Lie Groups II

Adjoint Representation and Campbell Identity¶

We define $\mathrm{Ad}_{X}(Y)$ by:

and define a matrix function $f(t)$ by:

Differentiate $f(t)$:

We apply it recursively for the second derivative:

We evaluate these derivatives at $t=0$:

We expand $f(t)$ using a Taylor series around the point $t=0$:

and we get:

We can set $t=1$:

This last equation is called the Campbell identity. We can also write it as:

Lie Algebra¶

Lie algebra $g$ for a Lie group $G$ that is a topologically closed subgroup of GL(n, R) is defined by:

The following can be proven:

1. $g$ is a vector space

2. $X, Y \in g \Rightarrow [X, Y] \in g$

3. $g$ is the tangent space to $G$ at $I$.

4. $\exp: g \rightarrow G$ is locally invertible

Invariant subgroup $a$ of a group $g$: for every $a$ and $g$ the following holds:

where $a'$ is a member of $a$.

Invariant abelian subgroup: is an invariant subgroup that is abelian: $a a' = a' a$, or $[a, a'] = 0$.

Lie group is:

• simple if it does not posses an invariant Lie subgroup.

• semisimple if it does not posses an abelian invariant Lie subgroup.

Typical examples of semisimple groups are the direct products of simple groups.

Using the structure constants $[L_i, L_j] = C_{ij}{}^k L_k$ we can define a metric tensor (also called Killing form):

The regular representation is:

From this it follows:

This fulfills all properties of a metric, e.g.:

but it is not necessarily positive definite (i.e., $(L_i, L_i) > 0$).

Theorem: a Lie algebra is semisimple if and only if $\det g_{ij} \neq 0$.

The center of a group G is a set of elements that commute with every element of G:

The center is a normal subgroup of G, so the quotient $G / Z(G)$ is a group.

To construct every other connected Lie group with the same Lie algebra su(2) as SU(2), such as SO(3), we take quotients of SU(2) by discrete subgroups of its center $Z(SU(2))=\{I,−I\}$. The possible quotients are:

Utilities¶

Computing $\exp(\theta J_3) P_1 \exp(-\theta J_3)$¶

Where $J_i$ are generators of the SO(3) group and $P_1$ are generators of the translation group. We do not know their matrices, we only know their commutators. We use the formula above with $t = \theta$, $X=J_3$ and $Y=P_1$:

We need to compute the commutators. The Lie algebra commutator is:

So:

And higher terms repeat. The commutators alternate between $P_2$ and $P_1$ and their signs:

Any vector operator has the same Lie bracket, for example for $J_i$ we have $[J_i, J_j] = \epsilon_{ijk} J_k$ and obtain:

Computing $\exp(\theta J_3) P_2 \exp(-\theta J_3)$¶

Similarly, let’s compute:

Computing $\exp(\theta J_3) P_3 \exp(-\theta J_3)$¶

Finally:

since all the commutators are zero.

Summary¶

We have thus shown that for any operator satisfying $[J_i, P_j] = \epsilon_{ijk} P_k$ we get:

We can thus write these as a single equation in index notation:

where the matrix $(R_3)^i{}_j$ is a rotation around the $z$-axis:

We are interested in computing ${\bf a} \cdot {\bf P} = a^j P_j$:

and using the below indentity:

or:

In an analogous way we can obtain:

Putting it all together we get:

In other words, we can see how a general element of the rotation group $e^{\theta_1 J_1} e^{\theta_2 J_2} e^{\theta_3 J_3}$ acts on a general element of the translation group $T({\bf a})$. The result is another translation $T(R_1(\theta_1) R_2(\theta_2) R_3(\theta_3) {\bf a})$ and the translation vector ${\bf a}$ gets rotated with the three rotation matrices ${\bf a} \rightarrow R_1(\theta_1) R_2(\theta_2) R_3(\theta_3) {\bf a}$.

For T=J¶

As a special case, we’ll use $P_i = J_i$, and we’ll use $\bf n$ instead of $\bf a$:

In an analogous way we can obtain:

Putting it all together we get:

This shows how to “encode” the 3D vector $\bf n$ into any $j$ representation. The RHS is encoding the rotated vector, the LHS is encoding the original vector and then transforming it using the “sandwich” formula. Thus showing that the “sandwich” transformation is equivalent to a rotation. It is thus showing how any SU(2) transformation corresponds to SO(3) transformation, the “minus” cancels out in the sandwich, so 360 degree rotation transforms $\bf n$ into itself. To obtain the 360 rotation equal to -1 we need to act with just the left part of the sandwich on some “spinor”, and it would transform it to -1.

Thus we have proven:

where:

Spin 1/2¶

For spin j=1/2 we have:

Now we would like to show that:

where ${\bf n}$ is a unit vector representing the axis of rotation and $\theta$ is the rotation around this axis. This follows from the fact that the LHS is a general element of the SU(2) group, and so is the RHS. Constructively, this is proven in a separate file.

TODO: We also want to prove that the three rotation matrices can be written as just one matrix:

Then we get:

Here $U$ can be expanded as:

We can now identify this relation with quaternions as follows:

And the rotation equation is:

In other words, the LHS is encoding the vector ${\bf a}$ into a quaternion as $a^1 i + a^2 j + a^3 k$, and the RHS gives us the rotated quaternion vector that we can decode back.

Identity $g e^X g^{-1} = e^{g X g^{-1}}$¶

The identity

can be proven using:

Computing $\exp(\theta_3 J_3) \exp(\theta_1 P_1) \exp(-\theta_3 J_3)$¶

Using the identity $g e^X g^{-1} = e^{g X g^{-1}}$ we get:

Rodrigues’ Rotation Formula¶

If $\mathbf{v}$ is a vector in R^3 and $\mathbf{k}$ is a unit vector (axis of rotation) and $\theta$ the angle of rotation, then the vector $\mathbf{v}$ gets rotated to $\mathbf{v}_\mathrm{rot}$.

We define a matrix $\mathbf{K}$ using:

Note that $|\mathbf{k}|^2 = k_1^2 + k_2^2 + k_3^3 = 1$. We get:

We will also need:

where we used $\delta_{ml}k^l k^m=1$.

Now we compute the characteristic polynomial:

From the last equation we get $\mathbf{K}^4=-\mathbf{K}^2$ and $\mathbf{K}^5=-\mathbf{K}^3=\mathbf{K}$ and so on.

Then the rotation matrix $\mathbf{R}$ is:

So we got:

Now we can write:

and get:

We use:

and get:

This is called the Rodrigues’ formula. In components:

We can simplify: