I will soon be teaching an introductory course on bilinear algebra and quadratic forms. I will likely spend most of the time and effort on positive definite quadratic forms and euclidean spaces. These have nice and concrete (i.e. not to theoretical physics) applications since they are basic for many optimisation problems. Are there any applications of this type (so not Einstein's relativity theory) which involve isotropic forms (and which hopefully would be useful beyond just motivating the material for bored undergraduates, e.g. that would be a basis for a problem or a series of exercises)?
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5$\begingroup$ "Isotropic" = forms that represent zero? Sure; they often arise as intersection forms on algebraic surfaces (e.g. an elliptic fibration yields a nontrivial divisor with self-intersection zero), and are often useful to describe the structure of positive-definite forms (quite a few examples in Conway and Sloane's "SPLAG" = *Sphere Packings, Lattices, and Groups"). $\endgroup$– Noam D. ElkiesCommented Dec 15, 2016 at 18:24
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3$\begingroup$ also (and possibly more suitable for undergraduates) the determinant of a 2-by-2 matrix, and trace(A^2) for square matrices of any order, are natural quadratic forms that represent zero. And a conic plane curve is basically a quadratic form in 3 real variables that represents zero (up to multiplying the form by a nonzero scalar). If it has rational coefficients then the existence of rational points is equivalent to the question of whether the form represents zero nontrivially over the rationals. $\endgroup$– Noam D. ElkiesCommented Dec 15, 2016 at 18:26
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$\begingroup$ Made CW according to the OP's wishes (people are encouraged to tweak or extend or correct it). $\endgroup$– Todd TrimbleCommented Dec 15, 2016 at 19:43
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$\begingroup$ Isotropic forms are easy and fun to deal with over finite fields. And you might connect the particular case of $U_4(\mathbb{F}_2)$ to 27 lines on cubic surfaces... $\endgroup$– Dima PasechnikCommented Dec 15, 2016 at 22:22
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2 Answers
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I've enjoyed finding out how to parametrize the integer null vectors of an indefinite ternary with integer coefficients. The observation that an isotropic ternary integrally represents an integer multiple of $y^2 - zx$ goes back to Fricke and Klein (1897 I think).
It is easy to find all ration vectors in the null cone. However, it is fairly difficult to pull out all the primitive integer vectors. If we just multiply through by a denominator (in symbols) from stereographic projection, we typically miss a fixed fraction of solutions.
All told, the outcome is that the integer null vectors are parametrized by a finite number of Pythagorean Triple type formulas, but not necessarily just one.
why not...
This is about ternary quadratic forms with integer coefficients. We use the ordering in http://zakuski.math.utsa.edu/~kap/Lehman_1992.pdf and http://zakuski.math.utsa.edu/~kap/Watson_Representation_1954.pdf I like how Watson writes it, writing $zx$ instead of $xz.$ The ordered sextuple $$ \langle a,b,c, r,s,t \rangle $$ refers to the form $$ f(x,y,z) = a x^2 + b y^2 + c y^2 + r y z + s z x + t x y $$ The Hessian matrix, of second partial derivatives, is
$$ H = \left( \begin{array}{rrr} 2 a & t & s \\ t & 2b & r \\ s & r & 2c \end{array} \right) $$
The discriminant is half the determinant of $H$ and therefore an integer, $$ \Delta = 4abc + rst - a r^2 - b s^2 - c t^2. $$
Oh, right. Given a column vector $$ U = \left( \begin{array}{r} x \\ y \\ z \end{array} \right), $$ we have $$ f(x,y,z) = \frac{1}{2} \; \; U^T H U. $$
We will say that $f$ represents another form $g,$ with Hessian $G,$ if there is an integer matrix $P$ with $\det P \neq 0,$ such that $$ P^T H P = G. $$ Note that $ \det G = \det H \left( \det P \right)^2$ is still nonzero.
