Kenneth Kuttler

A.4. VECTOR IDENTITIES AND NOTATION 335

The way to think of ε i jk is that ε123 = 1 and if you switch any two of the numbers in thelist i, j,k, it changes the sign. Thus ε i jk =−ε jik and ε i jk =−εk ji etc. You should check thatthis rule reduces to the above definition. For example, it immediately implies that if thereis a repeated index, the answer is zero. This follows because ε ii j =−ε ii j and so ε ii j = 0.

It is useful to use the Einstein summation convention when dealing with these symbols.Simply stated, the convention is that you sum over the repeated index. Thus aibi means∑i aibi. Also, δ i jx j means ∑ j δ i jx j = xi. When you use this convention, there is one veryimportant thing to never forget. It is this: Never have an index be repeated more than once.Thus aibi is all right but aiibi is not. The reason for this is that you end up getting confusedabout what is meant. If you want to write ∑i aibici it is best to simply use the summationnotation. There is a very important reduction identity connecting these two symbols.

Lemma A.4.3 The following holds.

ε i jkε irs = (δ jrδ ks−δ krδ js) .

Proof: If { j,k} ̸= {r,s} then every term in the sum on the left must have either ε i jkor ε irs contains a repeated index. Therefore, the left side equals zero. The right side alsoequals zero in this case. To see this, note that if the two sets of indices are not equal, thenthere is one of the indices in one of the sets which is not in the other set. For example, itcould be that j is not equal to either r or s. Then the right side equals zero.

Therefore, it can be assumed { j,k} = {r,s} . If i = r and j = s for s ̸= r, then thereis exactly one term in the sum on the left and it equals 1. The right also reduces to 1in this case. If i = s and j = r, there is exactly one term in the sum on the left which isnonzero and it must equal −1. The right side also reduces to −1 in this case. If there isa repeated index in { j,k} , then every term in the sum on the left equals zero. The rightalso reduces to zero in this case because then j = k = r = s and so the right side becomes(1)(1)− (−1)(−1) = 0. ■

Proposition A.4.4 Let u,v be vectors in Rp where the Cartesian coordinates of u are(u1, · · · ,up) and the Cartesian coordinates of v are (v1, · · · ,vp). Then u ·v = uivi. If u,vare vectors in R3, then

(u×v)i = ε i jku jvk.

Also, δ ikak = ai.

Proof: The first claim is obvious from the definition of the dot product. The second isverified by simply checking it works. For example,

u×v ≡

∣∣∣∣∣∣i j k

u1 u2 u3v1 v2 v3

∣∣∣∣∣∣and so

(u×v)1 = (u2v3−u3v2) .

From the above formula in the proposition,

ε1 jku jvk ≡ u2v3−u3v2,

the same thing. The cases for (u×v)2 and (u×v)3 are verified similarly. The last claimfollows directly from the definition. ■

With this notation, you can easily discover vector identities and simplify expressionswhich involve the cross product.

A.4.. VECTOR IDENTITIES AND NOTATION 335The way to think of €;;, is that €}23 = 1 and if you switch any two of the numbers in thelist i, j,k, it changes the sign. Thus €;;, = —€ jiz and €;;, = —€,;; etc. You should check thatthis rule reduces to the above definition. For example, it immediately implies that if thereis a repeated index, the answer is zero. This follows because &€; = —€j;; and so €;;; = 0.It is useful to use the Einstein summation convention when dealing with these symbols.Simply stated, the convention is that you sum over the repeated index. Thus aj;b; meansy, aib;. Also, 6; jx; Means Li 6; ‘jx; = x;. When you use this convention, there is one veryimportant thing to never forget. It is this: Never have an index be repeated more than once.Thus a;J; is all right but a;;b; is not. The reason for this is that you end up getting confusedabout what is meant. If you want to write )'; ajb;c; it is best to simply use the summationnotation. There is a very important reduction identity connecting these two symbols.Lemma A.4.3 The following holds.Ej jKEirs = (5 jr OKs _ dK js) :Proof: If {j,k} 4 {r,s} then every term in the sum on the left must have either €; jxor €jr; contains a repeated index. Therefore, the left side equals zero. The right side alsoequals zero in this case. To see this, note that if the two sets of indices are not equal, thenthere is one of the indices in one of the sets which is not in the other set. For example, itcould be that j is not equal to either r or s. Then the right side equals zero.Therefore, it can be assumed {j,k} = {r,s}. If i =r and j = s for s 4+, then thereis exactly one term in the sum on the left and it equals 1. The right also reduces to |in this case. If i= s and j =7, there is exactly one term in the sum on the left which isnonzero and it must equal —1. The right side also reduces to —1 in this case. If there isa repeated index in {j,k}, then every term in the sum on the left equals zero. The rightalso reduces to zero in this case because then j = k = r=s and so the right side becomes()()-(-1)(-1) =0.Proposition A.4.4 Let u,v be vectors in R? where the Cartesian coordinates of u are(u1,**+ ,Up) and the Cartesian coordinates of v are (v\,-++ ,Vp). Then u-v = ujvj. If u,vare vectors in R3, then(u x v); = Ej jKUjVK-Also, 6 jxax = Qj-Proof: The first claim is obvious from the definition of the dot product. The second isverified by simply checking it works. For example,i jkUXV=|Uu WwW WYVi V2 V3and so(u x v); = (u2Vv3 —_ U3V2) .From the above formula in the proposition,E 1 jKUjVk = U2V3 — U3ZV2,the same thing. The cases for (uw x uv), and (wu x uv), are verified similarly. The last claimfollows directly from the definition.With this notation, you can easily discover vector identities and simplify expressionswhich involve the cross product.