Question: Let $(M,g)$ be a Riemannian manifold and $g$ the Riemannian metric in coordinates $g=g\_{\alpha \beta}dx^{\alpha} \otimes dx^{\beta}$, where $x^{i}$ are local coordinates on $M$. Denote by $g^{\alpha \beta}$ the inverse components of the inverse metric $g^{-1}$. Let $\nabla$ be the Levi-Civita connection of the metric $g$. Consider, locally, the function $\det((g\_{\alpha \beta})\_{\alpha \beta})$. It is known that $\nabla \det((g\_{\alpha \beta})\_{\alpha \beta}) = 0$ by using normal coordinates etc...

I would like to show this fact without using normal coordinates. Just by computation. Here is what I have so far: 

$$\nabla \det((g\_{\alpha \beta})\_{\alpha \beta}) = \left [ g^{\gamma \delta} \partial\_{\delta} \det((g\_{\alpha \beta})\_{\alpha \beta}) \right ] \partial\_{\gamma} = \left [ \det((g\_{\alpha \beta})\_{\alpha \beta}) g^{\gamma \delta} g^{\beta \alpha} \partial\_{\delta} g\_{\alpha \beta}\right ] \partial\_{\gamma}.$$

Here: the first equality sign follows from the definition of the gradient of a function and the second equality sign is the derivative of the determinant.

**Question:** How do I continue from here without using normal coordinates? Or are there any mistakes? If yes, where and which?

Answer: 
Better to specify your direction vector with $\theta$ and $\varphi$ in [spherical coordinates](https://en.wikipedia.org/wiki/Spherical_coordinate_system) $(r,\theta,\varphi)$. Otherwise you have a problem with how to specify which of the three coordinates is omitted $-$ this requires another parameter.

But in fact a line only has four degrees of freedom, not five. This is because you can choose any point on the line to represent it. Actually finding four numbers to represent a line is a bit messy; start with the direction $(\theta,\varphi)$, and consider the plane $L$ through the origin that is perpendicular to it. Then the line is specified by its point of intersection with $L$. This requires only two coordinates, but you need some linear algebra to set up your coordinate system on $L$.

Perhaps somebody knows a more elegant parametrisation?
The specification of lines in three dimensional Euclidean space by four numbers is not trivial. Let us start with a Euclidean plane. We can construct a Cartesian coordinate system where each point is uniquely specified by two coordinates. What about lines? Without projective coordinates, there is a difficulty. For example, consider two reference lines $L\_0$ and $L\_1$ with $L\_0$ given by the first coordinate being $0$ and $L\_1$ by the first coordinate being $1$. Then, any line $L$ **not** parallel to $L\_0,L\_1$ will intersect the two reference lines and the second coordinates of the points of intersection will determine $L$ uniquely. The only exceptions are lines parallel to $L\_0,L\_1$. In that case, the first coordinate determines the line uniquely. That is, $L\_x$ is the line all of whose points have first coordiante $x$.

We can extend this to three dimensional Euclidean space. We construct a Cartesian coordinate system where each point is uniquely specified by three coordinates. Consider two reference planes $P\_0$ and $P\_1$ with $P\_0$ given by the first coordinate being $0$ and $P\_1$ by the first coordinate being $1$. Then any line $L$ **not** parallel to $P\_0,P\_1$ will intersect each of the two reference planes and ech of the two points of intersection are given by coordinates of which the first coordinate is already fixed, and the remaining two coordinates of each point uniquely specify the line $L$ with four numbers. The only exceptions are the lines parallel to $P\_0,P\_1$. In that case, the first coordinate specifies a plane that contains the line $L$ and in that plane, by the case for lines in a plane, that is uniquely determined by two numbers, with exceptions noted already. 

In summary, there seems to be no simple way to always uniquely specify every line in three dimensional Euclidean space with four real numbers without exceptional cases where less than four numbers are used. The fundamental reason for this situation is that the space of lines is topologically **not** the same as four dimensional Euclidean space. This is similar to why a circle is not topologically the same as a line, but if you remove one point on the circle, then it is homeomorphic to a line.