Isotropy of quadratic forms
V. Suresh University Of Hyderabad Hyderabad
31 October 2008
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2 2 For example X1 + X2 + X1X2
We say that (a1, ··· , an) is a zero of the polynomial f if f (a1, ··· , an) = 0.
One of the main problems in Mathematics is to determine whether the given polynomial has a (non-trivial) zero or not.
For example, let us recall the Fermat’s last theorem:
X n + Y n = Z n (or X n + Y n − Z n)
Polynomials
f (X1, ··· , Xn) − −− polynomial
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We say that (a1, ··· , an) is a zero of the polynomial f if f (a1, ··· , an) = 0.
One of the main problems in Mathematics is to determine whether the given polynomial has a (non-trivial) zero or not.
For example, let us recall the Fermat’s last theorem:
X n + Y n = Z n (or X n + Y n − Z n)
Polynomials
f (X1, ··· , Xn) − −− polynomial
2 2 For example X1 + X2 + X1X2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms One of the main problems in Mathematics is to determine whether the given polynomial has a (non-trivial) zero or not.
For example, let us recall the Fermat’s last theorem:
X n + Y n = Z n (or X n + Y n − Z n)
Polynomials
f (X1, ··· , Xn) − −− polynomial
2 2 For example X1 + X2 + X1X2
We say that (a1, ··· , an) is a zero of the polynomial f if f (a1, ··· , an) = 0.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms For example, let us recall the Fermat’s last theorem:
X n + Y n = Z n (or X n + Y n − Z n)
Polynomials
f (X1, ··· , Xn) − −− polynomial
2 2 For example X1 + X2 + X1X2
We say that (a1, ··· , an) is a zero of the polynomial f if f (a1, ··· , an) = 0.
One of the main problems in Mathematics is to determine whether the given polynomial has a (non-trivial) zero or not.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms X n + Y n = Z n (or X n + Y n − Z n)
Polynomials
f (X1, ··· , Xn) − −− polynomial
2 2 For example X1 + X2 + X1X2
We say that (a1, ··· , an) is a zero of the polynomial f if f (a1, ··· , an) = 0.
One of the main problems in Mathematics is to determine whether the given polynomial has a (non-trivial) zero or not.
For example, let us recall the Fermat’s last theorem:
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Polynomials
f (X1, ··· , Xn) − −− polynomial
2 2 For example X1 + X2 + X1X2
We say that (a1, ··· , an) is a zero of the polynomial f if f (a1, ··· , an) = 0.
One of the main problems in Mathematics is to determine whether the given polynomial has a (non-trivial) zero or not.
For example, let us recall the Fermat’s last theorem:
X n + Y n = Z n (or X n + Y n − Z n)
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms i.e. the polynomial X 2 + Y 2 − Z 2 has a zero
X = a, Y = 0, Z = a is a zero of X n + Y n − Z n
We call this zero a trivial zero
The famous theorem of Fermat asserts that if n ≥ 3, this polynomial has no non-trivial zero over integers.
i.e. if a, b, c are three integers such that an + bn = cn and n ≥ 3, then one of the a, b, c must be zero.
We have : 32 + 42 = 52.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms X = a, Y = 0, Z = a is a zero of X n + Y n − Z n
We call this zero a trivial zero
The famous theorem of Fermat asserts that if n ≥ 3, this polynomial has no non-trivial zero over integers.
i.e. if a, b, c are three integers such that an + bn = cn and n ≥ 3, then one of the a, b, c must be zero.
We have : 32 + 42 = 52. i.e. the polynomial X 2 + Y 2 − Z 2 has a zero
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We call this zero a trivial zero
The famous theorem of Fermat asserts that if n ≥ 3, this polynomial has no non-trivial zero over integers.
i.e. if a, b, c are three integers such that an + bn = cn and n ≥ 3, then one of the a, b, c must be zero.
We have : 32 + 42 = 52. i.e. the polynomial X 2 + Y 2 − Z 2 has a zero
X = a, Y = 0, Z = a is a zero of X n + Y n − Z n
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms The famous theorem of Fermat asserts that if n ≥ 3, this polynomial has no non-trivial zero over integers.
i.e. if a, b, c are three integers such that an + bn = cn and n ≥ 3, then one of the a, b, c must be zero.
We have : 32 + 42 = 52. i.e. the polynomial X 2 + Y 2 − Z 2 has a zero
X = a, Y = 0, Z = a is a zero of X n + Y n − Z n
We call this zero a trivial zero
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms i.e. if a, b, c are three integers such that an + bn = cn and n ≥ 3, then one of the a, b, c must be zero.
