When Does a Cross Product on R^{N} Exist?

Home , Cross product

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and ) x y e × x R 1 1 1 2 u v y n n + ) 1 x y e ( = , hr “ (here EXIST? 2 2 v 2 ) stereader the As 0). bd edr a not may readers u ( = if il sfamiliar is ticle u x x y e triat A eterminant. ( 0, 6= R 1 e pc can space ree u 3 3 e n h cross the end 3 A y x , (0 = n bilinear and

ywy we ways ny n 1 × × ol note hould sasquare a is y , however, , 2 denotes ” v e x , v 2 rl we eral, ) , y , 0, 6= 3 0 and ) , 3 0 y , , ils. 1) 4 v e ) , 2 PETER F. MCLOUGHLIN natural way to do this on R4 would be to consider the following determinant:

e1 e2 e3 e4

a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34 As the reader can verify, this determinant idea can be easily extended to Rn. Recall that if two rows of a square matrix repeat then the determinant is zero. This implies, for i=1,2 or 3, that:

e1 e2 e3 e4 ai1 ai2 ai3 ai4

a11 a12 a13 a14 a11 a12 a13 a14 (ai1e1 + ai2e2 + ai3e3 + ai4e4) = =0 a21 a22 a23 a24 · a21 a22 a23 a24

a31 a32 a33 a34 a31 a32 a33 a34 It follows our determinant product has the perpendicular property on its row vectors. Note, however, for n > 3 the determinant product acts on more than two vectors which implies it cannot be a candidate for a cross product on Rn. Surprisingly, a cross product can exist in Rn if and only if n is 0, 1, 3 or 7. The intention of this article is to provide a new constructive elementary proof (i.e. could be included in an linear algebra undergraduate text) of this classical result which is accessible to a wide audience. The proof provided in this paper holds for any ﬁnite-dimensional inner product space over R. It has recently been brought to my attention that the approach taken in this paper is similar to the approach taken in an article in The American Mathematical Monthly written back in 1967 (see [13]).

In 1943 Beno Eckmann, using algebraic topology, gave the ﬁrst proof of this result (he actually proved the result under the weaker condition that the cross product is only continuous (see [1] and [11])). The result was later extended to nondegenrate symmetric bilinear forms over ﬁelds of characteristic not equal to two (see [2], [3], [4], [5], and [6]). It turns out that there are intimate relationships between the cross product, quaternions and octonions, and Hurwitz’ theorem (also called the “1,2,4 8 Theorem” named after Adolf Hurwitz, who proved it in 1898). Throughout the years many authors have written on the history of these topics and their intimate relationships to each other (see [5], [6],[7], [8], and [9]).

1. basic idea of proof From this point on, the symbol “ ” will always denote cross product. Moreover, × 3 unless otherwise stated, ui will be understood to mean a unit vector. In R the relations e1 e2 = e3, e2 e3 = e1, and e1 e3 = e2 determine the right-hand- rule cross product.× The crux× of our proof will× be to− show that in general if a cross n product exists on R then we can always ﬁnd an orthonormal basis e1,e2 ...en where for any i = j there exists a k such that ei ej = aek with a =1 or 1. How could we generate6 such a set? Before answering× this question, we will need− some basic properties of cross products. These properties are contained in the following Lemma and Corollary: n WHEN DOES A CROSS PRODUCT ON R EXIST? 3

Lemma 1. Suppose u, v and w are vectors in Rn. If a cross product exists on Rn then it must have the following properties: (1.1) w (u v)= u (w v) (1.2) u · v =× v −u which· × implies u u =0 (1.3) v × (v −u) =× (v u)v (v v)u × (1.4) w× (v× u)= ((· w −v) ·u) + (u v)w + (w v)u 2(w u)v × × − × × · · − ·

For a proof of this Lemma see [12]. Recall that two nonzero vectors u and v are orthogonal if and only if u v = 0. · Corollary 1. Suppose u, v and w are orthogonal unit vectors in Rn. If a cross product exists on Rn then it must have the following properties: (1.5) u (u v)= v (1.6) w× (v × u)=− ((w v) u) × × − × ×

