Composition by Hurwitz

The composition set of Hurwitz or the composition set of quadratic forms of Adolf Hurwitz states in the mathematics that for n = 1, 2, 4 or 8, the product of two sums of n squared real numbers into a sum of also n may decompose squares of numbers , which are bilinear forms of the former.

With this statement the squares theorem can be proven.

From Hurwitz's theorem it follows that there are only for composition algebras , the real numbers, the complex numbers , the quaternions and octonions . He found a new characterization of the special position of these algebras . Composition algebras are algebras over the real numbers with Euclidean norm (whereby the algebra is understood as a vector space over the real numbers) for which applies. In 1943, Beno Eckmann gave a group-theoretical proof of Hurwitz's composition theorem. ${\displaystyle n=1,2,4,8}$ ${\displaystyle |x|\cdot |y|=|x\cdot y|}$

statement

Only for n = 1, 2, 4 or 8 can it be bilinear forms

${\displaystyle z_{i}=\sum _{j,k=1}^{n}\gamma _{ijk}x_{j}y_{k},\quad \gamma _{ijk}\in \mathbb {R} }$

for i = 1, ..., n , so that for all real numbers x ₁ , ..., x _n , y ₁ , ..., y _{n the} following applies:

${\displaystyle \left(\sum _{j=1}^{n}x_{j}^{2}\right)\left(\sum _{k=1}^{n}y_{k}^{2}\right)=\sum _{i=1}^{n}z_{i}^{2}}$

Evidence sketch

The case n = 1 satisfies the statement because a² b² = (ab) ² . For n > 1, the equations occurring in the sentence are converted into matrix equations . Examining the matrices first shows that n must be even. Furthermore, 2 ^n-2 linearly independent n × n matrices can be found, of which there are only n ². Therefore n ≤ 8. The case n = 6 can be excluded, because then there would have to be 16 linearly independent skew-symmetric 6 × 6 matrices, which cannot be. So n = 1, 2, 4 or 8.

