The universal bundle for a topological group $G$ is usually written as $\pi:EG\to BG$ Any principal $G$ bundle for which the total space is contractible is universal; this will help us to find universal bundles without worrying about Milnor's construction of $EG$ involving infinite joins.

• $G=\mathbb{Z}_2$ $E\mathbb{Z}_2=S^\infty$ and $B\mathbb{Z}_2=\mathbb{R}P^\infty$
• $G=\mathbb{Z}_n$ $E\mathbb{Z}_n=S^\infty$ and $B\mathbb{Z}_n=S^\infty/\mathbb{Z}_n$ Here $\mathbb{Z}_n$ acts on $S^\infty$ (considered as a subset of a separable complex Hilbert space) via multiplication with an $n$ th root of unity.
• $G=\mathbb{Z}^n$ $E\mathbb{Z}^n=\mathbb{R}^n$ and $B\mathbb{Z}^n=T^n$
• More generally, if $G$ is any discrete group then one can take $BG$ to be any Eilenberg-Mac Lane space $ K(G,1)$ and $EG$ to be its universal cover. Indeed $EG$ is simply connected, and it follows from the lifting theorem that $\pi_n(EG)=0$ for $n\geq 0$ This example includes the previous three and many more.
• $G=S^1$ $ES^1=S^\infty$ and $BS^1=\mathbb{C}P^\infty$
• $G=SU(2)$ $ESU(2)=S^\infty$ and $BSU(2)=\mathbb{H}P^\infty$
• $G=O(n)$ the $n$ th orthogonal group: $EO(n)=V(\infty,n)$ the manifold of frames of $n$ orthonormal vectors in $\mathbb{R}^\infty$ and $BO(n)=G(\infty, n)$ the Grassmanian of $n$ planes in $\mathbb{R}^\infty$ The projection map is taking the subspace spanned by a frame of vectors.