Chapter 4
Index Notation and the Summation
Convention
Syllabus covered:
3. Index notation and the Summation Convention; summation over repeated indices; Kronecker delta and
εi jk ; formula for εi jk εklm .
We now introduce a very useful notation. In particular it makes proving identities such as those in Chapter
2 much simpler. There are many other uses: an extended version of it is used in Relativity, and it can be widely
used in linear algebra and its applications (e.g. input-output models in economics).
Index notation
In this notation we abbreviate a vector a = (a1 , a2 , a3 ) ≡ a1 i + a2j + a3k to ai .
The special vector r will be written as xi , so x1 = x, x2 = y, x3 = z in the usual notation.
The name of the index (here, i) is irrelevant: ai and a j mean exactly the same thing. However, it becomes
relevant when we write an equation such as ai = bi (which implies 3 equations in fact). This means a1 = b1 ,
a2 = b2 and a3 = b3 . The same equations could be written a j = b j . But they cannot be written ai = b j ,
because we would have no idea whether, for example, a1 was equal to b1 , b2 or b3 .
Really, the notation means the ith component of a, or the i-th entry in a. The same idea works equally
well in any number of dimensions, or even for infinite sequences (a1 , a2 , . . . , an , . . .). However, we shall
assume that indices always run from 1 to 3.
We can have objects with more than one index1 . For example, we can use Ci j as a notation for a 3 × 3 (or
more generally n × n) matrix C. The equation a = Cb could then be written ai = ∑3j=1 Ci j b j .
If we want to mix the vector and index ways of writing things we must write (a)i = ai and so on.
Einstein summation convention
This further compression of notation allows us to drop summation signs. It is that if an index is repeated,
1 In
general these will be objects called “Cartesian tensors” (since we stick to Cartesian coordinates here) or more generally just
tensors. In this course we avoid a proper definition of tensors and discussions of their general properties.
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the repetition signals that one should sum over its allowed values (1 to 3 in our case). Thus
3
ai bi ≡ ∑ ai bi = a1 b1 + a2 b2 + a3b3 = a.b .
(4.1)
i=1
Again, the name of the repeated index is irrelevant: a j b j means exactly the same thing.
Using this convention, we can write the matrix product above, Cb, more briefly still as Ci j b j .
Indices that appear twice and are summed over are called dummy indices. Indices that appear once (more
exactly, once in every term in an equation or expression) are called free. Free indices tell us how many
equations have been compressed into one. For example,
ai b j ck = di jk
stands for 27 equations such as a2 b1 c3 = d213 .
To make the summation convention work we have to stick to one more rule: no index can appear more
than twice. In practice we have to be careful that every time we need an extra index, we only use the same
name as an existing index if we really mean to sum over the possible values of the pair.
Example 4.1. We can easily prove the matrix identity (AB)T = BT AT . Writing the matrix C = AB, we
have
CiTj = C ji = A jk Bki = ATk j BTik = BTik ATk j .
Thus, the rules for the Einstein summation convention are:
1. In any term (which may be a product), an index can appear at most twice.
2. If an index appears twice in the same term, it means we are to sum over all allowed values (dummy
index).
3. If an index appears once only, the same index must appear once only in all other terms in an equation
(free index).
If we are using the index convention but wish to violate these rules, we can do so by going back to putting
summation signs in explicitly, or by saying after an equation “no sum”, or whatever other change is needed
to indicate the violation.
Exercise 4.1. Which of the following have meaning in Einstein’s summation convention:
1. (ai bi )c j = (c j d j )bi
2. (ak bk )c j = (am bm )d j
3. ai b j ck dk e f = mi n j p f qk
4. ai b j = a j bi
p
5. am = dk bm / (ck bk )
2
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∇ in index notation
Since ∇V = (
∂V ∂V ∂V
,
,
), we can write the ith component of ∇V as
∂ x1 ∂ x2 ∂ x3
∂V
.
∂ xi
For brevity we introduce the shorthand notation
∂i =
and write the gradient as
∂
∂ xi
(∇V )i = ∂iV.
