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Is the composition of the isometries a rotation?

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
3,966
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? :unsure:

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? :unsure:


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 $a\in O_2$ ? :unsure:
 
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Klaas van Aarsen

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Mar 5, 2012
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Hey mathmari !!

Those are not geometric descriptions are they?
A geometric description is for instance that we have rotation with a specific angle around a specific point. 🧐
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
3,966
$\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1}$ is a translation back by $v$, a rotation aroung $\alpha$ and then again a translation by $v$.

$\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1}$ is a translation back by $v$, a refection about a line throught origin that has angle $\frac{\alpha}{2}$ with the $x$-axis and then again a translation by $v$.

Are these the geometric descriptions ? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,632
Leiden
$\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1}$ is a translation back by $v$, a rotation aroung $\alpha$ and then again a translation by $v$.

$\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1}$ is a translation back by $v$, a refection about a line throught origin that has angle $\frac{\alpha}{2}$ with the $x$-axis and then again a translation by $v$.

Are these the geometric descriptions ?
More or less, but I think we can do a bit better.
One step further from your description, we can say that the transformation $\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1}$ converts an absolute vector to a vector relative to $v$, which is then rotated by an angle $\alpha$, and then converted back into an absolute vector, can't we? 🤔

The proper geometric description would be that $\tau_v\circ \delta_{\alpha}\circ \tau_v^{-1}$ is a rotation by an angle $\alpha$ around the point $v$.
How does that sound? 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
3,966
What do you mean by "absolute vector" ?

And at the second case we have the same but instead of rotation we have reflection?

:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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What do you mean by "absolute vector" ?
I've learned that an "absolute vector" is simply a vector relative to the origin that identifies a point.
As opposed to a "relative vector" that indicates that it is relative to some point.
Now that I look it up, I see that it is terminology that seems to be specific for Computer Graphics. 🧐
Either way, that is where a main application of isometries is.

And at the second case we have the same but instead of rotation we have reflection?
Yes, but with a different angle, and it is not around a point, is it? 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
3,966
We convert the vector by $v$ then we reflect it by the line and then we convert the vector back by v. Or not? But which is the angle?
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,632
Leiden
We convert the vector by $v$ then we reflect it by the line and then we convert the vector back by v. Or not? But which is the angle?
Doesn't the problem statement say: "reflection about a line through the origin that has angle $\frac α2$ with the x-axis"? 🤔

I'd say: $\tau_v\circ \sigma_{\alpha}\circ \tau_v^{-1}$ is a reflection in a line through point $v$ that has angle $\frac α2$ with the x-axis. 🧐