Einstein summation convention: Question bank — Maths deep dive

Throughout, remember the two load-bearing rules from Einstein summation convention:

an index appearing exactly twice in one term is summed (a dummy index, and it vanishes from the result);
an index appearing exactly once is free, is not summed, and must appear identically on both sides.

We also lean on the helpers Kronecker delta $δ_{ij}$ and Levi-Civita symbol $ε_{ij k}$ .

The two pictures below are the whole mental model. The first shows how a dummy index closes into a loop (contraction = a sum that vanishes), while a free index dangles as an open leg (a survivor you must carry). The second traces the actual index flow through a matrix–vector product.

True or false — justify

The expression

a_{i} b_{i}

is a scalar.

True. The index

i

appears twice, so it is summed away, leaving zero free indices — and zero free indices means a single number (see the Dot product).

The expression

A_{ij} x_{j}

is a scalar.

False. Only

j

is repeated (summed);

i

appears once, so it is a free index. One free index means the result is a vector — one number per value of

i

a_{i} b_{j}

(no repeated index) has a well-defined meaning in the convention.

True, but it is not a sum — it is the outer product, an order-2 object with two free indices

i

and

j

(an

n \times n

grid of the products

a_{i} b_{j}

δ_{ii} = 1

in three dimensions.

False. Here

i

is repeated, so it is summed:

δ_{ii} = δ_{11} + δ_{22} + δ_{33} = 3

. In

n

dimensions

δ_{ii} = n

(with

n

the dimension). The "identity is

1

on the diagonal" fact is about a single entry, not the summed trace.

ε_{ij k} a_{j} a_{k}

is the cross product of

a

with itself, and it equals

0

True. It is

(a \times a)_{i}

, which is zero. Index-only reason: swapping the dummy names

j \leftrightarrow k

leaves

a_{j} a_{k}

unchanged but flips

ε

's sign, so the sum equals its own negative — hence

0

The equation

y_{k} = A_{ij} x_{j}

is valid notation.

False. The left side has free index

k

; the right side has free index

i

. Free indices must be identical on both sides, so this is meaningless — a built-in error detector caught it.

You may rename

i

k

a_{i} b_{i}

without changing its value.

True. A dummy index is just the name of a summation variable:

\sum_{i} a_{i} b_{i} = \sum_{k} a_{k} b_{k}

. So

a_{i} b_{i} = a_{k} b_{k}

You may rename the free index

i

y_{i} = A_{ij} x_{j}

k

on the right side only.

False. A free index isn't a private summation variable — it labels which component you mean, and it must match across the whole equation. Rename it on both sides or not at all.

ε_{ij k} ε_{ij k} = 6

in 3D.

True. All three indices are summed. It counts the non-zero permutation triples: there are

6

of them

(3!)

, each contributing

(\pm 1)^{2} = 1

, so the total is

6

δ_{ij} ε_{ij k} = 0

True. The

δ

forces

i = j

, but

ε

vanishes whenever two indices are equal. So every surviving term is zero.

Spot the error

What is wrong with writing

a_{i} b_{i} c_{i}

for "three vectors combined"?

The index

i

appears three times, which is forbidden: a repeated index means a pairwise sum, and three copies leave the pairing ambiguous. If you genuinely want

\sum_{i} a_{i} b_{i} c_{i}

, keep the explicit

\sum

A student writes

a_{i} b_{i} c_{i} d_{i}

to mean "two separate dot products". Why is this broken, and how do you fix it?

As written,

i

appears four times in one term — illegal. Dummy indices are local to their term but not reusable within it; rename one pair:

a_{i} b_{i} c_{k} d_{k}

Why is "

A_{ij} B_{ij}

is the matrix product

A B

" wrong?

Both

i

and

j

are repeated, so both are summed — the result has zero free indices, a scalar (the double contraction / Frobenius inner product). Genuine Matrix multiplication shares a single middle index:

C_{ik} = A_{ij} B_{j k}

A student claims

δ_{ij} δ_{ij} = δ_{ii} = n

. Spot the slip.

The first step secretly reuses

i

four times. Do it carefully:

δ_{ij} δ_{ij} = \sum_{i, j} δ_{ij}^{2} = n

(only the

i = j

diagonal terms are

1

; here

n

is the dimension). The answer is

n

, but "

= δ_{ii}

" as a written expression is illegal notation.

What is wrong with

(a \times b)_{i} = ε_{j k} a_{j} b_{k}

The Levi-Civita symbol in 3D needs three indices,

ε_{ij k}

. Dropping the free

i

also destroys the vector nature of the cross product — the correct form is

ε_{ij k} a_{j} b_{k}

Someone writes

δ_{ij} a_{j} = a_{j}

. Why is the right side wrong?

The

δ

's substitution rule renames the summed index to the free one:

δ_{ij} a_{j} = a_{i}

, not

a_{j}

. Writing

a_{j}

leaves a free

j

on the right but the left's free index is

i

— a mismatch.

Why is it dangerous to write

a_{i} b_{i}

with both indices "downstairs" in curved space or special relativity?

