Einstein summation convention: Exercises — Maths deep dive

Throughout, indices run $1, 2, 3$ (three dimensions) unless a problem says otherwise, and the two helper objects are the Kronecker delta $δ_{ij}$ and the Levi-Civita symbol $ε_{ij k}$ .

Recall Two facts you will reuse constantly

Substitution: $δ_{ij} a_{j} = a_{i}$ (the delta renames the repeated index). Epsilon–delta master identity: $ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l}$ (summed index is the shared first slot $i$ ).

Level 1 — Recognition

Exercise 1.1

For each expression, list the free indices and the dummy (summed) indices, and say what kind of object it is (scalar, vector, or matrix): (a) $a_{i} b_{i}$ (b) $A_{ij} x_{j}$ (c) $A_{ij} B_{j k}$ (d) $ε_{ij k} a_{j} b_{k}$ .

Recall Solution

Rule: an index appearing exactly twice in a term is summed (dummy); one appearing exactly once is free. The number of free indices = the rank = the kind of object (0→scalar, 1→vector, 2→matrix).

(a) $i$ appears twice → dummy. Free: none. → scalar.
(b) $j$ appears twice ( $A_{ij}$ and $x_{j}$ ) → dummy. $i$ appears once → free. → vector (one free index).
(c) $j$ twice → dummy. $i$ once, $k$ once → free. Two free indices → matrix.
(d) $j$ twice, $k$ twice → both dummy. $i$ once → free. → vector.

Exercise 1.2

Which of these are legal in summation convention, and why? (a) $a_{i} b_{i} c_{i}$ (b) $a_{i} b_{i} + c_{j} d_{j}$ (c) $y_{k} = A_{ij} x_{j}$ (d) $δ_{ii}$ .

Recall Solution

(a) Illegal. $i$ appears three times. A repeated index means a pairwise sum; three copies give no unique pairing. Write $\sum_{i} a_{i} b_{i} c_{i}$ if you truly mean that.
(b) Legal. Each term separately has a valid doubled index ( $i$ in the first, $j$ in the second). Dummy names are local to their own term, so no clash.
(c) Illegal. Left side has free index $k$ ; right side has free index $i$ . Free indices must be identical on both sides. Fix: $y_{i} = A_{ij} x_{j}$ .
(d) Legal. $δ_{ii} = \sum_{i} δ_{ii} = n = 3$ . A scalar.

Level 2 — Application

Exercise 2.1

Take $a = (1, 2, 3)$ and $b = (4, 5, 6)$ . Compute $a_{i} b_{i}$ by hand and state what it represents.

Recall Solution

$a_{i} b_{i} = \sum_{i = 1}^{3} a_{i} b_{i} = (1) (4) + (2) (5) + (3) (6) = 4 + 10 + 18 = 32$ . This is the Dot product $a \cdot b$ — no free indices, so a single number, as a dot product should be.

Exercise 2.2

With $a = (1, 2, 3)$ , $b = (4, 5, 6)$ , compute the first component of $a \times b$ using $(a \times b)_{i} = ε_{ij k} a_{j} b_{k}$ . Then compute all three components.

Recall Solution

Set $i = 1$ . The only nonzero $ε_{1 j k}$ are $ε_{123} = + 1$ and $ε_{132} = - 1$ : $(a \times b)_{1} = ε_{123} a_{2} b_{3} + ε_{132} a_{3} b_{2} = a_{2} b_{3} - a_{3} b_{2} = (2) (6) - (3) (5) = 12 - 15 = - 3 .$ Similarly $(a \times b)_{2} = a_{3} b_{1} - a_{1} b_{3} = (3) (4) - (1) (6) = 12 - 6 = 6$ , and $(a \times b)_{3} = a_{1} b_{2} - a_{2} b_{1} = (1) (5) - (2) (4) = 5 - 8 = - 3$ . So $a \times b = (- 3, 6, - 3)$ . See the figure below for what the sign of each $ε$ picks out.

Exercise 2.3

Simplify $δ_{ij} a_{i} b_{j}$ to a familiar expression.

Recall Solution

Use substitution on the delta. $δ_{ij}$ eats a repeated index: applying it to $b_{j}$ gives $δ_{ij} b_{j} = b_{i}$ . So $δ_{ij} a_{i} b_{j} = a_{i} (δ_{ij} b_{j}) = a_{i} b_{i} .$ That is the dot product $a \cdot b$ . (Equivalently, the delta forces $i = j$ , collapsing the double sum to a single one.)

Level 3 — Analysis

Exercise 3.1

Prove that $δ_{ij} δ_{j k} = δ_{ik}$ .

