## Lorentz group definition

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.
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 Blog Entries: 1 Recognitions: Science Advisor The orthogonal group is the set of all matrices Λ that preserve the symmetric metric η under η' = ΛηΛT. Here 3,1 is the signature of η, that is η = Diag(1,1,1,-1).
 Recognitions: Science Advisor If you look the SO(3,1) in 4*4 matrix notation you will find that in the matrix elements Lαβ 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 = x1y1 + x2y2 + x3y3 + ... is replaced with = -x0y0 + x1y1 + x2y2 + ...

## Lorentz group definition

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

 Quote by Bill_K The orthogonal group is the set of all matrices Λ that preserve the symmetric metric η under η' = ΛηΛT. Here 3,1 is the signature of η, that is η = Diag(1,1,1,-1).
 Quote by tom.stoer If you look the SO(3,1) in 4*4 matrix notation you will find that in the matrix elements Lαβ 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 = x1y1 + x2y2 + x3y3 + ... is replaced with = -x0y0 + x1y1 + x2y2 + ...
 Quote by 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.
Thanks, everyone. I finally got it.

 Quote by Ger http://www.math.sunysb.edu/~kirillov.../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.

 Tags abstract algebra, lie algebra, lie group, lorentz group, lorentz transform