3
$\begingroup$

On wikipedia, they say about $SU(2)$ group : "Since the group SU(2) is simply connected, every representation of its Lie algebra can be integrated to a group representation".

They give a reference to a book but I don't find the corresponding theorem inside.

I would like to basically understand the connection between representations of Lie algebras and Lie groups. I am a huge beginner in representation theory, lie group and lie algebra, I basically know the definitions.

So I would like a really simple answer (I'm actually doing physics and I need to basically understand the connection between Lie group and Lie algebra representation).

$\endgroup$
  • $\begingroup$ Basically, a map $\mathfrak g \to \mathfrak h$ will induce a map $G \to H$ where $G,H$ are simply connected Lie group with Lie algebra $\mathfrak g, \mathfrak h$. $\endgroup$ – user171326 May 10 '17 at 10:04
4
$\begingroup$

Since you're doing physics, almost every Lie algebra you will encounter can be defined as a Lie algebra of matrices, where the bracket is given by the matrix commutator. In these cases, the elements of the corresponding Lie group can be written as a matrix exponential $g = e^{iX} = \sum_{n = 0}^\infty \frac 1 {n!}(i X)^n$, for some $X$ in the Lie algebra. (I'm assuming the Lie group is connected. And yes, physicists tend to put an $i$ inside the exponential.)

Now, suppose that $X \mapsto \rho(X) \in {\rm M}(n \times n, \mathbb C)$ is a representation of the elements of the Lie algebra as $n \times n$ matrices. Then the corresponding representation of the Lie group is given by the map $ e^{iX} \mapsto e^{i \rho(X)} \in {\rm Gl}(n , \mathbb C)$.

So why is this useful in physics? Suppose you're dealing with a quantum particle of spin $1$. Then the action of the angular momentum operators (generators of the $SU(2)$ Lie algebra) on the wavefunction of your particle are represented by the $3 \times 3$ matrices, $$ J_x = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{array} \right), \ \ \ J_y = \left( \begin{array}{ccc} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{array} \right), \ \ \ J_z = \left( \begin{array}{ccc} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) .$$

The transformation of the particle's wavefunction under a spatial rotation about the axis $(n_x, n_y, n_z)$ through an angle $\theta$ is then given by action of the quantum operator $\exp[ i \theta (n_x J_x + n_y J_y + n_z J_z)]$, which is an element of the $SU(2)$ Lie group, and for our spin-one particle, this operator is represented by the matrix $$\exp[ i \theta (n_x J_x + n_y J_y + n_z J_z) ] = \exp \left( \begin{array}{ccc} 0 & n_z \theta & -n_y \theta \\ -n_z \theta & 0 & n_x\theta \\ n_y \theta & -n_x \theta & 0 \end{array} \right) .$$

There is a similar story for the Lorentz group in relativity, and for the gauge groups in the standard model: in each case, you exponentiate the representation of the Lie algebra to get the representation of the Lie group.

$\endgroup$
  • $\begingroup$ Great answer. Just a little question to add : I have heard that not all Lie Groups have the property that any element $g$ can be written as an exponential of an element of the Lie algebra. What are the conditions to satisfy to ensure that it is true ? Is it the "connected" property of the Lie group you wrote (to be sure) $\endgroup$ – StarBucK May 10 '17 at 12:25
  • $\begingroup$ In fact there is still something I misunderstand for the representation of the group. You say it is a map $ g=e^{iX} \mapsto e^{i \rho(X)}$. But it needs to know what is $X$ when I know an element of the group $g$. But the exponential map is not inversible in general if we work with matrices ? $\endgroup$ – StarBucK May 10 '17 at 12:45
  • 1
    $\begingroup$ Great questions. (1) You're right that the connected property is required. For instance, consider $O(3, \mathbb R)$, the group of $3 \times 3$ orthogonal matrices. This has two connected components: orthogonal matrices with determinant $+1$, and orthogonal matrices with determinant $-1$. The Lie algebra is the space of antisymmetric matrices. But $e^{iX}$ for any $X$ in the Lie algebra will always have determinant $+1$. (For example, you can find a continuous path $t \mapsto e^{itX}$ from the identity matrix to $e^{iX}$, and the determinant can't "jump", due to continuity.) $\endgroup$ – Kenny Wong May 10 '17 at 13:29
  • 1
    $\begingroup$ (2) You're also right that it is possible to find distinct $X$ and $X'$ such that $e^{iX} = e^{iX'}$. For example, in the $SU(2)$ case, $e^{i\theta J_z} = e^{i(\theta + 4\pi)J_z}$. Fortunately, it doesn't matter which $X$ you pick. If $e^{iX} = e^{iX'}$, and if $\rho$ is a representation, then it's guaranteed that $e^{i\rho(X)} = e^{i \rho(X')}$. $\endgroup$ – Kenny Wong May 10 '17 at 13:33
  • 1
    $\begingroup$ @arctictern Ah, that was a typo - thanks! Physicists write $e^{iX}$ for $e^{iX}$; my point was that physicists want the Lie algebra of $U(N)$ to contain hemitian matrices rather than antihermitian matrices (because hermitian matrices are significant in quantum mechanics). $\endgroup$ – Kenny Wong May 11 '17 at 8:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.