Representation of a Lie group

Remarks

Example: $G = \mathrm{SO}(3)$, $V = \R^3$ and $\pi(g) = g$. The map is the inclusion of a closed subgroup of $\mathrm{GL}_3(\R)$, so it is a continuous homomorphism.

Non-example: let $f \colon \R \to \R$ be an additive map that is not linear, which exists by choosing a basis of $\R$ over $\Q$. Then $\pi(t) = e^{f(t)}$ is a homomorphism $\R \to \mathrm{GL}_1(\R)$, but it is not continuous, so it is not a representation of the Lie group $\R$.

A continuous homomorphism of Lie groups is automatically smooth. So a finite-dimensional representation $\pi$ has a differential $d\pi$ at the identity, which is a representation of the Lie algebra of $G$. If $G$ is connected, $d\pi$ determines $\pi$. A unitary representation is one where each $\pi(g)$ preserves a fixed inner product on $V$.