We have : 32 + 42 = 52. i.e. the polynomial X 2 + Y 2 − Z 2 has a zero
X = a, Y = 0, Z = a is a zero of X n + Y n − Z n
We call this zero a trivial zero
The famous theorem of Fermat asserts that if n ≥ 3, this polynomial has no non-trivial zero over integers.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We have : 32 + 42 = 52. i.e. the polynomial X 2 + Y 2 − Z 2 has a zero
X = a, Y = 0, Z = a is a zero of X n + Y n − Z n
We call this zero a trivial zero
The famous theorem of Fermat asserts that if n ≥ 3, this polynomial has no non-trivial zero over integers. i.e. if a, b, c are three integers such that an + bn = cn and n ≥ 3, then one of the a, b, c must be zero.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2X + 3Y + 4Z has one trivial solution, e.g. X = Y = Z = 0.
This also has many other solutions : X = 2a, Y = 0, Z = −a is also a solution.
Consider a general linear polynomial (of degree 1): a1X1 + ··· + anXn.
X1 = ··· = Xn = 0 is a zero of this polynomial, called trivial zero.
It is easy to see that if n ≥ 2, the polynomial a1X1 + ··· + anXn
has a non-trivial zero.
Examples
3 2X − 3 has only one zero : X = 2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms This also has many other solutions : X = 2a, Y = 0, Z = −a is also a solution.
Consider a general linear polynomial (of degree 1): a1X1 + ··· + anXn.
X1 = ··· = Xn = 0 is a zero of this polynomial, called trivial zero.
It is easy to see that if n ≥ 2, the polynomial a1X1 + ··· + anXn
has a non-trivial zero.
Examples
3 2X − 3 has only one zero : X = 2
2X + 3Y + 4Z has one trivial solution, e.g. X = Y = Z = 0.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Consider a general linear polynomial (of degree 1): a1X1 + ··· + anXn.
X1 = ··· = Xn = 0 is a zero of this polynomial, called trivial zero.
It is easy to see that if n ≥ 2, the polynomial a1X1 + ··· + anXn
has a non-trivial zero.
Examples
3 2X − 3 has only one zero : X = 2
2X + 3Y + 4Z has one trivial solution, e.g. X = Y = Z = 0.
This also has many other solutions : X = 2a, Y = 0, Z = −a is also a solution.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms X1 = ··· = Xn = 0 is a zero of this polynomial, called trivial zero.
It is easy to see that if n ≥ 2, the polynomial a1X1 + ··· + anXn
has a non-trivial zero.
Examples
3 2X − 3 has only one zero : X = 2
2X + 3Y + 4Z has one trivial solution, e.g. X = Y = Z = 0.
This also has many other solutions : X = 2a, Y = 0, Z = −a is also a solution.
Consider a general linear polynomial (of degree 1): a1X1 + ··· + anXn.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms It is easy to see that if n ≥ 2, the polynomial a1X1 + ··· + anXn
has a non-trivial zero.
Examples
3 2X − 3 has only one zero : X = 2
2X + 3Y + 4Z has one trivial solution, e.g. X = Y = Z = 0.
This also has many other solutions : X = 2a, Y = 0, Z = −a is also a solution.
Consider a general linear polynomial (of degree 1): a1X1 + ··· + anXn.
X1 = ··· = Xn = 0 is a zero of this polynomial, called trivial zero.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms has a non-trivial zero.
Examples
3 2X − 3 has only one zero : X = 2
2X + 3Y + 4Z has one trivial solution, e.g. X = Y = Z = 0.
This also has many other solutions : X = 2a, Y = 0, Z = −a is also a solution.
Consider a general linear polynomial (of degree 1): a1X1 + ··· + anXn.
X1 = ··· = Xn = 0 is a zero of this polynomial, called trivial zero.
It is easy to see that if n ≥ 2, the polynomial a1X1 + ··· + anXn
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Examples
3 2X − 3 has only one zero : X = 2
2X + 3Y + 4Z has one trivial solution, e.g. X = Y = Z = 0.
This also has many other solutions : X = 2a, Y = 0, Z = −a is also a solution.
Consider a general linear polynomial (of degree 1): a1X1 + ··· + anXn.
X1 = ··· = Xn = 0 is a zero of this polynomial, called trivial zero.
It is easy to see that if n ≥ 2, the polynomial a1X1 + ··· + anXn
has a non-trivial zero.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Examples
3 2X − 3 has only one zero : X = 2
2X + 3Y + 4Z has one trivial solution, e.g. X = Y = Z = 0.
This also has many other solutions : X = 2a, Y = 0, Z = −a is also a solution.
Consider a general linear polynomial (of degree 1): a1X1 + ··· + anXn.
X1 = ··· = Xn = 0 is a zero of this polynomial, called trivial zero.
