In a system of curvilinear coordinates, $(q^1,q^2,q^3)$, basis vector can be derived by differentiating a generic position $M(q^1,q^2,q^3)$ with respect to the coordinates, ie $$\mathbf{e}_i=\frac{\partial M}{\partial q^i}$$ and the expression of the metric tensor is $$g_{ik}=\mathbf{e}_i\cdot \mathbf{e}_k$$ Can I find the expression of this metric tensor in a system of cylindrical coordinates without using any other system of coordinates?
[Math] Metric tensor in cylindrical coordinates
cylindrical coordinates
Related Solutions
The expression you've got for the acceleration in cylindrical coordinates (for $r$ constant) is correct. Check here for similar derivations in other coordinate systems.
Say we have a point $P = (r, \theta, z)\in \Bbb R^3$, using cylindrical coordinates. This gives rise to a natural $\vec e_r$ which is of unit length and lies in the $xy$-plane and has angle $\theta$ to the $x$-axis, and a vector $\vec e_z$ which is just the unit vector along the $z$-axis. This way we have $$ P = r\vec e_r + z\vec e_z $$ A different point $Q$ would give a different $\vec e_r$ (and theoretiacally also a different $\vec e_z$, but by a quirk of the cylindrical coordinate system, that vector happens to be the same for any point). Thus $\vec e_r, \vec e_z\in \Bbb R^3$ aren't really well-defined by themselves, but only when given a specific point to work from.
At the same time, at this point $P$, we have the tangent space $T_P\Bbb R^3$. This is also a three-dimensional vector space, like $\Bbb R^3$, but it is not the same $\Bbb R^3$ which $P$ lies in. In this vector space we have a natural basis derived from the coordinate system around $P$, and that basis is sometimes called $\vec e_r, \vec e_\theta$ and $\vec e_z$ (the exact construction of these vectors uses the definition of $T_P\Bbb R^3$ extensively, so it will vary from book to book, and there are also different conventions as to what their lengths should be).
If we have a different point $Q\in \Bbb R^3$, then $T_Q\Bbb R^3$ is also a three-dimensional vector space, but it is different from $\Bbb R^3$ and it is different from $T_P\Bbb R^3$.
A vector ("point") in $\Bbb R^3$ has no natural translation into a vector in $T_P\Bbb R^3$, and neither does a vector in $T_Q\Bbb R^3$. However, both $T_P\Bbb R^3$ and $T_Q\Bbb R^3$ have their own naturally defined basis called $\vec e_r, \vec e_\theta$ and $\vec e_z$, but they only share their name because they are constructed the same way. A vector in $T_P\Bbb R^3$ is completely uncomparable to a vector in $T_Q\Bbb R^3$ (this is most easily seen by just looking at your definition of the tangent space; there is no natural way of making a vector of one space into a vector of the other (well, there is by translation, but that's not applicable to non-flat spaces and this a terrible habit to start)).
Along with the $\vec e_r, \vec e_z$ of $\Bbb R^3$, this is ripe for confusion. This is what I suspect has happened to you, and made you write this question. My best tip is to always be very aware of which vector space any vector is a part of when you write them down and manipulate them. If you do this by actually giving them different names (instead of calling all of them $\vec e_r$), then it's a little bit easier to keep things straight.
Best Answer
It all depends on what you mean by the dot product in your defining equation for $g_{ik}$. If you are defining an abstract Riemannian metric, then your manifold is an abstract thing. But, ordinarily, when you refer to cylindrical coordinates, you're thinking of $\Bbb R^3$ as your ambient space and you're using "$\cdot$" to represent the dot product on $\Bbb R^3$. In that case, you're going to need to compute $\mathbf e_i$ as vectors in $\Bbb R^3$, and for that you'll need specific coordinates in $\Bbb R^3$ for $M(r,\theta,z)$.