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Which is in Isom?

mathmari

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Apr 14, 2013
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Hey!! :giggle:

Let $\alpha\in \mathbb{R}$. We define the following maps:

(a) For $p\in \mathbb{R}^2$ let $\delta_{p,\alpha}=\tau_p\circ \delta_{\alpha}\circ\tau_p^{-1}$.
(b) For $p\in \mathbb{R}f_{\alpha}$ let $\sigma_{p,\alpha}=\tau_p\circ \sigma_{\alpha}\circ\tau_p^{-1}$.
(c) For $p\in \mathbb{R}f_{\alpha}$, $q\in \mathbb{R}e_{\alpha}$ let $\gamma_{q,p,\alpha}=\tau_q\circ \sigma_{p,\alpha}$.



Let $\beta\in \text{Isom}(\mathbb{R}^2)$. Then one of the following statements is true.

(i) $\beta=\text{id}$

(ii) $\beta=r_v$, $0\neq v\in \mathbb{R}^2$.

(iii) $\beta=\delta_{p,\alpha}$,$0<\alpha<2\pi$

(iv) $\beta=\sigma_{p,\alpha}$, $\alpha\in \mathbb{R}$, $p\in \mathbb{R}f_{\alpha}$

(v) $\beta=\gamma_{q,p,\alpha}$, $\alpha\in \mathbb{R}$, $p\in \mathbb{R}f_{\alpha}$ , $q\in \mathbb{R}e_{\alpha}$ , $0\neq q$



So we have to check which of these cases for $\beta$ we have an isometry, or not?

The identity function is in $\text{Isom}(\mathbb{R}^2)$, isn't it?

For the other ones do we have to checkif these maps are bijective?


:unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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Hey mathmari !!

What are $\mathbb Rf_\alpha$, $\mathbb Re_\alpha$, and $r_v$? 🤔

We would need to check if the $\beta$ are isometries. That is whether distances are preserved. 🤔

Btw, it looks as if all statements are true. o_O
If $r_v$ is a typo and is actually $\tau_v$, then we have the list of all isometries in $\mathbb R^2$.
 

mathmari

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Apr 14, 2013
4,713
What are $\mathbb Rf_\alpha$, $\mathbb Re_\alpha$, and $r_v$? 🤔

We would need to check if the $\beta$ are isometries. That is whether distances are preserved. 🤔

Btw, it looks as if all statements are true. o_O
If $r_v$ is a typo and is actually $\tau_v$, then we have the list of all isometries in $\mathbb R^2$.
$\mathbb Re_\alpha$ is a multiple of the vector $ \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) \\ \sin \left (\frac{\alpha}{2}\right )\end{pmatrix}$ and $\mathbb Rf_\alpha$ is a a multiplie of the vector $ \begin{pmatrix}-\sin \left (\frac{\alpha}{2}\right ) \\ \cos \left (\frac{\alpha}{2}\right )\end{pmatrix}$.

Yes it should be $\tau_v$, the translation, $\tau_v(x)=x+v$.

How do we see that all of the list are isometries in $\mathbb R^2$ ? How do we check of the distances are preserved? Do we calculate $\beta (x-y)$ ?

:unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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mathmari

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Apr 14, 2013
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The definition of an isometry says that it's a map with $d(\beta (x),\beta (y))=d(x,y)$. 🤔
(i) For the identity we have that $\beta (x)=x$ and $\beta (y)=y$, so $d(\beta (x),\beta (y))=d(x,y)$ is satisfied.

(ii) If $\beta=\tau_v$ then $\beta(x)=\tau_v(x)=x+v$, and $\beta(y)=\tau_v(y)=y+v$. Then $d(\beta (x),\beta (y))=d(x+v,y+v)$ this is equal to $d(x,y)$, right?

(iii) $\beta (x)= \tau_p\circ \delta_{\alpha}\circ\tau_p^{-1} (x)= \tau_p( \delta_{\alpha}(x-p))=\tau_p( d_{\alpha}(x-p))=d_{\alpha}(x-p)+p$, right?

(iv) $\beta (x)= \tau_p\circ \sigma_{\alpha}\circ\tau_p^{-1} (x)= \tau_p( \sigma_{\alpha}(x-p))=\tau_p( s_{\alpha}(x-p))=s_{\alpha}(x-p)+p$, right?

(v) $\beta (x)= \tau_q\circ \sigma_{p,\alpha} (x)=\tau_q\circ \tau_p\circ \sigma_{\alpha}\circ\tau_p^{-1} (x)=s_{\alpha}(x-p)+p+q$, right?


:unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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(i) For the identity we have that $\beta (x)=x$ and $\beta (y)=y$, so $d(\beta (x),\beta (y))=d(x,y)$ is satisfied.

