Chapter 4.1 Introduction to Symmetry

In this chapter, we focus all on the principle of symmetry.

Principle of Symmetry

The principle of symmetry is to identify the transformations that do not change a thing in order to understand that thing.

The goal of this section is to develop the language so that we clearly state what we mean above. We begin with defining what a group is and then state some examples of symmetries.

Invariance of Sets under a Symmetry Group

To define what we mean by “do not change a thing”, we first define a binary operation.

Binary Operation

A binary operation \(\cdot\) on a set \(S\) takes pairs of elements of \(S\) to elements of \(S\).

If \(a\) and \(b\) are in \(S\), write \(a\cdot b\) for the element to which \(\cdot\) takes the pair \((a,b)\).

A binary operation is associative if for every \(x\), \(y\), and \(z\) in \(S\) \[x\cdot(y\cdot z) = (x\cdot y)\cdot z.\]

This is something that you have already seen before. For example, addition is a binary operation on the set of natural numbers \(\mathbb{N}.\) It is also an associative operation.

The next definition defines the “thing” that we are going to focus on.

Group

A group is a pair \((G, \cdot)\), where \(G\) is a set, \(\cdot\) is an associative binary operation, and \(G\) has two additional properties:

there is an element \(e\) in \(G\), the identity element, so that for every \(g\) in \(G\), \[e\cdot g = g\cdot e = g;\]
for every \(g\) in \(G\), \(g\) has an inverse \(h\) in \(G\), an element with the property \[g\cdot h = h\cdot g = e.\]

Note that there is no assumption that \(\cdot\) is commutative.

Property i just states there is an element in the group that doesn’t change the other elements when the binary operation is applied.

Property ii just states that for each element in the group, there is an element such that the binary operation on these two elements produces the identity element.

Although the word group may be unfamilar to you, the concept is familar. In fact, you have already seen a group: the integers under addition and the real numbers also under addition.

Example 1

The integers is a group, where the operation \(\cdot\) is \(+\), the element \(e\) is \(0\), and the inverse of an integer \(z\) is the integer \(-z\).
The real numbers is a group, where \(\cdot\) is the addition operation \(+\).
The unit circle is a group, where \(\cdot\) is given by \(\star\).

Another example of a group is an analog clock. Study this next example.

Example 2

Label the hours on an analog clock as \(0, 1, 2, 3, \dots, 11\). Take \(\mathbb Z_{12}\) to be these hours. Define \(+\) on \(\mathbb Z_{12}\) (or \(\cdot\) in order to agree with the notation for groups) in the following way: if \(a\) and \(b\) are in \(\mathbb Z_{12}\), take \(a+b\) to be the hour that the clock reads after \(a\) hours pass when the hour hand was originally at \(b\).

Add some points in \(\mathbb Z_{12}\). What is the identity element in \(\mathbb Z_{12}\)?
Find the inverse element of every element of \(\mathbb Z_{12}\).

Here are some examples
- \(1+11=0\), the clock reads \(0\) after \(1\) hour passes when the hour hand was originally at \(11\).
- \(2+5=7\), the clock reads \(7\) after \(2\) hours pass when the hour hand was originally at \(5\).
- \(3+10=1\), the clock reads \(1\) after \(3\) hours pass when the hour hand was originally at \(10\).
- \(0+6=6\), the clock reads \(6\) after \(0\) hours pass when the hour hand was originally at \(6\).
- The identity element is \(0\)
For each element \(a\) in \(\mathbb Z_{12}\), we need to find an element \(b\) so that \(a+b=0\).
- The inverse of \(0\) is \(0\) because \(0+0=0\).
- The inverse of \(1\) is \(11\) because \(1+11=0\).
- The inverse of \(2\) is \(10\) because \(2+10=0\).
- The inverse of \(3\) is \(9\) because \(3+9=0\).
- The inverse of \(4\) is \(8\) because \(4+8=0\).
- The inverse of \(5\) is \(7\) because \(5+7=0\).
- The inverse of \(6\) is \(6\) because \(6+6=0\).
- The inverse of \(7\) is \(5\) because \(7+5=0\).
- The inverse of \(8\) is \(4\) because \(8+4=0\).
- The inverse of \(9\) is \(3\) because \(9+3=0\).
- The inverse of \(10\) is \(2\) because \(10+2=0\).
- The inverse of \(11\) is \(1\) because \(11+1=0\).

Now that we know what a group is, let us define what a symmetry is.

Symmetry

A symmetry acting on a set \(X\) is an invertible function from \(X\) to \(X\) that preserves a certain structure.

Note: We are purposely imprecise in using the word “structure” because we want to be as flexible as possible.

Since any two symmetries \(f\) and \(g\) that act on the same set have domain \(X\) and range \(X\), the composite \(f\circ g\) is defined.

The composite of any two symmetries that preserve the same structure will still preserve that structure.

So we have the following example.

Example 3

Take \(G\) to be the set of all symmetries that preserve a given structure.

