# WHEN DOES A CROSS PRODUCT ON R n EXIST?

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2 2 PETER F. MCLOUGHLIN natural way to do this on R 4 would be to consider the following determinant: e 1 e 2 e 3 e 4 a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 As the reader can verify, this determinant idea can be easily extended to R n. Recall that if two rows of a square matrix repeat then the determinant is zero. This implies, for i=1,2 or 3, that: e 1 e 2 e 3 e 4 a i1 a i2 a i3 a i4 a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 (a i1 e 1 + a i2 e 2 + a i3 e 3 + a i4 e 4 ) = a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 = 0 a 31 a 32 a 33 a 34 a 31 a 32 a 33 a 34 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 R n. Surprisingly, a cross product can exist in R n 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, unless otherwise stated, u i will be understood to mean a unit vector. In R 3 the relations e 1 e 2 = e 3, e 2 e 3 = e 1, and e 1 e 3 = e 2 determine the right-handrule cross product. The crux of our proof will be to show that in general if a cross product exists on R n then we can always find an orthonormal basis e 1, e 2... e n where for any i j there exists a k such that e i e j = ae k with a = 1 or 1. How could we generate 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:

3 WHEN DOES A CROSS PRODUCT ON R n EXIST? 3 Lemma 1. Suppose u, v and w are vectors in R n. If a cross product exists on R n 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 R n. If a cross product exists on R n 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 {e 1, e 2, e 3 } = {e 1, e 2, e 1 e 2 }. Let s see how we could generalize and/or extend this idea. Recall u i always denotes a unit vector, the symbol stands for orthogonal, and the symbol denotes cross product. We will now recursively define a sequence of sets {S k } by S 0 := {u 0 } and S k := S k 1 {u k } (S k 1 u k ) where u k S k 1. Let s explicitly compute S 0, S 1, S 2, and S 3. S 0 = {u 0 }; S 1 = S 0 {u 1 } (S 0 u 1 ) = {u 0, u 1, u 0 u 1 }; S 2 = S 1 {u 2 } (S 1 u 2 ) = {u 0, u 1, u 0 u 1, u 2, u 0 u 2, u 1 u 2, (u 0 u 1 ) u 2 } ; S 3 = S 2 {u 3 } (S 2 u 3 ) = {u 0, u 1, u 0 u 1, u 2, u 0 u 2, u 1 u 2, (u 0 u 1 ) u 2, u 3, u 0 u 3,u 1 u 3, (u 0 u 1 ) u 3, u 2 u 3, (u 0 u 2 ) u 3, (u 1 u 2 ) u 3, ((u 0 u 1 ) u 2 ) u 3 } Note: S 1 corresponds to {e 1, e 2, e 1 e 2 }. We define S i S j := {u v where u S i and v S j }. We also define ±S i := S i ( S i ). In the following two Lemmas we will show that S n is an orthonormal set which is closed under the cross product. This in turn implies that a cross can only exist in R n if n = 0 or n = S k. Lemma 2. S 1 = {u 0, u 1, u 1 u 0 } is an orthonormal set. Moreover, S 1 S 1 = ±S 1. Proof. By definition of cross product we have: (1) u 0 (u 0 u 1 ) = 0 (2) u 1 (u 0 u 1 ) = 0 It follows S 1 is an orthogonal set. Next by (1.1), (1.2), and (1.3) we have: (1) u 1 (u 0 u 1 ) = u 0 (2) u 0 (u 0 u 1 ) = u 1