It is easy to see that if n ≥ 2, the polynomial a1X1 + ··· + anXn
has a non-trivial zero.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms A homogeneous polynomial q of degree 2 is called a quadratic form X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
what are these aij ?
they are in some field k
e.g. C, R, Q, Qp or C(X ). X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n with aij ∈ k
quadratic forms
Now we consider the next case, i.e. degree 2.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
what are these aij ?
they are in some field k
e.g. C, R, Q, Qp or C(X ). X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n with aij ∈ k
quadratic forms
Now we consider the next case, i.e. degree 2.
A homogeneous polynomial q of degree 2 is called a quadratic form
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms what are these aij ?
they are in some field k
e.g. C, R, Q, Qp or C(X ). X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n with aij ∈ k
quadratic forms
Now we consider the next case, i.e. degree 2.
A homogeneous polynomial q of degree 2 is called a quadratic form X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms they are in some field k
e.g. C, R, Q, Qp or C(X ). X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n with aij ∈ k
quadratic forms
Now we consider the next case, i.e. degree 2.
A homogeneous polynomial q of degree 2 is called a quadratic form X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
what are these aij ?
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms e.g. C, R, Q, Qp or C(X ). X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n with aij ∈ k
quadratic forms
Now we consider the next case, i.e. degree 2.
A homogeneous polynomial q of degree 2 is called a quadratic form X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
what are these aij ?
they are in some field k
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n with aij ∈ k
quadratic forms
Now we consider the next case, i.e. degree 2.
A homogeneous polynomial q of degree 2 is called a quadratic form X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
what are these aij ?
they are in some field k
e.g. C, R, Q, Qp or C(X ).
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms with aij ∈ k
quadratic forms
Now we consider the next case, i.e. degree 2.
A homogeneous polynomial q of degree 2 is called a quadratic form X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
what are these aij ?
they are in some field k
e.g. C, R, Q, Qp or C(X ). X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms quadratic forms
Now we consider the next case, i.e. degree 2.
A homogeneous polynomial q of degree 2 is called a quadratic form X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
what are these aij ?
they are in some field k
e.g. C, R, Q, Qp or C(X ). X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n with aij ∈ k
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms quadratic forms
Now we consider the next case, i.e. degree 2.
A homogeneous polynomial q of degree 2 is called a quadratic form X q(X1, ··· , Xn) = aij Xi Xj 1≤i≤j≤n
what are these aij ?
they are in some field k
e.g. C, R, Q, Qp or C(X ). X By a quadratic form q over k, we mean aij Xi Xj 1≤i≤j≤n with aij ∈ k
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms that means 2 6= 0 in k.
Then by a suitable linear change of variables, we can assume that
2 2 q = a1X1 + ··· + anXn .
We assume that all ai ’s non-zero.
n is called the dimension of q
We assume that characteristic of k is not equal 2.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Then by a suitable linear change of variables, we can assume that
2 2 q = a1X1 + ··· + anXn .
We assume that all ai ’s non-zero.
n is called the dimension of q
We assume that characteristic of k is not equal 2. that means 2 6= 0 in k.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2 2 q = a1X1 + ··· + anXn .
We assume that all ai ’s non-zero.
n is called the dimension of q
We assume that characteristic of k is not equal 2. that means 2 6= 0 in k.
Then by a suitable linear change of variables, we can assume that
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We assume that all ai ’s non-zero.
n is called the dimension of q
We assume that characteristic of k is not equal 2. that means 2 6= 0 in k.
Then by a suitable linear change of variables, we can assume that
2 2 q = a1X1 + ··· + anXn .
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms n is called the dimension of q
We assume that characteristic of k is not equal 2. that means 2 6= 0 in k.
Then by a suitable linear change of variables, we can assume that
2 2 q = a1X1 + ··· + anXn .
We assume that all ai ’s non-zero.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We assume that characteristic of k is not equal 2. that means 2 6= 0 in k.
Then by a suitable linear change of variables, we can assume that
2 2 q = a1X1 + ··· + anXn .
We assume that all ai ’s non-zero. n is called the dimension of q
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Witt, in the early 30’s introduced a new direction in the study of quadratic forms — now termed as algebraic theory of quadratic forms — by introducing the Witt group of quadratic forms to study the totality of quadratic forms over a general field k.
Historically, the study of quadratic forms was part of number theory; Minkowski, Siegel, Hasse, Eichler, Kneser and several other mathematicians created a rich arithmetic theory of quadratic forms.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Historically, the study of quadratic forms was part of number theory; Minkowski, Siegel, Hasse, Eichler, Kneser and several other mathematicians created a rich arithmetic theory of quadratic forms.
