# 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 ? – Frederic Thomas Oct 23 at 15:43

The second equation does not directly follow from the first. You are correct in that $$$$\tilde{\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}| M|=\tilde{\epsilon}_{\mu_{0}..\mu_{n}}M^{\mu_{0}}_{\ \ \ \ \bar{\mu}_{0}}...M^{\mu_{n}}_{\ \ \ \ \bar{\mu}_{n}},$$$$

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:

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

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 $$$${\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}=\sqrt{|g|}\tilde{\epsilon}_{\mu_{0}..\mu_{n}}.$$$$ so that it transforms like a true tensor

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.

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