Composition, denoted by \(\circ\), is the associative binary operation on \(G\).

The identity function (takes each element to itself) is a symmetry, as is the inverse of any symmetry.

The pair \((G, \circ)\) is a group

One example of a symmetry is the set of all vertical lines \(L\) so that the reflection of \(f\) across \(L\) is equal to \(f\). It may be helpful to graph \(f\) to understand what one needs to look for.

Example 4

Take \(f\) to be the function that is given by

\[f(x) = (x-3)^6.\]

Determine a vertical line \(L\) so that the reflection of \(f\) across \(L\) across \(L\) is equal to \(f\).

A sketch of \(f\) looks like this:

The function \(f\) is similar to the quadratic function \(g(x)=(x-3)^2\) in that it has an axis of symmetry. The axis of symmetry for \(g\) is \(x=3,\) meaning that the reflection of \(g\) across \(x=3\) is \(g\) again. Another way to state this is for any real number \(x\) in \(\mathbb{R}\)

\[g(6-x)=g(x).\]

So for \(f\), its axis of symmetry is \(x=3\) as well.

So \(L=3\) is the right vertical line.

Identifying the symmetry group of a set with a certain structure can help us to better understand and determine the properties of the set and the structure.

Example 5

The group \(H\) formed by the translations of the plane together with the non-zero scalings of the \(y\) axis is a group that preserves the set of quadratic polynomials.

The quadratic polynomials are symmetric under this (transformation) group.

Furthermore, given any two quadratic polynomials \(f\) and \(g\), there is \(h\) an in \(H\) so that \[h(f) = g.\]

So, in this sense, all quadratic polynomials look the same—We used this to simplify calculations involving quadratic polynomials.

Example 6

Take \(f\) and \(g\) to be the quadratic polynomials given by \[f(x) = 3x^2 - x + 1 \quad {\rm and} \quad g(x) = 6x^2 + 5x +2.\] Find an element of \(H\) that transforms \(f\) into \(g\).

Find the transformations by first writing each quadratic into the form \(f(x)=A(x-h)^2+k\). The function \(f\) can be rewritten as \[f(x)=3\left(x-\tfrac{1}{6}\right)^2+\tfrac{11}{2}\] and \(g\) can be rewritten as \[ g(x)=6\left(x+\tfrac{5}{12}\right)^2+\tfrac{23}{24}. \]

Here are two ways to do it.

Method 1. Transform \(f\) so it looks like \(\mathrm{pow}_2\) and then transform it to look like \(g\).

Move by the vector \(V=\langle -\tfrac{1}{16},-\tfrac{11}{2}\rangle\) to move to the origin.

Scale by \(y\)-coordinate by \(\tfrac{1}{3}\): \(Y_{1/3}\)

This transforms it to \(\mathrm{pow}_2\). Transform it to look like \(g\).

Move the origin by \(W=\langle \tfrac{5}{12},\tfrac{23}{24}\rangle\) so the vertex is at \(\left(\tfrac{5}{12},\tfrac{23}{24}\right)\)

Now scale \(y\)-coordinate by \(6\): \(Y_6\)

And this is \(g\).

Method 2. Transform \(f\) directly to \(g\).

Move by the vector \[V=(\tfrac{5}{12},\tfrac{23}{24})-(\tfrac{1}{6},\tfrac{11}{2})=\langle \tfrac{1}{4},-\tfrac{109}{24}\rangle\] to move vertex from \((\tfrac{1}{6},\tfrac{11}{2})\) to \(\left(\tfrac{5}{12},\tfrac{23}{24}\right)\).

Now scale \(y\)-coordinate by \(2\): \(Y_2\)

And this is \(g\).

In this example, we define the group of rigid motions of the plane.

Example 7

The set \(G\) of all reflections across any line in the plane, rotations of any angle about any point, and translations in the plane, together with all finite composites of these functions is a group with a binary operation given by composition.

Every element \(g\) in \(G\) takes a pair of points to another pair of points and preserves the distance between points, that is, if \(x\) and \(y\) are in \(\mathbb R^2\), then \[||x-y|| = ||g(x) - g(y)||.\]

This group is called the group of rigid motions.

This summarizes most of the transformations that we studied back in Chapter 2!

Functions with Involutive Symmetry

Reflection across a line are involutions as are rotation by half a circle around a given point.

Even Function

A function \(f\) is an even function if for all \(x\) in the domain of \(f\), \(-x\) is in the domain of \(f\) and \[f(-x) = f(x).\]

In other words, if \(f\) is symmetric under the reflection across the \(y\)-axis.

Here is an example.

Example 8

This is an even function.

The function that exhibit symmetry by rotation by half a circle are called odd functions.

Odd Function

A function \(f\) is an odd function if for all \(x\) in the domain of \(f\), \(-x\) is in the domain of \(f\) and \[f(-x) = -f(x).\]

In other words, the reflection of \(f\) across the \(y\)-axis is \(-f\), i,e., \[M_yf = -f.\]

Here is an example.

