# 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:

$$$$\tilde{\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}| M|=\tilde{\epsilon}_{\mu_{0}..\mu_{n}}M^{\mu_{0}}_{\ \ \ \ \bar{\mu}_{0}}...M^{\mu_{n}}_{\ \ \ \ \bar{\mu}_{n}},$$$$ where $$M^{\mu_{0}}_{\ \ \ \ \bar{\mu}_{0}}$$ is a transformation matrix. Then he relates $$g$$ to the tensor $${\epsilon}_{\mu_{0}..\mu_{n}}$$ as follows:

$$$${\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}=\sqrt{|g|}\tilde{\epsilon}_{\mu_{0}..\mu_{n}}.$$$$

Now, I know from Algebra that the determinant of a matrix ($$4 \times 4$$ in this case) can be written as:

$$$$g=\tilde{\epsilon}^{\bar{\mu}_{0}..\bar{\mu}_{3}}g_{0\mu_{0}}g_{1\mu_{1}}g_{2\mu_{2}}g_{3\mu_{3}},$$$$

• 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 ? Oct 23, 2020 at 15:43

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.

• actually, the proof is what i do not understand
– M91
Oct 23, 2020 at 18:20
• @K91 proof of what? I still do not understand the question or confusion you have. Oct 23, 2020 at 18:21
• The equation in the squared box.
– M91
Oct 23, 2020 at 22:57
• 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. Oct 23, 2020 at 23:02
• It is clear now, thanks
– M91
Oct 25, 2020 at 9:42