Register to reply

2 dimensional quadratic forms

by lavinia
Tags: dimensional, forms, quadratic
Share this thread:
lavinia
#1
Apr18-11, 05:54 AM
Sci Advisor
P: 1,722
I am told that the set of positive definite quadratic forms on R^2 has a metric that turns it into

H x R where H is the hyperbolic plane. Can you describe this metric?

* As a space the forms are viewed as GL(2,R)/O(2).
Phys.Org News Partner Science news on Phys.org
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
fzero
#2
Apr18-11, 02:38 PM
Sci Advisor
HW Helper
PF Gold
P: 2,602
Let's note that there's a Gauss-type decomposition of [tex]GL(n,\mathbb{R})[/tex] matrices as

[tex] \mathcal{M} = U D \mathcal{O}, [/tex]

where [tex]U[/tex] is upper-triangular (with 1s on the diagonal), [tex]D[/tex] is diagonal (with [tex]\det{D}>0[/tex]) and [tex] \mathcal{O}[/tex] is orthogonal (in [tex]O(n)[/tex]). I claim that

[tex] M = \begin{pmatrix} 1 & \sqrt{2} x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} r y & 0\\ 0 & r/y \end{pmatrix},[/tex]

is the general element of the coset [tex]GL(2,\mathbb{R})/O(2)[/tex] provided that [tex]r,y>0[/tex].

We can place the usual left-invariant metric on [tex]GL(2,\mathbb{R})[/tex], which will induce the metric

[tex]ds^2 = \frac{1}{2} \text{Tr} \left[ (M^{-1} dM) (M^{-1} dM) ^T\right],[/tex]

[tex] = \left( \frac{dr}{r}\right)^2 + \frac{1}{y^2} \left( dx^2 + dy^2 \right)[/tex]

on the coset. The space is therefore [tex] GL(2,\mathbb{R})/O(2) = H \times \mathbb{R}^+[/tex]. Note that this is consistent with [tex] H = SL(2,\mathbb{R})/SO(2)[/tex] and [tex]\mathbb{R}^+ = \mathbb{R}/\{ 1,-1\}[/tex].
lavinia
#3
Apr19-11, 10:11 PM
Sci Advisor
P: 1,722
Quote Quote by fzero View Post
Let's note that there's a Gauss-type decomposition of [tex]GL(n,\mathbb{R})[/tex] matrices as

[tex] \mathcal{M} = U D \mathcal{O}, [/tex]

where [tex]U[/tex] is upper-triangular (with 1s on the diagonal), [tex]D[/tex] is diagonal (with [tex]\det{D}>0[/tex]) and [tex] \mathcal{O}[/tex] is orthogonal (in [tex]O(n)[/tex]). I claim that

[tex] M = \begin{pmatrix} 1 & \sqrt{2} x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} r y & 0\\ 0 & r/y \end{pmatrix},[/tex]

is the general element of the coset [tex]GL(2,\mathbb{R})/O(2)[/tex] provided that [tex]r,y>0[/tex].

We can place the usual left-invariant metric on [tex]GL(2,\mathbb{R})[/tex], which will induce the metric

[tex]ds^2 = \frac{1}{2} \text{Tr} \left[ (M^{-1} dM) (M^{-1} dM) ^T\right],[/tex]

[tex] = \left( \frac{dr}{r}\right)^2 + \frac{1}{y^2} \left( dx^2 + dy^2 \right)[/tex]

on the coset. The space is therefore [tex] GL(2,\mathbb{R})/O(2) = H \times \mathbb{R}^+[/tex]. Note that this is consistent with [tex] H = SL(2,\mathbb{R})/SO(2)[/tex] and [tex]\mathbb{R}^+ = \mathbb{R}/\{ 1,-1\}[/tex].
what is the usual left invariant metric on [tex]GL(2,\mathbb{R})[/tex]?

fzero
#4
Apr19-11, 10:58 PM
Sci Advisor
HW Helper
PF Gold
P: 2,602
2 dimensional quadratic forms

Quote Quote by lavinia View Post
what is the usual left invariant metric on [tex]GL(2,\mathbb{R})[/tex]?
It's

[tex]
ds^2 = \frac{1}{2} \text{Tr} \left[ (\mathcal{M}^{-1} d\mathcal{M}) (\mathcal{M}^{-1} d\mathcal{M}) ^T\right].
[/tex]

