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Units, Constants and Useful Formulas
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Last modified: January 28, 2020
The Essential Mathematics of Lie Groups
---------------------------------------
Abstract Lie Groups
-------------------
Lie groups lie at the intersection of two fundamental
fields of mathematics: group theory and differential
geometry. They are finite dimensional DIFFERENTIABLE
MANIFOLDS, G, together with a group structure on G
such that the group law, * (+ or .), and inversion
are smooth maps:
μ: G x G -> G
(g1,g2) -> g1*g2
inverse: G -> G
g -> g-1
The axioms regarding the identity and associativity
also apply.
Lie Algebra
-----------
Differentiable manifolds have tangent spaces. As such
a Lie group G has a tangent space, g. At first this
is just a vector space but the group structures on
the manifold give rise to an additional structure
of the tangent space. In a vector space, we cannot,
in general, multiply vectors, but in the case of a
tangent space to a Lie group we can. Specifically,
the tangent space is equipped with a multiplication
defined by the Lie bracket, i.e. x.y -> [x,y]. This
additional structure of the tangent space satisfies:
Closure: [a,b] = c (= 0 for abelian case)
Jacobi Identity: [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0
All together this additional structure is called a
LIE ALGEBRA. More formally we can say a Lie algebra
over a Field, F, is a vector space over F endowed
with an operation [ , ]. In this context a field
is a set with additive and multiplicative operations
together with an additive and multiplicative inverse
(for non elements). A field as described. should not
be confused with a vector field.
Lie algebras are easier to calculate and locally
are a good approximation to the object they represent.
It is worth reiterating that the Lie algebra is
distinct from the group itself. Group operations
(. or +) pertain to G. In a group, you multiply
(or add) 2 elements together to get another element
of the group. In the corresponding algebra, which
lives in the tangent space, you take 2 elements of
the algebra, and commute them to get another element
of the algebra. The commutator is not defined on G
itself. In the additive case, G is abelian and
[,] = 0 since a + b = b + a.
Lie groups are CONTINUOUS groups. As a result,
multiplication tables don't work. Instead, it is
the commutator in the algebra that plays a role
similar to the multiplication law for the group.
Example:
Consider the group of real numbers under multiplication.
The real line of positive numbers is a manifold of
dimension 1. Equivalently it is a vector space over
the field R of real numbers (that is, over itself) of
dimension 1 (Note: A Vector Space is any set of
elements that can be added together and multiplied
by a scalar. The set of real numbers satisfy these
axioms and therefore forms a vector space.)
G = (ℝ\{0},.)
r1 = 2.03 and r2 = 3.06
Identity: 1
Multiplication: 2.03 x 3.06 = 5.09 ∈ G
Associativity: 2.03 x 3.06 = 3.06 x 2.03
Inverse: 2.03(1/2.03) = 3.06(1/3.06) = 1
It is easy to see that the commutator equals 0.
[2.03,3.06] = 2.03 x 3.06 - 3.06 x 2.03 = 0
Also,
[1.01,[2.03,3.06]] + [2.03,[3.06,1.01]] + [3.06,[1.01,2.03]] = 0
So G = (ℝ\{0},.) is an Abelian Lie group.
We will now look at Lie algebras in more detail.
We start by recognizing that manifolds are also
associated with vector fields.
Manifolds and Vector Fields
---------------------------
Prerequisite: 'The Essential Mathematics of
General Relativity'.
A vector field on a manifold can be defined as
a function that takes a point, P, in a manifold,
M, and gives back a vector, v, in TP in M. Thus,
v: M -> TPM such that vP ∈ TPM for all P ∈ M.
Since a vector field is an assignment of a tangent
space to each point in M, we can write the vector
fields, X and Y, at P in terms of tangent vectors:
X = xi∂i and Y = yi∂i
For example, in 2D we might have the vector field
X(x,y):
- - - -
X(x,y) = | xy || ∂x | ≡ xy∂x + (x + y)∂y
| x + y || ∂y |
- -
- -
At (2,1) and (2,2) we get the vectors | 2 | and
| 3 |
- -
- - (2,3) o (4,4) o
| 4 | which are ^ and ^ respectively.
| 4 | | /
- - (2,1) o (2,2) o
Dual Space
----------
dxj∂i = δji
Where dxj are the dual basis vectors. Therefore,
dxjX = dxjxi∂i = xidxj∂i = xiδji = xj
When evaluated on a vector field, X, the dual
basis vectors return the components of X.
The Directional Derivative on Vector Fields
-------------------------------------------
Tangent vector ≡ directional derivative, v = d/dλ
= (dxi/dλ)(∂/∂xi)
= xi∂i
Where xi = dxi/dλ are the components and ∂i = ∂/∂xi
are the bases.
Let f be a smooth differentiable scalar function
on the manifold, C∞(M) then,
Xf = xi∂if
Represents the change in f along a curve in the
direction of the tangent vector, v. This produces
a real number at the point P. SO THE VECTOR FIELD,
X, CAN BE INTERPRETED AS A DERIVATIVE OPERATOR.