Recall Solution

Write the implied sum: $δ_{ij} δ_{j k} = \sum_{j} δ_{ij} δ_{j k}$ . In the first factor only $j = i$ is nonzero, contributing $δ_{ik}$ ; all other $j$ give $0$ . Hence $δ_{ij} δ_{j k} = δ_{ik}$ . (Interpretation: multiplying the identity matrix by itself gives the identity.)

Exercise 3.2

Show $ε_{ij k} a_{j} a_{k} = 0$ for any vector $a$ . (This is the statement $a \times a = 0$ .)

Recall Solution

The key is that $ε_{ij k}$ is antisymmetric in $j, k$ (swapping them flips the sign), while $a_{j} a_{k}$ is symmetric (swapping them changes nothing). Rename the dummy pair $j \leftrightarrow k$ (allowed — dummy names are free): $ε_{ij k} a_{j} a_{k} = ε_{ik j} a_{k} a_{j} = - ε_{ij k} a_{j} a_{k},$ using $ε_{ik j} = - ε_{ij k}$ and $a_{k} a_{j} = a_{j} a_{k}$ . A quantity equal to its own negative must be $0$ .

Exercise 3.3

Evaluate $ε_{ij k} ε_{ij k}$ (all three indices summed).

Recall Solution

Use the master identity with the last two slots matched. Set $l = j$ , $m = k$ in $ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l}$ : $ε_{ij k} ε_{ij k} = δ_{j j} δ_{k k} - δ_{j k} δ_{k j} .$ Now $δ_{j j} = 3$ , $δ_{k k} = 3$ , so the first term is $9$ . The second: $δ_{j k} δ_{k j} = δ_{j k} δ_{j k} = δ_{j j} = 3$ . Therefore $ε_{ij k} ε_{ij k} = 9 - 3 = 6 .$ (Sanity check: there are exactly $6$ nonzero components of $ε$ , each $\pm 1$ , and squaring gives $+ 1$ apiece → $6$ .)

Level 4 — Synthesis

Exercise 4.1

Prove the BAC–CAB rule $a \times (b \times c) = b (a \cdot c) - c (a \cdot b)$ using index notation.

Recall Solution

Start from the $i$ -th component and introduce a fresh dummy for the inner cross product: $[a \times (b \times c)]_{i} = ε_{ij k} a_{j} (b \times c)_{k} = ε_{ij k} a_{j} ε_{k l m} b_{l} c_{m} .$ Cycle the second epsilon so its shared index is first: $ε_{k l m} = ε_{k ij}$ ? No — we cycle the first one instead. Since $ε_{ij k} = ε_{k ij}$ (cyclic shift is even), rewrite $ε_{ij k} = ε_{k ij}$ so both share $k$ in the first slot: $= ε_{k ij} ε_{k l m} a_{j} b_{l} c_{m} .$ Apply the master identity with shared index $k$ : $ε_{k ij} ε_{k l m} = δ_{i l} δ_{j m} - δ_{im} δ_{j l}$ : $= (δ_{i l} δ_{j m} - δ_{im} δ_{j l}) a_{j} b_{l} c_{m} .$ Distribute and use substitution. First term: $δ_{i l} δ_{j m} a_{j} b_{l} c_{m} = b_{i} (a_{j} c_{j}) = b_{i} (a \cdot c)$ . Second term: $δ_{im} δ_{j l} a_{j} b_{l} c_{m} = c_{i} (a_{j} b_{j}) = c_{i} (a \cdot b)$ . Hence $[a \times (b \times c)]_{i} = b_{i} (a \cdot c) - c_{i} (a \cdot b),$ which is exactly $b (a \cdot c) - c (a \cdot b)$ . $■$

Exercise 4.2

Prove the scalar identity $(a \times b) \cdot (c \times d) = (a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)$ .

Recall Solution

Write both cross products in index form (they must share the dotting index $i$ ): $(a \times b)_{i} (c \times d)_{i} = (ε_{ij k} a_{j} b_{k}) (ε_{i l m} c_{l} d_{m}) = ε_{ij k} ε_{i l m} a_{j} b_{k} c_{l} d_{m} .$ The two epsilons already share the first slot $i$ , so apply the master identity directly: $= (δ_{j l} δ_{k m} - δ_{j m} δ_{k l}) a_{j} b_{k} c_{l} d_{m} .$ First term: $δ_{j l} δ_{k m} a_{j} b_{k} c_{l} d_{m} = (a_{l} c_{l}) (b_{m} d_{m}) = (a \cdot c) (b \cdot d)$ . Second term: $δ_{j m} δ_{k l} a_{j} b_{k} c_{l} d_{m} = (a_{m} d_{m}) (b_{l} c_{l}) = (a \cdot d) (b \cdot c)$ . Therefore the result is $(a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)$ . $■$ (Setting $c = a, d = b$ recovers Lagrange's identity $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$ .)

