Understanding the Lorentz Group: What does O(1,3) mean?

[dimension10]

I am totally confused about the Lorentz Group at the moment. According to wikipedia, the Lorentz group can be defined as the General Orthogonal Lie Group$O(1,3)$. However, the definition of the GO Lie Group that I know only works when there is a single number inside the bracket, not 2, e.g. $O(1)$. So, what does $O(1,3)$ mean? Thanks in advance.

 

[Bill_K]

The orthogonal group is the set of all matrices Λ that preserve the symmetric metric η under η' = ΛηΛ^T. Here 3,1 is the signature of η, that is η = Diag(1,1,1,-1).

 

[tom.stoer]

If you look the SO(3,1) in 4*4 matrix notation you will find that in the matrix elements L_αβ with α=0 and/or β=0 the sin(θ) or cos(θ) for rotations is replaced by sinh(η) or cosh(η) for boosts. So you can interpret the the "1" in SO(3,1) as one direction with imaginary rotations transforming sin, cos to sinh, cosh.

The invariant scalar product

<x,y> = x₁y₁ + x₂y₂ + x₃y₃ + ...

is replaced with

<x,y> = -x₀y₀ + x₁y₁ + x₂y₂ + ...

 

[Ger]

http://www.math.sunysb.edu/~kirillov/mat552/liegroups.pdf. Uses also two indices, parameters in the brackets in O(n,ℝ)

 

[vanhees71]

The name $\mathrm{O}(1,3)$ means (pseudo-)orthogonal group wrt. the fundamental bilinear form with one positive and three negative principle values in $\mathbb{R}^4$. In components with respect to (pseudo-)orthonormal vectors this scalar produkt reads
$$x \cdot y=\eta_{\mu \nu} x^{\mu} y^{\nu}=x^0 y^0-x^1 y^1-x^2 y^2-x^3 y^3.$$
Then the Lorentz transformations are represented by such matrices that leave this bilinear form invariant for all vectors.

There are important subgroups. The most important one is the special orthochronous Lorentz group, $\mathrm{SO}(1,3)^{\uparrow}$, which is continously connected with the identity matrix. That's the symmetry group of the special relativistic spacetime manifold. The special orthonormal Lorentz group consists of all matrices, leaving the above explained Minkowski product invariant for any pair of vectors, have determinant 1, and for which ${\Lambda^0}_0 \geq +1$. Since the zeroth component of the four vectors denote time this latter restriction means that the transformation doesn't flip the direction of time.

 

[dimension10]

The orthogonal group is the set of all matrices Λ that preserve the symmetric metric η under η' = ΛηΛ^T. Here 3,1 is the signature of η, that is η = Diag(1,1,1,-1).

If you look the SO(3,1) in 4*4 matrix notation you will find that in the matrix elements L_αβ with α=0 and/or β=0 the sin(θ) or cos(θ) for rotations is replaced by sinh(η) or cosh(η) for boosts. So you can interpret the the "1" in SO(3,1) as one direction with imaginary rotations transforming sin, cos to sinh, cosh.

The invariant scalar product

<x,y> = x₁y₁ + x₂y₂ + x₃y₃ + ...

is replaced with

<x,y> = -x₀y₀ + x₁y₁ + x₂y₂ + ...

The name $\mathrm{O}(1,3)$ means (pseudo-)orthogonal group wrt. the fundamental bilinear form with one positive and three negative principle values in $\mathbb{R}^4$. In components with respect to (pseudo-)orthonormal vectors this scalar produkt reads
$$x \cdot y=\eta_{\mu \nu} x^{\mu} y^{\nu}=x^0 y^0-x^1 y^1-x^2 y^2-x^3 y^3.$$
Then the Lorentz transformations are represented by such matrices that leave this bilinear form invariant for all vectors.

There are important subgroups. The most important one is the special orthochronous Lorentz group, $\mathrm{SO}(1,3)^{\uparrow}$, which is continously connected with the identity matrix. That's the symmetry group of the special relativistic spacetime manifold. The special orthonormal Lorentz group consists of all matrices, leaving the above explained Minkowski product invariant for any pair of vectors, have determinant 1, and for which ${\Lambda^0}_0 \geq +1$. Since the zeroth component of the four vectors denote time this latter restriction means that the transformation doesn't flip the direction of time.

Thanks, everyone. I finally got it.

http://www.math.sunysb.edu/~kirillov/mat552/liegroups.pdf. Uses also two indices, parameters in the brackets in O(n,ℝ)

Actually, I think this is a different thing. $O(n,\mathbb R)$ means $O(n)$ over the real numbers $\mathbb R$, I think. Correct me if I'm wrong.