# Lie group as manifold

A Lie group is a manifold $G$ that is a group s.t. the multiplication map $G\rightarrow G$ and its inverse is differentiable. Further suppose that $G$ is compact, connected and is a $2n$ manifold in a $2n+1$ dimensional vector space $V$.

1. How do I prove that the set $U$ of quaternions of unit length is a 3d compact Lie group in $\mathbb{R}^ 4$?
2. How do I prove that $G=U\times S^1\subset\mathbb{R}^6$ is such a group?

What I thought:
1. Quaternions are elements of the form $a+bi+cj+dk$ with $a,b,c,d$ real numbers. I know what the multiplication is here. But how can I show that the multiplication is differentiable? And what is the inverse of the multiplication?
2. So we want to prove that $U\times S^1$ is compact, connected and a $2n$ manifold in a $2n+1$ dimensional vector space $V$. It seems to me that this is a $6$-manifold in $\mathbb{R}^7$, but how do I show the rest?

• 1. See the answers here. $U$ is isomorphic to $SU(2)$, and hence a Lie group. – Dietrich Burde Dec 21 '16 at 13:41
• @DietrichBurde Could you also help with 2.? – user389900 Dec 21 '16 at 13:44
• 2. follows from 1. Since $SU(2)$ and $S^1$ are compact, so is the direct product $U\times S^1\cong SU(2)\times S^1$. Similarly for the other claims. – Dietrich Burde Dec 21 '16 at 13:45
• @DietrichBurde Is $U\times S^1$ indeed a $6$-manifold? Is it not 4 dimensional? – user389900 Dec 21 '16 at 16:29
• $S^1$ is a manifold of dimension $1$, which can be embedded into $\mathbb{R}^2$, and $U$ has dimension $3$, and can be embedded into $\mathbb{R}^4$. Hence $U\times S^1$ can be embedded into $\mathbb{R}^6$. Its dimension, though, is $4$. – Dietrich Burde Dec 21 '16 at 16:37