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I was watching a video lecture on Algebraic Geometry, but I do not have their book. Something came up that confused me.

What is a non-degenerate homogeneous quadratic polynomial? Please with few examples of degenerate polynomials and why they fail.

The reason I am asking this is because apparently any non-degenerate homogeneous quadratic polynomial in projective space can be written as $f(x,y,z)=xz-y^2$ via a transformation. Does someone know how to prove this claim?

EDIT, Will Jagy: the result is Theorem I.9 on page 15 of PLESKEN

Will Jagy
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Kori
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  • Is this video freely available? If so, it may help to provide a link. – Michael Albanese Jan 10 '16 at 01:24
  • Any homogeneous quadratic polynomial in $n$ variables $x_1, \dots, x_n$ over $\mathbb C$ can be brought into the form $x_1^2 + \cdots + x_r^2$ for some $0 \leq r \leq n$ after a change of variables. The form is degenerate if $r < n$. Maybe it helps to think of this as the rank of a representing matrix, if you know something about that correspondence. – Hoot Jan 10 '16 at 01:31
  • It cannot be all polynomials, because $zx - y^2$ is isotropic over $\mathbb Q$ and therefore $\mathbb Z.$ This is one pages 507-508 of Fricke and Klein for isotropic (indefinite) ternaries https://books.google.com/books?id=H5kLAAAAYAAJ&printsec=frontcover&dq=Vorlesungen+%C3%BCber+die+Theorie+der+automorphen++Functionen&hl=en&sa=X&ved=0CCAQ6AEwAGoVChMIscb4rNvYxgIVSCiICh3qMwQX#v=onepage&q=muesli&f=false I can explain more but need more detail – Will Jagy Jan 10 '16 at 01:59
  • This is briefly alluded to in Magnus, Noneuclidean Tesselations and Their Groups, but explicit in an article, W. Plesken https://books.google.com/books?id=4TAxOSzxODUC&pg=PA15&lpg=PA15&dq=Let+g+be+an+integral+ternary+quadratic+form,+which+is+indefinite+and+nondegenerate&source=bl&ots=ETCH01cGlX&sig=kx0OSGeCR185rDg_f0N4DWG_ewU&hl=en&sa=X&ved=0ahUKEwiYlMTgn_XJAhUWwWMKHTjqDXUQ6AEIHTAA#v=onepage&q=Let%20g%20be%20an%20integral%20ternary%20quadratic%20form%2C%20which%20is%20indefinite%20and%20nondegenerate&f=false – Will Jagy Jan 10 '16 at 02:04
  • Yes exactly the proof from W. Plesken, I have hard time understanding it – Kori Jan 10 '16 at 02:14

1 Answers1

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Just an example. We begin with an isotropic ternary quadratic form, $$ f(x,y,z) = 24 x^2 + 24 y^2 + 24 z^2 -43 yz - 43 z x - 43 x y. $$ Its Hessian matrix of second partial derivatives is

$$ H = \left( \begin{array}{rrr} 48 & -43 & -43 \\ -43 & 48 & -43 \\ -43 & -43 & 48 \end{array} \right) $$

Isotropic means that there is at least one triple $(x,y,z)$ of integers, not all equal to zero, with $f(x,y,z) = 0.$ Nondegenerate means $\det H \neq 0.$

The theorem in question guarantees an integer matrix,

$$ R = \left( \begin{array}{rrr} 58 & 61 & 18 \\ 18 & -25 & 15 \\ 15 & 55 & 58 \end{array} \right) $$

Back to that in a minute. The indicated form, $y^2 - z x,$ has Hessian matrix $$ G = \left( \begin{array}{rrr} 0 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 0 \end{array} \right) $$

Alright, the relationship is $R^T H R = 157339 G.$ That is, $$ \left( \begin{array}{rrr} 58 & 18 & 15 \\ 61 & -25 & 55 \\ 18 & 15 & 58 \end{array} \right) \left( \begin{array}{rrr} 48 & -43 & -43 \\ -43 & 48 & -43 \\ -43 & -43 & 48 \end{array} \right) \left( \begin{array}{rrr} 58 & 61 & 18 \\ 18 & -25 & 15 \\ 15 & 55 & 58 \end{array} \right) = \left( \begin{array}{rrr} 0 & 0 & -157339 \\ 0 & 314678 & 0 \\ -157339 & 0 & 0 \end{array} \right) $$

The rows of $R$ tell us that, for any integers $(u,v),$ we always have $$ f( 58 u^2 + 61 uv + 18 v^2, 18 u^2 -25 uv + 15 v^2, 15 u^2 + 55 uv + 58 v^2 ) = 0 $$

Will Jagy
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