# Levi Civita identity

I've been trying to use the following identity (provided by wikipedia)

$$\epsilon^{i_1...i_n}\epsilon_{j_1...j_n} = \delta_{j_1...j_n}^{i_1...i_n} \equiv n!\delta_{[j_1}^{i_1}...\delta_{j_n]}^{i_n}$$

To prove the following relation in Peskin & Schroeder (p.134) $$\epsilon^{\alpha\beta\mu\nu}\epsilon_{\alpha\beta\rho\sigma} = -2({\delta^\mu}_\rho {\delta^\nu}_\sigma - {\delta^\mu}_\sigma {\delta^\nu}_\rho)$$

This was stated without proof, as you do. They did however say that it could be proven by "appealing to symmetry arguments, then evaluating one special case to determine the overall constant".

Ideally I would like to find a general formula for the product of two $$d$$ dimensional Levi Civita symbols with $$n$$ contracted indices in addition to the proof of the first equation, but any help is greatly appreciated.

If anyone knows any sources that go into this in detail and actually provide proofs, I would be very grateful.

• A starting point would be to notice that the only non-zero terms will be the ones where none of the indices repeat. – Superfast Jellyfish Feb 27 at 19:11

Following the method in L&L vol. 2 ch. 1, the left-hand side $$\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma}$$ is a product of a pseudo-tensor, which is invariant under Lorentz transformations up to a factor of a determinant, and a pseudo-tensor which transforms under the inverse determinant.

This tells us that the right-hand side has to be an invariant tensor and so must be constructed from Kronecker delta's.

The right hand side must also be anti-symmetric in $$\mu,\nu$$ and anti-symmetric in $$\rho,\sigma$$ so that on general grounds $$\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma} = - A (\delta^{\mu}_{\rho} \delta^{\nu}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\nu}_{\rho})$$ must hold, where the minus sign is due to the Minkowski metric, and the factor $$A=2$$ is fixed by the requirement that $$\varepsilon^{\mu \nu \rho \sigma} \varepsilon_{\mu \nu \rho \sigma} = - 4!$$ holds, so setting $$- A (\delta^{\mu}_{\mu} \delta^{\nu}_{\nu} - \delta^{\mu}_{\nu} \delta^{\nu}_{\mu}) = - 4!$$ gives $$A(16 - 4) = 24$$ or $$A = 2$$.

More generally, the same thinking tells us that the product $$\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma}$$ must be an anti-symmetric combination of Kronecker delta's up to an overall normalization factor, where the normalization factor must respect the fact that $$\varepsilon^{\mu \nu \rho \sigma} \varepsilon_{\mu \nu \rho \sigma} = - 4!$$ should hold, but $$\delta^{\mu \nu \rho \sigma}_{\alpha \beta \gamma \delta}$$ is an anti-symmetric combination of Kronecker delta's such that $$\delta^{\mu \nu \rho \sigma}_{\mu \nu \rho \sigma} = 4!$$, and so $$\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma} = - \delta^{\mu \nu \rho \sigma}_{\alpha \beta \rho \sigma}$$ must hold. Thus we can work out contractions like $$\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma} = - \delta^{\alpha \beta \mu \nu}_{\alpha \beta \rho \sigma}$$ directly \begin{align*} \varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma} &= - \delta^{\alpha \beta \mu \nu}_{\alpha \beta \rho \sigma} \\ &= - (\delta^{\alpha}_{\alpha} \delta^{\beta \mu \nu}_{\beta \rho \sigma} - \delta^{\alpha}_{\beta} \delta^{\beta \mu \nu}_{\alpha \rho \sigma} + \delta^{\alpha}_{\rho} \delta^{\beta \mu \nu}_{\alpha \beta \sigma} - \delta^{\alpha}_{\sigma}\delta^{\beta \mu \nu}_{\alpha \beta \rho}) \\ &= - (4\delta^{\beta \mu \nu}_{\beta \rho \sigma} - \delta^{\alpha \mu \nu}_{\alpha \rho \sigma} + \delta^{\beta \mu \nu}_{\rho \beta \sigma} - \delta^{\beta \mu \nu}_{\sigma \beta \rho}) \\ &= - [4(\delta^{\beta}_{\beta} \delta^{\mu \nu}_{\rho \sigma} - \delta^{\beta}_{\rho} \delta^{\mu \nu}_{\beta \sigma} + \delta^{\beta}_{\sigma}\delta^{\mu \nu}_{\beta \rho}) - ( \delta^{\alpha }_{\alpha} \delta^{\mu \nu}_{\rho \sigma} - \delta^{\alpha}_{\rho} \delta^{\mu \nu}_{\alpha \sigma} + \delta^{\alpha}_{\sigma} \delta^{\mu \nu}_{\alpha \rho}) - \delta^{\beta \mu \nu}_{\beta \rho \sigma} + \delta^{\beta \mu \nu}_{\beta \sigma \rho}] \\ &= - [4(4\delta^{\mu \nu}_{\rho \sigma} - \delta^{\mu \nu}_{\rho \sigma} + \delta^{\mu \nu}_{\sigma \rho}) - ( 4 \delta^{\mu \nu}_{\rho \sigma} - \delta^{\mu \nu}_{\rho \sigma} + \delta^{\mu \nu}_{\sigma \rho}) - \delta^{\beta \mu \nu}_{\beta \rho \sigma} + \delta^{\beta \mu \nu}_{\beta \sigma \rho}] \\ &= - [4(2\delta^{\mu \nu}_{\rho \sigma} ) - 2 \delta^{\mu \nu}_{\rho \sigma} - 2 \delta^{\mu \nu}_{\rho \sigma} + 2 \delta^{\mu \nu}_{\sigma \rho}] \\ &= - [8 \delta^{\mu \nu}_{\rho \sigma} - 6 \delta^{\mu \nu}_{\rho \sigma} ] \\ &= - 2 \delta^{\mu \nu}_{\rho \sigma} \\ &= - 2 (\delta^{\mu}_{\rho} \delta^{\nu}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\nu}_{\rho}). \end{align*} Similarly, in $$d$$ dimensions we have $$\varepsilon^{\mu_1 .. \mu_d} \varepsilon_{\mu_1 .. \mu_d} = - d!$$ From this one immediately sees that $$\varepsilon^{\mu_1 .. \mu_d} \varepsilon_{\nu_1 .. \mu_d} = - \delta^{\mu_1 .. \mu_d}_{\nu_1 .. \nu_d}$$ and so contractions obey identities like $$\varepsilon^{\mu_1 \ldots \mu_r \mu_{r+1} .. \mu_d} \varepsilon_{\mu_1 .. \mu_r \nu_{r+1} \ldots \nu_d} = - A \delta^{\mu_{r+1} .. \mu_d}_{\nu_{r+1} .. \nu_d}$$ must hold in $$d$$ dimensions, where $$A$$ can be fixed by expanding $$\varepsilon^{\mu_1 \ldots \mu_r \mu_{r+1} .. \mu_d} \varepsilon_{\mu_1 .. \mu_r \nu_{r+1} \ldots \nu_d} = - \delta^{\mu_1 .. \mu_r \mu_{r+1} .. \mu_d}_{\mu_1 .. \mu_r \nu_{r+1} .. \nu_d}$$ one step at a time as in the example above, and obviously $$A = r!$$ should hold so that $$A = - d!$$ when $$r = d$$, as can be proven by induction.