Proof. Follows from previous Lemma by substituting (u v) = (u w) = (w v)=0 and (w w) = (v v) = (u u) = 1. · · · · · · Now that we have our basic properties of cross products, let’s start to answer our question. Observe that e1,e2,e3 = e1,e2,e1 e2 . Let’s see how we could generalize and/or extend this{ idea. } { × }

Recall ui always denotes a unit vector, the symbol “ ” stands for orthogonal, and the symbol “ ” denotes cross product. We will now⊥ recursively define a se- × quence of sets Sk by S := u and Sk := Sk uk (Sk uk) where { } 0 { 0} −1 ∪{ } ∪ −1 × uk Sk . ⊥ −1 Let’s explicitly compute S0, S1, S2, and S3. S0 = u0 ; S1 = S0 u1 (S0 u1)= u0,u1,u0 u1 ; S2 = S1 u2 (S1 u2)= u ,u{,u} u ,u ,u∪{ }∪u ,u × u , (u{ u ) u× ; S} = S ∪{u }∪(S ×u )= { 0 1 0 × 1 2 0 × 2 1 × 2 0 × 1 × 2} 3 2 ∪{ 3} ∪ 2 × 3 u0,u1,u0 u1,u2,u0 u2,u1 u2, (u0 u1) u2,u3,u0 u3,u1 u3, (u0 u1) {u ,u u ×, (u u ) ×u , (u × u ) u×, ((u× u ) u ×) u × × × 3 2 × 3 0 × 2 × 3 1 × 2 × 3 0 × 1 × 2 × 3} Note: S corresponds to e ,e ,e e . 1 { 1 2 1 × 2} We define Si Sj := u v where u Si and v Sj . × { × ∈ ∈ } We also define Si := Si ( Si). ± ∪ − In the following two Lemmas we will show that Sn is an orthonormal set which is closed under the cross product. This in turn implies that a cross can only exist in n R if n =0 or n = Sk . | | Lemma 2. S = u ,u ,u u is an orthonormal set. Moreover, S S = S . 1 { 0 1 1 × 0} 1× 1 ± 1 Proof. By definition of cross product we have: (1) u0 (u0 u1)=0 (2) u · (u × u )=0 1 · 0 × 1 It follows S1 is an orthogonal set. Next by (1.1), (1.2), and (1.3) we have: (1) u1 (u0 u1)= u0 (2) u × (u × u )= u 0 × 0 × 1 − 1 4 PETER F. MCLOUGHLIN

(3) (u0 u1) (u0 u1)= u0 (u1 (u0 u1)) = u0 u0 =1 It follows× S is· an× orthonormal· set× and S× S = S· . 1 1 × 1 ± 1

k+1 Lemma 3. Sk is an orthonormal set. Moreover, Sk Sk = Sk and Sk =2 1 . × ± | | −

Proof. We proceed by induction. When k = 1 claim follows from previous Lemma. k Suppose Sk is an orthonormal set, Sk Sk = Sk , and Sk = 2 1. −1 −1 × −1 ± −1 | −1| − Let y1,y2 Sk−1. By deﬁnition any element of Sk is of the form y1, uk, or y1 uk. By (1.1), (1.2),∈ and (1.6) we have: × (1) y (y uk)= uk (y y ) = 0 since uk Sk and y y Sk . 1 · 2 × · 1 × 2 ⊥ −1 1 × 2 ∈ −1 (2) (y uk) (y uk)= uk ((uk y ) y )= uk (uk (y y ))=0 1 × · 2 × · × 1 × 2 − · × 1 × 2 It follows Sk is an orthogonal set. Next by (1.1), (1.2), (1.5), and (1.6) we have: (1) (y uk) (y uk)= y (uk (y uk)) = y y 1 × × 2 × − 1 × × 2 × − 1 × 2 (2) y (y uk)= ((y y ) uk) and y (y uk)= uk 1 × 2 × − 1 × 2 × 1 × 1 × − (3) (y uk) (y uk)= y (uk (y uk)) = y y =1 1 × · 1 × 1 · × 1 × 1 · 1 It follows Sk is an orthonormal set and Sk Sk = Sk. Lastly, note Sk = k+1 × ± | | 2 Sk +1=2 1 | −1| −