Witt, in the early 30’s introduced a new direction in the study of quadratic forms — now termed as algebraic theory of quadratic forms — by introducing the Witt group of quadratic forms to study the totality of quadratic forms over a general field k.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We say that a quadratic form q is isotropic over k if q has a non-trivial zero.
n i.e. there exists a non-zero vector x = (x1, ··· xn) ∈ k
such that q(x) = 0.
If q is not isotropic, then it is called anisotropic
A typical isotropic quadratic form is the hyperbolic plane X 2 − Y 2.
(1, 1) is a zero of this quadratic form
Let q(X1, ··· , Xn) be a quadratic form over a field k. Clearly X1 = ··· = Xn = 0 is a zero of q.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms n i.e. there exists a non-zero vector x = (x1, ··· xn) ∈ k
such that q(x) = 0.
If q is not isotropic, then it is called anisotropic
A typical isotropic quadratic form is the hyperbolic plane X 2 − Y 2.
(1, 1) is a zero of this quadratic form
Let q(X1, ··· , Xn) be a quadratic form over a field k. Clearly X1 = ··· = Xn = 0 is a zero of q.
We say that a quadratic form q is isotropic over k if q has a non-trivial zero.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms such that q(x) = 0.
If q is not isotropic, then it is called anisotropic
A typical isotropic quadratic form is the hyperbolic plane X 2 − Y 2.
(1, 1) is a zero of this quadratic form
Let q(X1, ··· , Xn) be a quadratic form over a field k. Clearly X1 = ··· = Xn = 0 is a zero of q.
We say that a quadratic form q is isotropic over k if q has a non-trivial zero.
n i.e. there exists a non-zero vector x = (x1, ··· xn) ∈ k
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms If q is not isotropic, then it is called anisotropic
A typical isotropic quadratic form is the hyperbolic plane X 2 − Y 2.
(1, 1) is a zero of this quadratic form
Let q(X1, ··· , Xn) be a quadratic form over a field k. Clearly X1 = ··· = Xn = 0 is a zero of q.
We say that a quadratic form q is isotropic over k if q has a non-trivial zero.
n i.e. there exists a non-zero vector x = (x1, ··· xn) ∈ k such that q(x) = 0.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms A typical isotropic quadratic form is the hyperbolic plane X 2 − Y 2.
(1, 1) is a zero of this quadratic form
Let q(X1, ··· , Xn) be a quadratic form over a field k. Clearly X1 = ··· = Xn = 0 is a zero of q.
We say that a quadratic form q is isotropic over k if q has a non-trivial zero.
n i.e. there exists a non-zero vector x = (x1, ··· xn) ∈ k such that q(x) = 0.
If q is not isotropic, then it is called anisotropic
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms (1, 1) is a zero of this quadratic form
Let q(X1, ··· , Xn) be a quadratic form over a field k. Clearly X1 = ··· = Xn = 0 is a zero of q.
We say that a quadratic form q is isotropic over k if q has a non-trivial zero.
n i.e. there exists a non-zero vector x = (x1, ··· xn) ∈ k such that q(x) = 0.
If q is not isotropic, then it is called anisotropic
A typical isotropic quadratic form is the hyperbolic plane X 2 − Y 2.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Let q(X1, ··· , Xn) be a quadratic form over a field k. Clearly X1 = ··· = Xn = 0 is a zero of q.
We say that a quadratic form q is isotropic over k if q has a non-trivial zero.
n i.e. there exists a non-zero vector x = (x1, ··· xn) ∈ k such that q(x) = 0.
If q is not isotropic, then it is called anisotropic
A typical isotropic quadratic form is the hyperbolic plane X 2 − Y 2.