• +1 just for typing out all the indices and indices of indices! – JamalS Feb 27 at 21:02
• This is pure gold. Thank you so much for such a detailed answer. – fosheimdet Feb 27 at 23:28

I'll give an alternative derivation. Let me just say that I will work entirely in Euclidean signature for simplicity. Generalizing to Lorentzian just requires an extra $$\det(\eta_{\mu\nu}) = (-1)^{n-1}$$ overall.

By definition of determinant, given an $$n\times n$$ matrix $$A_{ij}$$ one has $$\det A = \epsilon_{i_1\cdots i_n} A_{1,i_1}\cdots A_{n,i_n}\,.\label{1}\tag{1}$$ Here we will use Einstein convention throughout. This is also equivalent to $$\det A = \frac{1}{n!}\epsilon_{i_1\cdots i_n}\epsilon_{j_1\cdots j_n} A_{j_1,i_1}\cdots A_{j_n,i_n}\,.\tag{2}\label{2}$$ See the part at the botton if you want a proof of it. At the same time we have \begin{aligned} \epsilon_{k_1\cdots k_n}\det A &= \epsilon_{k_1\cdots k_n}\epsilon_{i_1\cdots i_n} A_{1,i_1}\cdots A_{n,i_n} =\\&= \epsilon_{i_1\cdots i_n} A_{k_1,i_1}\cdots A_{k_n,i_n}\,. \end{aligned}\tag{3}\label{3} The way to show \eqref{3} is by sorting the $$A_{a,i_a}$$ so that they appear with the row indices matching $$k_a$$, then we apply the inverse permutation to the $$i_a$$ to put them back in the initial order (call this permutation $$\sigma$$). This comes at a cost of introducing a $$\mathrm{sgn}(\sigma)$$, i.e. the parity of the permutation, which is precisely $$\epsilon_{k_1\cdots k_n}$$, thus cancelling that factor (since either $$1^2$$ or $$(-1)^2$$ is $$1$$).

