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[SOLVED] Symmetry map and rotation

mathmari

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Apr 14, 2013
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Hey!! :eek:

Let $F\subseteq \mathbb{R}^2$. A map $\pi:\mathbb{R}^2\rightarrow \mathbb{R}$ is called symmetry map of $F$, if $\pi(F)=F$. A symmetry map of $F$ is a map where the figures $F$ and $\pi (F)$ are congruent.

  • Let $\pi_1:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $x\mapsto \begin{pmatrix}0 &-1 \\ 1 & 0\end{pmatrix}\cdot x$ and $\pi_2:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $x\mapsto \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\cdot x$ be symmetry maps of the below square:

    maps.JPG
    1. Determine the mapping formula of $\pi_1\circ \pi_2$ and $\pi_2\circ\pi_1$. Give the geometric interpretation of $\pi_1, \ \pi_2, \ \pi_1\circ \pi_2, \ \pi_2\circ \pi_1$.
    2. Describe all other symmetry maps of the square geometrically and give the mapping formula in each case.
    3. Show that all symmetry maps of the square can be represented by executing $\pi_1$ and $\pi_2$ one after the other.
  • Let $\delta_a:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $x\mapsto d_a\cdot x$ be a rotation by $a$ anticlockwise around the origin.
    1. Determine $\delta_a\left (\begin{pmatrix}1 \\ 0\end{pmatrix}\right )$ and $\delta_a\left (\begin{pmatrix}0 \\ 1\end{pmatrix}\right )$. Make a graph for that.
    2. Determine $d_a$.
    3. Show that $\delta_a\circ \delta_b=\delta_{a+b}$.


I have done the following:


    1. For $\pi_1\circ \pi_2$ we have:
      \begin{equation*}\left (\pi_1\circ \pi_2\right )(x)=\pi_1 \left (\pi_2(x)\right )=\pi_1 \left (\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\cdot x\right )=\begin{pmatrix}0 &-1 \\ 1 & 0\end{pmatrix}\cdot \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\cdot x=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\cdot x\end{equation*}

      For $\pi_2\circ \pi_1$ we have:
      \begin{equation*}\left (\pi_2\circ \pi_1\right )(x)=\pi_2 \left (\pi_1(x)\right )=\pi_2 \left (\begin{pmatrix}0 &-1 \\ 1 & 0\end{pmatrix}\cdot x\right )=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\cdot \begin{pmatrix}0 &-1 \\ 1 & 0\end{pmatrix}\cdot x=\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\cdot x\end{equation*}

      For the geometric interpretation we have to see graphically what the map does at the unit vectors, or not? (Wondering)

    2. Could you give a hint for that? (Wondering)

    3. Could you give me a hint how we could show that? (Wondering)

    1. How can we do that without knowing $d_a$ ? (Wondering)

    2. How can we determine $d_a$ ? (Wondering)


 

Klaas van Aarsen

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Mar 5, 2012
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1. For $\pi_1\circ \pi_2$ we have:
\begin{equation*}\left (\pi_1\circ \pi_2\right )(x)=\pi_1 \left (\pi_2(x)\right )=\pi_1 \left (\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\cdot x\right )=\begin{pmatrix}0 &-1 \\ 1 & 0\end{pmatrix}\cdot \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\cdot x=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\cdot x\end{equation*}

For $\pi_2\circ \pi_1$ we have:
\begin{equation*}\left (\pi_2\circ \pi_1\right )(x)=\pi_2 \left (\pi_1(x)\right )=\pi_2 \left (\begin{pmatrix}0 &-1 \\ 1 & 0\end{pmatrix}\cdot x\right )=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\cdot \begin{pmatrix}0 &-1 \\ 1 & 0\end{pmatrix}\cdot x=\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\cdot x\end{equation*}

For the geometric interpretation we have to see graphically what the map does at the unit vectors, or not?
Hey mathmari !!