proof
Conversion into a matrix equation
The numbers x i , y i and z i are entered in column vectors of dimension n . Then the bilinear forms can be written as a matrix equation where G is an n × n matrix whose coefficients are homogeneous linear functions of the vector . Then ${\displaystyle {\vec {x}},{\vec {y}},{\vec {z}}}$ ${\displaystyle {\vec {z}}=G{\vec {y}}}$ ${\displaystyle G_{ik}=\sum _{j=1}^{n}\gamma _{ijk}x_{j}}$ ${\displaystyle {\vec {x}}}$ ${\displaystyle {\begin{aligned}z_{1}^{2}+z_{2}^{2}+\dots +z_{n}^{2}={\vec {z}}^{\top }{\vec {z}}=&(G{\vec {y}})^{\top }(G{\vec {y}})={\vec {y}}^{\top }G^{\top }G{\vec {y}}\\{\stackrel {\displaystyle !}{=}}&(x_{1}^{2}+x_{2}^{2}+\dots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\dots +y_{n}^{2})\end{aligned}}}$ The superscript T denotes the transposed matrix . Because there are no mixed terms y i y k with i ≠ k, G T G is a diagonal matrix , and because all prefactors of are identical, G T G is even a multiple of the identity matrix E n . ${\displaystyle y_{i}^{2}}$
Construction of the skew-symmetric n × n matrices B i
The G matrix is ​​divided into matrices A j with the constant coefficients ( A j ) ik = γ ijk . Then we get: ${\displaystyle G=\sum _{j=1}^{n}x_{j}A_{j}}$ ${\displaystyle G^{\top }G=\left(\sum _{i=1}^{n}x_{i}A_{i}\right)^{\top }\left(\sum _{j=1}^{n}x_{j}A_{j}\right)=\sum _{i,j=1}^{n}x_{i}x_{j}A_{i}^{\top }A_{j}{\stackrel {\displaystyle !}{=}}({\vec {x}}^{\top }{\vec {x}})E_{n}}$ Comparison of the two terms with shows . Multiplication of the above equation from the left with and from the right with gives: ${\displaystyle x_{n}^{2}}$ ${\displaystyle A_{n}^{\top }A_{n}=E_{n}=A_{n}A_{n}^{\top }}$ ${\displaystyle A_{n}}$ ${\displaystyle A_{n}^{\top }}$ ${\displaystyle \sum _{i,j=1}^{n}x_{i}x_{j}A_{n}A_{i}^{\top }A_{j}A_{n}^{\top }=({\vec {x}}^{\top }{\vec {x}})A_{n}E_{n}A_{n}^{\top }\quad \rightarrow \quad \sum _{i,j=1}^{n}x_{i}x_{j}B_{i}^{\top }B_{j}=({\vec {x}}^{\top }{\vec {x}})E_{n}}$ with the matrices and B n = E n . The comparison of coefficients in the last equation shows for i, j <n : ${\displaystyle B_{i}:=A_{i}A_{n}^{\top }}$ ${\displaystyle {\begin{aligned}B_{i}^{\top }B_{n}+B_{n}^{\top }B_{i}=&B_{i}^{\top }+B_{i}=O_{n}\\i=j\quad \rightarrow \quad &B_{i}^{\top }B_{i}=E_{n}\\i\neq j\quad \rightarrow \quad &B_{i}^{\top }B_{j}+B_{j}^{\top }B_{i}=O_{n}\end{aligned}}}$ with the zero matrix O n . That is synonymous with ${\displaystyle B_{i}^{\top }=-B_{i}\,,\quad B_{i}B_{i}=-E_{n}\,,\quad B_{i}B_{j}=-(B_{i}B_{j})^{\top }=-B_{j}B_{i}}$     (*)   for and . Well allowed ${\displaystyle i,j\in \{1,2,\dots ,n-1\}}$ ${\displaystyle i\neq j}$ ${\displaystyle \operatorname {det} (B_{i})^{2}=\operatorname {det} (B_{i}B_{i})=\operatorname {det} (-E_{n})=(-1)^{n}}$ to restrict the analysis to even n based on the determinant product theorem.
Construction of 2 n-1 transformations
A combination of one to n - 1 matrices (*) and the identity matrix results in transformations ${\displaystyle E_{n},\quad B_{i_{1}},\quad B_{i_{1}}B_{i_{2}},\quad B_{i_{1}}B_{i_{2}}B_{i_{3}},\quad \dots ,\quad B_{1}B_{2}\dots B_{n-1}}$     (Δ)   in which the natural indices in all products should be increasing according to 0 < i 1 < i 2 <... < n . The transformations (Δ) are referred to below as Δ transformations. Each of the Δ transformations except E n is created by “pulling” k matrices ( k = 1,…, n-1) from the set { B 1 ,…, B n-1 } without replacing, sorting the drawn matrices according to ascending order Indexes and formation of the matrix product. The binomial coefficient gives the number of variants for a given k and the number of all Δ transformations including E n is 2 n-1 because of the sum formula of the binomial coefficients . ${\displaystyle \textstyle {\binom {n-1}{k}}}$
Symmetry of transformations
The Δ transformations can be checked for symmetry : ${\displaystyle (B_{i_{1}}B_{i_{2}}\dots B_{i_{r}})^{\top }=B_{i_{r}}^{\top }\dots B_{i_{2}}^{\top }B_{i_{1}}^{\top }=(-1)^{r}B_{i_{r}}\dots B_{i_{2}}B_{i_{1}}}$ because of B i T = -B i (*). In order to sort the factors in the product according to ascending indices, the last factor ( r - 1) has to be swapped with its predecessor ( r - 1) times, whereby the product according to B i B j = -B j B i (*) has the sign ( r - 1) changes. The now last factor must be swapped ( r - 2) times with its predecessor, the product ( r - 2) times changing the sign, etc. Overall, the product has the sign after sorting ${\displaystyle (-1)^{r}(-1)^{(r-1)}(-1)^{(r-2)}\dots (-1)^{1}=(-1)^{r+(r-1)+(r-2)+\dots +1}=(-1)^{\frac {r(r+1)}{2}}}$ Symmetry exists with an even exponent, i.e. with r (r + 1) divisible by four , which is why r then has to leave the remainder zero or three when dividing by four. In the other cases, remainder one or two when dividing r by four, the transformation is skew symmetric.