Example 9

This is an odd function.

In this next example, use the graphs to identify whether the function is even or odd.

Example 10

Label the graphs below as being either even or odd functions.

The function \(f_1\) is even,
The function \(f_2\) is neither.
The function \(f_3\) is odd.
The function \(f_4\) is even.

Determining whether a function is even or odd can be determined by analyzing its formula.

Example 11

Show that the function \({\rm pow_2}\) is an even function.

Notice that \(f(-x)={\rm{pow}}_2(-x)=(-x)^2=x^2=f(x)\). So \(f\) is even.

In this next example, we show the following function is odd.

Example 12

Show that the function \({\rm pow_3}\) is an odd function.

Notice that \(f(-x)={\rm{pow}}_3(-x)=(-x)^3=-x^3=-[x^3]=-f(x)\). So \(f\) is odd.

In this example, we must now identify whether it is even or odd.

Example 13

Determine whether or not the function \(f\) is an odd or an even function, where \[f(x) = x|x|.\]

Notice that \(f(-x)=(-x)|-x|=-x|x|=-[x|x|]=-f(x)\). So \(f\) is odd.

There are some special properties that even and odd functions have.

Even and Odd Function Theorem

Suppose that \(f\) and \(g\) are functions that are defined on the same domain.

If \(f\) and \(g\) are both even, then \(f+g\) and \(f\cdot g\) are both even.
If \(f\) and \(g\) are both odd, then \(f+g\) is odd and \(f\cdot g\) is even.
If \(f\) is odd and \(g\) is even, then \(f\cdot g\) is odd.

Do you see why the above are true?

However, not all functions are even or odd. They can be neither.

To show that a function \(f\) is neither even nor odd, it is sufficient to show that:

there is a point \(a\) in \(\mathcal D(f)\) such that \(-a\) is not in \(\mathcal D(f)\)

there are points \(a\) and \(b\) in \(\mathcal D(f)\) such that \(-a\) and \(-b\) are in \(\mathcal D(f)\) and \[f(-a) \ne f(a) \quad {\rm and}\quad f(-b) \ne -f(b).\]

In this example explain why it is neither.

Example 14

Determine whether or not the function \(f\) is an odd or an even function, where \[f(x) = x + |x|.\]

Notice that \(f(-x)=(-x)+|-x|=-x+|x|\).Rewrite it to look like \(f(x)\) nor \(-f(x)=-x-|x|\), so \(f\) is neither even nor odd.

Explain why the square root function is neither.

Example 15

Determine whether or not the function \(f\) is an odd or an even function, where \[f(x) = \sqrt{x}.\]

Neither since \(f(x)=\sqrt{x}\) is not defined when \(x<0\). Specifically, \(x=1\) is in the domain of \(f\) but \(-x=-1\) is not in the domain of \(f\).

Although not every function is even or odd, some functions can be translated by the right vector to make them even or odd.

Example 16

Take \(f\) to be the function that is given by \[f(x)=\cos(x+4).\] Identify a vector \(\langle a,0\rangle\) so that \(\langle a,0\rangle+f\) is an even function.

A sketch of \(f\) looks like this:

The function \(f\) is the cosine function translated by \(4\) to the left: \[f=\cos\circ T_4.\]

Cosine is an even function, so translate \(f\) by \(V=\langle 4,0\) to get that

\[\begin{align*} V+f&=(x,\cos(x)). \end{align*}\]

In the next example, we find the right vector \(V\) to transform back a non-odd function to an odd function.

Example 17

Take \(f\) to be the function that is given by \[f(x)=(x-2)^5+4.\] Identify a vector \(V\) so that \(V+f\) is an odd function.

A sketch of \(f\) looks like this:

The function \(f\) is the \(\rm{pow}_5\) translated by \(2\) to the right and up by \(4\): \[f=T_4\circ\rm{pow}_5\circ T_{-2}.\]

The \(\rm{pow}_5\) is an odd function, so translate \(f\) by \(V=\langle -2,-4\rangle\) to get that

\[\begin{align*} V+f&=(x,x^5). \end{align*}\]

In the final example, we look at a symmetry that odd functions exhibit, which is that rotation by half a circle around \((0,0)\) of \(f\) produces \(f\). We find the corresponding symmetry for an odd function that has been translated.

Example 18

Take \(f\) to be the function that is given by \[f(x)=\sin(x+1)+2.\] Determine a point \(p\) so that the rotation of \(f\) around \(p\) by half a circle is equal to \(f\).

A sketch of \(f\) looks like this:

The function \(f\) is the sine function translated by \(1\) to the left and upward by \(2\): \[f=T_2\circ\sin\circ T_{1}.\]

The sine is an odd function, rotation of sine about \((0,0)\) by half a circle equals sine. So rotation of \(f\) about \((-1,2)\) by half a circle equals \(f:\)

\[\begin{align*} \langle -1,2\rangle+R_{(-1,0)}(\langle 1,-2\rangle +f)&=f. \end{align*}\]

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