can be reached in a unique way (i.e. is the exponential map a bijection). By looking at the matrix logarithm however we conclude that this is not the case for real matrices, but it is for complex matrices.

Examples:

TODO example it can be seen that the Lie algebra is not closed matrix multiplication, even though the corresponding group is by definition. But it is closed under the Lie bracket operation.

Lie bracket of a matrix Lie group

0 0

[X, Y] = X Y - Y X

(1)

This makes it clear how the Lie bracket can be seen as a "measure of non-commutativity"

Because the Lie bracket has to be a bilinear map, all we need to do to specify it uniquely is to specify how it acts on every pair of some basis of the Lie algebra.

Then, together with the Baker-Campbell-Hausdorff formula and the Lie group-Lie algebra correspondence, this forms an exceptionally compact description of a Lie group.

One parameter subgroup

0 0

The one parameter subgroup of a Lie group for a given element

M

of its Lie algebra is a subgroup of

G

given by:

e^{t M} \in G ∣ t \in R

(1)

Intuitively,

M

is a direction, and

t

is how far we move along a given direction. This intuition is especially vivid in for example in the case of the Lie algebra of

S O (3)

, the rotation group.

One parameter subgroups can be seen as the continuous analogue to the cycle of an element of a group.

Classical group

0 0

Symplectic group ( $S p (n, F)$ )

0 0

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:

O^{T} I O = I

(1)

and its generalization the indefinite orthogonal group has:

O^{T} S O = I

(2)

where S is symmetric. So for the symplectic group we have matrices Y such as:

Y^{T} A Y = I

(3)

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.

Symplectic matrix

1 0

Unitary symplectic group ( $S p (n)$ )

0 0

General linear group ( $G L (n)$ , $G L (n, F)$ )

1 0

Invertible matrices. Or if you think a bit more generally, an invertible linear map.

When the field

F

is not given, it defaults to the real numbers.

Non-invertible are excluded "because" otherwise it would not form a group (every element must have an inverse). This is therefore the largest possible group under matrix multiplication, other matrix multiplication groups being subgroups of it.

Finite general linear group ( $G L (n, F_{m})$ , $G L (n, m)$ )

0 0

general linear group over a finite field of order

m

. Remember that due to the classification of finite fields, there is one single field for each prime power

m

Exactly as over the real numbers, you just put the finite field elements into a

n \times n

matrix, and then take the invertible ones.

Lie algebra of $G L (n)$

0 0

Is the set of all n-by-y square matrices.

Because

G L (n)

is a Lie group we can use Section "Lie algebra of a matrix Lie group".

For every matrix

x

in the set of all n-by-y square matrices

M_{n}

e^{x}

has inverse

e^{-} x

Note that this works even if

x

is not invertible, and therefore not in

G L (n)

Therefore, the Lie algebra of

G L (n)

is the entire

M_{n}

Special linear group ( $S L (n)$ )

1 0

Specials sub case of the general linear group when the determinant equals exactly 1.

Special linear group of dimension 2 ( $S L (2)$ )

0 0

Lie algebra of $S L (n)$

0 0

Lie algebra of $S L (2)$

0 0

This is a good first concrete example of a Lie algebra. Shown at Lie Groups, Physics, and Geometry by Robert Gilmore (2008) Chapter 4.2 "How to linearize a Lie Group" has an example.

We can use use the following parametrization of the special linear group on variables

x

y

and

z

M = [1 + x z y (1 + y z) / (1 + x)]

(1)

Every element with this parametrization has determinant 1:

d e t (M) = (1 + x) (1 + y z) / (1 + x) - y z = 1

(2)

Furthermore, any element can be reached, because by independently settting

x

y

and

z

M_{00}

M_{01}

and

M_{10}

can have any value, and once those three are set,

M_{11}

is fixed by the determinant.

