# Levi civita symbol and kronecker delta identities in 4 dimensions

1. Jul 17, 2014

### Emil

I'm trying to explicitly show that

$$\varepsilon^{0 i j k} \varepsilon_{0 i j l} = - 2 \delta^k_l$$

I sort of went off the deep end and tried to express everything instead of using snazzy tricks and ended up with

$$\begin{eqnarray*} \delta^{\mu \rho}_{\nu \sigma} & = & \delta^{\mu}_{\nu} \delta^{\rho}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\rho}_{\nu}\\ & & \\ \delta^{\mu \rho_1 \rho_2}_{\nu \sigma_1 \sigma_2} & = & \delta^{\mu}_{\nu} \delta^{\rho_1 \rho_2}_{\sigma_1 \sigma_2} - \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2}_{\nu \sigma_2} + \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2}_{\sigma_2 \nu}\\ & & \\ \delta^{\mu \rho_1 \rho_2 \rho_3}_{\nu \sigma_1 \sigma_2 \sigma_3} & = & \delta^{\mu}_{\nu} \delta^{\rho_1 \rho_2 \rho_3}_{\sigma_1 \sigma_2 \sigma_3} - \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2 \rho_3}_{\nu \sigma_2 \sigma_3} + \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2 \rho_3}_{\sigma_2 \nu \sigma_3} - \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2 \rho_3}_{\sigma_2 \sigma_3 \nu}\\ & & \\ \varepsilon^{0 i j k} \varepsilon_{0 i j l} = \delta^{0 i j k}_{0 i j l} & = & \delta^0_0 \delta^{i j k}_{i j l} - \delta^0_i \delta^{i j k}_{0 j l} + \delta^0_i \delta^{i j k}_{j 0 l} - \delta^0_i \delta^{i j k}_{j l 0}\\ & & \\ & = & \delta^0_0 \left( \delta^i_i \delta^{j k}_{j l} - \delta^i_j \delta^{j k}_{i l} + \delta^i_i \delta^{j k}_{l j} \right) \ldots\\ & & - \delta^0_i \left( \delta^i_0 \delta^{j k}_{j l} - \delta^i_j \delta^{j k}_{0 l} + \delta^0_j \delta^{j k}_{l 0} \right) \ldots\\ & & + \delta^0_i \left( \delta^i_j \delta^{j k}_{0 l} - \delta^i_0 \delta^{j k}_{j l} + \delta^0_0 \delta^{j k}_{l j} \right) \ldots\\ & & - \delta^0_i \left( \delta^i_j \delta^{j k}_{l 0} - \delta^i_l \delta^{j k}_{j 0} + \delta^0_l \delta^{j k}_{0 j} \right)\\ & & \\ & = & \delta^0_0 \left( \delta^i_i \left( \delta^j_j \delta^k_l - \delta^j_l \delta^k_j \right) - \delta^i_j \left( \delta^j_i \delta^k_l - \delta^j_l \delta^k_i \right) + \delta^i_i \left( \delta^j_l \delta^k_j - \delta^j_j \delta^k_l \right) \right) \ldots\\ & & - \delta^0_i \left( \delta^i_0 \left( \delta^j_j \delta^k_l - \delta^j_l \delta^k_j \right) - \delta^i_j \left( \delta^j_0 \delta^k_l - \delta^j_l \delta^k_0 \right) + \delta^0_j \left( \delta^j_l \delta^k_0 - \delta^j_0 \delta^k_l \right) \right) \ldots\\ & & + \delta^0_i \left( \delta^i_j \left( \delta^j_0 \delta^k_l - \delta^j_l \delta^k_0 \right) - \delta^i_0 \left( \delta^j_j \delta^k_l - \delta^j_l \delta^k_j \right) + \delta^0_0 \left( \delta^j_l \delta^k_j - \delta^j_j \delta^k_l \right) \right) \ldots\\ & & - \delta^0_i \left( \delta^i_j \left( \delta^j_l \delta^k_0 - \delta^j_0 \delta^k_l \right) - \delta^i_l \left( \delta^j_j \delta^k_0 - \delta^j_0 \delta^k_j \right) + \delta^0_l \left( \delta^j_0 \delta^k_j - \delta^j_j \delta^k_0 \right) \right)\\ & & \\ & & 0 = i = j\\ & & \\ & = & \delta^0_0 \delta^i_i \delta^j_j \delta^k_l - \delta^0_0 \delta^i_j \delta^j_i \delta^k_l - \delta^0_0 \delta^i_i \delta^j_j \delta^k_l \ldots\\ & & - \delta^0_i \delta^i_0 \delta^j_j \delta^k_l + \delta^0_i \delta^i_j \delta^j_0 \delta^k_l + \delta^0_i \delta^0_j \delta^j_0 \delta^k_l \ldots\\ & & + \delta^0_i \delta^i_j \delta^j_0 \delta^k_l - \delta^0_i \delta^i_0 \delta^j_j \delta^k_l - \delta^0_0 \delta^j_j \delta^k_l \ldots\\ & & + \delta^0_i \delta^i_j \delta^j_0 \delta^k_l\\ & & \\ & = & \delta^k_l - \delta^k_l\\ \end{eqnarray*}$$