X also obeys the product rule (fg)' = f'g + fg'.
Therefore,
X(fg) = (Xf)g + f(Xg)
Example.
- -
Suppose we have the tangent vector | 1 | = V and
| 2 |
- -
we want to find the value of f = x2 + y3 at the
point (1,0).
Xf = 1∂xf + 2∂yf
= (1)2x + (2)3y2
At (1,0) we get:
Xf = 2
So the DD of f in the direction of the tangent
vector, V, at the point (1,0) is 2.
A concrete example of this would be the following.
Suppose X is a wind velocity field across the North
American continent and f is a temperature described
by a scalar function defined over the same area. An
airborne temperature probe caught in the wind field
would measure how the temperature changes as it drifts
across the continent.
Compare the DD with the divergence and curl:
divF = ∂F1/∂x + ∂F2/∂y = y + 1
curlF = (∂F2/∂x - ∂F1/∂y)k = (1 - x)k
The Lie Derivative on Vector Fields
-----------------------------------
Let X be a collection of all fields on M (i.e.
X ∈ Γ(M)). Let f be a function on M (i.e.
f ∈ C∞(M)).
The Lie derivative of f, LXf, is given by:
LXf = xi∂if
So the Lie derivative for a scalar is the directional
derivative discussed above.
The Lie derivative of a vector can be be found as
follows:
γ(λ) is called an INTEGRAL CURVE of the vector field.
Integral curves for an electric field or magnetic
field are known as field lines, and integral curves
for the velocity field of a fluid are known as
streamlines.
Consider:
xi' = xi + dxi
= xi + xidλ since xi = dxi/dλ
A vector, vi, transforms as:
vi'(x') = (∂xi'/∂xj)vj(x)
= [(∂xi'/∂xj) + (∂xi'/∂xj)dλ]vj(x)
= [δij + (∂jxi)dλ]vj(x)
= vi(x) + (∂jxi)vi(x)dλ
We now use the expand vi(x + dx) as a Taylor series
f(x + dx) ~ f(x) + f'(x)dx
vi(x + dx) ~ vi(x) + dxj∂jvi(x)
But, dxi = xidλ. Therefore,
vi(x + dx) ~ vi(x) + xj∂jvi(x)dλ
LXvi = lim (vi(x + dx) - vi(x))/dλ
dλ->0
= lim [[vi(x) + xj∂jvi(x)dλ]
dλ->0 - [vi(x) + vi(x)(∂jxi)dλ]]/dλ
Therefore,
LXvi = xj∂jvi(x) - vj(x)(∂jxi)
The Killing Vector
------------------
A vector field, K, is a Killing field if:
LXg = 0
Where g is the metric.
This means that when you move the metric g a small
amount along K, g doesn't change.
A typical use of the Killing Field is to express a
symmetry in General Relativity (in which the geometry
of spacetime as distorted by gravitational fields is
viewed as a 4-dimensional Riemannian manifold). In a
static configuration, in which nothing changes with
time, the time vector will be a Killing vector, and
thus the Killing field will point in the direction of
the forward motion in time (i.e. g does mot change
with time).
Covariant vs Lie Derivative
---------------------------
The covariant derivative allows vectors in different
tangent spaces to be compared and relies on the
manifold being equipped with a connection. If you
have a vector X at one point, the connection tells
you what are the parallel vectors at nearby points.
In contrast, the Lie derivative evaluates the change
of one vector field along the flow of another vector
field.
The Lie Bracket on 2 Vector Fields
----------------------------------
Consider 2 vector fields and the aforementioned
scalar function, f:
LX(LYf) = xi∂i(yj∂jf)
= xi∂iyj∂jf + xiyj∂2ijf
And,
LY(LXf) = yi(∂i(xj∂jf))
= yi∂ixj∂jf + yixj∂2ijf
The pesky term in each case is the ∂2ijf term
because coordinate transformations never contain
second derivatives. However, if we subtract the
2 results we get:
[LX(LYf) - LY(LXf)] = {xi(∂iyj) - yi(∂ixj)}∂jf ... 1.
The {} term is a new vector field with components
made from the 2 old vector fields (i.e. Z = Zj∂j).
We can write this is as:
[LX(LYf) - LY(LXf)] = {(xi∂i)(yj∂j) - (yi∂i)(xj∂j)}f
... 2.
Knowing that X = xi∂i and Y = yi∂i enables us to
write:
LX(LYf) - LY(LXf) = [X,Y]f ≡ X(Yf) - Y(Xf)
We can also write 1. in the form:
[X,Y] = [X,Y]j∂j where [ ]j ≡ { ... }
This is the LIE BRACKET. Because it is a vector
field it takes a point, P, in M and gives back a
tangent vector, X. Indeed the RHS has the structure
of a tangent vector.
The Lie bracket is often written as:
[X,Y] = JYX - JXY ... 3.
where JY and JX are the JACOBEAN MATRICES of Y
and X respectively.
NOTE THAT THE LIE BRACKET OF 2 VECTOR FIELDS IS THE
SAME AS THE LIE DERIVATIVE OF 2 VECTOR FIELDS. i.e.