4 4 PETER F. MCLOUGHLIN (3) (u 0 u 1 ) (u 0 u 1 ) = u 0 (u 1 (u 0 u 1 )) = u 0 u 0 = 1 It follows S 1 is an orthonormal set and S 1 S 1 = ±S 1. Lemma 3. S k is an orthonormal set. Moreover, S k S k = ±S k and S k = 2 k+1 1. Proof. We proceed by induction. When k = 1 claim follows from previous Lemma. Suppose S k 1 is an orthonormal set, S k 1 S k 1 = ±S k 1, and S k 1 = 2 k 1. Let y 1, y 2 S k 1. By definition any element of S k is of the form y 1, u k, or y 1 u k. By (1.1), (1.2), and (1.6) we have: (1) y 1 (y 2 u k ) = u k (y 1 y 2 ) = 0 since u k S k 1 and y 1 y 2 S k 1. (2) (y 1 u k ) (y 2 u k ) = u k ((u k y 1 ) y 2 ) = u k (u k (y 1 y 2 )) = 0 It follows S k is an orthogonal set. Next by (1.1), (1.2), (1.5), and (1.6) we have: (1) (y 1 u k ) (y 2 u k ) = y 1 (u k (y 2 u k )) = y 1 y 2 (2) y 1 (y 2 u k ) = ((y 1 y 2 ) u k ) and y 1 (y 1 u k ) = u k (3) (y 1 u k ) (y 1 u k ) = y 1 (u k (y 1 u k )) = y 1 y 1 = 1 It follows S k is an orthonormal set and S k S k = ±S k. Lastly, note S k = 2 S k = 2 k+1 1 Lemma 3 tells us how to construct a multiplication table for the cross product on S k. Let s take a look at the multiplication table in R 3. In order to simplify the notation for the basis elements we will make the following assignments: e 1 := u 0, e 2 := u 1, e 3 := u 0 u 1 e 1 e 2 e 3 e 1 0 e 3 e 2 e 2 e 3 0 e 1 e 3 e 2 e 1 0 In the table above we used (1.2), (1.5) and (1.6) to compute each product. In particular, e 1 e 3 = u 0 (u 0 u 1 ) = u 1 = e 2. The other products are computed in a similar way. Now when v = x 1 e 1 + x 2 e 2 + x 3 e 3 and w = y 1 e 1 + y 2 e 2 + y 3 e 3 we have v w = (x 1 e 1 +x 2 e 2 +x 3 e 3 ) (y 1 e 1 +y 2 e 2 +y 3 e 3 ) = x 1 y 1 (e 1 e 1 )+x 1 y 2 (e 1 e 2 )+x 1 y 3 (e 1 e 3 ) + x 2 y 1 (e 2 e 1 ) + x 2 y 2 (e 2 e 2 ) + x 2 y 3 (e 2 e 3 ) + x 3 y 1 (e 3 e 1 ) + x 3 y 2 (e 3 e 2 ) + x 3 y 3 (e 3 e 3 ). By the multiplication table above we can simplify this expression. Hence, v w = (x 2 y 3 x 3 y 2 )e 1 + (x 3 y 1 x 1 y 3 )e 2 + (x 1 y 2 x 2 y 1 )e 3 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 R 7. In order to simplify the notation for the basis elements we will make the following assignments: e 1 := u 0, e 2 := u 1, e 3 := u 0 u 1, e 4 := u 2, e 5 := u 0 u 2, e 6 := u 1 u 2, e 7 := (u 0 u 1 ) u 2.