Lemma 3 tells us how to construct a multiplication table for the cross product 3 on Sk. Let’s take a look at the multiplication table in R . In order to simplify the notation for the basis elements we will make the following assignments: e1 := u0, e := u , e := u u 2 1 3 0 × 1 e e e × 1 2 3 e 0 e e 1 3 − 2 e e 0 e 2 − 3 1 e e e 0 3 2 − 1 In the table above we used (1.2), (1.5) and (1.6) to compute each product. In particular, e1 e3 = u0 (u0 u1) = u1 = e2. The other products are computed in a similar× way. × × − − Now when v = x e + x e + x e and w = y e + y e + y e we have v w = 1 1 2 2 3 3 1 1 2 2 3 3 × (x1e1 +x2e2 +x3e3) (y1e1 +y2e2 +y3e3)= x1y1(e1 e1)+x1y2(e1 e2)+x1y3(e1 e )+ x y (e e )+×x y (e e )+ x y (e e )+×x y (e e )+×x y (e e )+× 3 2 1 2 × 1 2 2 2 × 2 2 3 2 × 3 3 1 3 × 1 3 2 3 × 2 x3y3(e3 e3). By the multiplication table above we can simplify this expression. Hence, × v w = (x2y3 x3y2)e1 + (x3y1 x1y3)e2 + (x1y2 x2y1)e3 The× reader should− note that this cross− product is the− standard right-hand-rule cross product. Let’s next take a look at the multiplication table in R7. In order to simplify the notation for the basis elements we will make the following assignments: e1 := u0, e := u , e := u u , e := u , e := u u , e := u u , e := (u u ) u . 2 1 3 0 × 1 4 2 5 0 × 2 6 1 × 2 7 0 × 1 × 2 n WHEN DOES A CROSS PRODUCT ON R EXIST? 5

e e e e e e e × 1 2 3 4 5 6 7 e 0 e e e e e e 1 3 − 2 5 − 4 − 7 6 e e 0 e e e e e 2 − 3 1 6 7 − 4 − 5 e e e 0 e e e e 3 2 − 1 7 − 6 5 − 4 e e e e 0 e e e 4 − 5 − 6 − 7 1 2 3 e e e e e 0 e e 5 4 − 7 6 − 1 − 3 2 e e e e e e 0 e 6 7 4 − 5 − 2 3 − 1 e e e e e e e 0 7 − 6 5 4 − 3 − 2 1 In the table above we used (1.2), (1.5) and (1.6) to compute each product. In particular, e e = (u u ) (u u )= (u u ) (u u ) = ((u u ) u ) u = 5 × 6 0 × 2 × 1 × 2 − 0 × 2 × 2 × 1 0 × 2 × 2 × 1 (u2 (u0 u2)) u1 = u0 u1 = e3. The other products are computed in a similar− × way.× × − × −

Let’s look at v w when v = (x1, x2, x3, x4, x5, x6, x7) and w = (y1,y2,y3,y4,y5,y6,y7). Using the bilinear× property of the cross product and the multiplication table for R7 we have: v w = ( x3y2 +x2y3 x5y4 +x4y5 x6y7 +x7y6)e1 +( x1y3 +x3y1 x6y4 +x4y6 x×y + x −y )e + ( x −y + x y x−y + x y x y +−x y )e + ( −x y + x y − 7 5 5 7 2 − 2 1 1 2 − 7 4 4 7 − 5 6 6 5 3 − 1 5 5 1 − x2y6 +x6y2 x3y7 +x7y3)e4 +( x4y1 +x1y4 x2y7 +x7y2 x6y3+x3y6)e5 +( x7y1 + x y x y −+ x y x y + x −y )e + ( x −y + x y x−y + x y x y +−x y )e 1 7 − 4 2 2 4 − 3 5 5 3 6 − 5 2 2 5 − 4 3 3 4 − 1 6 6 1 7 Lemma 4. If u = (u0 u1) + (u1 u3) and v = (u1 u2) (((u0 u1) u2) u3) then u v =0 and u ×v. × × − × × × × ⊥ Proof. Using (1.2), (1.5), (1.6), and the bilinear property it can be shown that u v = 0. Next, note that u u , u u , u u , and ((u u ) u ) u are × 0 × 1 1 × 3 1 × 2 0 × 1 × 2 × 3 all elements of Si when i 3. By Lemma 3 all these vectors form an orthonormal set. This implies u v. ≥ ⊥

