Lorentz group ($SO(1,3)$)

Subgroup of the Poincaré group without translations. Therefore, in those, the spacetime origin is always fixed.

Or in other words, it is as if two observers had their space and time origins at the exact same place. However, their space axes may be rotated, and one may be at a relative speed to the other to create a Lorentz boost. Note however that if they are at relative speeds to one another, then their axes will immediately stop being at the same location in the next moment of time, so things are only valid infinitesimally in that case.

This group is made up of matrix multiplication alone, no need to add the offset vector: space rotations and Lorentz boost only spin around and bend things around the origin.

One definition: set of all 4x4 matrices that keep the Minkowski inner product, mentioned at Physics from Symmetry by Jakob Schwichtenberg (2015) page 63. This then implies:

Physics from Symmetry by Jakob Schwichtenberg (2015) page 66 shows one in terms of 4x4 complex matrices.

More importantly though, are the representations of the Lie algebra of the Lorentz group, which are generally also just also called "Representation of the Lorentz group" since you can reach the representation from the algebra via the exponential map.

Bibliography:

One of the representations of the Lorentz group that show up in the Representation theory of the Lorentz group.

TODO understand a bit more intuitively.

Two observers travel at fixed speed relative to each other. They synchronize origins at x=0 and t=0, and their spacial axes are perfectly aligned. This is a subset of the Lorentz group. TODO confirm it does not form a subgroup however.

Generalization of orthogonal group to preserve different bilinear forms. Important because the Lorentz group is $SO(1,3)$.

Given a matrix $A$ with metric signature containing $m$ positive and $n$ negative entries, the indefinite orthogonal group is the set of all matrices that preserve the associated bilinear form, i.e.: Note that if $A=I$, we just have the standard dot product, and that subcase corresponds to the following definition of the orthogonal group: Section "The orthogonal group is the group of all matrices that preserve the dot product".

As shown at all indefinite orthogonal groups of matrices of equal metric signature are isomorphic, due to the Sylvester's law of inertia, only the metric signature of $A$ matters. E.g., if we take two different matrices with the same metric signature such as: and: both produce isomorphic spaces. So it is customary to just always pick the matrix with only +1 and -1 as entries.

Following the definition of the indefinite orthogonal group, we want to show that only the metric signature matters.

First we can observe that the exact matrices are different. For example, taking the standard matrix of $O(2)$: and: both have the same metric signature. However, we notice that a rotation of 90 degrees, which preserves the first form, does not preserve the second one! E.g. consider the vector $x=(1,0)$, then $x\cdot x=1$. But after a rotation of 90 degrees, it becomes $x_2=(0,1)$, and now $x_2\cdot x_2=2$! Therefore, we have to search for an isomorphism between the two sets of matrices.

For example, consider the orthogonal group, which can be defined as shown at the orthogonal group is the group of all matrices that preserve the dot product can be defined as:

Like the special orthogonal group is to the orthogonal group, $SO(m,n)$ is the subset of $O(m,n)$ with determinant equal to exactly 1.