(1, 1) is a zero of this quadratic form
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms consider the quadratic form q(X , Y ) = X 2 + Y 2
2 2 if x, y ∈ R then x ≥ 0 and y ≥ 0
x2 + y 2 = 0 implies x = 0 and y = 0
q is anisotropic
2 2 2 More generally X1 + X2 + ··· + Xn is an anisotropic quadratic form over R
Consider the field of real numbers R
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2 2 if x, y ∈ R then x ≥ 0 and y ≥ 0
x2 + y 2 = 0 implies x = 0 and y = 0
q is anisotropic
2 2 2 More generally X1 + X2 + ··· + Xn is an anisotropic quadratic form over R
Consider the field of real numbers R consider the quadratic form q(X , Y ) = X 2 + Y 2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms x2 + y 2 = 0 implies x = 0 and y = 0
q is anisotropic
2 2 2 More generally X1 + X2 + ··· + Xn is an anisotropic quadratic form over R
Consider the field of real numbers R consider the quadratic form q(X , Y ) = X 2 + Y 2
2 2 if x, y ∈ R then x ≥ 0 and y ≥ 0
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms q is anisotropic
2 2 2 More generally X1 + X2 + ··· + Xn is an anisotropic quadratic form over R
Consider the field of real numbers R consider the quadratic form q(X , Y ) = X 2 + Y 2
2 2 if x, y ∈ R then x ≥ 0 and y ≥ 0 x2 + y 2 = 0 implies x = 0 and y = 0
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2 2 2 More generally X1 + X2 + ··· + Xn is an anisotropic quadratic form over R
Consider the field of real numbers R consider the quadratic form q(X , Y ) = X 2 + Y 2
2 2 if x, y ∈ R then x ≥ 0 and y ≥ 0 x2 + y 2 = 0 implies x = 0 and y = 0 q is anisotropic
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Consider the field of real numbers R consider the quadratic form q(X , Y ) = X 2 + Y 2
2 2 if x, y ∈ R then x ≥ 0 and y ≥ 0 x2 + y 2 = 0 implies x = 0 and y = 0 q is anisotropic
2 2 2 More generally X1 + X2 + ··· + Xn is an anisotropic quadratic form over R
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms q(X , Y ) = X 2 + Y 2 √ √ q(1, −1) = 1 − 1 = 0 ( −1 ∈ C)
q is isotropic
2 2 2 q(X1, ··· , Xn) = a1X1 + a2X + ··· + anXn
n ≥ 2
p q(1, −a1/a2, 0, ··· , 0) = 0
q is isotropic over C
C - field of complex numbers
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms √ √ q(1, −1) = 1 − 1 = 0 ( −1 ∈ C)
q is isotropic
2 2 2 q(X1, ··· , Xn) = a1X1 + a2X + ··· + anXn
n ≥ 2
p q(1, −a1/a2, 0, ··· , 0) = 0
q is isotropic over C
C - field of complex numbers q(X , Y ) = X 2 + Y 2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms q is isotropic
2 2 2 q(X1, ··· , Xn) = a1X1 + a2X + ··· + anXn
n ≥ 2
p q(1, −a1/a2, 0, ··· , 0) = 0
q is isotropic over C
C - field of complex numbers q(X , Y ) = X 2 + Y 2 √ √ q(1, −1) = 1 − 1 = 0 ( −1 ∈ C)
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2 2 2 q(X1, ··· , Xn) = a1X1 + a2X + ··· + anXn
n ≥ 2
p q(1, −a1/a2, 0, ··· , 0) = 0
q is isotropic over C
C - field of complex numbers q(X , Y ) = X 2 + Y 2 √ √ q(1, −1) = 1 − 1 = 0 ( −1 ∈ C) q is isotropic
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms n ≥ 2
p q(1, −a1/a2, 0, ··· , 0) = 0
q is isotropic over C
C - field of complex numbers q(X , Y ) = X 2 + Y 2 √ √ q(1, −1) = 1 − 1 = 0 ( −1 ∈ C) q is isotropic
2 2 2 q(X1, ··· , Xn) = a1X1 + a2X + ··· + anXn
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms p q(1, −a1/a2, 0, ··· , 0) = 0
q is isotropic over C
C - field of complex numbers q(X , Y ) = X 2 + Y 2 √ √ q(1, −1) = 1 − 1 = 0 ( −1 ∈ C) q is isotropic
2 2 2 q(X1, ··· , Xn) = a1X1 + a2X + ··· + anXn n ≥ 2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms q is isotropic over C
C - field of complex numbers q(X , Y ) = X 2 + Y 2 √ √ q(1, −1) = 1 − 1 = 0 ( −1 ∈ C) q is isotropic
2 2 2 q(X1, ··· , Xn) = a1X1 + a2X + ··· + anXn n ≥ 2
p q(1, −a1/a2, 0, ··· , 0) = 0
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms C - field of complex numbers q(X , Y ) = X 2 + Y 2 √ √ q(1, −1) = 1 − 1 = 0 ( −1 ∈ C) q is isotropic
2 2 2 q(X1, ··· , Xn) = a1X1 + a2X + ··· + anXn n ≥ 2
p q(1, −a1/a2, 0, ··· , 0) = 0 q is isotropic over C
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms For example Z/pZ, p a prime number
q(X1, ··· , Xn) − quadratic form over F
Theorem. If n ≥ 3, then q isotropic.
∗ ∗2 Proof : Exercise (use the fact that F /F is a group of order 2.)
Q − the field of rational numbers
2 2 2 X1 + X2 + ··· + Xn is anisotropic over Q
F − a finite field
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms q(X1, ··· , Xn) − quadratic form over F
Theorem. If n ≥ 3, then q isotropic.
