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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).

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  • $\begingroup$ 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, 2017 at 10:04

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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.

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  • $\begingroup$ 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, 2017 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, 2017 at 12:45
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    $\begingroup$ 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, 2017 at 13:29
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    $\begingroup$ (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, 2017 at 13:33
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    $\begingroup$ @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, 2017 at 8:32

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