(4.2)
The divergence of a vector field Fi is
∂ F1 /∂ x1 + ∂ F2/∂ x2 + ∂ F3 /∂ x3 ,
so we can write
∇·F =
Now ∂i x j =
∂ Fi
= ∂i Fi .
∂ xi
(4.3)
∂xj
. If i = j, this is 1, but if i 6= j, it is 0. We introduce a notation for this, δi j .
∂ xi
Kronecker delta
We define an object called the Kronecker delta:
1
δi j =
0
if i = j
.
if i 6= j
If written as a matrix, Kronecker delta is just the unit matrix. From the definitions,
δ1 j a j =
3
∑ δ1 j a j = a1
j=1
and sinilarly for i = 2 and i = 3; hence
δi j a j =
3
∑ δi j a j = ai.
(4.4)
j=1
Thinking of δi j as the unit matrix I this just says Ia = a.
What is δii ? Don’t forget that we must sum over a repeated index; so
3
δii = ∑ δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3.
i=1
Again, if δi j is the unit matrix, δii is its trace.
We now have all the ingredients to work with index notation on any vector equation which does not
involve a cross-product.
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Example 4.2. Show that
∇(a.r) = a,
where r is the position vector and a is a constant vector.
The index notation form of the left side is ∂i (a j x j ). This gives
∂i (a j x j ) = a j ∂i x j + x j ∂i a j = a j δi j = ai
since a being constant implies ∂i a j = 0. The right side is just the index notation for a, of course.
Finally, we add an another new object, the Levi-Civita epsilon, to enable us to handle cross-products.
Levi-Civita epsilon
This is defined by
εi jk =


1 if (i, j, k) is a cyclic permutation of (1, 2, 3),
−1 if (i, j, k) is an anticyclic permutation of (1, 2, 3),

0 otherwise.
[Aside: in n dimensions, we would have to replace cyclic and anticyclic in the above definition by even and
odd.] Thus
ε123 = ε231 = ε312 = 1,
ε132 = ε321 = ε213 = −1,
and all other possibilities (e.g. ε112 , ε333 , ε232 ) are zero.
Consider then
Likewise
ε1 jk a j bk = ε123 a2 b3 + ε132 a3 b2 = a2 b3 − a3 b2 .
ε2 jk a j bk = a3 b1 − a1b3 ,
Thus
ε3 jk a j bk = a1 b2 − a2 b1 .
εi jk a j bk = (a × b)i .
(4.5)
The determinant of a matrix A can be written εi jk A1i A2 j A3k . ∇ × F can be expressed as
εi jk
∂ Fk
= εi jk ∂ j Fk
∂xj
(4.6)
since this is the ith component of the curl.
Example 4.3. Show that if T jk = Tk j for all values of j and k then εi jk T jk must be zero.
εi jk T jk
So 2εi jk T jk
= −εik j Tk j swapping indices and using the symmetries
= −εimn Tmn renaming dummy indices k → m and j → n.
= −εi jk T jk
= 0.
again renaming dummy indices, now m → j and n → k.
Here εi jk could be replaced with anything skew in jk: the outcome is referred to as “skew summed with
symmetric is zero”.
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This looks like a trick. To understand it better, it may help to look at a two-dimensional case with a skew
object Ai j . Then
Ai j Ti j = A11 T11 + A12T12 + A21T21 + A22 T22 .
Since A is skew, A11 = A22 = 0. Then substituting A21 = −A12 and T21 = T12 the remaining terms cancel.
This argument can be repeated in any number of dimensions but is very laborious to write out. The index
form achieves the same much more economically.
An important identity is
εi jk εilm = δ jl δkm − δ jm δkl .
(∗∗)
This can be proved by comparing both sides for all possible choices of ( j, k, l, m): e.g. if ( j, k, l, m) =
(1, 2, 1, 2) the left-hand side of (∗∗) is
εi12 εi12 = ε312 ε312 = 1,
and the right-hand side is
δ11 δ22 − δ12δ12 = 1 − 0 = 1.