∗ ∗2 Proof : Exercise (use the fact that F /F is a group of order 2.)
Q − the field of rational numbers
2 2 2 X1 + X2 + ··· + Xn is anisotropic over Q
F − a finite field
For example Z/pZ, p a prime number
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Theorem. If n ≥ 3, then q isotropic.
∗ ∗2 Proof : Exercise (use the fact that F /F is a group of order 2.)
Q − the field of rational numbers
2 2 2 X1 + X2 + ··· + Xn is anisotropic over Q
F − a finite field
For example Z/pZ, p a prime number q(X1, ··· , Xn) − quadratic form over F
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms ∗ ∗2 Proof : Exercise (use the fact that F /F is a group of order 2.)
Q − the field of rational numbers
2 2 2 X1 + X2 + ··· + Xn is anisotropic over Q
F − a finite field
For example Z/pZ, p a prime number q(X1, ··· , Xn) − quadratic form over F
Theorem. If n ≥ 3, then q isotropic.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Q − the field of rational numbers
2 2 2 X1 + X2 + ··· + Xn is anisotropic over Q
F − a finite field
For example Z/pZ, p a prime number q(X1, ··· , Xn) − quadratic form over F
Theorem. If n ≥ 3, then q isotropic.
∗ ∗2 Proof : Exercise (use the fact that F /F is a group of order 2.)
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2 2 2 X1 + X2 + ··· + Xn is anisotropic over Q
F − a finite field
For example Z/pZ, p a prime number q(X1, ··· , Xn) − quadratic form over F
Theorem. If n ≥ 3, then q isotropic.
∗ ∗2 Proof : Exercise (use the fact that F /F is a group of order 2.)
Q − the field of rational numbers
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms F − a finite field
For example Z/pZ, p a prime number q(X1, ··· , Xn) − quadratic form over F
Theorem. If n ≥ 3, then q isotropic.
∗ ∗2 Proof : Exercise (use the fact that F /F is a group of order 2.)
Q − the field of rational numbers
2 2 2 X1 + X2 + ··· + Xn is anisotropic over Q
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms q(X1, ··· , Xn) − a quadratic form over k
If n ≥ 5, then q is isotropic.
This follows from the case of finite fields and Hensel’s Lemma. √ k - totally imaginary field, e.g. Q( −1)
q(X1, ··· , Xn) − a quadratic form over k
The classical Hasse - Minkowski Theorem asserts that if n ≥ 5, then q is isotropic
k − p-adic field (i.e. k/Qp a finite extension)
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms If n ≥ 5, then q is isotropic.
This follows from the case of finite fields and Hensel’s Lemma. √ k - totally imaginary field, e.g. Q( −1)
q(X1, ··· , Xn) − a quadratic form over k
The classical Hasse - Minkowski Theorem asserts that if n ≥ 5, then q is isotropic
k − p-adic field (i.e. k/Qp a finite extension) q(X1, ··· , Xn) − a quadratic form over k
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms This follows from the case of finite fields and Hensel’s Lemma. √ k - totally imaginary field, e.g. Q( −1)
q(X1, ··· , Xn) − a quadratic form over k
The classical Hasse - Minkowski Theorem asserts that if n ≥ 5, then q is isotropic
k − p-adic field (i.e. k/Qp a finite extension) q(X1, ··· , Xn) − a quadratic form over k
If n ≥ 5, then q is isotropic.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms √ k - totally imaginary field, e.g. Q( −1)
q(X1, ··· , Xn) − a quadratic form over k
The classical Hasse - Minkowski Theorem asserts that if n ≥ 5, then q is isotropic
k − p-adic field (i.e. k/Qp a finite extension) q(X1, ··· , Xn) − a quadratic form over k
If n ≥ 5, then q is isotropic.
This follows from the case of finite fields and Hensel’s Lemma.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms q(X1, ··· , Xn) − a quadratic form over k
The classical Hasse - Minkowski Theorem asserts that if n ≥ 5, then q is isotropic
k − p-adic field (i.e. k/Qp a finite extension) q(X1, ··· , Xn) − a quadratic form over k
If n ≥ 5, then q is isotropic.
This follows from the case of finite fields and Hensel’s Lemma. √ k - totally imaginary field, e.g. Q( −1)
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms The classical Hasse - Minkowski Theorem asserts that if n ≥ 5, then q is isotropic
k − p-adic field (i.e. k/Qp a finite extension) q(X1, ··· , Xn) − a quadratic form over k
If n ≥ 5, then q is isotropic.