LXY = [X,Y]
-----------------------------------------------------
Aside:
A slightly simpler explanation of the above is:
XY = Xμ∂μ(Yν∂ν)
Using the product rule:
= XμYν(∂2μν) + Xμ(∂μYν)∂ν
There is no particular interpretation for the second
derivative, and as we saw above, it does not transform
nicely. But suppose we take the commutator:
[X,Y] = XY - YX = Xμ(∂μYν)∂ν - Yμ(∂μXν)∂ν
The second derivatives cancel leaving the directional
derivative.
[X,Y] = (Xμ(∂μYν) - Yμ(∂μXν))∂ν
= [X,Y]ν∂ν
Where [X,Y]ν = (Xμ(∂μYν) - Yμ(∂μXν)) are the components
of a new vector field made from the 2 old vector
fields.
-----------------------------------------------------
Example:
Consider the following 2 vector fields (tangent
vectors) on the 2-sphere. The 2 sphere is a 2D
surface that forms the boundary of a sphere in 3
dimensions.
- - - - - - - -
| y || ∂x | | 0 || ∂x |
X = | -x || ∂y | Y = | z || ∂y |
| 0 || ∂z | | -y || ∂z |
- - - - - - - -
Or,
X = x1∂1 + x2∂2 + x3∂3
Y = y1∂1 + y2∂2 + y3∂3
Therefore,
X = y∂1 - x∂2 and Y = z∂2 - y∂3
Using equation 2. from above:
ΣΣ[xi∂i,yj∂j]
ij
= [y∂x,z∂y] + [y∂x,-y∂z] + [-x∂y,z∂y] + [-x∂y,-y∂z]
= y∂x(z)∂y - z∂y(y)∂x = -z∂x
+
y∂x(-y)∂z - (-y)∂z(y)∂x = 0
+
(-x)∂y(z)∂y - z∂y(-x)∂y = 0
+
(-x)∂y(-y)∂z - (-y)∂z(-x)∂y = x∂z
= -z∂x + x∂z
- - - -
| -z || ∂x |
= | 0 || ∂y |
| x || ∂z |
- - - -
Using equation 3. from above:
JX:
- -
| ∂x(y) ∂y(y) ∂z(y) |
| ∂x(-x) ∂y(-x) ∂z(-x) |
| ∂x(0) ∂y(0) ∂z(0) |
- -
JXY:
- - - - - -
| 0 1 0 || 0 | | z |
| -1 0 0 || z | = | 0 |
| 0 0 0 || -y | | 0 |
- - - - - -
Which is NOT a tangent vector.
JY:
- -
| ∂x(0) ∂y(0) ∂z(0) |
| ∂x(z) ∂y(z) ∂z(z) |
| ∂x(-y) ∂y(-y) ∂z(-y) |
- -
JYX:
- - - - - -
| 0 0 0 || y | | 0 |
| 0 0 1 || -x | = | 0 |
| 0 -1 0 || 0 | | x |
- - - - - -
Which is also NOT a tangent vector.
However, JYX - JXY gives:
- - - - - -
| 0 | | z | | -z |
| 0 | - | 0 | = | 0 |
| x | | 0 | | x |
- - - - - -
Which IS a tangent vector.
So the 2 approaches give the same result.
We can envisage X and Y as 2 velocity fields.
Suppose we now move along X for a brief time t,
then along Y for another brief interval t. Next
we switch back to X, but with a minus sign for
time t, and then to Y with a minus sign for time
t. We have tried to retrace our path but we will
fail to return to our exact starting point.
Cross Product
-------------
If we consider the special case of vector fields
on the manifold R3, the set of all smooth
vector fields forms an infinite-dimensional vector
space. In this instance the Lie bracket relates a
3-dimensional subspace of vector fields on R3 to
the cross product.
Note: This is a different case from the previous
example where 2 specific vector fields, X and Y,
on the specific manifold S2 illustrates
the general property that the Lie bracket of any
two vector fields on a manifold is itself a vector
field on the manifold. In that instance the
resulting vector has nothing to do with the cross
product of two vectors.
