I have been reading up on the permutation tensor, and have come across the following expression (in 'Generalized Calculus with Applications to Matter and Forces' by L.M.B.C Campos page 709): $$e_{i_1,\ldots,i_n}=e^{i_1,\ldots,i_n}=\begin{cases} 0 & \text{if repeated indices} \\ 1 & \text{if ($i_1,\ldots,i_n$) is an even permutation} \\ -1 & \text{if ($i_1,\ldots,i_n$) is an odd permutation} \end{cases}$$ However when I try and prove that: $$e_{i_1,\ldots,i_n}=e^{i_1,\ldots,i_n}$$ I instead get: $$e_{i_1,\ldots,i_n}=\textrm{det}(g)e^{i_1,\ldots,i_n}$$ why is there a difference between my result and that given above? Is it to do with the way $e_{i_1,\ldots,i_n}$ has been defined?

My working $$e_{i_1,\ldots, i_n}=\sum_\sigma g_{i_{\sigma(1)}i_1}\ldots g_{i_{\sigma(n)}i_n}e^{i_{\sigma(1)}\ldots i_{\sigma(n)}}$$ where the summation is over all permutations, $\sigma$, of $(1,\ldots,n)$ $$\begin{align}e_{i_1...i_n} &=e^{i_1...i_n}\sum_\sigma \textrm{sgn}(\sigma) g_{i_{\sigma(1)}i_1}\ldots g_{i_{\sigma(n)}i_n}\\ &=\textrm{det}(g)e^{i_1,\ldots,i_n}\end{align}$$

  • $\begingroup$ Indeed, you are right. The statement of the book is false as soon as $e^{i_1\ldots i_n} = g^{i_1j_1}\cdots g^{i_nj_n} e_{j_1\cdots j_n}$ and your identity is correct. Perhaps the book uses different conventions... $\endgroup$ – Valter Moretti Oct 4 '16 at 8:42
  • $\begingroup$ Related: physics.stackexchange.com/q/123231/2451 $\endgroup$ – Qmechanic Oct 4 '16 at 8:44

The symbol defined as \begin{align} e_{i_1i_2\ldots i_n} &= e^{i_1i_2\ldots i_n} = \begin{cases} 0 &\text{if repeated indices} \\ 1 &\text{if even permutation} \\ -1 &\text{if odd permutation} \end{cases}\end{align} is indeed not a tensor. It is called the Levi-Civita symbol (and is a pseudo-tensor density), but we can turn it into a pseudo-tensor, by defining the Levi-Civita tensor \begin{align} \epsilon_{i_1i_2\ldots i_n} \equiv \sqrt{|\det(g)|}e_{i_1i_2\ldots i_n}, \\ \epsilon^{i_1i_2\ldots i_n} \equiv \frac{1}{\sqrt{|\det(g)|}}e^{i_1i_2\ldots i_n}. \end{align} Since you have already found how the symbol reacts to index lowering, you can immediately verify that \begin{align} g_{i_1j_1}g_{i_2j_2}\cdots g_{i_nj_n}\epsilon^{j_1j_2\ldots j_n} = (-1)^s\epsilon_{i_1i_2\ldots i_n}, \end{align} where $s$ is the number of negatives in the metric signature. The $(-1)^s$-factor is why we call it a pseudo-tensor. Note that the definition carries two consequences: \begin{align} \epsilon_{i_1i_2\ldots i_n}\epsilon^{j_1j_2\ldots j_n} = e_{i_1i_2\ldots i_n}e^{j_1j_2\ldots j_n}, \end{align} and crucially, under some frame transformation $\Lambda_k^\ell$ we have \begin{align} \Lambda_{i_1}^{j_1}\Lambda_{i_2}^{j_2}\cdots\Lambda_{i_n}^{j_n}\epsilon_{j_1j_2\ldots j_n} = \mathrm{sgn}(\Lambda)\sqrt{|\det(\Lambda^2g)|}e_{i_1i_2\ldots i_n} = \mathrm{sgn}(\Lambda)\widetilde{\epsilon}_{i_1i_2\ldots i_n}, \end{align} where $\widetilde{\epsilon}_{i_1i_2\ldots i_n}$ is the Levi-Civita tensor of the transformed frame, and $\mathrm{sgn}(\Lambda)$ is the sign of the determinant. We can also consider this the reason to call it a pseudo-tensor, since both the $(-1)^s$ factor and the $\mathrm{sgn}(\Lambda)$ factor are consequences of the same property. In particular, note that $(-1)^s = \mathrm{sgn}(g)$.

A finaly word of caution: in the literature it is common to use $\epsilon$ or $\varepsilon$ for either the symbol or the tensor, and sometimes the other for the other, and sometimes without clarifying which one is used. In such cases it can typically be inferred from the context.


Another aspect.. I will not touch about pseudotensor but I will try to conclude what we use in basic GR. We can interpret it as the volume form of Euclidean space. It is a tensor in flat but not in curved spacetime. In curved spacetime the Levi-Civita symbol is tensor density not tensor but Levi-Civita indeed is the a tensor they read \begin{align} \epsilon_{i_1i_2\ldots i_n} \equiv \sqrt{|\det(g)|}\varepsilon_{i_1i_2\ldots i_n}, \\ \epsilon^{i_1i_2\ldots i_n} \equiv \frac{sgn(g)}{\sqrt{|\det(g)|}}\varepsilon^{i_1i_2\ldots i_n}\equiv \frac{(-1)^s}{\sqrt{|\det(g)|}}\varepsilon^{i_1i_2\ldots i_n}. \end{align}

You can check ther are transform as the tensor \begin{align} g_{i_1j_1}g_{i_2j_2}\cdots g_{i_nj_n}\epsilon^{j_1j_2\ldots j_n} = \epsilon_{i_1i_2\ldots i_n}, \end{align} The contraction reads \begin{align} (-1)^s\epsilon_{i_1i_2\ldots i_n}\epsilon^{j_1j_2\ldots j_n} = \varepsilon_{i_1i_2\ldots i_n}\varepsilon^{j_1j_2\ldots j_n}=n!, \end{align} As the other tensor the L-C tensor change frame by $$ \Lambda^{i_1}_{j'_1} ... \Lambda^{i_n}_{j'_n} \epsilon_{i_1i_2\ldots i_n}=\epsilon_{j'_1j'_2\ldots j'_n} $$ So $$ \Lambda^{i_1}_{j'_1} ... \Lambda^{i_n}_{j'_n} \sqrt{|g|}\varepsilon_{i_1i_2\ldots i_n}=\sqrt{|g'|}\varepsilon_{j'_1j'_2\ldots j'_n} $$ and $$ \varepsilon_{j'_1j'_2\ldots j'_n} = \sqrt{\frac{|g|}{|g'|}}\Lambda^{i_1}_{j'_1} ... \Lambda^{i_n}_{j'_n} \varepsilon_{i_1i_2\ldots i_n} $$ $$ \quad\quad =J\;\Lambda^{i_1}_{j'_1} ... \Lambda^{i_n}_{j'_n} \varepsilon_{i_1i_2\ldots i_n} $$ where $J$ is Jacobian of transforming coordinate $ x \mapsto x'$ . So $\epsilon_{i....j}$ is a tensor density of weight 1 (-1 for some author)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.