5 WHEN DOES A CROSS PRODUCT ON R n EXIST? 5 e 1 e 2 e 3 e 4 e 5 e 6 e 7 e 1 0 e 3 e 2 e 5 e 4 e 7 e 6 e 2 e 3 0 e 1 e 6 e 7 e 4 e 5 e 3 e 2 e 1 0 e 7 e 6 e 5 e 4 e 4 e 5 e 6 e 7 0 e 1 e 2 e 3 e 5 e 4 e 7 e 6 e 1 0 e 3 e 2 e 6 e 7 e 4 e 5 e 2 e 3 0 e 1 e 7 e 6 e 5 e 4 e 3 e 2 e 1 0 In the table above we used (1.2), (1.5) and (1.6) to compute each product. In particular, e 5 e 6 = (u 0 u 2 ) (u 1 u 2 ) = (u 0 u 2 ) (u 2 u 1 ) = ((u 0 u 2 ) u 2 ) u 1 = (u 2 (u 0 u 2 )) u 1 = u 0 u 1 = e 3. The other products are computed in a similar way. Let s look at v w when v = (x 1, x 2, x 3, x 4, x 5, x 6, x 7 ) and w = (y 1, y 2, y 3, y 4, y 5, y 6, y 7 ). Using the bilinear property of the cross product and the multiplication table for R 7 we have: v w = ( x 3 y 2 +x 2 y 3 x 5 y 4 +x 4 y 5 x 6 y 7 +x 7 y 6 )e 1 +( x 1 y 3 +x 3 y 1 x 6 y 4 +x 4 y 6 x 7 y 5 + x 5 y 7 )e 2 + ( x 2 y 1 + x 1 y 2 x 7 y 4 + x 4 y 7 x 5 y 6 + x 6 y 5 )e 3 + ( x 1 y 5 + x 5 y 1 x 2 y 6 +x 6 y 2 x 3 y 7 +x 7 y 3 )e 4 +( x 4 y 1 +x 1 y 4 x 2 y 7 +x 7 y 2 x 6 y 3 +x 3 y 6 )e 5 +( x 7 y 1 + x 1 y 7 x 4 y 2 + x 2 y 4 x 3 y 5 + x 5 y 3 )e 6 + ( x 5 y 2 + x 2 y 5 x 4 y 3 + x 3 y 4 x 1 y 6 + x 6 y 1 )e 7 Lemma 4. If u = (u 0 u 1 ) + (u 1 u 3 ) and v = (u 1 u 2 ) (((u 0 u 1 ) u 2 ) u 3 ) 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 0 u 1, u 1 u 3, u 1 u 2, and ((u 0 u 1 ) u 2 ) u 3 are all elements of S i when i 3. By Lemma 3 all these vectors form an orthonormal set. This implies u v. Lemma 5. If u = (u 0 u 1 ) + (u 1 u 3 ) and v = (u 1 u 2 ) (((u 0 u 1 ) u 2 ) u 3 ) then (u u)(v v) (u v) (u v) + (u v) 2 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 R n if and only if n=0, 1, 3 or 7. Moreover, there exists orthonormal bases S 1 and S 2 for R 3 and R 7 respectively such that S i S i = ±S i for i = 1 or 2. Proof. By Lemma 3 we have that a cross product can only exist in R n if n = 2 k+1 1. Moreover, Lemmas 4 and 5 tells us that if we try to define a cross product in R 2k+1 1 then the Pythagorean property does not always hold when k 3. It follows R 0, R 1, R 3, and R 7 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 R 7 satisfies all the properties of a cross product and the zero map defines a valid cross product on R 0, and R 1. Lastly, Lemma 3 tells us how to generate orthonormal bases S 1 and S 2 for R 3 and R 7 respectively such that S i S i = ±S i for i = 1 or 2.

6 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), [2] R.B. Brown and A. Gray, Vector cross products, Comment. Math. Helv. 42 (1967), [3] M. Rost, On the dimension of a composition algebra, Doc. Math. 1 (1996), No. 10, (electronic). [4] K. Meyberg, Trace formulas in vector product algebras, Commun. Algebra 30 (2002), no. 6, [5] Alberto Elduque, Vector Cross Products, Talk presented at the Seminario Rubio de Francia of the Universidad de Zaragoza on April 1, Electronic copy found at: [6] W.S. Massey, Cross Products of Vectors in Higher Dimensional Euclidean Spaces, The American Mathematical Monthly (1983), Vol. 90, No. 10, [7] John, Baez, The Octonions, Bull. Amer. Math. Soc. 39 (02): Electronic copy found at: [8] D. B. Shapiro, Compositions of Quadratic Forms, W. de Gruyter Verlag, [9] Bruce W. Westbury, Hurwitz Theorem, Electronic copy found at: [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: [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: [13] Bertram Walsh, The Scarcity of Cross Products on Euclidean Spaces, The American Mathematical Monthly (Feb. 1967), Vol. 74, No. 2, address:

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