Index notation handled the dot product with a single repeated index. But the cross product, the determinant and the curl all involve something the Kronecker delta cannot express: orientation. They care whether an ordering is right-handed or left-handed, and they flip sign when you swap two things. To write them with indices you need a companion to \delta_{ij} that is antisymmetric — the Levi-Civita symbol \varepsilon_{ijk}.

With this one symbol, the cross product, the scalar triple product, the determinant and the curl all collapse into single index expressions — and the tangle of vector identities you meet in electromagnetism becomes routine algebra.

The definition: +1, −1, or 0

\varepsilon_{ijk} has three indices, each running over 1, 2, 3, and takes just three values:

Concretely, the cyclic orders (1,2,3), (2,3,1), (3,1,2) give +1; the anticyclic orders (3,2,1), (2,1,3), (1,3,2) give -1; and any repeat, like (1,1,2), gives 0. The figure is the mnemonic: read the circle clockwise for +1, against it for -1.

The defining feature is total antisymmetry: swapping any two indices flips the sign,

which instantly explains the last case — if two indices are equal, swapping them changes nothing yet must flip the sign, so the value has to be 0.

The cross product

The i-th component of \mathbf{a}\times\mathbf{b} is one clean expression — i is free, while j and k are summed:

Worked example — recover the familiar first component. Set the free index i = 1 and sum over j, k \in \{1,2,3\}. Only terms with all three indices distinct survive (the rest carry a repeated index and vanish), so the only survivors have \{j,k\} = \{2,3\}:

Exactly the first row of the determinant formula — but derived by pure bookkeeping. The other two components follow by cycling 1 \to 2 \to 3.

The determinant and the scalar triple product

Contract three vectors against \varepsilon and you get the scalar triple product — which is exactly the 3\times 3 determinant:

The antisymmetry of \varepsilon is the antisymmetry of the determinant: swap two rows and the sign flips; repeat a row and it is zero. More generally, for any 3\times 3 matrix M,

so the Levi-Civita symbol is the engine inside every determinant you have ever expanded.

The curl

Replace one vector by the derivative operator \partial_j \equiv \partial/\partial x_j and the cross-product formula becomes the curl:

Setting i = 3, say, and summing gives \varepsilon_{312}\,\partial_1 F_2 + \varepsilon_{321}\,\partial_2 F_1 = \partial_1 F_2 - \partial_2 F_1 — the familiar \partial F_y/\partial x - \partial F_x/\partial y. One formula now encodes all three messy components of the curl, and identities like \nabla\cdot(\nabla\times\mathbf{F}) = 0 become one-line consequences of antisymmetry meeting symmetry.

The epsilon–delta identity — the master tool

When two Levi-Civita symbols meet and share one summed index, they collapse into Kronecker deltas. This single identity dispatches almost every vector identity in physics:

The shared index here is the first one, i (summed). The surviving indices pair up: the "in-order" pairing (j\text{–}l,\ k\text{–}m) comes with a plus, the "crossed" pairing (j\text{–}m,\ k\text{–}l) with a minus. Two useful special cases follow by contracting further:

Worked identity: the "BAC−CAB" rule

The vector triple product \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) is a nightmare to expand geometrically, but the epsilon–delta identity makes it fall out in a few lines. This is the calculation that convinces people to learn index notation.

Step 1 — write the i-th component with two epsilons. The inner cross product has component (\mathbf{b}\times\mathbf{c})_k = \varepsilon_{klm} b_l c_m, so

Step 2 — line up the shared index. The identity needs the summed index (k) in the first slot of both symbols. Cycle the first one, \varepsilon_{ijk} = \varepsilon_{kij}, so that \varepsilon_{ijk}\varepsilon_{klm} = \varepsilon_{kij}\varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}.

Step 3 — let the deltas sift. In the first term \delta_{il} forces l = i and \delta_{jm} forces m = j; in the second, m = i and l = j:

Step 4 — read off the vectors. The repeated j in a_j c_j is the dot product \mathbf{a}\cdot\mathbf{c}, and a_j b_j = \mathbf{a}\cdot\mathbf{b}. So

The famous "BAC−CAB" rule, proved with nothing but the definition of \varepsilon and one identity — no diagrams, no case-checking. Every step is mechanical, which is exactly the point.