To find the elements of the Lie algebra, we evaluate the derivative on each parameter at 0:

M_{x} M_{y} M_{z} = \frac{d M}{d x} ∣ ∣ ∣ ∣ ∣_{(x, y, z) = (0, 0, 0)} = \frac{d M}{d y} ∣ ∣ ∣ ∣ ∣_{(x, y, z) = (0, 0, 0)} = \frac{d M}{d z} ∣ ∣ ∣ ∣ ∣_{(x, y, z) = (0, 0, 0)} = [10 0 - (1 + y z) / (1 + x)^{2}] ∣ ∣ ∣ ∣ ∣_{(x, y, z) = (0, 0, 0)} = [00 1 z / (1 + x)] ∣ ∣ ∣ ∣ ∣_{(x, y, z) = (0, 0, 0)} = [01 0 y / (1 + x)] ∣ ∣ ∣ ∣ ∣_{(x, y, z) = (0, 0, 0)} = [10 0 - 1] = [0010] = [0100]

(3)

Remembering that the Lie bracket of a matrix Lie group is really simple, we can then observe the following Lie bracket relations between them:

[M_{x}, M_{y}] [M_{x}, M_{z}] [M_{y}, M_{z}] = M_{x} M_{y} - M_{y} M_{x} = M_{x} M_{z} - M_{z} M_{x} = M_{y} M_{z} - M_{z} M_{y} = [0010] = [0 - 1 00] = [1000] - [00 - 1 0] - [0100] - [0001] = [0020] = [0 - 2 00] = [10 0 - 1] = 2 M_{y} = - 2 M_{z} = M_{x}

(4)

One key thing to note is that the specific matrices

M_{x}

M_{y}

and

M_{z}

are not really fundamental: we could easily have had different matrices if we had chosen any other parametrization of the group.

TODO confirm: however, no matter which parametrization we choose, the Lie bracket relations between the three elements would always be the same, since it is the number of elements, and the definition of the Lie bracket, that is truly fundamental.

Lie Groups, Physics, and Geometry by Robert Gilmore (2008) Chapter 4.2 "How to linearize a Lie Group" then calculates the exponential map of the vector

x M_{x} + y M_{y} + z M_{z}

as:

I c o s h (θ) + M_{x} s i n h (θ) / θ

(5)

with:

θ^{2} = x^{2} + b c

(6)

TODO now the natural question is: can we cover the entire Lie group with this exponential? Lie Groups, Physics, and Geometry by Robert Gilmore (2008) Chapter 7 "EXPonentiation" explains why not.

Finite special general linear group ( $S L (n, m)$ )

0 0

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.

Isometry group

1 0

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

orthogonal group preserves the inner product
unitary group preserves a Hermitian form
Lorentz group preserves the Minkowski inner product

Lie algebra of a isometry group

0 0

We can almost reach the Lie algebra of any isometry group in a single go. For every

X

in the Lie algebra we must have:

\forall v, w \in V, t \in R (e^{t X} v ∣ e^{t X} w) = (v ∣ w)

(1)

because

e^{t X}

has to be in the isometry group by definition as shown at Section "Lie algebra of a matrix Lie group".

Then:

\frac{d ( e ^{t X} v ∣ e ^{t X} w )}{d t} ∣ ∣ ∣ ∣ ∣_{0} = 0 ⟹ (X e^{t X} v ∣ e^{t X} w) + (e^{t X} v ∣ X e^{t X} w) ∣ ∣ ∣ ∣ ∣_{0} = 0 ⟹ (X v ∣ w) + (v ∣ X w) = 0

(2)

so we reach:

\forall v, w \in V (X v ∣ w) = - (v ∣ X w)

(3)

With this relation, we can easily determine the Lie algebra of common isometries:

Lie algebra of $O (n)$

Bibliography:

An Introduction to Tensors and Group Theory for Physicists by Nadir Jeevanjee (2011) page 151

Orthogonal group ( $O (n)$ )

0 0

Definition of the orthogonal group

0 0

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.

The orthogonal group is the group of all matrices that preserve the dot product

0 0

When viewed as matrices, it is the group of all matrices that preserve the dot product, i.e.:

O (n) = O \in M (n) ∣ \forall x, y, x^{T} y = (O x)^{T} (O y)

(1)

This implies that it also preserves important geometric notions such as norm (intuitively: distance between two points) and angles.

This is perhaps the best "default definition".

What happens to the definition of the orthogonal group if we choose other types of symmetric bilinear forms

0 0

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.:

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

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".

The orthogonal group is the group of all invertible matrices where the inverse is equal to the transpose

0 0

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:

x^{T} y = (O x)^{T} (O y) = x^{T} O^{T} O y

(1)

and for that to be true for all possible

x

and

y

then we must have:

O^{T} O = I

(2)

i.e. the matrix inverse is equal to the transpose.

Conversely, if:

O^{T} O = I

(3)

is true, then

(O x)^{T} (O y) = x^{T} (O^{T} O) y = x^{T} y

(4)

These matricese are called the orthogonal matrices.

