Ciro Santilli @cirosantilli 34

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

The group of all transformations that preserve some bilinear form, notable examples:

Like everything else in Lie groups, first start with the matrix as discussed at Section "Lie algebra of a matrix Lie group".

Intuitively, a Lie algebra is a simpler object than a Lie group. Without any extra structure, groups can be very complicated non-linear objects. But a Lie algebra is just an algebra over a field, and one with a restricted bilinear map called the Lie bracket, that has to also be alternating and satisfy the Jacobi identity.

Another important way to think about Lie algebras, is as infinitesimal generators.

Because of the Lie group-Lie algebra correspondence, we know that there is almost a bijection between each Lie group and the corresponding Lie algebra. So it makes sense to try and study the algebra instead of the group itself whenever possible, to try and get insight and proofs in that simpler framework. This is the key reason why people study Lie algebras. One is philosophically reminded of how normal subgroups are a simpler representation of group homomorphisms.

To make things even simpler, because all vector spaces of the same dimension on a given field are isomorphic, the only things we need to specify a Lie group through a Lie algebra are:Note that the Lie bracket can look different under different basis of the Lie algebra however. This is shown for example at Physics from Symmetry by Jakob Schwichtenberg (2015) page 71 for the Lorentz group.

As mentioned at Lie Groups, Physics, and Geometry by Robert Gilmore (2008) Chapter 4 "Lie Algebras", taking the Lie algebra around the identity is mostly a convention, we could treat any other point, and things are more or less equivalent.

Bibliography:

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.

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

Basically, a "representation" means associating each group element as an invertible matrices, i.e. a matrix in (possibly some subset of) $GL(n)$, that has the same properties as the group.

Or in other words, associating to the more abstract notion of a group more concrete objects with which we are familiar (e.g. a matrix).

Each such matrix then represents one specific element of the group.

This is basically what everyone does (or should do!) when starting to study Lie groups: we start looking at matrix Lie groups, which are very concrete.

Or more precisely, mapping each group element to a linear map over some vector field $V$ (which can be represented by a matrix infinite dimension), in a way that respects the group operations:

As shown at Physics from Symmetry by Jakob Schwichtenberg (2015)

Motivation:

Bibliography:

I like relativistic quantum mechanics.

Best mathematical explanation: Section "Spin comes naturally when adding relativity to quantum mechanics".

Physics from Symmetry by Jakob Schwichtenberg (2015) chapter 3.9 "Elementary particles" has an amazing summary of the preceding chapters the spin value has a relation to the representations of the Lorentz group, which encodes the spacetime symmetry that each particle observes. These symmetries can be characterized by small integer numbers:

Bibliography:

The main interest of this theorem is in classifying the indefinite orthogonal groups, which in turn is fundamental because the Lorentz group is an indefinite orthogonal groups, see: all indefinite orthogonal groups of matrices of equal metric signature are isomorphic.

It also tells us that a change of basis does not the alter the metric signature of a bilinear form, see matrix congruence can be seen as the change of basis of a bilinear form.

The theorem states that the number of 0, 1 and -1 in the metric signature is the same for two symmetric matrices that are congruent matrices.

For example, consider:

The eigenvalues of $A$ are $1$ and $4$, and the associated eigenvectors are:symPy code:and from the eigendecomposition of a real symmetric matrix we know that:

Now, instead of $P$, we could use $PE$, where $E$ is an arbitrary diagonal matrix of type:With this, would reach a new matrix $B$:Therefore, with this congruence, we are able to multiply the eigenvalues of $A$ by any positive number $e_1^2$ and $e_2^2$. Since we are multiplying by two arbitrary positive numbers, we cannot change the signs of the original eigenvalues, and so the metric signature is maintained, but respecting that any value can be reached.

Note that the matrix congruence relation looks a bit like the eigendecomposition of a matrix:but note that $D$ does not have to contain eigenvalues, unlike the eigendecomposition of a matrix. This is because here $S$ is not fixed to having eigenvectors in its columns.

But because the matrix is symmetric however, we could always choose $S$ to actually diagonalize as mentioned at eigendecomposition of a real symmetric matrix. Therefore, the metric signature can be seen directly from eigenvalues.

Also, because $D$ is a diagonal matrix, and thus symmetric, it must be that:

What this does represent, is a general change of basis that maintains the matrix a symmetric matrix.

Related: