Ciro Santilli @cirosantilli 34

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:

The orthogonal group has 2 connected components:

It is instructive to visualize how the $\pm 1$ looks like in $SO(3)$:

As a result it is isomorphic to the direct product of the special orthogonal group by the cyclic group of order 2:

A low dimensional example:because you can only do two things: to flip or not to flip the line around zero.

Note that having the determinant plus or minus 1 is not a definition: there are non-orthogonal groups with determinant plus or minus 1. This is just a property. E.g.:has determinant 1, but:so $M$ is not orthogonal.

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.

Just like for the finite general linear group, the definition of special also works for finite fields, where 1 is the multiplicative identity!

Note that the definition of orthogonal group may not have such a clear finite analogue on the other hand.

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

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.

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

Members of the orthogonal group.

Group of rotations of a rigid body.

Like orthogonal group but without reflections. So it is a "special case" of the orthogonal group.

This is a subgroup of both the orthogonal group and the special linear group.

Intuition, please? Example? mathoverflow.net/questions/278641/intuition-for-symplectic-groups The key motivation seems to be related to Hamiltonian mechanics. The two arguments of the bilinear form correspond to each set of variables in Hamiltonian mechanics: the generalized positions and generalized momentums, which appear in the same number each.

Seems to be set of matrices that preserve a skew-symmetric bilinear form, which is comparable to the orthogonal group, which preserves a symmetric bilinear form. More precisely, the orthogonal group has:and its generalization the indefinite orthogonal group has:where S is symmetric. So for the symplectic group we have matrices Y such as:where A is antisymmetric. This is explained at: www.ucl.ac.uk/~ucahad0/7302_handout_13.pdf They also explain there that unlike as in the analogous orthogonal group, that definition ends up excluding determinant -1 automatically.

Therefore, just like the special orthogonal group, the symplectic group is also a subgroup of the special linear group.

Topology is the plumbing of calculus.

The key concept of topology is a neighbourhood.

Just by havin the notion of neighbourhood, concepts such as limit and continuity can be defined without the need to specify a precise numerical value to the distance between two points with a metric.

As an example. consider the orthogonal group, which is also naturally a topological space. That group does not usually have a notion of distance defined for it by default. However, we can still talk about certain properties of it, e.g. that the orthogonal group is compact, and that the orthogonal group has two connected components.

Group of the unitary matrices.

Complex analogue of the orthogonal group.

One notable difference from the orthogonal group however is that the unitary group is connected "because" its determinant is not fixed to two disconnected values 1/-1, but rather goes around in a continuous unit circle. $U(1)$ is the unit circle.