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

Elements of the orthogonal group have determinant plus or minus one

0 0

the orthogonal group is the group of all invertible matrices where the inverse is equal to the transpose

The orthogonal group is the group of all matrices with orthonormal rows and orthonormal columns

0 0

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.

Topology of the orthogonal group

0 0

The orthogonal group is compact

0 0

Connected components of the orthogonal group (The orthogonal group has two connected components)

0 0

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

\pm 1

looks like in

S O (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) ≅ S O (n) \times C_{2}

(1)

A low dimensional example:

O (1) ≅ S O (2) \times C_{2}

(2)

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.:

M = [2132]

(3)

has determinant 1, but:

M^{T} M = [58811]

(4)

M

is not orthogonal.

Lie algebra of $O (n)$

0 0

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

Special orthogonal group ( $S O (n)$ , Rotation group)

0 0

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.

Lie algebra of $S O (3)$

0 0

We can reach it by taking the rotations in three directions, e.g. a rotation around the z axis:

R_{z} (θ) = ⎣ ⎢ ⎡ c o s (θ) s i n (θ) 0 - s i n (θ) c o s (θ) 0 001 ⎦ ⎥ ⎤

(1)

then we derive and evaluate at 0:

L_{z} = \frac{d R _{z} ( θ )}{d θ} ∣ ∣ ∣ ∣ ∣_{0} = ⎣ ⎢ ⎡ - s i n (θ) c o s (θ) 0 - c o s (θ) - s i n (θ) 0 001 ⎦ ⎥ ⎤ ∣ ∣ ∣ ∣ ∣_{0} = ⎣ ⎢ ⎡ 010 - 1 00 000 ⎦ ⎥ ⎤

(2)

L_{z}

therefore represents the infinitesimal rotation.

Note that the exponential map reverses this and gives a finite rotation around the Z axis back from the infinitesimal generator

L_{z}

e^{θ L_{z}} = R_{z} (θ)

(3)

Repeating the same process for the other directions gives:

L_{x} = T O D O L_{y} = T O D O

(4)

We have now found 3 linearly independent elements of the Lie algebra, and since

S O (3)

has dimension 3, we are done.

Lie bracket of the rotation group

0 0

Based on the

L_{x}

L_{y}

and

L_{z}

derived at Lie algebra of

S O (3)

we can calculate the Lie bracket as:

T O D O

(1)

3D rotation group ( $S O (3)$ )

0 0

Has

S U (2)

as a double cover.

Unitary group ( $U (n)$ )

0 0

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.

Unitary group of degree 1 ( $U (1)$ )

0 0

Unitary group of degree 2 ( $U (2)$ )

0 0

Diffeomorphic to the 3 sphere.

Unit circle

1 0

The

U (1)

unitary group is one very over-generalized way of looking at it :-)

Special unitary group ( $S U (n)$ )

1 0

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.

Special unitary of degree 2 ( $S U (2)$ )

0 0

en.wikipedia.org/wiki/Representation_theory_of_SU(2)

Double cover of

S O (3)

Isomorphic to the quaternions.

Representations of $S U (2)$

0 0

Lie algebra of $S U (2)$

0 0

Bibliography:

Physics from Symmetry by Jakob Schwichtenberg (2015) page 54.

2D representation of $S U (2)$

0 0

Pauli matrix.

Projective linear group

1 0

TODO motivation. Motivation. Motivation. Motivation. The definitin with quotient group is easy to understand.

Finite projective linear group ( $P G L (q, p)$ )

0 0

Projective special linear group

0 0

Finite projective special linear group ( $P S L (p, q)$ )

0 0

$P S L (2, p)$

0 0

PSL(2,7)

1 0

The second smallest non-Abelian finite simple group after the alternating group of degree 5.

Poincaré group

1 0

Full set of all possible special relativity symmetries:

translations in space and time
rotations in space
Lorentz boosts

In simple and concrete terms. Suppose you observe N particles following different trajectories in Spacetime.

There are two observers traveling at constant speed relative to each other, and so they see different trajectories for those particles:

space and time shifts, because their space origin and time origin (time they consider 0, i.e. when they started their timers) are not synchronized. This can be modelled with a 4-vector addition.
their space axes are rotated relative to one another. This can be modelled with a 4x4 matrix multiplication.
and they are moving relative to each other, which leads to the usual spacetime interactions of special relativity. Also modelled with a 4x4 matrix multiplication.