Example:
- - - - - - - - - - - -
| 0 || ∂x | | z || ∂x | | -y || ∂x |
X = | -z || ∂y | Y = | 0 || ∂y | Z = | x || ∂y |
| y || ∂z | | -x || ∂z | | 0 || ∂z |
- - - - - - - - - - - -
JX:
- -
| ∂x(0) ∂y(0) ∂z(0) |
| ∂x(-z) ∂y(-z) ∂z(-z) |
| ∂x(y) ∂y(y) ∂z(y) |
- -
JXY:
- - - - - -
| 0 0 0 || z | | 0 |
| 0 0 -1 || 0 | = | x |
| 0 1 0 || -x | | 0 |
- - - - - -
JY:
- -
| ∂x(z) ∂y(z) ∂z(z) |
| ∂x(0) ∂y(0) ∂z(0) |
| ∂x(-x) ∂y(-x) ∂z(-x) |
- -
JYX:
- - - - - -
| 0 0 1 || 0 | | y |
| 0 0 0 || -z | = | 0 |
| -1 0 0 || y | | 0 |
- - - - - -
JYX - JXY gives:
- - - - - -
| y | | 0 | | y |
| 0 | - | x | = | -x |
| 0 | | 0 | | 0 |
- - - - - -
The cross product is defined as:
(ux,uy,uz) ^ (vx,vy,vz) = (uyvz - uzvy,uzvx
- uxvz,uxvy - uyvx)
= |u||v|sinθ
- - - - - -
| 0,-z,y | ^ | z,0,-x | = | zx,zy,y2 |
- - - - - -
Now,
x ^ y = z
y ^ x = -z
y ^ z = x
z ^ y = -x
z ^ x = y
x ^ z = -y
x2 = y2 = z2 = 0
Therefore,
- - - - - -
| 0,-z,y | ^ | z,0,-x | = | y,-x,0 |
- - - - - -
Left Invariant Vector Field
---------------------------
Any Lie group, G, acts on itself by left multiplication.
(recall that the only group operation is .). If h ∈
G is fixed, and g ∈ G, we denote this action by:
Lg: h -> gh
or,
Lg(h) = gh
This is interpreted as the translation of a point
h by g. We will show the justification for this
interpretation later when we discuss one-parameter
subgroups.
A vector field, X, is a function that takes a
point, h, in G (which in this case is also a
manifold) and gives back a vector, vh, in ThG.
Consider a vector field, X, at h, (X)h We can
do 2 things:
1. We can evaluate X at another point g.
Therefore,
X(Lgh) is (X)h evaluated at gh, (X)gh.
2. We can take the tangent vector of X at h,
xh and apply the pushforward, Lg*, to xg
to get the vector at gh. Therefore,
Lg*xh = xgh
Let's look at this in more detail. From the
discussion 'The Essential Mathematics of
General Relativity' we had that the pushforward
of a tangent vector from a manifold M to a
manifold N is given by:
(φ*v)α∂α = (φ*)αμvμ∂α
Where (φ*)αμ is the Jacobean.
If M and N are the same manifold then:
(φ*)vα∂α = vμ∂μ
We now replace φ by L (for left), vα∂α with xa
and multiply by g to get:
Lg*xa = xga
Therefore, Lg*, is a linear map from the
tangent space at h to the tangent space at
the point gh.
Now we can define:
Lg*(xh) := (Lg*(X))gh
^ ^
| |
vector in vector
ThG field at gh
≡ gdLgX (see later)
This is true because a vector field assigns a
tangent vector to every point on the manifold
and vice-versa. Therefore, we can write 2 as:
Lg*Xh = Xgh ... 3.
If 1. and 3. yield the same result we say that
X is LEFT INVARIANT VECTOR FIELD, i.e.
Lg*Xh = X(Lg(h)) = (X)gh
Therefore, the vector field at gh is the SAME
as the vector field h.
A LIVF satisfies the commutation diagram:
vh Lg* vgh
ThG -----------> TghG
^ ^
| |
Xh | | Xgh
| |
| |
G -------------> G
h Lg gh
If we now let h = the element at the identity we
get:
Lg*XI = X(Lg(I)) = XgI
Therefore, X at g is the same as the same as X
at the identity.
Note: We picked a tangent vector at the identity
and then translated that vector to another point.
However, we could have equally well picked an
element that is not at the identity but working
with the vector at the identity is much easier.
Summarizing:
The Lie algebra of a Lie group is defined to
be the left invariant vector fields on the group.
Integral Curves (aka Flow) of a Vector Field
--------------------------------------------
An integral curve for the vector field, X, with
an initial condition, P0, is a curve such that:
dγ(t)/dt = Xγ(t)
and,
γ(0) = P0
What this says is that an integral curve, γ(t),
is a curve with initial condition P0 such that
for every point, P = γ(t), on the curve, the
tangent vector, vγ(t), is equal to the value of
the vector field, X, at P. That is, the vector
field gives the value of the tangent to the path
at every point.
One-Parameter Subgroups
-----------------------
Consider a left invariant vector field, XL. Let
us denote the integral curve of XL as γL(t). By
definition, XL assigns a tangent vector to each
point on γL(t). γL(t) is referred to as a ONE
PARAMETER SUBGROUP where an element of the subgroup
is γL(t) for each t.
dγL(t)/dt = vγL(t)
But we know for a LIVF that,
LγL(t)*v = vγL(t)
Therefore, we can say,
dγL(t)/dt ≡ LγL(t)*
There are 2 ways we can move along the curve,
translation by s and pushforward by γL(s).
Translation by s gives γL(s + t). Therefore,
dγL(s + t)/dt = vγL(s + t)
Pushforward by γL(s) gives:
LγL(s)*γL(t) = γL(s)vγL(t) = vγL(s)γL(t)
Clearly, these 2 ways have to be equal. Therefore,
γL(s + t) = γL(s)γL(t)
THIS CAN ONLY BE SATISFIED IF γ(t) IS AN EXPONENTIAL
FUNCTION.
Note that:
. γ(0) = 1 γ(t) at the identity.