(ii) If $\beta=\tau_v$ then $\beta(x)=\tau_v(x)=x+v$, and $\beta(y)=\tau_v(y)=y+v$. Then $d(\beta (x),\beta (y))=d(x+v,y+v)$ this is equal to $d(x,y)$, right?
The standard Euclidean metric is $d(x,y)=\|x-y\|$.
So $d(\beta (x),\beta (y))=d(x+v,y+v)=\|(x+v)-(y+v)\|=\|x-y\|=d(x,y)$. So yes. (Nod)

(iii) $\beta (x)= \tau_p\circ \delta_{\alpha}\circ\tau_p^{-1} (x)= \tau_p( \delta_{\alpha}(x-p))=\tau_p( d_{\alpha}(x-p))=d_{\alpha}(x-p)+p$, right?

(iv) $\beta (x)= \tau_p\circ \sigma_{\alpha}\circ\tau_p^{-1} (x)= \tau_p( \sigma_{\alpha}(x-p))=\tau_p( s_{\alpha}(x-p))=s_{\alpha}(x-p)+p$, right?

(v) $\beta (x)= \tau_q\circ \sigma_{p,\alpha} (x)=\tau_q\circ \tau_p\circ \sigma_{\alpha}\circ\tau_p^{-1} (x)=s_{\alpha}(x-p)+p+q$, right?
Yep.

So if we have that $\|d_\alpha y\|=\|y\|$ and $\|s_\alpha y\|=\|y\|$, then each of these will follow. 🤔
 

mathmari

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Apr 14, 2013
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The standard Euclidean metric is $d(x,y)=\|x-y\|$.
So $d(\beta (x),\beta (y))=d(x+v,y+v)=\|(x+v)-(y+v)\|=\|x-y\|=d(x,y)$. So yes. (Nod)


Yep.

So if we have that $\|d_\alpha y\|=\|y\|$ and $\|s_\alpha y\|=\|y\|$, then each of these will follow. 🤔

I got it now!! 🤩


One question... At the question statement is it meant that $\beta$ can be ONLY one of the given maps or that these maps have the property of $\beta$, i.e . that they are isometries? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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One question... At the question statement is it meant that $\beta$ can be ONLY one of the given maps or that these maps have the property of $\beta$, i.e . that they are isometries?
Makes sense yes.
I think we're supposed to prove that if $\beta$ is an isometry, that exactly 1 of the statements is true. 🤔
 

mathmari

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Apr 14, 2013
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Makes sense yes.
I think we're supposed to prove that if $\beta$ is an isometry, that exactly 1 of the statements is true. 🤔
But how can exactly one be true if all maps are isometries? I got stuck right now. :unsure:
 

Klaas van Aarsen

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But how can exactly one be true if all maps are isometries? I got stuck right now.
An isometry cannot be both a reflection and a rotation can it? :rolleyes:
 

mathmari

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Apr 14, 2013
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An isometry cannot be both a reflection and a rotation can it? :rolleyes:
I got stuck right now, you mean that not all maps can be an isometry although the distances are preserved? :unsure:
 

Klaas van Aarsen

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I got stuck right now, you mean that not all maps can be an isometry although the distances are preserved?
All those maps are isometries.
We need to prove that any given isometry must be one of them.
That is that we didn't 'miss' any isometries. And moreover that if an isometry matches one of them, that it won't simultaneously match another one as well. 🤔
 

mathmari

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Apr 14, 2013
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All those maps are isometries.
We need to prove that any given isometry must be one of them.
That is that we didn't 'miss' any isometries. And moreover that if an isometry matches one of them, that it won't simultaneously match another one as well. 🤔
Ah ok!

To show that we didn't 'miss' any isometries, do we suppose that we miss some and we get a contradiction, or how do we show that? :unsure:
 

Klaas van Aarsen

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To show that we didn't 'miss' any isometries, do we suppose that we miss some and we get a contradiction, or how do we show that?
I think we can state the list from wiki as the list of the possible euclidean isometries.
Then we should probably just prove that the given statements correspond to them and that they are mutually exclusive. 🤔
 

mathmari

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Apr 14, 2013
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I think we can state the list from wiki as the list of the possible euclidean isometries.
Then we should probably just prove that the given statements correspond to them and that they are mutually exclusive. 🤔
Ok!

(i) is the identity
(ii) is the translation nas for $v$
(iii) is the rotation around the point $p$ and with angle $a$
(iv) is the reflection about aline through the point $p$ and with angle $a$
(v) is a glide reflection


Is that correct? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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(i) is the identity
(ii) is the translation nas for $v$
(iii) is the rotation around the point $p$ and with angle $a$
(iv) is the reflection about a line through the point $p$ and with angle $a$
(v) is a glide reflection
Is that correct?
Yep. That is, it is in one direction. (Nod)
I think we still need to verify that every rotation can be written with a $p$ in $\mathbb Rf_\alpha$.
And similarly for reflections and glide reflections. 🤔

And also that a $\gamma$ as we defined it, is really a glide reflection and not a regular reflection. 🤔