Let $(V,q )$ be a quadratic space over $ \mathbb{Q} $. A subspace $ U $ is called totally isotropic if $ q(x) = 0 $ for all $ x \in U $ and a subspace $ U $ is called an anisotropic subspace if $ q(x) \neq 0 $ for all non zero $ x \in U $. Let us consider two quadratic forms $ q_{1}$, $q_{2} $, defined by $$ q_{1}(a,b,c,d) = ab + 2 ac - 2 ad - 2 bc + 2 bd - cd $$ and $$ q_{2}(a,b,c,d) = 2 a^{2} - 13 ab + 4 ac + 4 ad + 2 b^{2} + 4 bc + 4 bd + 2 c^{2} - 13 cd + 2 d^{2} .$$ Then, what is the maximum dimension of a common anisotropic subspace of $ q_{1} $ and $ q_{2} $?
$\begingroup$
$\endgroup$
8
-
1$\begingroup$ Context would help: how did these particular ones arise, or are you hoping for a general algorithm? Do you have info on the local behavior? (Thinking of Hasse-Minkowski...) $\endgroup$– paul garrettCommented Jan 3, 2022 at 19:41
-
$\begingroup$ I have tried , I computed some anisotropic subspace of 1st one but some vector is isotropic for 2nd one. $\endgroup$– SkyCommented Jan 4, 2022 at 5:26
-
$\begingroup$ Is there any algorithm? $\endgroup$– SkyCommented Jan 4, 2022 at 5:27
-
$\begingroup$ Where do these particular forms come from? $\endgroup$– Antoine LabelleCommented Jan 4, 2022 at 21:53
-
$\begingroup$ These forms come from lot of calculations , that don't help much. Does there any trick to guess the anisotropic subspace we can also think what is the maximum dimension of totally isotropic subspace. $\endgroup$– SkyCommented Jan 5, 2022 at 6:52
|
Show 3 more comments
1 Answer
$\begingroup$
$\endgroup$
3
Dimension 4 is clearly impossible since your quadratic forms are isotropic, but dimension 3 is possible.
The following SageMath code generate random 3-dimensional subspaces and check whether they are simultaneously isotropic for both forms. It finds quite a lot of such subspaces, as an example you can take the subspace generated by the vectors
$$\pmatrix{2\cr0\cr0\cr1}, \pmatrix{-1\cr0\cr1\cr1}, \pmatrix{1\cr1\cr12\cr0}$$
M1 = Matrix([[ 0, 1, 2,-2],
[ 1, 0,-2, 2],
[ 2,-2, 0,-1],
[-2, 2,-1, 0]])
M2 = Matrix([[ 4,-13, 4, 4],
[-13, 4, 4, 4],
[ 4, 4, 4,-13],
[ 4, 4,-13, 4]])
Q1 = QuadraticForm(QQ, M1)
Q2 = QuadraticForm(QQ, M2)
for k in range(50):
basis = random_matrix(ZZ, 3, 4)
N1 = matrix(QQ, 3, 3, lambda i, j: Q1.bilinear_map(basis[i], basis[j]))
N2 = matrix(QQ, 3, 3, lambda i, j: Q2.bilinear_map(basis[i], basis[j]))
q1 = QuadraticForm(QQ, N1)
q2 = QuadraticForm(QQ, N2)
if q1.anisotropic_primes() and q2.anisotropic_primes():
print(basis)
print()
answered Jan 5, 2022 at 21:50
-
$\begingroup$ What does mean anisotropic primes here? So in general for any quadratic form we can compute an anisotropic subspace of a quadratic form or common anisotropic subspace for both with the help of sage math? $\endgroup$– SkyCommented Jan 12, 2022 at 20:10
-
$\begingroup$ q.anisotropic_primes() gives the list of all primes $p$ (including $\infty$) such that $q$ is anisotropic over $\mathbb{Q}_p$. So by Hasse-Minkowski q is anisotropic over $\mathbb{Q}$ if and only if q.anisotropic_primes() is nonempty (note that in python for a list L, "if L" evaluates to False if L is empty and True otherwise). $\endgroup$ Commented Jan 12, 2022 at 21:42
-
$\begingroup$ Thanks @Antoine Labelle , now I understand $\endgroup$– SkyCommented Jan 13, 2022 at 13:23