This follows from the case of finite fields and Hensel’s Lemma. √ k - totally imaginary field, e.g. Q( −1) q(X1, ··· , Xn) − a quadratic form over k
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms k − p-adic field (i.e. k/Qp a finite extension) q(X1, ··· , Xn) − a quadratic form over k
If n ≥ 5, then q is isotropic.
This follows from the case of finite fields and Hensel’s Lemma. √ k - totally imaginary field, e.g. Q( −1) q(X1, ··· , Xn) − a quadratic form over k
The classical Hasse - Minkowski Theorem asserts that if n ≥ 5, then q is isotropic
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms called the u-invariant, denoted by u(k), as
u(k) = sup { dim(q) | q anisotropic quadratic form over k }
Examples
1) u(C) = 1
2) u(R) = ∞
3) F - finite field u(F) = 2
Kaplansky (1953) defined an invariant attached to a field k,
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms u(k) = sup { dim(q) | q anisotropic quadratic form over k }
Examples
1) u(C) = 1
2) u(R) = ∞
3) F - finite field u(F) = 2
Kaplansky (1953) defined an invariant attached to a field k, called the u-invariant, denoted by u(k), as
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Examples
1) u(C) = 1
2) u(R) = ∞
3) F - finite field u(F) = 2
Kaplansky (1953) defined an invariant attached to a field k, called the u-invariant, denoted by u(k), as
u(k) = sup { dim(q) | q anisotropic quadratic form over k }
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 1) u(C) = 1
2) u(R) = ∞
3) F - finite field u(F) = 2
Kaplansky (1953) defined an invariant attached to a field k, called the u-invariant, denoted by u(k), as
u(k) = sup { dim(q) | q anisotropic quadratic form over k }
Examples
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2) u(R) = ∞
3) F - finite field u(F) = 2
Kaplansky (1953) defined an invariant attached to a field k, called the u-invariant, denoted by u(k), as
u(k) = sup { dim(q) | q anisotropic quadratic form over k }
Examples
1) u(C) = 1
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 3) F - finite field u(F) = 2
Kaplansky (1953) defined an invariant attached to a field k, called the u-invariant, denoted by u(k), as
u(k) = sup { dim(q) | q anisotropic quadratic form over k }
Examples
1) u(C) = 1
2) u(R) = ∞
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Kaplansky (1953) defined an invariant attached to a field k, called the u-invariant, denoted by u(k), as
u(k) = sup { dim(q) | q anisotropic quadratic form over k }
Examples
1) u(C) = 1
2) u(R) = ∞
3) F - finite field u(F) = 2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Kaplansky (1953) defined an invariant attached to a field k, called the u-invariant, denoted by u(k), as
u(k) = sup { dim(q) | q anisotropic quadratic form over k }
Examples
1) u(C) = 1
2) u(R) = ∞
3) F - finite field u(F) = 2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 5) k - totally imaginary number field, u(k) = 4
6) If u(k) = n, then u(k((t))) = 2n.
4) k - p-adic field u(k) = 4
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 6) If u(k) = n, then u(k((t))) = 2n.
4) k - p-adic field u(k) = 4
5) k - totally imaginary number field, u(k) = 4
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 4) k - p-adic field u(k) = 4
5) k - totally imaginary number field, u(k) = 4
6) If u(k) = n, then u(k((t))) = 2n.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 4) k - p-adic field u(k) = 4
5) k - totally imaginary number field, u(k) = 4
6) If u(k) = n, then u(k((t))) = 2n.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms and define k(T1, ··· Tn) by induction.
We know the following:
i u(C(X1, ··· , Xi )) = 2
i+1 u(F(X1, ··· , Xi )) = 2
where F is a finite field.
f (t) For a field k, let k(T ) = { g(t) | g(t) 6= 0}
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We know the following:
i u(C(X1, ··· , Xi )) = 2
i+1 u(F(X1, ··· , Xi )) = 2
where F is a finite field.
f (t) For a field k, let k(T ) = { g(t) | g(t) 6= 0} and define k(T1, ··· Tn) by induction.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms i u(C(X1, ··· , Xi )) = 2
i+1 u(F(X1, ··· , Xi )) = 2
where F is a finite field.
f (t) For a field k, let k(T ) = { g(t) | g(t) 6= 0} and define k(T1, ··· Tn) by induction.
We know the following:
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms i+1 u(F(X1, ··· , Xi )) = 2
where F is a finite field.
f (t) For a field k, let k(T ) = { g(t) | g(t) 6= 0} and define k(T1, ··· Tn) by induction.
We know the following:
i u(C(X1, ··· , Xi )) = 2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms where F is a finite field.
f (t) For a field k, let k(T ) = { g(t) | g(t) 6= 0} and define k(T1, ··· Tn) by induction.
