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Hurwitz problem

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In mathematics, theHurwitz problem (named afterAdolf Hurwitz) is the problem of finding multiplicative relations betweenquadratic forms which generalise those known to exist between sums of squares in certain numbers of variables.

Description

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There are well-known multiplicative relationships between sums of squares in two variables

(x2+y2)(u2+v2)=(xuyv)2+(xv+yu)2 ,{\displaystyle (x^{2}+y^{2})(u^{2}+v^{2})=(xu-yv)^{2}+(xv+yu)^{2}\ ,}

(known as theBrahmagupta–Fibonacci identity), and alsoEuler's four-square identity andDegen's eight-square identity. These may be interpreted as multiplicativity for the norms on thecomplex numbers (C{\displaystyle \mathbb {C} }),quaternions (H{\displaystyle \mathbb {H} }), andoctonions (O{\displaystyle \mathbb {O} }), respectively.[1]: 1–3 [2]

The Hurwitz problem for the fieldK is to find general relations of the form

(x12++xr2)(y12++ys2)=(z12++zn2) ,{\displaystyle (x_{1}^{2}+\cdots +x_{r}^{2})\cdot (y_{1}^{2}+\cdots +y_{s}^{2})=(z_{1}^{2}+\cdots +z_{n}^{2})\ ,}

with thez being bilinear forms in thex andy: that is, eachz is aK-linear combination of terms of the formxi yj.[3]: 127 

We call a triple(r,s,n){\displaystyle \;(r,s,n)\;}admissible forK if such an identity exists.[1]: 125  Trivial cases of admissible triples include(r,s,rs).{\displaystyle \;(r,s,rs)\;.} The problem is uninteresting forK ofcharacteristic 2, since over such fields every sum of squares is a square, and we exclude this case. It is believed that otherwise admissibility is independent of the field of definition.[1]: 137 

The Hurwitz–Radon theorem

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Hurwitz posed the problem in 1898 in the special caser=s=n{\displaystyle \;r=s=n\;} and showed that, when coefficients are taken inC{\displaystyle \mathbb {C} }, the only admissible values(n,n,n){\displaystyle \,(n,n,n)\,} weren{1,2,4,8}.{\displaystyle \;n\in \{1,2,4,8\}\;.}[3]: 130  His proof extends to a field of anycharacteristicexcept 2.[1]: 3 

The "Hurwitz–Radon" problem is that of finding admissible triples of the form(r,n,n).{\displaystyle \,(r,n,n)\;.} Obviously(1,n,n){\displaystyle \;(1,n,n)\;} is admissible. TheHurwitz–Radon theorem states that(ρ(n),n,n){\displaystyle \;\left(\rho (n),n,n\right)\;} is admissible over any field whereρ(n){\displaystyle \,\rho (n)\,} is the function defined forn=2uv,{\displaystyle \;n=2^{u}v\;,}v odd,u=4a+b,{\displaystyle \;u=4a+b\;,} with0b3,{\displaystyle \;0\leq b\leq 3\;,} andρ(n)=8a+2b.{\displaystyle \;\rho (n)=8a+2^{b}\;.}[1]: 137 [3]: 130 

Other admissible triples include(3,5,7){\displaystyle \,(3,5,7)\,}[1]: 138  and(10,10,16).{\displaystyle \,(10,10,16)\;.}[1]: 137 

See also

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References

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  1. ^abcdefgRajwade, A.R. (1993).Squares. London Mathematical Society Lecture Note Series. Vol. 171.Cambridge University Press.ISBN 0-521-42668-5.Zbl 0785.11022.
  2. ^Curtis, C.W. (1963). "The four and eight square problem and division algebras". In Albert, A.A. (ed.).Studies in Modern Algebra.Mathematical Association of America. pp. 100–125, esp. 115. — Solution of Hurwitz's Problem on page 115.
  3. ^abcLam, Tsit-Yuen (2005).Introduction to Quadratic Forms over Fields.Graduate Studies in Mathematics. Vol. 67.American Mathematical Society.ISBN 0-8218-1095-2.MR 2104929.Zbl 1068.11023.
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