Yep. (Nod)

2. Could you give a hint for that?
The symmetries of the square is the dihedral group $D_4$.
It consists of identity, 3 rotations, and 4 reflections.
Can we find a matrix for each of them? (Wondering)

3. Could you give me a hint how we could show that?
How are the elements of $D_4$ represented? (Wondering)

  • Let $\delta_a:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $x\mapsto d_a\cdot x$ be a rotation by $a$ anticlockwise around the origin.
1. How can we do that without knowing $d_a$ ?
Isn't $d_a$ given? It is a rotation over an angle $a$ in anticlockwise direction. (Thinking)

2. How can we determine $d_a$ ?
What is the matrix for a rotation around angle $a$?
Or perhaps more specifically, which vector do we get if we rotate $\binom 10$ over an angle $a$? (Wondering)
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,004
The symmetries of the square is the dihedral group $D_4$.
It consists of identity, 3 rotations, and 4 reflections.
Can we find a matrix for each of them? (Wondering)
The identity is the identity matrix.

The rotation is $\begin{pmatrix}\cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta)\end{pmatrix}$. We already have the matrix for $\theta=\frac{\pi}{2}$ (map $\pi_1$).

The remaining rotations are for $\theta=\pi$ and $\theta=\frac{3\pi}{2}$, right? (Wondering)



The reflection is $\begin{pmatrix}\cos (2\theta) & \sin (2\theta) \\ \sin (2\theta) & -\cos (2\theta)\end{pmatrix}$. We already have the matrix for $\theta=\pi$ (map $\pi_2$).

The remaining reflections are for $\theta=\frac{\pi}{2}$ and $\theta=\frac{3\pi}{2}$. Which is $4$th one? (Wondering)



How are the elements of $D_4$ represented? (Wondering)
Using the matrices above? (Wondering)


Isn't $d_a$ given? It is a rotation over an angle $a$ in anticlockwise direction. (Thinking)


What is the matrix for a rotation around angle $a$?
Or perhaps more specifically, which vector do we get if we rotate $\binom 10$ over an angle $a$? (Wondering)
So do we take here the matrices as above? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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The identity is the identity matrix.

The rotation is $\begin{pmatrix}\cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta)\end{pmatrix}$. We already have the matrix for $\theta=\frac{\pi}{2}$ (map $\pi_1$).

The remaining rotations are for $\theta=\pi$ and $\theta=\frac{3\pi}{2}$, right? (Wondering)



The reflection is $\begin{pmatrix}\cos (2\theta) & \sin (2\theta) \\ \sin (2\theta) & -\cos (2\theta)\end{pmatrix}$. We already have the matrix for $\theta=\pi$ (map $\pi_2$).

The remaining reflections are for $\theta=\frac{\pi}{2}$ and $\theta=\frac{3\pi}{2}$. Which is $4$th one? (Wondering)
What about $\theta=0$? (Wondering)

Using the matrices above? (Wondering)

So do we take here the matrices as above? (Wondering)
Yep. The matrices above are good. (Nod)
 

mathmari

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Apr 14, 2013
4,004

mathmari

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Apr 14, 2013
4,004
Let's consider the map $\pi_1$ and see how this changes the square with vertices $(1,1), \ (-1,1), \ (1,-1), \ (-1,-1)$.

The square is:

Quadrat.JPG

From the map $\pi_1$ we get the new vertices:
\begin{align*}&\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}-1 \\ 1\end{pmatrix} \\
&\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}-1 \\ 1\end{pmatrix}=\begin{pmatrix}-1 \\ -1\end{pmatrix} \\
&\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}1 \\ -1\end{pmatrix}=\begin{pmatrix}1 \\ 1\end{pmatrix} \\
&\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}-1 \\ -1\end{pmatrix}=\begin{pmatrix}1 \\ -1\end{pmatrix}\end{align*}

And so the square becomes:

quadrat_p1.JPG

So $\pi_1$ is a rotation by $\frac{\pi}{2}$ anticlockwise.


From the map $\pi_2$ we get the new vertices:
\begin{align*}&\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ -1\end{pmatrix} \\
&\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}-1 \\ 1\end{pmatrix}=\begin{pmatrix}-1 \\ -1\end{pmatrix} \\
&\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}1 \\ -1\end{pmatrix}=\begin{pmatrix}1 \\ 1\end{pmatrix} \\
&\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}-1 \\ -1\end{pmatrix}=\begin{pmatrix}-1 \\ 1\end{pmatrix}\end{align*}

And so the square becomes:

quadrat_p2.JPG

So $\pi_2$ is a reflection in respect to the $x$-axis.