The normalization is put in for convenience. [tex]\mathcal{M}^{-1} d\mathcal{M}[/tex] is the Maurer-Cartan form.
lavinia
#5
Apr20-11, 08:10 AM
Sci Advisor
P: 1,722
Quote Quote by fzero View Post
It's

[tex]
ds^2 = \frac{1}{2} \text{Tr} \left[ (\mathcal{M}^{-1} d\mathcal{M}) (\mathcal{M}^{-1} d\mathcal{M}) ^T\right].
[/tex]

The normalization is put in for convenience. [tex]\mathcal{M}^{-1} d\mathcal{M}[/tex] is the Maurer-Cartan form.
Thanks. I will do some reading about this.

Do you have a simple geometric picture of this metric? Start with an ellipse centered at the origin of R^2. Map it onto the Poincare disk somehow. Then maybe the factor of R comes from uniformly scaling the ellipses.
fzero
#6
Apr20-11, 09:40 AM
Sci Advisor
HW Helper
PF Gold
P: 2,602
Quote Quote by lavinia View Post
Thanks. I will do some reading about this.

Do you have a simple geometric picture of this metric? Start with an ellipse centered at the origin of R^2, map it onto the Poincare disk somehow. Then maybe the factor of R comes from uniformly scaling the ellipses.
The factor of [tex]\mathbb{R}^+[/tex] corresponds to the determinant of a matrix in the component [tex]GL(n,\mathbb{R})^+[/tex] connected to the identity. The metric we found is consistent with the well known fact that [tex]GL(n,\mathbb{R})^+=SL(n,\mathbb{R})\times \mathbb{R}^+[/tex].
lavinia
#7
Apr22-11, 04:27 AM
Sci Advisor
P: 1,722
Quote Quote by fzero View Post
The factor of [tex]\mathbb{R}^+[/tex] corresponds to the determinant of a matrix in the component [tex]GL(n,\mathbb{R})^+[/tex] connected to the identity. The metric we found is consistent with the well known fact that [tex]GL(n,\mathbb{R})^+=SL(n,\mathbb{R})\times \mathbb{R}^+[/tex].
Ok. This is helpful.

Does this picture work?

Each positive definite quadratic form determines an ellipse, its ellipse of unit vectors.

An ellipse has three parameters, the lengths of its two axes and its angle of tilt to the xy-coordinate axes.
The subset that are rotations of ellipses with tilt angle zero and fixed x-axis length, (They are rotations of ellipses with equation, x^2 + y^2/b^2 = K) form a two parameter family.
(By convention, angles of rotation can be chosen to be between 0 and pi (the interval, [0,pi), and the first axis can be chosen to be the one that lines up with the x-axis when the rotation is undone. )

For each ellipse in this 2 parameter family, map it to the point in the Poincare disk whose angle is twice the tilt angle and whose Euclidean distance to the origin is |k1 - k2|/k1 + k2 where k1 and k2 are the lengths of its axes.

This two parameter family looks like the Poincare disk under this mapping. As the x-axis length is varied to account for all positive definite quadratic forms, the entire set looks like the Cartesian product of the Poincare disk with the open half line of positive real numbers.

Is this the correct picture?

What is the importance of this metric for understanding the space of positive definite quadratic forms?
fzero
#8
Apr23-11, 03:36 PM
Sci Advisor
HW Helper
PF Gold
P: 2,602
I would identify your K with [tex]\mathbb{R}^+[/tex] in the coset, then the ratio of axis lengths b and tilt angle form the UHP. I think you're still mapping the ratio and tilt angle to the Poincare disk, so you might find that the metric on your space is in the same conformal class as the one on the UHP.

As far as the importance of the metric, I have no specific applications in mind for doing geometry on the space of positive definite quadratic forms. That is not to say that there aren't any. I just happened to be aware of some Lie group geometry that could shed some light on your original question.


Register to reply

Related Discussions
Quadratic Forms (Sum) Calculus & Beyond Homework 2
Quadratic Forms for SL(2;R) Advanced Physics Homework 1
Quadratic forms Linear & Abstract Algebra 2
Quadratic forms Calculus & Beyond Homework 3
Quadratic Forms Calculus & Beyond Homework 0