# Relation between representations of Lie Group and Lie Algebra

If $G$ is a Lie Group, a representation of $G$ is a pair $(\rho,V)$ where $V$ is a vector space and $\rho : G\to GL(V)$ is a group homomorphism.

Similarly, if $\mathfrak{g}$ is a Lie Algebra, a representation of $\mathfrak{g}$ is a Lie Algebra homomorphism $\rho : \mathfrak{g}\to \mathfrak{gl}(V)$ to the Lie Algebra of endomorphisms.

Now, many Physics books treating Quantum Field Theory, immediately relate the representations of Lie Groups and Lie Algebras without citing the result being used nor explaining how is it used really.

This is quite common in order to find the representations of the Lorentz group $SO(1,3)$ in terms of elements of its Lie Algebra.

Now since physicists don't clear this in the books, I'm asking here. What is actually the relation between representations of Lie Groups and Lie Algebras that allows one to find the representations of the Lie Group in terms of the representations of the Lie Algebra?

## 2 Answers

I think the answer on your question is given probably from a geometric point of view. There is a beautiful theorem from Lie himself and is usually referred as Lie's 3rd Theorem, and states something which nowadays is rephrased as follows (over the complex numbers)

Theorem: There is an equivalence between the category of complex simply connected Lie groups and category of complex Lie algebras.

The above theorem isn't difficult to get proved and what it says in fact is that any complex Lie Algebra can be thought as the Lie Algebra of some Lie group. Think of a complex Lie Algebra $\mathfrak{g}$, and get the group generated by $\{ exp(A) \thinspace | \thinspace A \in \mathfrak{g} \}$, inside $GL_n(\mathbf{C})$; with some topological reasonable assumptions such that admits a cover, take its universal covering space (hence simply connected). Then this new manifold can be endowed with a group structure, hence it can be proven that this new Lie group has $\mathfrak{g}$ as its associated algebra.

Moreover, because of the equivalence you get a correspondence $$Hom_{\mathbb{C}}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C})) = Hom_{\mathsf{Lie}}(G, GL_n(\mathbb{C})),$$ so the representations of one structure correspond to the representations of the other.

Hope the above helps! Cheers!

• Thanks for the answer. In do understand that by differentiation, if $\pi : G \to GL(V)$ is a representation of $G$ then $d\pi : \mathfrak{g}\to \mathfrak{gl}(V)$ is a representation of the Lie algebra $\mathfrak{g}$. So a Lie group representation induces a Lie algebra representation, and we know how to construct it. But why the opposite is true? I believe we need to use the exponential map, but not all of $G$ is rebuilt from the exponential map if I'm not mistaken. – Gold Jun 15 '17 at 17:45
• You're welcome, well if I understand correctly your question, then the answer is exactly the equivalence of categories in the answer. The associated representation of $\mathfrak{g}$ is on the Lie group provided by the equivalence of categories. And by construction you can see that means the covering space of the Lie group generated by the set of $exp(A)$. The other direction of the equivalence is given by differentiation as you pointed out. – user321268 Jun 15 '17 at 18:12
• What I mean is that if you have a representation of an arbitrary lie algebra $\mathfrak{g}$, then the associated Lie Group is a very specific group constructed by this Lie Algebra, and I've delineated this construction in my answer. – user321268 Jun 15 '17 at 18:20

Each finite-dimensional linear representation of a Lie group $$G$$ on a vector space $$V$$ induces a representation of its Lie algebra $$\mathfrak g$$ on $$V$$. If $$X\in\mathfrak g$$ and if $$v\in V$$, define$$X.v=\left.\frac{d}{dt}\exp(tX).v\right|_{t=0}$$Note that not all representations of $$\mathfrak g$$ can be obtained by this process, although this is actually true if $$G$$ turns out to be simply connected.