In non-Euclidean spaces the convention pairs an upper index with a lower one:

a^{i} b_{i}

. A sum of two lower indices,

a_{i} b_{i}

, is not coordinate-independent — you first lower an index with the metric

g_{ij}

a_{i} = g_{ij} a^{j}

. In flat Euclidean space with Cartesian axes

g_{ij} = δ_{ij}

, so up/down coincide and the all-downstairs shorthand is safe.

Why questions

Why must a summed index appear exactly twice, never three or more times?

Summation pairs factors,

\sum_{i} (this_{i}) (that_{i})

. Two copies specify a unique pairing; three copies give no unique way to group the factors, so the notation would be ambiguous and is therefore banned.

Why does the number of free indices tell you the order of the object (not its matrix rank)?

Each free index independently runs over

1 \dots n

, generating a separate component: 0 → scalar, 1 → vector, 2 → matrix/order-2 Tensors. This count is the tensor's order (number of index slots), which is unrelated to linear-algebra rank (the number of independent rows of a matrix) — same English word, different concept.

Why is the Kronecker delta called a "substitution operator"?

Because

δ_{ij} a_{j} = a_{i}

: contracting a

δ

against an index deletes the summed index and renames the other object's index to the delta's surviving one — it swaps one label for another.

Why does the Levi-Civita symbol naturally encode cross products and Determinants, while

δ

encodes dot products?

ε

carries a sign that flips under swaps (orientation), which is exactly the antisymmetry a cross product or determinant needs.

δ

is symmetric and just matches indices, which is what the symmetric Dot product needs.

Why does swapping any two indices of

ε_{ij k}

flip its sign, and why is that useful in Vector calculus identities?

Swapping two entries turns an even permutation into an odd one (and vice versa), flipping

+ 1 \leftrightarrow - 1

. This antisymmetry lets you cyclically reorder indices (

ε_{ij k} = ε_{k ij}

) to line up the shared index before applying the epsilon–delta identity.

Why can the single epsilon–delta identity

ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l}

collapse so many messy proofs?

It converts a product of two orientation objects (hard, sign-laden) into plain index-matching

δ

's (easy substitutions). Almost every double-cross-product identity reduces to this one line — e.g. the BAC–CAB rule.

Edge cases

In 1D, what happen to

δ

and

ε

With

n = 1

every index can only be

1

, so

δ_{ii} = δ_{11} = 1

and any

ε

with

\geq 2

slots is all zeros (it needs distinct values to be non-zero). A 1D "vector" has a single component, so the convention degenerates to ordinary multiplication.

In 2D, is

ε_{ij k}

meaningful?

No — the 3-index

ε_{ij k}

is intrinsically 3-dimensional. In 2D you use the two-index

ε_{ij}

(with

ε_{12} = + 1, ε_{21} = - 1

). Match the symbol's index count to the dimension

n

What does the convention give for

δ_{ij} a_{i} b_{j}

The delta forces

i = j

, collapsing it to

a_{i} b_{i} = a \cdot b

. So

δ_{ij}

inside a contraction just glues two vectors into their dot product.

A_{ii}

a scalar or a matrix?

A scalar — the repeated

i

is summed, giving

A_{11} + A_{22} + \dots + A_{nn}

, the trace of

A

. Zero free indices remain.

If a term has no repeated index at all, is the convention "doing nothing"?

Correct — with no repeated index there is no implied sum. Every index is free, so the term is a genuine multi-component object (like the outer product

a_{i} b_{j}

), evaluated component-by-component.

What is

ε_{ij k}

when

i = j

Zero. Any repeated index makes

ε = 0

, because a permutation cannot repeat a value. This single fact is what forces

a \times a = 0

and encodes antisymmetry.

Can a valid equation have a free index on one side but none on the other, e.g.

c = a_{i} b_{i}

Yes here, because both sides have zero free indices (

i

is summed on the right,

c

is a scalar). The rule is only that free indices must match — and there are none to mismatch.

Does the order of factors matter in

a_{i} b_{i}

versus

b_{i} a_{i}

No — each term is an ordinary product of numbers, so

a_{i} b_{i} = b_{i} a_{i}

; the sum is commutative. (Order does matter for

A_{ij} B_{j k}

, because there the shared-index positions define which is row and which is column.)

In flat Euclidean space, does raising or lowering an index change any numbers?

No — with

g_{ij} = δ_{ij}

a^{i} = a_{i}

component-for-component, so upper/lower placement is cosmetic. It only starts to matter once the metric differs from

δ_{ij}

(curved space or a non-trivial signature).

Recall One-line self-test

If you can instantly say (a) how many free indices a term has, (b) which index is summed, and (c) whether any index illegally repeats three+ times — you have internalised the whole convention. "Twice = sum, Once = run, Thrice = crime." ::: The three-word summary of every trap on this page.

Connections

Parent: Einstein summation convention.
Traps drawn from Dot product, Cross product, Matrix multiplication, Determinants, Tensors, and Vector calculus identities — with the helpers Kronecker delta and Levi-Civita symbol.