The bottom line is that all I want for christmas is to get $$- 2 \delta^k_l$$ from

$$\varepsilon^{0 i j k} \varepsilon_{0 i j l} = \delta^{0 i j k}_{0 i j l} = \left|\begin{array}{cccc} \delta^0_0 & \delta^0_i & \delta^0_j & \delta^0_l\\ \delta^i_0 & \delta^i_i & \delta^i_j & \delta^i_l\\ \delta^j_0 & \delta^j_i & \delta^j_j & \delta^j_l\\ \delta^k_0 & \delta^k_i & \delta^k_j & \delta^k_l \end{array}\right| =$$

in a way that doesn't involve 100000 kronecker deltas. THAAAAANKS :rofl:

2. Jul 17, 2014

### Fredrik

Staff Emeritus
The formula doesn't hold if k=l=0.

I have never tried to brute-force this sort of thing. It's so much easier to make observations that simplify the problem. Let k be an arbitrary element of {1,2,3}. $\varepsilon^{0 i j k} \varepsilon_{0 i j l}$ is a sum with 4×4=16 terms, but most of them are zero. Clearly all terms with i=j, all terms with i=0 or j=0, and all terms with i=k or j=k, are zero. This only leaves two terms!

Let a,b be the two elements of {1,2,3} that aren't equal to k. The only terms that we haven't proved are zero are (no summation) $\varepsilon^{0 a b k}\varepsilon_{0 a b l}$ and $\varepsilon^{0 b a k}\varepsilon_{0 b a l}$. If $l\neq k$, then $l\in\{a,b\}$, and both terms are zero. If $l=k$, then one of the terms is 1×1=1, and the other is (-1)×(-1)=1.

Hm, I didn't get a minus sign. I'm guessing that your convention isn't that $\varepsilon^{0123}$ and $\varepsilon_{0123}$ are both 1. One of them is defined to be -1, right?

3. Jul 18, 2014

### Emil

Not understanding how contra/covariance comes in, and what to sum over

I know that the convention in use is $$\varepsilon^{\alpha \beta \gamma \delta} = - \varepsilon_{\alpha \beta \gamma \delta}$$ I'm not quite comfortable on how it produces the minus signs.
Does one of the terms become $$\varepsilon^{0 a b k} \varepsilon_{0 a b l} = \left( - 1 \right) \left( 1 \right) = - 1$$ and the other $$\varepsilon^{0 b a k} \varepsilon_{0 b a l} = \left( 1 \right) \left( - 1 \right) = - 1$$ If so, why explicitly? Then I want
$$\varepsilon^{0 i j k} \varepsilon_{0 i j l} =EinsteinSummation?OverWhat?= \varepsilon^{0 a b k} \varepsilon_{0 a b l} + \varepsilon^{0 b a k} \varepsilon_{0 b a l} = \left( - 1 \right) \left( 1 \right) + \left( 1 \right) \left( - 1 \right) = - 2$$
for all $$k = l$$ i.e. $$\varepsilon^{0 i j k} \varepsilon_{0 i j l} = - 2 \delta^k_l$$