Determinant metric tensor I do not understand the relation between the determinant of the metric tensor $g$ and the non-tensorial symbol $\tilde{\epsilon}_{\mu_{0}..\mu_{n}}$. This is explained in Carrol's book as followed:
\begin{equation}
\tilde{\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}| M|=\tilde{\epsilon}_{\mu_{0}..\mu_{n}}M^{\mu_{0}}_{\ \ \ \ \bar{\mu}_{0}}...M^{\mu_{n}}_{\ \ \ \ \bar{\mu}_{n}},
\end{equation}
where $M^{\mu_{0}}_{\ \ \ \ \bar{\mu}_{0}}$ is a transformation matrix. Then he relates $g$ to the tensor ${\epsilon}_{\mu_{0}..\mu_{n}}$ as follows:
\begin{equation}
{\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}=\sqrt{|g|}\tilde{\epsilon}_{\mu_{0}..\mu_{n}}.
\end{equation}
Now, I know from Algebra that the determinant of a matrix ($4 \times 4$ in this case) can be written as:
\begin{equation}
g=\tilde{\epsilon}^{\bar{\mu}_{0}..\bar{\mu}_{3}}g_{0\mu_{0}}g_{1\mu_{1}}g_{2\mu_{2}}g_{3\mu_{3}},
\end{equation}
but cannot follow his approach
 A: Your question is a little confusing, so I'm going to explain what I think it's asking. Please let me know if I misunderstood it.
Let us first define the object ${\tilde \epsilon}_{a_1\cdots a_n}$ as follows
$$
{\tilde \epsilon}_{a_1 \cdots a_i a_{i+1} \cdots a_n} = - {\tilde \epsilon}_{a_1 \cdots a_{i+1}  a_i \cdots a_n} , \qquad {\tilde \epsilon}_{12\cdots n} = 1.
\tag{1}
$$
In other words, ${\tilde \epsilon}$ is completely antisymmetric in all its indices and is normalized as shown above.
We start by proving that this is not a tensor. First, recall the definition of the determinant of an $n\times n$ matrix
$$
\det M \equiv {\tilde \epsilon}_{a_1 \cdots a_n} M^{a_1}{}_1 \cdots M^{a_n}{}_n
$$
Using the two formulae above, we can deduce the following identity
$$
\boxed{ {\tilde \epsilon}_{a_1 \cdots a_n} M^{a_1}{}_{b_1} \cdots M^{a_n}{}_{b_n} = \det M {\tilde \epsilon}_{b_1 \cdots b_n} } 
$$
I leave its proof as an exercise.
Now, consider the transformation of ${\tilde \epsilon}_{a_1 \cdots a_n}$ under a coordinate transformation, $x^a \to x'^a$. Under this, we have
$$\tag{2}
{\tilde \epsilon}_{a_1 \cdots a_n} \to {\tilde \epsilon}'_{a_1 \cdots a_n} =  {\tilde \epsilon}_{b_1 \cdots b_n} J^{b_1}{}_{a_1} \cdots J^{b_n}{}_{a_n}  = \det J {\tilde \epsilon}_{a_1 \cdots a_n} , \qquad (J^{-1})^a{}_b = \frac{\partial x'^a}{\partial x^b} . 
$$
This proves that ${\tilde \epsilon}$ is NOT a tensor since in the new coordinates it should satisfy (1) and it doesn't.
However, we can now construct a tensor from this object by defining
$$
\epsilon_{a_1\cdots a_n} \equiv \sqrt{|\det g|} {\tilde \epsilon}_{a_1\cdots a_n}
$$
To prove that this a tensor we simply need to determine the new metric determinant. This is easy since
$$\tag{3}
g'_{ab} = g_{cd} J^c{}_a J^d{}_b \implies g' = J^T g J \implies \det g' = \det g (\det J)^2 .
$$
We now consider the transformation of $\epsilon$ under coordinate transformations, we have
\begin{align}
\epsilon_{a_1\cdots a_n} \to \epsilon'_{a_1\cdots a_n} &= \epsilon_{b_1 \cdots b_n} J^{b_1}{}_{a_1} \cdots J^{b_n}{}_{a_n} \\
&= \sqrt{|\det g|} {\tilde \epsilon}_{b_1 \cdots b_n} J^{b_1}{}_{a_1} \cdots J^{b_n}{}_{a_n} \\
&= \sqrt{|\det g|} \det J {\tilde \epsilon}_{a_1 \cdots a_n} \\
&= \sqrt{|\det g'|} \text{sign}(\det J) {\tilde \epsilon}_{a_1 \cdots a_n} \\
&=  \text{sign}(\det J) \epsilon_{a_1 \cdots a_n}
\end{align}
Thus, we see that this object transforms exactly like a tensor apart from the $\text{sign}(\det J)$ term. This sign represents the parity of the coordinate transformations (i.e. whether $x'^a$ and $x^a$ have the same orientation or not). The object $\epsilon$ is a tensor under orientation preserving coordinate transformations.
${\tilde \epsilon}$ is called the Levi-Civita symbol and $\epsilon$ is called the Levi-Civita tensor.