. γ(-t) = (γ(t))-1
This now also justifies the previous interpretation
of Lg(a) = ga as a translation, i.e.
γL(g)γL(a) ≡ γL(a + g)
-----------------------------------------------------
Digression:
What we have described is called a GROUP
HOMOMORPHISM. A homomorphism is a way to compare
2 groups. Consider 2 groups, (G,*) and (H,◆), a
group homomorphism from G to H is a function
φ: G -> H such that for all x and v ∈ in G it
holds that:
φ(x*y) = φ(x)◆φ(y)
Where * and ◆ are the group operations.
Example:
G = ℝ under +
H = ℝ' under .
φ: G -> H
φ(x + y) = φ(x).φ(y)
φ = exponential function.
So φ is a homomorphism.
A one-parameter subgroup of a Lie group is a
group homomorphism i.e. a smooth map:
γ: ℝ -> G where (ℝ,+) and (G,.)
For which:
γ(s + t) = γ(s)γ(t)
-----------------------------------------------------
Recovering the Lie Group from its Lie Algebra
---------------------------------------------
The tangent vector to the integral curve, γ(t),
at the identity is given by:
v = dγ(t)/dt|t=0
Where we have simplied the notation such that
vγL ≡ v.
Which has the solution:
γ(t) = exp(tv)
This is the ONE-PARAMETER SUBGROUP from before
which can be written as a group element, X.
So we can recover the Lie algebra from the Lie
group by taking the differential of the group
element evaluated at t = 0.
One can also recover the Lie algebra by simply
taking the logarithm of the group element.
To show that these operations are equivalent,
we need to prove that dX/dt|t=0 = lnX.
d(exp(tv))/dt|t=0 = ln(exp(tv))
xexp(tv)|t=0 = v
Therefore, v = v.
The exponential and the logarithm give a 1-to-1
correspondence, continuous in both directions,
between a neighborhood of 1 in any Lie group
and a neighborhood of 0 in its Lie algebra.
These operations are illustrated in the following
diagram.
We can linearize the one-parameter Lie subgroup
γ(t) = exp(tx) by considering infinitessimal
changes of the form:
γ(t) ~ I + tx + O((tx)2)
Which is just the Taylor series expansion of the
exponential function.
Matrix Lie Groups
-----------------
Matrix Lie groups are a special class of abstract
Lie groups which are also smooth differentiable
manifolds. For a matrix Lie group the vector
fields associated with manifolds are represented
by matrices. The most interesting Lie groups
turn out to be matrix groups.
For matrices, one can regard the smooth manifold
as the General Linear group GL(n,ℝ) which is the
set of n x n invertible matrices with det ≠ 0.
Indeed, sometimes this is written as M(n,ℝ).
Having det ≠ 0 means that the matrices have an
inverse and satisfy the other axioms of Lie groups.
Hence, GL(n,ℝ) is a Lie group.
Matrix Exponential
------------------
To move from the matrix Lie algebra to the Lie group
we use the MATRIX EXPONENTIAL. The power series
expansion of this is:
∞
M = exp(tm) = Σ(tm)k/k! X ∈ G, m ∈ g,
k=0
Where M and m are now matrices.
One can also move from the Lie group back to
the Lie algebra by differentiating the matrix
exponential:
m = dM/dt|t=0 M ∈ G, m ∈ g,
Alternatively, we can use the MATRIX LOGARITHM
which is the inverse map of the matrix exponential
(i.e. log = exp-1). The power series expansion is:
∞
m = ln(1 + M) = Σ(-1)k+1Mk/k M ∈ G, x ∈ g,
k=1
This is a difficult calculation for most
practical purposes. Fortunately, if M can be
diagonalized to a matrix MD, then:
M = PMDP-1
lnM = P(lnMD)P-1
Where the diagonal elements are ln(d11), ln(d22)
etc. and all other entries are 0.
For matrices that cannot be diagonalized one
needs to find its JORDAN DECOMPOSITION and
calculate the logarithm of the JORDAN BLOCKS.
Jacobi Formula
-------------
Jacobi's formula states:
det(exp(x)) = exp(Tr(x))
Where x is a real or complex matrix.
The Lie Bracket of Matrix Lie Groups
------------------------------------
Since for a matrix Lie groups the vector fields
are represented by matrices, the Lie bracket
corresponds to the usual commutator for a matrix
group. We can demonstrate this by using the
BAKER-CAMPBELL-HAUSDORFF formula:
exp(z) = exp(x)exp(y) = exp(x + y + (1/2)[x,y] + ... )
Let W = [X,Y]
Therefore:
exp(tw) = [exp(tx),exp(ty)]
= exp(tx)exp(ty) - exp(ty)exp(tx)
= exp(tx + ty + (1/2)[tx,ty] + ... )
- exp(tx + ty + (1/2)[ty,tx] + ... )
= exp(tx + ty + (1/2)[tx,ty] + ... )
- exp(tx + ty - (1/2)[tx,ty] + ... )
From here we can either take the log of both sides
to get (setting t = 1):
w = (x + y + (1/2)[x,y] + ... )
- (x + y - (1/2)[x,y] + ... )
= [x,y]
or,
d{exp(tw))/dt|t=0 = [x,y]
Examples
--------
G = (SO(2),.):
- - - -
AB = | cosα -sinα || cosβ -sinβ |
| sinα cosα || sinβ cosβ |
- - - -
- -
= | cosαcosβ - sinαsinβ -cosαsinβ - sinαcosβ |
| sinαcosβ + cosαsinβ -sinαsinβ + cosαcosβ |
- -
- -
= | cos(α + β) -sin(α + β) | ∈ G
| sin(α + β) cos(α + β) |
- -
With generator,
Xg = dA(α)/dα|α = 0
- -
= | 0 -1 |
| 1 0 |
- -
[Xg,Xg] = 0
Therefore (SO(2),.) is an Abelian Lie group.
