Is there a hole in the Galilean group? According to Wikipedia > If ${\displaystyle G}$ is a compact simply connected Lie group, then it is determined by its Lie algebra, so it should be possible to calculate its cohomology from the Lie algebra. So, if Galilean group were compact and simply connected, then it would have one two-dimensional hole inside because the second cohomology group of the Galilei algebra is one-dimensional, and because the dimension of the $k$-th cohomology group means the number of the $k$-dimensional holes in $G$ as topological space. But, unfortunately, the Galilean group isn't compact, so I'm not sure that the cohomology of its Lie-algebra reflects its homology. But perhaps yes. So, my question is that how this group looks like. Is there a hole inside, or not?

Every connected Lie group $G$ has a maximal compact subgroup $K$ to which it is homotopy equivalent; in particular $G$ and $K$ have the same cohomology (with any coefficients). The maximal compact subgroup of the Galilean group is the subgroup $SO(3)$ of rotations. Topologically this is real projective space $\mathbb{RP}^3$, whose real (and even rational) cohomology is the same as that of the $3$-sphere $S^3$: we have $H^0 \cong H^3 \cong \mathbb{R}$ and all other real cohomology vanishes.

Hence, if you like, there is a "3-dimensional hole." But IMO one shouldn't take this hole idea too seriously as a description of cohomology.