How to derive Levi-civita identity: $\varepsilon_{kj}\delta_{il}-\delta_{ij}\varepsilon_{kl}=-\varepsilon_{jl}\delta_{ik}$ in 2D

Question

I came across the following identity in 2D (and I believe in 2D only):

$$\varepsilon_{kj}\delta_{il}-\delta_{ij}\varepsilon_{kl}=-\varepsilon_{jl}\delta_{ik},$$

with $\varepsilon$ the Levi-Civita tensor and $\delta$ the Kronecker delta. I can verify it by plugging in component values, but I'm struggling to see where it could come from.

I’m no expert with indices myself, but what helped me most was simply working through a few examples by hand. When I first studied differential geometry, doing these computations manually did helped me develop an intuition for these scary looking creatures. I think it’s one of those things you just have to do by hand until you get a feel for it. It’s not very “sexy” and can be pretty tedious, but I don’t think there’s much more to it. — userא0, Dec 04 '25 at 15:23
@userא0 You are probably right :/ I was hoping for something using the known identity (things relating $\varepsilon_{ij}\varepsilon_{jk}$ to $\delta$s for example, but I guess it will be more complicated than just checking by hand even if there is a solution..) — Syrocco, Dec 04 '25 at 15:27

score 3 · Accepted Answer · answered Dec 04 '25 at 16:59

We may use the identity \begin{equation*} \varepsilon_{jl}\varepsilon_{mi}=\delta_{jm}\delta_{li}-\delta_{ji}\delta_{lm}. \end{equation*} Now introduce a new index $k$ and multiply by $\varepsilon_{mk}$ on both sides (summing over $m$), we will get \begin{align*} \varepsilon_{mk}\varepsilon_{jl}\varepsilon_{mi}&=\varepsilon_{mk}\delta_{jm}\delta_{li}-\varepsilon_{mk}\delta_{ji}\delta_{lm} \\ \varepsilon_{jl}\delta_{ki}&=\varepsilon_{jk}\delta_{li}-\varepsilon_{lk}\delta_{ji}, \end{align*} where we have used that $\varepsilon_{mi}\varepsilon_{mk}=\delta_{ik}$, your result is a few permutations of indices away. (note that $\delta_{ij}=\delta_{ji}$ and $\varepsilon_{ij}=-\varepsilon_{ji}$)

I was missing the (2D) $ \varepsilon_{mi}\varepsilon_{mk}=\delta_{ik}$, thanks! — Syrocco, Dec 04 '25 at 17:10

score 3 · Answer 2 · answered Dec 04 '25 at 18:55

Eskil answers the question well, so here I will give a little tease. This can actually be realized as an identity on complex numbers; why this is equivalent will be left as an exercise. What we want to show is $$ \Im(k\bar j)\Re(il) + \Im(l\bar k)\Re(ij) + \Im(j\bar l)\Re(ik) = 0 $$ for all complex $i,j,k,l$ (which I have cheekily named after the indices in the original problem) and where $\Im$ and $\Re$ are the imaginary and real parts. We just expand and compute: $$ (k\bar j - j\bar k)(il + \bar l\bar i) + (l\bar k - k\bar l)(ij + \bar j\bar i) + (j\bar l - l\bar j)(ik + \bar k\bar i) $$$$ = k\bar jil + k\bar j\bar l\bar i - j\bar kil - j\bar k\bar l\bar i + l\bar kij + l\bar k\bar j\bar i - k\bar lij - k\bar l\bar j\bar i + k\bar lik + j\bar l\bar k\bar i - l\bar jik - l\bar j\bar k\bar i. $$ As you can see, all terms cancel.

Very pretty (and useful, as I was just thinking of using complex numbers instead of tensors for the 2D problem in which this identity arose)!! Thanks a lot — Syrocco, Dec 04 '25 at 21:53

How to derive Levi-civita identity: $\varepsilon_{kj}\delta_{il}-\delta_{ij}\varepsilon_{kl}=-\varepsilon_{jl}\delta_{ik}$ in 2D

2 Answers2