Suppose that $f$ is isotropic over the integers. This means that there is some $(u_1, u_2, u_3),$ not all zero, with $f(u_1, u_2, u_3)= 0.$
Theorem: if $f$ is isotropic with nonzero discriminant, then there is a nonzero integer $n$ such that $f$ represents $n (y^2 - zx).$
It took me three steps to prove this, so I will present it that way. The first statement of this that I know is pages 507-508 of
Fricke Klein
This is also on page 303 of Cassels
but is not part of the online preview.
First, we have $$ U = \left( \begin{array}{r} u_1 \\ u_2 \\ u_3 \end{array} \right). $$ The hypothesis is that $$ U^T H U = 0. $$ On the other hand, we are told that the determinant of $H$ is nonzero, which tells us that $HU \neq \vec{0}.$ It follow that there is a nonzero integer vector $$ V = \left( \begin{array}{r} v_1 \\ v_2 \\ v_3 \end{array} \right) $$ such that $$ V^T H U = 0. $$ For example, using the ordinary cross product, we could simply use $V = (HU) \times U.$
We may take any third vector as the final column of a matrix
$$ P = \left( \begin{array}{rrr} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{array} \right) $$ as long as the determinant is nonzero. The outcome is that, with some integers $i,j,k,l,$ $$ P^T H P = \left( \begin{array}{rrr} 0 & 0 & l \\ 0 & 2i & j \\ l & j & 2k \end{array} \right). $$ That is, $f$ represents $$ \langle 0,i,k,j,l,0 \rangle $$
Next, we take a new matrix $$ Q = \left( \begin{array}{rrr} 1 & 0 & j^2 - 4ik \\ 0 & 1 & -2jl \\ 0 & 0 & 4il \end{array} \right). $$ It took me a long time to find $Q.$ We now have, with some nonzero integers $m,n,$ $$ Q^T P^T H P Q = \left( \begin{array}{rrr} 0 & 0 & n \\ 0 & 2m & 0 \\ n & 0 & 0 \end{array} \right). $$ Finally, we take a diagonal matrix $R,$ $$ R = \left( \begin{array}{rrr} m & 0 & 0 \\ 0 & -n & 0 \\ 0 & 0 & -n \end{array} \right). $$ We reach $$ R^T Q^T P^T H P Q R = \left( \begin{array}{rrr} 0 & 0 & -m n^2 \\ 0 & 2mn^2 & 0 \\ -m n^2 & 0 & 0 \end{array} \right) \; \; = \; \; m n^2 \; \; \left( \begin{array}{rrr} 0 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 0 \end{array} \right) $$
That is, $f$ represents an integer multiple of $y^2 - zx.$ Let us call that $N(y^2 - z x),$ so this $N = m n^2.$
The matrix entries I came up with for the general case may be much larger than required. On page 28, Cassels asks about $3 x^2 - 2 y^2 - z^2.$ I have a computer program to find these expressions with small numbers. I will not bother doubling the diagonal entries,
% jagy@phobeusjunior:~$ ./homothety_indef 3 -2 -1 0 0 0 0 -24 0 0 24 0 4
$$ \left( \begin{array}{rrr} 2 & 2 & -2 \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 1 \\ 2 & 2 & -1 \\ -2 & 4 & 1 \end{array} \right) = \left( \begin{array}{rrr} 0 & 0 & 12 \\ 0 & -24 & 0 \\ 12 & 0 & 0 \end{array} \right) $$ That is, $$ \langle 3,-2,-1,0,0,0 \rangle $$ represents $$ \langle 0, -24,0,0,24,0 \rangle $$ or $$ -24 \cdot \langle 0, 1,0,0,-1,0 \rangle. $$
This shows the six Pythagorean Triple type parametrizations needed for $$ 66 (x^2 + y^2 + z^2) - 115(yz + zx + xy) =0 $$ The first matrix means $$ x = 120 u^2 + 229 uv + 118 v^2; \; \; y = 118 u^2 + 7 u v + 9 v^2; \; \; z = 9 u^2 + 11 u v + 120 v^2. $$ This is up to order (lots of symmetry) and negating all three $xyz.$ $u,v$ are not required of the same $\pm$ sign. Indeed, we rename and negate so that $x \geq |y| \geq |z|.$ Otherwise many more recipes (such matrices) might be needed. Oh, there is something very unusual caused by the extreme symmetry for this problem: in all cases, we may demand $u,v \geq 0.$ Rare behavior. The other oddity, that we always get $x,y,z \geq 0,$ is simply that $2 \cdot 66 > 115,$ and says only that the null cone happens to lie entirely inside the positive octant (and the negative octant).
