TM}(v)}$}
\end{align}
Likewise, on the cotangent bundle, we have a coordinate chart $(T^*U,T^*\alpha)$, where we label the coordinates as $T^*\alpha= (x^1\circ \pi_{T^*M},\dots, x^n\circ \pi_{T^*M}, \zeta_1,\dots, \zeta_n)$, and for completeness, the definition of $\zeta_i$ is that for any $\theta\in T^*U$,
\begin{align}
\zeta_i(\theta)&=\theta\left(\frac{\partial}{\partial x^i}\bigg|_{\pi_{T^*M}(\theta)}\right)\\
&=\text{$i^{th}$ coefficient in the expansion $\theta=\sum_{j=1}^n\theta_j\,dx^j_{\pi_{T^*M}(\theta)}$}
\end{align}
Now, if we suppose $g$ is a pseudo-Riemannian metric on $M$, then we have the vector-bundle isomorphism $g^{\flat}:TM\to T^*M$ (the musical isomorphism). Since we have this isomorphism, we can "pull-back" any chart on the cotangent bundle and make it into a chart on the tangent bundle. This is just one of many instances of "transport of structure". So, $\bigg((g^{\flat})^{-1}(T^*U), (T^*\alpha) \circ g^{\flat}\bigg)=\bigg(TU, (T^*\alpha)\circ g^{\flat}\bigg)$ is a chart on $TM$. Note that $(x^i\circ \pi_{T^*M})\circ g^{\flat}=x^i\circ \pi_{TM}$ (this just says that if we're at a point $p\in M$, then $g^{\flat}$ sets up an isomorphism between $T_pM$ and $T_p^*M$, the tangent and cotangent spaces over the same base point). Next, we have that
\begin{align}
\zeta_i\circ g^{\flat}&=(g_{ij}\circ \pi_{TM})\cdot \xi^j\tag{$*$}
\end{align}
This equation is saying that if we take a vector $v\in T_pM$, and we look at the corresponding covector $\theta:= g^{\flat}(v)\in T_p^*M$, then their components are related as $\theta_i=g_{ij}(p)v^j$. The equation $(*)$ looks a little scary, but it's the correct way of writing things. Both sides are functions on $TU$.
Therefore, with $(*)$, we know exactly how the two charts for the tangent bundle, $(TU,T\alpha)$ and $(TU, (T^*\alpha)\circ g^{\flat})$ are related to one another.

Step 1 of Abusing Notation.
Typically, we're too lazy to write things "the correct way", so we omit the various compositions in $(*)$, and simply write it as
\begin{align}
\zeta_i&=g_{ij}\xi^j\tag{$**$}
\end{align}
The equal sign in $(**)$ means "the two sides aren't technically equal, but they're equal if the reader knows how to fill in the blanks".
As a side remark, you may have seen this exact equation in a different form in classical mechanics. There, one calls the configuration manifold $Q$ instead of $M$, and denotes coordinates on the base manifold as $(q^1,\dots, q^n)$, and by abuse of notation (suppressing the composition with $\pi_{TQ}$) the coordinates on the tangent bundle as $(q^1,\dots, q^n,\dot{q}^1,\dots, \dot{q}^n)$, and the coordinates on the cotangent bundle as $(q^1,\dots, q^n, p_1,\dots, p_n)$. Then, given the pseudo-Riemannian metric, one writes
\begin{align}
p_i=g_{ij}\dot{q}^j
\end{align}
(and then, since this equation relates momenta to velocities, one gives an interpretation to the metric as a "generalized mass matrix", but now I'm just rambling).

Step 2 of Abusing Notation.
Suppose now that we're in $\Bbb{R}^n$. Then, we have the identity chart (i.e Cartesian coordinates) $(\Bbb{R}^n,\text{id}_{\Bbb{R}^n}=(x^1,\dots, x^n))$. Using the identity chart, one can provide a very natural diffeomorphism $T(\Bbb{R}^n)\cong \Bbb{R}^n\times \Bbb{R}^n$ (ultimately, this boils down to the fact that the tangent space at any point of a real/complex vector space is canonically isomorphic to the vector space itself). So, the coordinates on the tangent bundle are $(x^1,\dots, x^n,\xi^1,\dots \xi^n)$, where $x^i$ takes a point in $\Bbb{R}^{n}\times \Bbb{R}^n$ and spits out the $i^{th}$ entry, while $\xi^i$ takes a point in $\Bbb{R}^{n}\times \Bbb{R}^n$ and spits out the $(n+i)^{th}$ entry.
As I mentioned, the fact that we have such a simple diffeomorphism of the tangent bundle onto $\Bbb{R}^n\times\Bbb{R}^n$ means that for many purposes, we don't actually have to bother keeping track of the first copy of $\Bbb{R}^n$ (i.e the base point), we only care about the second copy, which is the "vector-part". Therefore, to keep in line with this philosophy of not caring about the distinction between a point in $\Bbb{R}^n$ versus a vector tangent to it, we simply by abuse of notation, call $\xi^i:= x^i$ again. So, now given $a\in \Bbb{R}^n$ the number $x^i(a)=a^i$ can either mean the $i^{th}$ coordinate of a point $a$ or of a tangent vector. Likewise, on the cotangent bundle, we call $\zeta_i:=x_i$. Then, with these abuses of notation, we never have to talk about tangent/cotangent bundles or seriously distinguish $\Bbb{R}^n$ from its dual or anything. So given a constant pesudo-Riemannian metric $g$ on $\Bbb{R}^n$ (i.e simply a pseudo-inner product on the vector space $\Bbb{R}^n$), the equation $(**)$ above becomes
\begin{align}
x_i&=g_{ij}x^j.
\end{align}
Hopefully the stages of abuse of notation makes it clear what's going on.
A: 
Maybe the vector space structure of $M$ (say, $M=\mathbb R^{3,1}$) is relevant.

Yes, indeed, because flat spacetime is also an affine space, where displacements between points behave in the same way as vectors. Once a point is chosen to be the origin, a one-to-one correspondence between points and displacements from the origin is established. This enables position vectors to be defined. Moreover, the tangent space at any point is globally canonically isomorphic to the space itself. This enables tensors at different points to be compared via a trivial connection, and consequently, the position vector to be placed in the tangent space at the origin.
For more information, see this post.