A more abstract argument is to note that (a) the product of the epsilons is zero unless the pairs jk and lm
contain the same pair of indices, (b) if the order in the pairs is the same then both epsilons are 1 or both are
−1, and the right side is 1, and (c) a similar argument for when jk and lm have opposite orders.
Exercise 4.2. Pick another set of values for ( j, k, l, m) and verify that (∗∗) holds.
2
The previous lecturers on this course said:
“Identity (∗∗) is important – write it on your bathroom mirror so you see it every morning!”
Note that due to the cyclic symmetry
εi jk εilm = εi jk εlmi = εi jk εmil = −εi jk εmli etc. ,
so the key point is that one of the indices on the first epsilon is the same as one of the indices on the second
epsilon: then it can always be rearranged to the standard form quoted. Moreover, the names of the indices
themselves can be changed – only the pattern of their occurrence is significant, e.g. it also follows that
ε pik εm j p = δim δk j − δi j δkm .
Example 4.4. Expand out (a × b).(c × d) using index notation.
(a × b).(c × d) =
=
=
εi jk a j bk εilm cl dm [using (4.1) and (4.5)]
(δ jl δkm − δ jm δkl )a j bk cl dm [using (∗∗)]
al cl bm dm − am dm bk ck [using (4.4)]
so (a × b).(c × d) = (a.c)(b.d) − (a.d)(b.c)
Example 4.5. Expand ∇ × (F × G) using index notation. (Note this is section 2.5, identity 6.)
∇ × (F × G)i
εi jk ∂ j (εklm Fl Gm )
= (δil δ jm − δim δ jl )∂ j (Fl Gm )
= ∂m (Fi Gm ) − ∂l (Fl Gi )
= Fi (∂m Gm ) + Gm ∂m Fi − Gi (∂m Fm ) − Fm ∂m Gi
=
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so ∇ × (F × G) = F(∇ · G) + (G.∇)F − G(∇ · F) − (F.∇)G.
Explanation: Here we are substituting F × G for F in (4.6), using (∗∗), differentiating the products, and
converting back to vector notation using (4.2) and (4.3).
Exercise 4.3. Using index notation, prove the identity
∇ × ( f v) = f ∇ × v + (∇ f ) × v.
(This is, with different names, identity 5 in section 2.5.) Now repeat this exercise, but without using index
notation (i.e. writing out the left-hand sides of the identity in full, and showing that it can be rearranged to
give the right-hand side).
2
Exercise 4.4. Show that
Hence prove that
∂i (εi jk u j vk ) = vk εki j ∂i u j − u j ε jik ∂i vk .
∇ · (u × v) = v.(∇ × u) − u.(∇ × v).
(∗)
(This is identity 4 in section 2.5.) Now prove identity (∗) without using index notation (i.e. writing out the
left-hand side of the identity in full and showing that this can be rearranged to give the right-hand side). 2
One can prove the remaining identities for derivatives of products or products of derivatives similarly.
Even the harder standard product identities, e.g. the one stated as number 2 in the list in section 2.5,
∇ (u.v) = u × (∇ × v) + v × (∇ × u) + u.∇v + v.∇u
(4.7)
are not very difficult: for this one, we start with the first term on the right, using (4.5) and (4.6), and work
analogously to Examples 4.4 and 4.5 to get
εi jk u j (εklm ∂l vm ) = u j ∂i v j − u j ∂ j vi
(4.8)
and then swap u and v and add.
Example 4.6. Use index notation to evaluate ∇ × (∇ × F). (Note the result is the identity given at the
end of section 2.5.)
∇ × (∇ × F)i
so
εi jk ∂ j (εklm ∂l Fm )
= (δil δ jm − δim δ jl )∂ j ∂l Fm
= ∂i ∂m Fm − ∂l ∂l Fi
=
∇ × (∇ × F) = ∇(∇ · F) − ∇2F
using (4.6) for the first equality; the second follows from (∗∗).
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