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

  • $\begingroup$ Your question is not clear. What is the question: to get the determinant of the metric tensor by the 3. formula ? Or is it about the whole approach using the anti-symmetric Levi-Civita-(pseudo)-tensor all 3 equations of the post ? $\endgroup$ – Frederic Thomas Oct 23 at 15:43

The second equation does not directly follow from the first. You are correct in that \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}

which is the full algebraic expression for the determinant of a matrix. Your third equation also seems correct to this end, but the first equation writes out the determinant without making any explicit references to specific indices.

Anyway, if you take $M$ as $\frac{\partial{x^\mu}}{\partial{x^{\mu'}}}$, i.e a coordinate transformation, you can see that $\tilde{\epsilon}$ doesn't quite transform as a tensor:

\begin{equation} \tilde{\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}\lvert \frac{\partial{x^\mu}}{\partial{x^{\mu'}}} \rvert =\tilde{\epsilon}_{\mu_{0}..\mu_{n}}\frac{\partial{x^\mu}}{\partial{x^{\mu'_0}}} ...\frac{\partial{x^\mu}}{\partial{x^{\mu'_n}}} \end{equation}

Note that it would transform like a tensor if only that pesky $\vert\frac{\partial{x^\mu}}{\partial{x^{\mu'}}}\rvert$ wasn't there. Now, the metric determinant $g$ transforms in a similar way:

$$ g(x^{\mu'}) = \lvert\frac{\partial{x^{\mu'}}}{\partial{x^{\mu}}}\rvert^{-2}\, g(x^\mu) $$

which you can verify by taking the determinant of the metric transformation equation. You'll see how multiplying an equation with $\sqrt{|g|}$ will contribute a $|\frac{\partial{x^{\mu'}}}{\partial{x^{\mu}}}|$ during transformations, which is exactly what is needed to cancel the pesky factor at the front in the Levi-Civita symbol transformation.

So, the Levi-Civita (epsilon) tensor is defined by \begin{equation} {\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}=\sqrt{|g|}\tilde{\epsilon}_{\mu_{0}..\mu_{n}}. \end{equation} so that it transforms like a true tensor

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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 $$ {\tilde \epsilon}_{a_1 \cdots a_i \cdots a_j \cdots a_n} = - {\tilde \epsilon}_{a_1 \cdots a_j \cdots a_i \cdots a_n} , \qquad {\tilde \epsilon}_{12\cdots n} = 1. \tag{1} $$ 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.

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  • $\begingroup$ actually, the proof is what i do not understand $\endgroup$ – M91 Oct 23 at 18:20
  • $\begingroup$ @K91 proof of what? I still do not understand the question or confusion you have. $\endgroup$ – Prahar Oct 23 at 18:21
  • $\begingroup$ The equation in the squared box. $\endgroup$ – M91 Oct 23 at 22:57
  • $\begingroup$ I'll help you with hints: 1) Prove LHS is completely antisymmetric in indices $b_1\cdots b_n$. 2) What is most general possible structure of a totally antisymmetric tensor with $n$ indices? Show that there is a unique choice up to a normalization. 3) Fix the normalization by setting $b_1\cdots b_n = 1\cdots n$ and use definition of determinant. $\endgroup$ – Prahar Oct 23 at 23:02
  • $\begingroup$ It is clear now, thanks $\endgroup$ – M91 Oct 25 at 9:42

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