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- Apr 14, 2013

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Hey!!

Let $\delta_a:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the rotation around the origin with angle $\alpha$ and let $\sigma_{\alpha}:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the reflection about a line through the origin that has angle $\frac{\alpha}{2}$ with the $x$-axis.

Let $v\in V$ and $\alpha\in \mathbb{R}$.

I want to determine the geometric description of the isometries $\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1}$ and $\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1}$.

$\tau_v$ is the translation about $v$, i.e. $\tau_v(x)=x+v$.

After that I want to show that for $a\in O_2$ and $v\in V$ it holds that $\phi_a\circ \tau_v\circ\phi_a^{-1}=\tau_{av}$.

Let's consider the isometry $\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1}$:

$$(\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1})(x)=\tau_v(\delta_{\alpha}(\tau_v^{-1}(x)))=\tau_v(\delta_{\alpha}(x-v))=\delta_{\alpha}(x-v)+v$$ Is the total result a rotation?

Let's consider now the isometry $\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1}$:

$$(\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1})(x)=\tau_v(\sigma_{\alpha}(\tau_v^{-1}(x)))=\tau_v(\sigma_{\alpha}(x-v))=\sigma_{\alpha}(x-v)+v$$ Is the total result again a rotation?

For the other question: We have that $\phi_a:V\rightarrow V, \ v\mapsto av$.

$O_2$ is the set of orthogonal $2\times 2$ matrices.

We have that \begin{align*}(\phi_a\circ \tau_v\circ\phi_a^{-1})(x)&=\phi_a( \tau_v(\phi_a^{-1}(x)))=\phi_a( \tau_v(a^{-1}x))=\phi_a(a^{-1}x+v) \\ & =a(a^{-1}x+v)=x+av=\tau_{av}(x)\end{align*} So it is done. Is this correct? Where did we use here that $a\in O_2$ ?

Let $\delta_a:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the rotation around the origin with angle $\alpha$ and let $\sigma_{\alpha}:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the reflection about a line through the origin that has angle $\frac{\alpha}{2}$ with the $x$-axis.

Let $v\in V$ and $\alpha\in \mathbb{R}$.

I want to determine the geometric description of the isometries $\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1}$ and $\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1}$.

$\tau_v$ is the translation about $v$, i.e. $\tau_v(x)=x+v$.

After that I want to show that for $a\in O_2$ and $v\in V$ it holds that $\phi_a\circ \tau_v\circ\phi_a^{-1}=\tau_{av}$.

Let's consider the isometry $\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1}$:

$$(\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1})(x)=\tau_v(\delta_{\alpha}(\tau_v^{-1}(x)))=\tau_v(\delta_{\alpha}(x-v))=\delta_{\alpha}(x-v)+v$$ Is the total result a rotation?

Let's consider now the isometry $\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1}$:

$$(\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1})(x)=\tau_v(\sigma_{\alpha}(\tau_v^{-1}(x)))=\tau_v(\sigma_{\alpha}(x-v))=\sigma_{\alpha}(x-v)+v$$ Is the total result again a rotation?

For the other question: We have that $\phi_a:V\rightarrow V, \ v\mapsto av$.

$O_2$ is the set of orthogonal $2\times 2$ matrices.

We have that \begin{align*}(\phi_a\circ \tau_v\circ\phi_a^{-1})(x)&=\phi_a( \tau_v(\phi_a^{-1}(x)))=\phi_a( \tau_v(a^{-1}x))=\phi_a(a^{-1}x+v) \\ & =a(a^{-1}x+v)=x+av=\tau_{av}(x)\end{align*} So it is done. Is this correct? Where did we use here that $a\in O_2$ ?

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