We know the following:
i u(C(X1, ··· , Xi )) = 2
i+1 u(F(X1, ··· , Xi )) = 2
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms f (t) For a field k, let k(T ) = { g(t) | g(t) 6= 0} and define k(T1, ··· Tn) by induction.
We know the following:
i u(C(X1, ··· , Xi )) = 2
i+1 u(F(X1, ··· , Xi )) = 2 where F is a finite field.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms f (t) For a field k, let k(T ) = { g(t) | g(t) 6= 0} and define k(T1, ··· Tn) by induction.
We know the following:
i u(C(X1, ··· , Xi )) = 2
i+1 u(F(X1, ··· , Xi )) = 2 where F is a finite field.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Does u(k) < ∞ imply u(k(T )) < ∞?
This problem is open and seems to be very difficult for general fields.
This remained open till late 90’s even for the p-adic field Qp.
We have the following
Conjecture : If k is a p-adic field, then u(k(T )) = 8.
Since u(k) = 4, it is easy to see that u(k(T )) ≥ 8
This led to the following natural question
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms This problem is open and seems to be very difficult for general fields.
This remained open till late 90’s even for the p-adic field Qp.
We have the following
Conjecture : If k is a p-adic field, then u(k(T )) = 8.
Since u(k) = 4, it is easy to see that u(k(T )) ≥ 8
This led to the following natural question
Does u(k) < ∞ imply u(k(T )) < ∞?
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms This remained open till late 90’s even for the p-adic field Qp.
We have the following
Conjecture : If k is a p-adic field, then u(k(T )) = 8.
Since u(k) = 4, it is easy to see that u(k(T )) ≥ 8
This led to the following natural question
Does u(k) < ∞ imply u(k(T )) < ∞?
This problem is open and seems to be very difficult for general fields.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We have the following
Conjecture : If k is a p-adic field, then u(k(T )) = 8.
Since u(k) = 4, it is easy to see that u(k(T )) ≥ 8
This led to the following natural question
Does u(k) < ∞ imply u(k(T )) < ∞?
This problem is open and seems to be very difficult for general fields.
This remained open till late 90’s even for the p-adic field Qp.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Conjecture : If k is a p-adic field, then u(k(T )) = 8.
Since u(k) = 4, it is easy to see that u(k(T )) ≥ 8
This led to the following natural question
Does u(k) < ∞ imply u(k(T )) < ∞?
This problem is open and seems to be very difficult for general fields.
This remained open till late 90’s even for the p-adic field Qp.
We have the following
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Since u(k) = 4, it is easy to see that u(k(T )) ≥ 8
This led to the following natural question
Does u(k) < ∞ imply u(k(T )) < ∞?
This problem is open and seems to be very difficult for general fields.
This remained open till late 90’s even for the p-adic field Qp.
We have the following
Conjecture : If k is a p-adic field, then u(k(T )) = 8.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms This led to the following natural question
Does u(k) < ∞ imply u(k(T )) < ∞?
This problem is open and seems to be very difficult for general fields.
This remained open till late 90’s even for the p-adic field Qp.
We have the following
Conjecture : If k is a p-adic field, then u(k(T )) = 8.
Since u(k) = 4, it is easy to see that u(k(T )) ≥ 8
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms We have the following :
Theorem(Parimala - Suresh)(1998): Let k be a p-adic field. If p 6= 2, then u(k(X )) ≤ 10.
Till late 90’s it was not even known whether u(k(T )) is finite.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Theorem(Parimala - Suresh)(1998): Let k be a p-adic field. If p 6= 2, then u(k(X )) ≤ 10.
Till late 90’s it was not even known whether u(k(T )) is finite.
We have the following :
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Till late 90’s it was not even known whether u(k(T )) is finite.
We have the following :
Theorem(Parimala - Suresh)(1998): Let k be a p-adic field. If p 6= 2, then u(k(X )) ≤ 10.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 1999
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2000
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2001
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2002
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2003
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2004
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2005
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms 2006
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Theorem(Parimala - Suresh)(2007) : Let k be a p-adic field. If p 6= 2, then u(k(X )) = 8.
The proof of the above theorems uses many techniques from cohomology, arithematic geometry and Bruaer groups.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms The proof of the above theorems uses many techniques from cohomology, arithematic geometry and Bruaer groups.
Theorem(Parimala - Suresh)(2007) : Let k be a p-adic field. If p 6= 2, then u(k(X )) = 8.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms Theorem(Parimala - Suresh)(2007) : Let k be a p-adic field. If p 6= 2, then u(k(X )) = 8.
The proof of the above theorems uses many techniques from cohomology, arithematic geometry and Bruaer groups.
V. Suresh University Of Hyderabad Hyderabad Isotropy of quadratic forms