From the map $\pi_1\circ\pi_2$ we get the new vertices:
\begin{align*}&\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ 1\end{pmatrix} \\
&\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}-1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ -1\end{pmatrix} \\
&\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}1 \\ -1\end{pmatrix}=\begin{pmatrix}-1 \\ 1\end{pmatrix} \\
&\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}-1 \\ -1\end{pmatrix}=\begin{pmatrix}-1 \\ -1\end{pmatrix}\end{align*}

And so the square becomes:

quadrat_p1p2.JPG

So $\pi_1\circ\pi_2$ is a rotation by $\frac{\pi}{2}$ anticlockwise and then a reflection.



From the map $\pi_2\circ\pi_1$ we get the new vertices:
\begin{align*}&\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}-1 \\ -1\end{pmatrix} \\
&\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}-1 \\ 1\end{pmatrix}=\begin{pmatrix}-1 \\ 1\end{pmatrix} \\
&\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}1 \\ -1\end{pmatrix}=\begin{pmatrix}1 \\ -1\end{pmatrix} \\
&\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}-1 \\ -1\end{pmatrix}=\begin{pmatrix}1 \\ 1\end{pmatrix}\end{align*}

And so the square becomes:

quadrat_p2p1.JPG

So $\pi_2\circ\pi_1$ is a reflection and then a rotation by $\frac{\pi}{2}$.



Is everything correct so far? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Do we not have the identity with $\theta=0$ ?
That is the case for a rotation, but not for a reflection. (Worried)

Still, a reflection in a line with angle $0$ is the same as a reflection in a line with angle $\pi$, isn't it?
The reflection in the line with angle $\frac\pi 2$ is also the same as in the line with angle $\frac{3\pi}{2}$.
How about $\frac\pi 4$? (Wondering)

Let's consider the map $\pi_1$ and see how this changes the square with vertices $(1,1), \ (-1,1), \ (1,-1), \ (-1,-1)$.

...

Is everything correct so far?
Looks good. (Nod)
 

mathmari

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Apr 14, 2013
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Looks good. (Nod)
How can we show that all symmetry maps of the square can be represented by executing $\pi_1$ and $\pi_2$ one after the other? (Wondering)


All the symmetry maps of the square are:

- Rotation by $i\cdot \frac{\pi}{2}$ with $i\in \{1,2,3\}$.
- Reflection by the axes $x$ and $y$.
- Reflection by the diagonal and the antidiagonal.
- Identity map.

The map $\pi_1$ is the rotation by $\frac{\pi}{2}$, right? So $\pi_1\circ\pi_1$ is the rotation by $2\cdot \frac{\pi}{2}=\pi$ and $\pi_1\circ\pi_1\circ$ is the rotation by $3\cdot \frac{\pi}{2}$.

The map $\pi_2$ is the reflection by the $x$-axis, right? The $\pi_1\circ\pi_2$ is the reflection by the diagonal, or not? The $\pi_2\circ\pi_1$ is the reflection by the antidiagonal, right? (Wondering)

So we miss the reflection by the $y$ axis and the identity, or not? (Wondering)
 
Last edited:

Klaas van Aarsen

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Staff member
Mar 5, 2012
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How can we show that all symmetry maps of the square can be represented by executing $\pi_1$ and $\pi_2$ one after the other? (Wondering)


All the symmetry maps of the square are:

- Rotation by $i\cdot \frac{\pi}{2}$ with $i\in \{1,2,3\}$.
- Reflection by the axes $x$ and $y$.
- Reflection by the diagonal and the antidiagonal.
- Identity map.

The map $\pi_1$ is the rotation by $\frac{\pi}{2}$, right? So $\pi_1\circ\pi_1$ is the rotation by $2\cdot \frac{\pi}{2}=\pi$ and $\pi_1\circ\pi_1\circ$ is the rotation by $3\cdot \frac{\pi}{2}$.