Level 5 — Mastery

Exercise 5.1

Using index notation and the master identity, prove the vector-calculus identity $\nabla \times (\nabla \times F) = \nabla (\nabla \cdot F) - \nabla^{2} F .$ Here $\nabla$ acts as the operator $\partial_{i} = \frac{\partial}{\partial x _{i}}$ , and $(\nabla \times F)_{i} = ε_{ij k} \partial_{j} F_{k}$ . See Vector calculus identities.

Recall Solution

Write the $i$ -th component, with a fresh dummy set for the inner curl: $[\nabla \times (\nabla \times F)]_{i} = ε_{ij k} \partial_{j} (\nabla \times F)_{k} = ε_{ij k} \partial_{j} (ε_{k l m} \partial_{l} F_{m}) .$ Pull constants (the $ε$ are just numbers, and derivatives commute with them): $= ε_{ij k} ε_{k l m} \partial_{j} \partial_{l} F_{m} .$ Cycle the first epsilon to share $k$ up front: $ε_{ij k} = ε_{k ij}$ , so $= ε_{k ij} ε_{k l m} \partial_{j} \partial_{l} F_{m} = (δ_{i l} δ_{j m} - δ_{im} δ_{j l}) \partial_{j} \partial_{l} F_{m} .$ First term: $δ_{i l} δ_{j m} \partial_{j} \partial_{l} F_{m} = \partial_{m} \partial_{i} F_{m} = \partial_{i} (\partial_{m} F_{m}) = \partial_{i} (\nabla \cdot F) = [\nabla (\nabla \cdot F)]_{i} .$ Second term: $δ_{im} δ_{j l} \partial_{j} \partial_{l} F_{m} = \partial_{j} \partial_{j} F_{i} = (\partial_{j} \partial_{j}) F_{i} = \nabla^{2} F_{i} .$ Combining: $[\nabla \times (\nabla \times F)]_{i} = [\nabla (\nabla \cdot F)]_{i} - \nabla^{2} F_{i}$ , i.e. $\nabla \times (\nabla \times F) = \nabla (\nabla \cdot F) - \nabla^{2} F . ■$

Exercise 5.2

Show that $ε_{ij k} \partial_{j} \partial_{k} f = 0$ for any twice-differentiable scalar field $f$ (i.e. $\nabla \times \nabla f = 0$ ).

Recall Solution

Same symmetry argument as Exercise 3.2, now with derivatives. The symbol $ε_{ij k}$ is antisymmetric under $j \leftrightarrow k$ ; the double derivative $\partial_{j} \partial_{k} f$ is symmetric (mixed partials commute: $\partial_{j} \partial_{k} f = \partial_{k} \partial_{j} f$ ). Rename the dummies $j \leftrightarrow k$ : $ε_{ij k} \partial_{j} \partial_{k} f = ε_{ik j} \partial_{k} \partial_{j} f = - ε_{ij k} \partial_{j} \partial_{k} f .$ Equal to its own negative ⇒ $0$ . So the curl of any gradient vanishes.

Exercise 5.3

Express the determinant of a $3 \times 3$ matrix $A$ in index form and verify it for $A = diag (2, 3, 4)$ . See Determinants.

Recall Solution

The determinant is $det A = ε_{ij k} A_{1 i} A_{2 j} A_{3 k}$ (sum over $i, j, k$ ; the $ε$ supplies the signed permutation orientation, exactly what Determinants need). Only permutations of $(1, 2, 3)$ survive, and for a diagonal matrix the only nonzero entries are $A_{11}, A_{22}, A_{33}$ , forcing $i = 1, j = 2, k = 3$ : $det A = ε_{123} A_{11} A_{22} A_{33} = (+ 1) (2) (3) (4) = 24 .$ Matches the product of the diagonal. ✔

Recall Self-test checklist

Ran the ladder? ::: L1 recognise · L2 compute · L3 manipulate deltas/epsilons · L4 prove named vector identities · L5 prove calculus identities and determinants — all with index notation alone.

Connections

Parent: Einstein summation convention.
Built on Dot product, Cross product, Kronecker delta, Levi-Civita symbol.
Extends to Matrix multiplication, Determinants, Tensors, and Vector calculus identities.