Note that the first two types of transformation are exactly the non-relativistic Galilean transformations.

The Poincare group is the set of all matrices such that such a relationship like this exists between two frames of reference.

Galilean transformation

0 0

Translation (geometry)

1 0

Subset of Galilean transformation with speed equals 0.

Translation group

0 0

This is a good and simple first example of Lie algebra to look into.

The derivative is the generator of the translation group

0 0

Take the group of all Translation in

R^{1}

Let's see how the generator of this group is the derivative operator:

\frac{\partial}{\partial x}

(1)

The way to think about this is:

the translation group operates on the argument of a function $f (x)$
the generator is an operator that operates on $f$ itself

So let's take the exponential map:

e^{x_{0} \frac{\partial}{\partial x}} f (x) = (1 + x_{0} \frac{\partial}{\partial x} + x_{0}^{2} \frac{\partial ^{2}}{\partial x ^{2}} + \dots) f (x)

(2)

and we notice that this is exactly the Taylor series of

f (x)

around the identity element of the translation group, which is 0! Therefore, if

f (x)

behaves nicely enough, within some radius of convergence around the origin we have for finite

x_{0}

e^{x_{0} \frac{\partial}{\partial x}} f (x) = f (x + x_{0})

(3)

This example shows clearly how the exponential map applied to a (differential) operator can generate finite (non-infinitesimal) Translation!

Galilean invariance

1 0

A law of physics is Galilean invariant if the same formula works both when you are standing still on land, or when you are on a boat moving at constant velocity.

For example, if we were describing the movement of a point particle, the exact same formulas that predict the evolution of

x_{l a n d} (t)

must also predict

x_{b o a t} (t)

, even though of course both of those

x (t)

will have different values.

It would be extremely unsatisfactory if the formulas of the laws of physics did not obey Galilean invariance. Especially if you remember that Earth is travelling extremelly fast relative to the Sun. If there was no such invariance, that would mean for example that the laws of physics would be different in other planets that are moving at different speeds. That would be a strong sign that our laws of physics are not complete.

The consequence/cause of that is that you cannot know if you are moving at a constant speed or not.

Lorentz invariance generalizes Galilean invariance to also account for special relativity, in which a more complicated invariant that also takes into account different times observed in different inertial frames of reference is also taken into account. But the fundamental desire for the Lorentz invariance of the laws of physics remains the same.

Covariance

0 0

Generally means that he form of the equation

f (x)

does not change if we transform

x

This is generally what we want from the laws of physics.

E.g. a Galilean transformation generally changes the exact values of coordinates, but not the form of the laws of physics themselves.

Lorentz covariance is the main context under which the word "covariant" appears, because we really don't want the form of the equations to change under Lorentz transforms, and "covariance" is often used as a synonym of "Lorentz covariance".

TODO some sources distinguish "invariant" from "covariant": invariant vs covariant.

Invariant vs covariant

0 0

Some sources distinguish "invariant" from "covariant" such that under some transformation (typically Lie group):

invariant: the value of $f (x)$ does not change if we transform $x$
covariant: the form of the equation $f (x)$ does not change if we transform $x$ .

TODO examples.

Bibliography:

Lorentz group ( $S O (1, 3)$ )

1 0

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:

Λ^{T} η Λ = η

(1)

Representation theory of the Lorentz group

1 0

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:

Physics from Symmetry by Jakob Schwichtenberg (2015) chapter 3.7 "The Lorentz Group O (1, 3)"

Representation of the Lorentz group

0 0

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

Lie algebra of the Lorentz group

0 0

Spinor

1 0

TODO understand a bit more intuitively.

Lorentz boost

0 0

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.

Indefinite orthogonal group ( $O (m, n)$ )

0 0

Generalization of orthogonal group to preserve different bilinear forms. Important because the Lorentz group is

S O (1, 3)

Definition of the indefinite orthogonal group

0 0

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.:

O (m, n) = O \in M (m + n) ∣ \forall x, y x^{T} A y = (O x)^{T} A (O y)

(1)

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:

[100 - 1]

(2)

and:

[200 - 3]

(3)

both produce isomorphic spaces. So it is customary to just always pick the matrix with only +1 and -1 as entries.

All indefinite orthogonal groups of matrices of equal metric signature are isomorphic

0 0

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)

[1001]

(1)

and:

[2001]

(2)

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