The map $\pi_2$ is the reflection by the $x$-axis, right? The $\pi_1\circ\pi_2$ is the reflection by the diagonal, or not? The $\pi_2\circ\pi_1$ is the reflection by the antidiagonal, right? (Wondering)

So we miss the reflection by the $y$ axis and the identity, or not?
Usually the dihedral group $D_4$ is represented by $\{\operatorname{id}, \rho, \rho^2, \rho^3, \sigma, \rho\sigma, \rho^2\sigma, \rho^3\sigma\}$, where $\rho$ is the anticlockwise rotation by $\frac\pi 2\,\text{rad}$, and $\sigma$ is the reflection in the x-axis.

In this case we have $\rho=\pi_1$ and $\sigma=\pi_2$ don't we? (Thinking)
 

mathmari

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Apr 14, 2013
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Usually the dihedral group $D_4$ is represented by $\{\operatorname{id}, \rho, \rho^2, \rho^3, \sigma, \rho\sigma, \rho^2\sigma, \rho^3\sigma\}$, where $\rho$ is the anticlockwise rotation by $\frac\pi 2\,\text{rad}$, and $\sigma$ is the reflection in the x-axis.

In this case we have $\rho=\pi_1$ and $\sigma=\pi_2$ don't we? (Thinking)
Ahh yes!!

But how can we that all the symmetry maps of the square can be represented by executing $\pi_1$ and $\pi_2$ one after the other? (Wondering)

Is it enough to give the above ones? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Ahh yes!!

But how can we that all the symmetry maps of the square can be represented by executing $\pi_1$ and $\pi_2$ one after the other?

Is it enough to give the above ones?
We can construct identity by applying $\pi_2$ twice (or by not applying anything at all).
We can construct the reflection in the y-axis with $\pi_2$ in combination with a repetition of $\pi_1$ can't we? (Wondering)
 

mathmari

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Apr 14, 2013
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We can construct identity by applying $\pi_2$ twice (or by not applying anything at all).
We can construct the reflection in the y-axis with $\pi_2$ in combination with a repetition of $\pi_1$ can't we? (Wondering)
Ah ok! So do we have the following? (Wondering)

All the symmetry maps of the square are:

- Rotation by $i\cdot \frac{\pi}{2}$ with $i\in \{1,2,3\}$.
- Reflection by the axes $x$ and $y$.
- Reflection by the diagonal and the antidiagonal.
- Identity map.

The map $\pi_1$ is the rotation by $\frac{\pi}{2}$. So $\pi_1\circ\pi_1$ is the rotation by $2\cdot \frac{\pi}{2}=\pi$ and $\pi_1\circ\pi_1\circ$ is the rotation by $3\cdot \frac{\pi}{2}$.

The map $\pi_2$ is the reflection by the $x$-axis. The $\pi_1\circ\pi_2$ is the reflection by the diagonal. The $\pi_2\circ\pi_1$ is the reflection by the antidiagonal.

The map $\pi_2\circ\pi_2$ is the identity map.

The map $\pi_1\circ \pi_2\circ\pi_1$ is the reflection in the y-axis.


Is this enough to show the desired result? Or do wwe have to prove something further? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Ah ok! So do we have the following? (Wondering)

All the symmetry maps of the square are:

- Rotation by $i\cdot \frac{\pi}{2}$ with $i\in \{1,2,3\}$.
- Reflection by the axes $x$ and $y$.
- Reflection by the diagonal and the antidiagonal.
- Identity map.

The map $\pi_1$ is the rotation by $\frac{\pi}{2}$. So $\pi_1\circ\pi_1$ is the rotation by $2\cdot \frac{\pi}{2}=\pi$ and $\pi_1\circ\pi_1\circ$ is the rotation by $3\cdot \frac{\pi}{2}$.

The map $\pi_2$ is the reflection by the $x$-axis. The $\pi_1\circ\pi_2$ is the reflection by the diagonal. The $\pi_2\circ\pi_1$ is the reflection by the antidiagonal.

