What is the relation between representations of Lie Groups and Lie Algebras? 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?
 A: 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.
A: 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!
