Intuitive definition: real group of rotations + reflections.

Mathematical definition that most directly represents this: the orthogonal group is the group of all matrices that preserve the dot product.

When viewed as matrices, it is the group of all matrices that preserve the dot product, i.e.:
This implies that it also preserves important geometric notions such as norm (intuitively: distance between two points) and angles.

$O(n)=O∈M(n)∣∀x,y,x_{T}y=(Ox)_{T}(Oy)$

This is perhaps the best "default definition".

We looking at the definition the orthogonal group is the group of all matrices that preserve the dot product, we notice that the dot product is one example of positive definite symmetric bilinear form, which in turn can also be represented by a matrix as shown at: Section "Matrix representation of a symmetric bilinear form".

By looking at this more general point of view, we could ask ourselves what happens to the group if instead of the dot product we took a more general bilinear form, e.g.:The answers to those questions are given by the Sylvester's law of inertia at Section "All indefinite orthogonal groups of matrices of equal metric signature are isomorphic".

- $I_{2}$: another positive definite symmetric bilinear form such as $(x_{1},x_{2})_{T}(y_{1},y_{2})=2x_{1}y_{1}+x_{2}y_{2}$?
- $I_{−}$ what if we drop the positive definite requirement, e.g. $(x_{1},x_{2})_{T}(y_{1},y_{2})=−x_{1}y_{1}+x_{2}y_{2}$?

Let's show that this definition is equivalent to the orthogonal group is the group of all matrices that preserve the dot product.

Note that:
and for that to be true for all possible $x$ and $y$ then we must have:
i.e. the matrix inverse is equal to the transpose.

$x_{T}y=(Ox)_{T}(Oy)=x_{T}O_{T}Oy$

$O_{T}O=I$

These matricese are called the orthogonal matrices.

TODO is there any more intuitive way to think about this?

Or equivalently, the set of rows is orthonormal, and so is the set of columns. TODO proof that it is equivalent to the orthogonal group is the group of all matrices that preserve the dot product.

The orthogonal group has 2 connected components:

- one with determinant +1, which is itself a subgroup known as the special orthogonal group. These are pure rotations without a reflection.
- the other with determinant -1. This is not a subgroup as it does not contain the origin. It represents rotations with a reflection.

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

- you take the first basis vector and move it to any other. You have therefore two angular parameters.
- you take the second one, and move it to be orthogonal to the first new vector. (you can choose a circle around the first new vector, and so you have another angular parameter.
- at last, for the last one, there are only two choices that are orthogonal to both previous ones, one in each direction. It is this directio, relative to the others, that determines the "has a reflection or not" thing

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

$O(n)≅SO(n)×C_{2}$

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

$O(1)≅SO(2)×C_{2}$

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.

$M=[21 32 ]$

$M_{T}M=[58 811 ]$

From Section "Lie algebra of a isometry group" we reach:

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.

We can reach it by taking the rotations in three directions, e.g. a rotation around the z axis:
then we derive and evaluate at 0:
$L_{z}$ therefore represents the infinitesimal rotation.

$R_{z}(θ)=⎣⎢⎡ cos(θ)sin(θ)0 −sin(θ)cos(θ)0 001 ⎦⎥⎤ $

$L_{z}=dθdR_{z}(θ) ∣∣∣∣∣ _{0}=⎣⎢⎡ −sin(θ)cos(θ)0 −cos(θ)−sin(θ)0 001 ⎦⎥⎤ ∣∣∣∣∣ _{0}=⎣⎢⎡ 010 −100 000 ⎦⎥⎤ $

Note that the exponential map reverses this and gives a finite rotation around the Z axis back from the infinitesimal generator $L_{z}$:

$e_{θL_{z}}=R_{z}(θ)$

Repeating the same process for the other directions gives:
We have now found 3 linearly independent elements of the Lie algebra, and since $SO(3)$ has dimension 3, we are done.

$L_{x}=TODOL_{y}=TODO$

Based on the $L_{x}$,$L_{y}$ and $L_{z}$ derived at Lie algebra of $SO(3)$ we can calculate the Lie bracket as:

$TODO$

Has $SU(2)$ as a double cover.

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.Diffeomorphic to the 3 sphere.

The $U(1)$ unitary group is one very over-generalized way of looking at it :-)

The complex analogue of the special orthogonal group, i.e. the subgroup of the unitary group with determinant equals exactly 1 instead of an arbitrary complex number with absolute value equal 1 as is the case for the unitary group.

Double cover of $SO(3)$.

Isomorphic to the quaternions.

Bibliography: