Skip to main content
edited body
Source Link
Alain Valette
  • 11.1k
  • 44
  • 62

To complement Igor's answer. Consider the semi-direct product $G=\mathbb{C}\rtimes\mathbb{R}$ associated with the action of $\mathbb{R}$ on $\mathbb{C}$ by rotations; so the product in $G$ is given by $(z,s)(z's')=(z+e^{is}z',s+s')$. Note that $G$ is just the universal cover of the motion group of $\mathbb{R}^2$. Take the standard euclidean structure on $\mathbb{C}\times\mathbb{R}$: from the formula for the product, it is clearly left-invariant but nornot right invariant.

To complement Igor's answer. Consider the semi-direct product $G=\mathbb{C}\rtimes\mathbb{R}$ associated with the action of $\mathbb{R}$ on $\mathbb{C}$ by rotations; so the product in $G$ is given by $(z,s)(z's')=(z+e^{is}z',s+s')$. Note that $G$ is just the universal cover of the motion group of $\mathbb{R}^2$. Take the standard euclidean structure on $\mathbb{C}\times\mathbb{R}$: from the formula for the product, it is clearly left-invariant but nor right invariant.

To complement Igor's answer. Consider the semi-direct product $G=\mathbb{C}\rtimes\mathbb{R}$ associated with the action of $\mathbb{R}$ on $\mathbb{C}$ by rotations; so the product in $G$ is given by $(z,s)(z's')=(z+e^{is}z',s+s')$. Note that $G$ is just the universal cover of the motion group of $\mathbb{R}^2$. Take the standard euclidean structure on $\mathbb{C}\times\mathbb{R}$: from the formula for the product, it is clearly left-invariant but not right invariant.

Source Link
Alain Valette
  • 11.1k
  • 44
  • 62

To complement Igor's answer. Consider the semi-direct product $G=\mathbb{C}\rtimes\mathbb{R}$ associated with the action of $\mathbb{R}$ on $\mathbb{C}$ by rotations; so the product in $G$ is given by $(z,s)(z's')=(z+e^{is}z',s+s')$. Note that $G$ is just the universal cover of the motion group of $\mathbb{R}^2$. Take the standard euclidean structure on $\mathbb{C}\times\mathbb{R}$: from the formula for the product, it is clearly left-invariant but nor right invariant.