Lemma 5. If u = (u0 u1) + (u1 u3) and v = (u1 u2) (((u0 u1) u2) u3) then (u u)(v v) = (u × v) (u v×) + (u v)2 × − × × × · · 6 × · × ·

Proof. Using the previous Lemma and Lemma 3, we have (u v)=0,(u u) = 2, (v v)=2, and (u v) = 0. Hence, claim follows. × · · · Theorem 1. A cross product can exist in Rn if and only if n=0, 1, 3 or 7. More- 3 7 over, there exists orthonormal bases S1 and S2 for R and R respectively such that Si Si = Si for i =1 or 2. × ± Proof. By Lemma 3 we have that a cross product can only exist in Rn if n = 2k+1 1. Moreover, Lemmas 4 and 5 tells us that if we try to define a cross − k+1 product in R2 −1 then the Pythagorean property does not always hold when k 3. It follows R0, R1, R3, and R7 are the only spaces on which a cross product can≥ exist. It is left as an exercise to show that the cross product that we defined above for R7 satisfies all the properties of a cross product and the zero map defines a valid cross product on R0, and R1. Lastly, Lemma 3 tells us how to generate 3 7 orthonormal bases S1 and S2 for R and R respectively such that Si Si = Si for i = 1 or 2. × ± 6 PETER F. MCLOUGHLIN

acknowledgments

I would like to thank Dr. Daniel Shapiro, Dr. Alberto Elduque, and Dr. John Baez for their kindness and encouragement and for reading over and commenting on various drafts of this paper. I would also like to thank the editor and all the anonymous referees for reading over and commenting on various versions of this paper. This in turn has lead to overall improvements in the mathematical integrity, readability, and presentation of the paper. References

[1] B. Eckmann, “Stetige Losungen linearer Gleichungssysteme”, Comment. Math. Helv. 15 (1943), 318-339. [2] R.B. Brown and A. Gray, “Vector cross products”, Comment. Math. Helv. 42 (1967), 222-236. [3] M. Rost, “On the dimension of a composition algebra”, Doc. Math. 1 (1996), No. 10, 209-214 (electronic). [4] K. Meyberg, “Trace formulas in vector product algebras”, Commun. Algebra 30 (2002), no. 6, 2933-2940. [5] Alberto Elduque, “Vector Cross Products”, Talk presented at the Seminario Rubio de Francia of the Universidad de Zaragoza on April 1, 2004. Electronic copy found at: http://www.unizar.es/matematicas/algebra/elduque/Talks/crossproducts.pdf [6] W.S. Massey, “Cross Products of Vectors in Higher Dimensional Euclidean Spaces”, The American Mathematical Monthly (1983), Vol. 90, No. 10, 697-701. [7] John, Baez, “The Octonions”, Bull. Amer. Math. Soc. 39 (02): 145205. Electronic copy found at: http://www.ams.org/journals/bull/2002-39-02/S0273-0979-01-00934-X/S0273-0979-01-00934-X.pdf [8] D. B. Shapiro, Compositions of Quadratic Forms, W. de Gruyter Verlag, 2000. [9] Bruce W. Westbury, “Hurwitz’ Theorem”, 2009. Electronic copy found at: http://arxiv.org/abs/1011.6197 [10] A. Hurwitz, (1898). “Ueber die Composition der quadratischen Formen von beliebig vielen Variabeln (On the composition of quadratic forms of arbitrary many variables)” (in German). Nachr. Ges. Wiss. Gttingen: 309-316. [11] Beno Eckmann, “Continuous solutions of linear equations An old problem, its history, and its solution”, Exposition Math, 9 (1991) pp. [12] Peter F. Mcloughlin, “Basic Properties of Cross Products”, Electronic copy found at: http://www.math.csusb.edu/faculty/pmclough/BP.pdf [13] Bertram Walsh, “The Scarcity of Cross Products on Euclidean Spaces”, The American Math- ematical Monthly (Feb. 1967), Vol. 74, No. 2, 188-194. E-mail address: [email protected]