Einstein summation convention: Visual walkthrough

Before we touch the identity, a promise: every symbol below is defined the first time it appears, and anchored to a picture. If you've never seen an index in your life, start at Step 1 and keep reading.

Step 1 — What an "index" even is, on a picture

WHAT we're doing: naming the three drawers of a chest of drawers. WHY: the whole game is about which labels repeat. To talk about repeats we first need labels that are crystal-clear. PICTURE: three drawers, each with a number painted on it. Below, the vector $a$ is an arrow whose shadow onto each axis fills one drawer.

Recall Why does "repeated index = sum" matter here?

Because the identity we're proving has an index $i$ that appears twice on the left. ::: That double- $i$ is a hidden instruction "add over $i = 1, 2, 3$ ", per the convention.

Step 2 — Building the Levi-Civita symbol as a "spin detector"

WHAT: a tiny machine that scores an ordering of $(1, 2, 3)$ as "clockwise / counter-clockwise / degenerate". WHY this object and not a plain number? Because cross products and determinants care about orientation — which way you circle around. A ratio or a sum can't remember orientation; a sign can. $ε_{ij k}$ is the smallest thing that stores "which way you're spinning". PICTURE: a circle with $1 \to 2 \to 3 \to 1$ arrows. Read the three labels in the direction of the arrows → $+ 1$ . Read them against → $- 1$ . Any doubled label can't sit on the wheel → $0$ .

Step 3 — What the product $ε_{ij k} ε_{i l m}$ is asking

WHAT: we're multiplying two spin-detectors that share their first slot $i$ , then adding over the three values of that shared slot. WHY: sharing $i$ and summing it is the algebraic fingerprint of "how much do these two oriented triples overlap once we forget the first direction?" That overlap is a number that depends only on the leftovers $j, k, l, m$ . PICTURE: two spin-wheels standing side by side, joined by a chalk bridge labelled $i$ — the bridge is the summed index. Four free stubs $j, k$ and $l, m$ poke out.

Recall How many free indices → what kind of object is the answer?

Four free indices $j, k, l, m$ . ::: So the result is a rank-4 array — but we'll show it's built purely from products of $δ$ 's.

Step 4 — The key fact: $ε$ is a determinant of $δ$ 's

WHAT: we rewrite the sign-machine as a determinant whose entries are just the Kronecker delta $δ_{ab}$ ( $= 1$ if $a = b$ , else $0$ ). WHY this move? A determinant already flips sign when you swap two rows and vanishes when two rows are equal — exactly the two rules of $ε$ . So this determinant is $ε_{ij k}$ , not merely a lookalike. It converts an orientation question into ordinary matrix algebra. PICTURE: the grid of $δ$ -entries. Row $i$ is a unit row with its $1$ in column $i$ ; three such rows stacked. Swapping rows $j$ and $k$ flips the picture's handedness — and the determinant's sign.

Step 5 — Multiply two such determinants (the product-of-determinants trick)

WHAT: we glued the two $δ$ -grids by matrix-multiplying them, so every new entry is "row-of-first dotted with row-of-second" $= δ$ of the two labels. WHY: $det (M) det (N) = det (M N)$ turns a product of two orientation-numbers into one $3 \times 3$ determinant of overlaps. All the geometry is now inside plain $δ$ 's. PICTURE: two label-strips $(i, j, k)$ and $(i, l, m)$ crossed like a grid; each cell lights up chalk-blue only when its row-label equals its column-label.

Step 6 — Sum the shared index $i$ : the matrix shrinks $3 \to 2$

WHAT: we ran the hidden sum over $i$ ; the whole first row and column (all involving $i$ ) collapse, and the determinant drops from $3 \times 3$ to $2 \times 2$ in the leftovers $j, k, l, m$ . WHY: $δ_{ii} = 3$ (the dimension) and the substitution rule $δ_{ai} δ_{ib} = δ_{ab}$ are the only tools — no new machinery. The result must be free of $i$ because $i$ was summed away. PICTURE: the $3 \times 3$ grid with its $i$ -row and $i$ -column greyed out and folded away, leaving a $2 \times 2$ block whose determinant is the two diagonal minus two off-diagonal products.

Step 7 — Edge & degenerate cases (never leave a scenario unshown)

WHY show these: they pin down each piece of the formula — Case A confirms the "any repeat → 0" behaviour, Case B confirms the $+$ term, Case C confirms the $-$ term. Every combination of matches is now witnessed.

The one-picture summary

Recall Feynman retelling — the whole walkthrough in plain words

Picture two little spinning wheels, each labelled with three directions. A wheel outputs $+ 1$ if you read its labels the "correct" way round, $- 1$ the wrong way, and $0$ if it's stuck (two labels the same). Now tie the two wheels together at their first spoke — that shared spoke is the letter $i$ we're going to average over. Question: after we average over that shared direction, how much do the two wheels agree? The trick is that each wheel is secretly a determinant of tiny yes/no switches (the deltas, which say "these two labels match: yes=1 / no=0"). Two determinants multiplied become one bigger determinant of match-switches. When we finally add over the shared spoke $i$ , that whole shared row and column fold up and vanish, leaving a small $2 \times 2$ box. Its value is "straight matches minus crossed matches": $δ_{j l} δ_{k m} - δ_{j m} δ_{k l}$ . The plus is honest agreement; the minus is the leftover memory of spin. That single line is what flattens the triple cross product into the BAC–CAB rule.

Connections

Levi-Civita symbol and Kronecker delta — the two objects glued together here.
Determinants — the engine (Steps 4–6): $ε$ is a determinant of $δ$ 's.
Matrix multiplication — how the two grids fuse in Step 5.
Cross product and Vector calculus identities — where this identity is spent.
Dot product — each entry of the fused grid is a dot of unit rows.
Parent: Einstein summation convention.