G = (SO(3),.):
- -
| 1 0 0 |
Rx = | 0 cosθ -sinθ |
| 0 sinθ cosθ |
- -
- -
| cosθ 0 sinθ |
Ry = | 0 1 0 |
| -sinθ 0 cosθ |
- -
- -
| cosθ -sinθ 0 |
Rz = | sinθ cosθ 0 |
| 0 0 1 |
- -
With generators:
- -
| 0 0 0 |
Xx = | 0 0 -1 |
| 0 1 0 |
- -
- -
| 0 0 1 |
Xy = | 0 0 0 |
| -1 0 0 |
- -
- -
| 0 -1 0 |
Xz = | 1 0 0 |
| 0 0 0 |
- -
Rx.Ry is another rotation (∴ ∈ G) but it is not equal
to Rz. Whereas [Xx,Xy] = Xz
G = (SU(3),.):
An SU(3) group element is obtained by exponentiating
the 3 × 3 traceless Hermitian Gell-Mann matrices.
Unfortunately the resulting expression is rather
complicated. However, for λ1 - λ7 we can use the
following formula:
exp(itθλi/2) = I + iλisin(tθ/2) + λi2(cos(tθ/2) - 1)
i.e. exp(itλ1/2)
- - - -
| 1 0 0 | | 0 1 0 |
exp(itλ1/2) = | 0 1 0 | + i| 1 0 0 |sin(tθ/2)
| 0 0 1 | | 0 0 0 |
- - - -
- -
| 1 0 0 |
+ | 0 1 0 |(cos(tθ/2) - 1)
| 0 0 0 |
- -
- -
| cos(tθ/2) isin(tθ/2) 0 |
= | isin(tθ/2) cos(tθ/2) 0 | = X
| 0 0 1 |
- -
Note that det(exp(iλ1/2)) = exp(Tr(λ1)) = 1
(Jacobi's identity) and XX† = I
The eigenvalues/eigenvectors of this matrix are:
- - - - - -
| 0 | | -1 | | 1 |
1| 0 | exp(-itθ/2)| 1 | exp(itθ/2)| 1 |
| 1 | | 0 | | 0 |
- - - - - -
The eigenvectors are linearly independent and
therefore we can diagonalize X.
- - - -
| 0 -1 1 | | 0 0 1 |
P = | 0 1 1 | P-1 = | -1/2 1/2 0 |
| 1 0 0 | | 1/2 1/2 0 |
- - - -
After diagonalization (XD = P-1XP):
- -
| 1 0 0 |
| 0 cos(tθ/2) - isin(tθ/2) 0 |
| 0 0 cos(tθ/2) + isin(tθ/2) |
- -
- -
| 1 0 0 |
≡ | 0 exp(-itθ/2) 0 |
| 0 0 exp(itθ/2) |
- -
Take the logarithm:
- -
| 0 0 0 |
| 0 -itθ/2 0 | = γ
| 0 0 itθ/2 |
- -
Apply the similarity transform:
λ1 = PγP-1
- - - - - -
| 0 -1 1 || 0 0 0 || 0 0 1 |
= | 0 1 1 || 0 -itθ/2 0 || -1/2 1/2 0 |
| 1 0 0 || 0 0 itθ/2 || 1/2 1/2 0 |
- - - - - -
- -
| 0 itθ/2 0 |
= | itθ/2 0 0 |
| 0 0 0 |
- -
= itθλ1/2
We get the same result if we find the derivative
at t = 0. Thus,
- -
| -(θ/2)sin(tθ/2) i(θ/2)cos(tθ/2) 0 |
dX/dt|t=0 = | i(θ/2)cos(tθ/2) -(θ/2)sin(tθ/2) 0 |
| 0 0 1 |t=0
- -
- -
| 0 1 0 |
= (iθ/2)| 1 0 0 |
| 0 0 0 |
- -
LIVF:
Lg*XI = gdX/dt|t=0
= gd(exp(tλ1))/dt|t=0
= gλ1
= (λ1)g
where g ∈ G, v ∈ TgG
- -
So the LIVF is: | 0 1 0 |
| 1 0 0 |
| 0 0 0 |
- -
Finally, lets show that algebraic operations can
only be performed in the tangent space and not
on the manifold itself. In other words, elements
have to be 'referred' back to the tangent space
for alebraic operations. For example,
[λ1,λ2] = iλ3 (f123 = 1)
- -
| costθ + isintθ 0 0 |
and X3 = exp(itθλ3) = | 0 costθ - isintθ 0 |
| 0 0 1 |
- -
- -
| iθ 0 0 |
dX3/dt|t=0 = | 0 -iθ 0 | = iλ3
| 0 0 0 |
- -
If we try and perform the same operation on
the manifold after exponentiation we get:
- -
| -2isin2tθ 0 0 |
[exp(itθλ1),exp(itθλ2)] = | 0 2isin2tθ 0 |
| 0 0 1 |
- -
With dX3/dt|t=0 = 0
Which clearly doesn't work!