Just as a sanity check: contract \eqref{3} with $$\epsilon_{k_1\cdots k_n}$$ and compare with \eqref{2} to see that, indeed $$\epsilon_{k_1\cdots k_n}\epsilon_{k_1\cdots k_n} = n!\,.$$ But back to business. I will now do something weird. Let me take the entries of $$A$$ to be symbols, namely $$A_{i,j} \equiv \delta_{i, K_j}\,.$$ By that I do not mean the identity matrix but rather $$A = \left(\begin{matrix} \delta_{1,K_1} &\delta_{1,K_2} &\ldots&\delta_{1,K_n}\\ \delta_{2,K_1} &\delta_{2,K_2} \\ \vdots && \ddots\\ \delta_{n,K_1}&&&\delta_{n,K_n} \end{matrix}\right)\,.$$ The symbols still commute, so everything goes through, but now \eqref{2} and \eqref{3} say $$\epsilon_{i_1\cdots i_n} \delta_{k_1,K_{i_1}}\cdots \delta_{k_n,K_{i_n}} = \frac1{n!} \epsilon_{k_1\cdots k_n}\epsilon_{i_1\cdots i_n}\epsilon_{j_1\cdots j_n}\delta_{j_1,K_{i_1}}\cdots \delta_{j_n,K_{i_n}}\,.$$ That's a lot of indices, I apologize. But we can work our way to a more readable expression. First let's just contract the $$\delta$$'s in the right hand side $$\epsilon_{i_1\cdots i_n} \delta_{k_1,K_{i_1}}\cdots \delta_{k_n,K_{i_n}} = \frac1{n!} \epsilon_{k_1\cdots k_n}\epsilon_{i_1\cdots i_n}\epsilon_{K_{i_1}\cdots K_{i_n}}\,.$$ Now notice that the sum over the $$i_a$$'s in the RHS produces $$n!$$ times the same term, because I can either have an even or odd permutation, but the sign of the $$\epsilon_{i_1\ldots}$$ tensor is the same as that of the $$\epsilon_{K_{i_1}\cdots}$$ tensor. So $$\epsilon_{i_1\cdots i_n} \delta_{k_1,K_{i_1}}\cdots \delta_{k_n,K_{i_n}} = \epsilon_{k_1\cdots k_n}\epsilon_{K_1\cdots K_n}\,.$$ Finally, the left hand side is by definition the antisymmetrization of the $$\delta$$'s over the second index. By convention the antisymmetrization has weight $$1$$ (see here or here), meaning that it doesn't overcount. Since here we instead have $$n!$$ terms we have to multiply by that. $$n!\,\delta_{k_1,[K_1}\cdots \delta_{k_n,K_n]} = \epsilon_{k_1\cdots k_n}\epsilon_{K_1\cdots K_n}\,.$$ I also replaced $$i_a$$ by $$a$$ because they are antisymmetrized so the order they had initially doesn't matter and I can re-sort them as I please (paying signs of course).

### Proof of the equality \eqref{1} \eqref{2}.

Take the expression \eqref{1} and exchange the position of $$A_{p, i_p}$$ and $$A_{q, i_q}$$. For the sake of concreteness let's say $$p. This does nothing because the entries of the matrix are just numbers! \begin{aligned} \det A &= \epsilon_{i_1\cdots i_n} A_{1,i_1}\cdots A_{p, i_p}\cdots A_{q, i_q}\cdots A_{n,i_n} \\& = \epsilon_{i_1\cdots i_n} A_{1,i_1}\cdots A_{q, i_q}\cdots A_{p, i_p}\cdots A_{n,i_n}\,. \end{aligned} Ok, nothing happened, but let me now swap $$i_p$$ and $$i_q$$ in the $$\epsilon$$: $$\epsilon_{i_1\cdots i_p\cdots i_q\cdots i_n} = -\epsilon_{i_1\cdots i_q\cdots i_p\cdots i_n}\,.$$ Obvious! We get a minus sign. So therefore I'll just rename $$i_p$$ to $$i_q$$ and $$i_q$$ to $$i_p$$ (I'm always free to do so since there are summed over) \begin{aligned} \det A &= \epsilon_{i_1\cdots i_n} A_{1,i_1}\cdots A_{p, i_p}\cdots A_{q, i_q}\cdots A_{n,i_n} \\& = -\epsilon_{i_1\cdots i_n} A_{1,i_1}\cdots A_{q, i_p}\cdots A_{p, i_q}\cdots A_{n,i_n}\,. \end{aligned} We have just proven that in the product $$A_{1,i_1}\cdots A_{n,i_n}$$ we can antisymmetrize over the row indices as well. That's because, as we saw, swapping any two row indices gives the same contribution up to a sign. We can then say $$\det A = \frac{1}{n!}\sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)\,\epsilon_{i_1\cdots i_n} A_{\sigma(1),i_1}\cdots A_{\sigma(n),i_n}\,,$$ where $$S_n$$ is the permutation group of $$n$$ elements and $$\mathrm{sgn}(\sigma)$$ is the parity of the permutation. I divided by $$n!$$ because every term is now counted $$|S_n|= n!$$ times. You might know that, in general, for any tensor $$\sum_{\sigma \in S_n}\mathrm{sgn}(\sigma) \,T_{\sigma(1)\cdots \sigma(n)} = \epsilon_{i_1\ldots i_n} T_{i_1\ldots i_n}\,.$$ This is basically by definition of $$\epsilon_{i_1\ldots i_n}$$. Looking back, we just proved \eqref{2}.