The map $\pi_2\circ\pi_2$ is the identity map.

The map $\pi_1\circ \pi_2\circ\pi_1$ is the reflection in the y-axis.
Don't we have $(\pi_1\circ \pi_2\circ\pi_1)(\mathbf e_1) = \pi_1(\pi_2(\pi_1(\mathbf e_1))) = \pi_1(\pi_2(\mathbf e_2)) = \pi_1(-\mathbf e_2) = \mathbf e_1$?
Shouldn't a reflection in the y-axis result in $-\mathbf e_1$?
I think we have the wrong composition. (Worried)

Is this enough to show the desired result? Or do we have to prove something further?
I believe this covers question 3 of the first bullet - if we have the correct reflection in the y-axis. (Nod)
 

mathmari

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Don't we have $(\pi_1\circ \pi_2\circ\pi_1)(\mathbf e_1) = \pi_1(\pi_2(\pi_1(\mathbf e_1))) = \pi_1(\pi_2(\mathbf e_2)) = \pi_1(-\mathbf e_2) = \mathbf e_1$?
Shouldn't a reflection in the y-axis result in $-\mathbf e_1$?
I think we have the wrong composition. (Worried)
Ahh ok! So it must be $\pi_2\circ \pi_1\circ \pi_1$, right? (Wondering)



I believe this covers question 3 of the first bullet - if we have the correct reflection in the y-axis. (Nod)
Great!! But what about the question 2 of the first bullet? Isn't this similar to the question 3? (Wondering)



About the second bullet:

For question 1 do we draw an unit circle and take an arbitrary angle and then we use the trigonometric functions to get the image of the map? (Wondering)

Using the images of $\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $\begin{pmatrix}0 \\ 1\end{pmatrix}$ can we determine $d_a$ ? (Wondering)
 

Klaas van Aarsen

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Ahh ok! So it must be $\pi_2\circ \pi_1\circ \pi_1$, right?
Yep. (Nod)

Great!! But what about the question 2 of the first bullet? Isn't this similar to the question 3?
You have listed all symmetries geometrically yes.
Question 2 also asks for their mappings. If I'm not mistaken you've only listed 4 of them. (Thinking)

About the second bullet:

For question 1 do we draw an unit circle and take an arbitrary angle and then we use the trigonometric functions to get the image of the map?

Using the images of $\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $\begin{pmatrix}0 \\ 1\end{pmatrix}$ can we determine $d_a$ ?
Yes. You have already listed the matrix for a rotation by an angle $\theta$. Replace the angle by $a$ and we're good to go. (Happy)
 

mathmari

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Apr 14, 2013
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You have listed all symmetries geometrically yes.
Question 2 also asks for their mappings. If I'm not mistaken you've only listed 4 of them. (Thinking)
You mean the compositions? Or do we use the compositions of question 3 and calculate the matrices?

But can we do that in that way? Because I list the compositions in question 3,can I use these also in question 2 to calculate the matrices? Or is there also an other way? (Wondering)
 

Klaas van Aarsen

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You mean the compositions? Or do we use the compositions of question 3 and calculate the matrices?

But can we do that in that way? Because I list the compositions in question 3,can I use these also in question 2 to calculate the matrices? Or is there also an other way? (Wondering)
We can write down the matrix for each geometric transformation directly.
For instance the reflection in the y-axis maps $\binom 10$ to $\binom{-1}0$.
And it maps $\binom 01$ to $\binom 01$.
Therefore the matrix is $$\begin{pmatrix}-1&0\\0&1\end{pmatrix}.$$
The columns in the matrix are the images of the 2 unit vectors.
(Thinking)
 

mathmari

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Apr 14, 2013
4,004
We can write down the matrix for each geometric transformation directly.
For instance the reflection in the y-axis maps $\binom 10$ to $\binom{-1}0$.
And it maps $\binom 01$ to $\binom 01$.
Therefore the matrix is $$\begin{pmatrix}-1&0\\0&1\end{pmatrix}.$$
The columns in the matrix are the images of the 2 unit vectors.
(Thinking)
Ok!! Thank you!! (Mmm)