Linear combinations of the transformations
Let R , R 1 , R 2 ... be linear combinations of the Δ transformations. Then a relation R = O n shows a linear dependency of the Δ-transformations which are entered with a coefficient not equal to zero. This love than on the relation involved hot. If R 1 = O n and R 2 = O n are two relations, then they should be called foreign to one another if there is no Δ-transformation that is involved in both relations. A reducible relation is one that can be represented as the sum of two alien relations according to R = R 1 + R 2 with R 1 = R 2 = O n . The irreducible are not reducible. In the following only the irreducible relations are considered, because the reducible ones are made up of them.
Linear independent transformations
A linear combination of the Δ transformations containing symmetric and skew-symmetric transformations is reducible. Because because the zero matrix is ​​the only both symmetrical and skew-symmetrical matrix, both the symmetrical and skew-symmetrical transformations add to the zero matrix. Conversely, every irreducible relation contains only symmetrical or only skew-symmetrical transformations. An irreducible relation remains irreducible if it is multiplied by a Δ transformation. By means of such a multiplication, for example with the first summand involved, it can be achieved that E n is also involved. Be ${\displaystyle E_{n}=\sum c_{i}B_{i}+\sum c_{ij}B_{i}B_{j}+\sum c_{i_{1}i_{2}i_{3}}B_{i_{1}}B_{i_{2}}B_{i_{3}}+\sum c_{i_{1}i_{2}i_{3}i_{4}}B_{i_{1}}B_{i_{2}}B_{i_{3}}B_{i_{4}}+\dots }$ equivalent to an irreducible relation. On the left side there is a symmetrical matrix and therefore all the summands involved are symmetrical on the right side. The number of factor matrices in a summand must therefore leave the remainder zero or three when divided by four. Thus all c i and c ij vanish . Multiplying both sides by any B i gives: ${\displaystyle B_{i}=\sum c_{i_{1}i_{2}i_{3}}B_{i_{1}}B_{i_{2}}B_{i_{3}}B_{i}+\sum c_{i_{1}i_{2}i_{3}i_{4}}B_{i_{1}}B_{i_{2}}B_{i_{3}}B_{i_{4}}B_{i}+\dots }$ where there is now a skew-symmetric matrix on the left as well as on the right-hand side and all summands are skew-symmetric due to irreducibility. The number of factor matrices in a summand must therefore leave the remainder one or two when divided by four. It must therefore be if i is not among the indices i 1 , i 2 or i 3 . Because i can be chosen arbitrarily, all vanish at n > 4. Analogously it can be shown that all coefficients vanish in the sums and only ${\displaystyle c_{i_{1}i_{2}i_{3}}=0}$ ${\displaystyle c_{i_{1}i_{2}i_{3}}}$ ${\displaystyle E_{n}=cB_{1}B_{2}\dots B_{n-1}}$     (□)   remains - even if n ≤ 4. In addition, it turns out that there is a symmetrical matrix on the right-hand side, which is why n must be divisible by four so that a relation (□) can exist.
The case of the six bilinear forms
Because six is ​​not divisible by four, the relation (□) is excluded and all Δ transformations should be linearly independent. Of these, the five are with one factor matrix, the ten with two factor matrices and the one with five factor matrices, so a total of 16 are skew symmetrical. Of these 6 × 6 matrices, however, only a maximum of 5 + 4 + 3 + 2 + 1 = 15 can be linearly independent. Therefore a composition with n = 6 cannot exist.
Enough
The 2 n-1 Δ-transformations are either linearly independent, or if n is divisible by four, there are relations between them, which arise from equation (□) and multiplication with one of the first 2 n-2 Δ-transformations. The first 2 n-2 Δ transformations are therefore linearly independent. But now more than n ² n × n matrices are always linearly dependent on what ${\displaystyle 2^{n-2}\leq n^{2}}$ and therefore n ≤ 8 must be. The case n = 1 is possible because there are no matrix equations here. Otherwise, n has to be different from six. Thus, a composition can only exist if n = 1, 2, 4 or 8.
Quod erat demonstrandum .

literature

1. ↑ Guido Walz (Ed.): Lexicon of Mathematics . tape 4 (Moo-Sch). Springer Spektrum Verlag, Mannheim 2017, ISBN 978-3-662-53499-1 , doi : 10.1007 / 978-3-662-53500-4 .
2. ↑ Hurwitz: About the composition of the square forms of any number of variables . In: News from the k. Society of Sciences in Göttingen, mathematical-physical class . 1898, p. 309–316 ( Computer Science University of Toronto [PDF; accessed June 18, 2017]).
3. ^ M. Koecher, R. Remmert, Kompositionsalgebren, theorem by Hurwitz, in: D. Ebbinghaus u. a., Numbers, Springer 1983, Chapter 9