For SU(3), the group elements represent rotations
of complex 3 component (u, d, s) vectors in an 8
dimensional space.
Note that if we had performed the same calculation
using Rz from SO(3) we would get the
following LIVF.
- - - - - -
| 0 -1 0 || x | | -y |
| 1 0 0 || y | = | x |
| 0 0 0 || z | | 0 |
- - - - - -
Which is:
XL = -y∂x + x∂y
For SO(3), the group elements represent rotations
of real 3 component (x, y, z) vectors in a 3
dimensional space.
Group Actions
-------------
Let G be a group and S be a set. Let g ∈ G
and s ∈ S. One can construct a left action
of g on s denoted as g.s as follows:
g.s = gs
Likewise, one can construct a right action as
follows:
s.g = g.s
One can construct a left action from a right
action by composing with the inverse operation
of the group. Therefore,
g.s = gs ≡ sg-1
We can think of the multiplication as the act
of moving points around inside of S. For example,
in D4, the symmetry group of the square, the
elements move the set of vertices to achieve
rotations and reflections. (see see D4 here)
We will focus on left actions. A left group
action has to satisfy the following 3 axioms.
Let s ∈ S and g,h ∈ G then:
1. g.s ∈ S
2. e.s = s
3. g.(h.s) = (gh).s
For example, from D4 it is easy to see that
axiom 3. is satisfied i.e.
d.[r.s] = [dr].s
where s = {1,2,3,4}.
G Acting on Itself
------------------
To make G act on itself we let S = G. For example,
in D4 the element d, performs a reflection about the
vertical axis and the element, r, performs a 90°
rotation. The left action d.r = dr combines these
2 operations to achieve a reflection about the
right diagonal. All possible operations can be
seen in the Cayley table.
For example, from the Cayley table for D4 it is
easy to see that d.[r.r] = [dr].r.
In addition, to the left action, a group also
acts on itself is by conjugation. Therefore,
g.x = gxg-1 x, g ∈ G
Proof of axioms 2 and 3 for conjugation:
2. e.x = ese-1
3. g1.(g2.x) = g1.(g2xg2-1) since g2.x = g2xg2-1
= g1.p where p = g2xg2-1
= g1pg1-1 since g1.p = g1pg1-1
= g1(g2xg2-1)g1-1
= (g1g2)x(g2-1g1-1)
= (g1g2)x(g1g2)-1
= (g1g2).x Q.E.D
The conjugacy equation also describes matrix
similarity. However, conjugacy in this context
is more restrictive than similarity in that g,
g-1 and x have to be in G.
The Adjoint Representation
--------------------------
We start with the conjugacy class equation and
change the notation. We use upper case letters for
elements of the group, G, and lower case letters
for elements of the Lie algebra. Therefore,
Z = YXY-1 X, Y, Z ∈ G
The differential of the conjugation action,
evaluated at the identity, is called the adjoint
action. Therefore,
AdYX = dZ/dt|t=0
X is obtained by exponentiating the generator x.
Therefore, we can write:
AdYX = d(YXY-1)/dt|t=0
= d(Yexp(tx)Y-1)/dt|t=0
= Yxexp(tx)Y-1|t=0
= YxY-1
Where Y, X ∈ G and x ∈ the Lie algebra.
Therefore, the adjoint representation is a
representation where the group elements in G
act on a basis which is comprised of the group
generators in the tangent space at the identity
and not the column vectors that appears in the
fundamental representation. In other words, in
the adjoint representation, the Lie algebra itself
forms the basis upon which G acts.
Derivative of Ad = ad
---------------------
We can pass from a representation of a Lie group
G, to a representation of its Lie algebra by
taking the derivative at the identity.
adYx = d(AdYX)/dt|t=0
= d(YxY-1)/dt|t=0
= d(exp(ty)xexp(-ty))/dt|t=0
= yexp(ty)xexp(-ty) + exp(ty)x(-y)exp(-ty)|t=0
= yx - xy
= [y,x]
The ad operator is the adjoint action of the group
generators on themselves.
As one would expect, AdY and adY are related
through the exponential map:
AdY = exp(adY)
Diagramatically, we can summarize these relationships
in a commutative diagram.
ad
g --------> gl(g)
: :
: exp : exp
v v
G --------> GL(g)
Ad
Non-Abelian Gauge Theory
------------------------
Lie groups and Lie algebras play an important
role in the various dynamical theories which
govern the behaviour of particles - the gauge
theories. We now present the relationship
between Lie algebras and gauge theories.