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parisize = 4000000, primelimit = 500509
? x = 120 * u^2 + 229 * u * v + 118 * v^2
%1 = 120*u^2 + 229*v*u + 118*v^2
?
? y = 118 * u^2 + 7 * u * v + 9 * v^2
%2 = 118*u^2 + 7*v*u + 9*v^2
?
? z = 9 * u^2 + 11 * u * v + 120 * v^2
%3 = 9*u^2 + 11*v*u + 120*v^2
? 66 * ( x^2 + y^2 + z^2) - 115 * ( y * z + z * x + x * y)
%4 = 0
?
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jagy@phobeusjunior:~$ ./isotropy 66 115
A = 66 B = 115
120 229 118
118 7 9
9 11 120
129 227 108
108 -11 10
10 31 129
145 211 84
84 -43 18
18 79 145
150 199 73
73 -53 24
24 101 150
153 187 64
64 -59 30
30 119 153
154 181 60
60 -61 33
33 127 154
end of A = 66 B = 115
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120 118 9 < 120, 229, 118 > 1 0
129 108 10 < 129, 227, 108 > 1 0
145 84 18 < 145, 211, 84 > 1 0
150 73 24 < 150, 199, 73 > 1 0
153 64 30 < 153, 187, 64 > 1 0
154 60 33 < 154, 181, 60 > 1 0
395 314 32 < 154, 181, 60 > 1 1
404 302 35 < 153, 187, 64 > 1 1
422 275 44 < 150, 199, 73 > 1 1
440 242 59 < 145, 211, 84 > 1 1
464 170 107 < 129, 227, 108 > 1 1
467 140 134 < 120, 229, 118 > 1 1
880 783 66 < 153, 187, 64 > 1 2
1038 540 151 < 154, 181, 60 > 2 1
1056 495 178 < 120, 229, 118 > 2 1
1086 375 268 < 145, 211, 84 > 2 1
1764 1290 157 < 153, 187, 64 > 1 3
1782 1264 165 < 129, 227, 108 > 1 3
1800 1237 174 < 154, 181, 60 > 1 3
1869 1122 220 < 120, 229, 118 > 1 3
2020 669 522 < 150, 199, 73 > 3 1
2022 645 544 < 145, 211, 84 > 3 1
2310 2211 172 < 153, 187, 64 > 2 3
2602 1851 240 < 145, 211, 84 > 2 3
2654 2333 200 < 145, 211, 84 > 1 4
2712 1675 306 < 154, 181, 60 > 3 2
2836 1422 435 < 150, 199, 73 > 3 2
2916 1182 595 < 120, 229, 118 > 2 3
2924 1973 290 < 120, 229, 118 > 1 4
2955 946 792 < 129, 227, 108 > 3 2
3005 1838 344 < 154, 181, 60 > 1 4
3260 1109 818 < 153, 187, 64 > 4 1
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One standard construction of $L^2$ is to start from a space of square-integrable functions, which is only positive semidefinite (and thus contains many isotropic vectors), and form the quotient by the kernel of the inner product (which consists of those isotropic vectors). Of course $L^2$ is the arena for much important mathematics, both pure and applied.