Do you know a Lie group that has no faithful finite-dimensional representation?
Solution
A standard example is the universal covering group \[ G=\widetilde{\operatorname{SL}_2(\mathbb R)}. \] Let \[ p:G\to\operatorname{SL}_2(\mathbb R) \] be the covering homomorphism. Its kernel is infinite cyclic and central: \[ \ker p\cong\pi_1(\operatorname{SL}_2(\mathbb R))\cong\mathbb Z. \]
Let \(\rho:G\to\operatorname{GL}(V)\) be a finite-dimensional continuous representation. Its differential \[ d\rho:\mathfrak{sl}_2(\mathbb R)\to\mathfrak{gl}(V) \] complexifies to a finite-dimensional representation of \(\mathfrak{sl}_2(\mathbb C)\). Every such representation integrates to the simply connected complex group \(\operatorname{SL}_2(\mathbb C)\). Restricting the resulting group representation to the real subgroup \(\operatorname{SL}_2(\mathbb R)\) and then pulling back along \(p\) gives a representation of \(G\) with differential \(d\rho\).
Because \(G\) is connected and simply connected, a finite-dimensional Lie-group representation is uniquely determined by its differential. Hence \(\rho\) is exactly that pullback. Therefore \[ \ker p\subseteq\ker\rho. \] Since \(\ker p\ne1\), no finite-dimensional representation of \(G\) is faithful.
Thus \[ \boxed{\widetilde{\operatorname{SL}_2(\mathbb R)}} \] is a connected Lie group with no faithful finite-dimensional representation.