We have seen that the elements of a group are
related to the Lie algebra via the exponential
map, i.e.
U = exp(tTa)
We have used U in place of the group element,
X, as a reminder that the matrices are unitary.
We now want to look at the situation where t
is not a constant but becomes a function of
the spacetime coordinates, xμ. This means
that the above formula becomes:
U(xμ) = exp(t(xμ)Ta)
Now, the group elements can act on a field, φ,
where φ denotes an array of different fields
written as a column vector (e.g. quark fields).
The fields transform among themselves, just as
the components of a three-dimensional vector
transform among themselves under the group of
rotations. In other words the fields form a
basis and the action of the group on this basis
constitutes a group representation.
These fields transform as (dropping the μ
and (x) for brevity):
φ -> φ' = Uφ
∂μφ -> (∂μφ)' = U∂μφ + (∂μU)φ
However, the second term is a problem. What
we want is a derivative such that:
Dμφ -> (Dμφ)' = UDμφ
Try:
Dμφ = ∂μφ - Wμφ
Rearranging,
Wμφ = ∂μφ - Dμφ
Consider the transformation:
Wμφ -> (Wμφ)' = (∂μφ)' - (Dμφ)'
= U∂μφ + (∂μU)φ - UDμφ
= {U∂μ - UDμ + (∂μU)}φ
= {U(∂μ - Dμ) + (∂μU)}φ
= {UWμ + (∂μU)}φ
But φ = U-1φ'. Therefore,
(Wμφ)' = {UWμU-1 + (∂μU)U(x)-1}φ'
So Wμ transforms as:
Wμ -> Wμ' = UWμU-1 + (∂μU)U-1
Dμφ = ∂μφ - igWμφ is called the GAUGE COVARIANT
DERIVATIVE. Wμ is a matrix of the type generated
by an infinitesimal gauge transformation and
carries information regarding the group from
one spacetime point to another. Wμ can be
decomposed as:
Wμ = WμaTa
Where Wμa are the GAUGE FIELDS that act as
coefficients to the generators, Ta. When
these gauge fields are quantized, their quanta are
called GAUGE BOSONS.
The gauge covariant derivative differs from the
'ordinary' covariant derivative ∇μvρ = ∂μvρ + Γρμνvν
in that the latter applies to transformations
that behave properly under a change of basis.
Now lets look at 2 successive transformations.
Wμ -> Wμ' = U1WμU1-1 + (∂μU1)U1-1
Wμ -> Wμ'' = U2{U1WμU1-1 + (∂μU1)U1-1}U2-1 + (∂μU2)U2-1
= (U2U1)Wμ(U2U1)-1 + (∂μU2U1)(U2U1)-1
= U3WμU3-1 + (∂μU3)U3-1
Therefore, performing 2 gauge transformations is
also a gauge transformation. In addition, the
transformation with U = 1 corresponds to the
identity transformation
Wμ -> Wμ' = 1Wμ1-1 + (∂μ1)1-1 = Wμ
And a transformation with U followed by a
transformation with U-1 yields the identity.
Wμ -> Wμ' = 1.1-1Wμ(1.1-1)-1 + (∂μ1.1-1)(1.1-1)-1
= 1Wμ
This fits the definition of a group whose elements
are the gauge transformations. This is called the
GAUGE GROUP.
Specifically, the first term in the transformation,
Wμ -> Wμ' = U1WμU1-1 + (∂μU1)U1-1
is the conjugacy class equation that gives the
adjoint representation of a Lie group.
In the the adjoint representation U(1) has 1 one
dimensional generator corresponding to the photon.
SU(2) has 3 three dimensional generators that
correspond to the W+/- and Z0 bosons. SU(3) has
8 eight dimensional generators corresponding to
the gluons. The adjoint representation contrasts
with the defining (fundamental) representations
that describe the Fermionic particles (e-1, u, d
and s quarks).
For the second term U(x) displaced over an
infinitesimally small distance dx will differ
from U(x) by an infinitesimal gauge transformation.
Therefore, we can use a Taylor series expansion
to get U(x + dx) ~ U(x) + U'(x)dx + .... Thus,
U(x + dx) = U(x) + ε(x)TaU(x)dx + ...
Therefore, by comparison,
∂μU(x) = ε(x)TaU(x)
So,
∂μU(x)U(x)-1 = ε(x)Ta
As such it is also part of the Lie algebra.
Field Strength, Gμν
-------------------
We define:
Gμν := [Dμ,Dν]
= ∂μWν - ∂νWμ - [Wμ,Wν]
Or, setting Gμν = GμνaTa, we get:
GμνaTa = ∂μWνaTa - ∂νWμaTa - WμaWνb[Ta,Tb]
= ∂μWνaTa - ∂νWμaTa - WμaWνbfabcTc
In the abelian case the commutator = 0 so
Gμν = = ∂μWν - ∂νWμ