Symmetry

Defn.

A symmetry of \(t:T\) is an invertible function \(f:T\to T\) where \(f(t)=_T t\).

Symmetries form a group

• The identity function is a symmetry.
• If \(f,g:T\to T\) are symmetries then \(fg:T\to T\) is a symmetry.  Note: composition is associative.
• \(f:T\to T\) is a symmetry implies \(f^{-1}:T\to T\) is a symmetry.

Why insist on invertible \(f:T\to T\)?

• Actually we don't always, but we call those instead "transformation monoids".
• Invertible has key role in connecting to equality...

Equivalence

• Transitive law \[\begin{aligned}(x\equiv y) \& (y\times z) & \Longrightarrow (x\equiv z)\\ (x\equiv y) \times (y\times z) & \to (x\equiv z)\end{aligned}\] becomes a (parital) product between the evidence.
• Reflexive law \[\begin{aligned}(x\equiv x) \& (x\times y) & \Longrightarrow (x\equiv y)\\ (x\equiv x) \times (x\times y) & \to (x\equiv y)\end{aligned}\] shows there is an identity for the product.
• Symmetric law \[\begin{aligned}(x\equiv y) \& (y\times x) & \Longrightarrow (x\equiv x)\\ (x\equiv y) \times (y\times x) & \to (x\equiv x)\end{aligned}\] ensures inverses exists.

Officially...

• Symmetries form a groupoid as you vary what point is being considered.
• Equivalence is a groupoid as well.  We get a group when we fix a point.

So why are groups so studied?

They are the algebra of equality.

Important Groups & Loops

Derived form composition algebras

Hurwitz' Theorem.

Take invertible elements!

• Field, units are the nonzeros.
• Better yet, take group of length 1:
  • \[U(A)=\{z\in A\mid |z|=1\}\]
  • \[z,w\in U(A)\Rightarrow |zw|=|z||w|=1\Rightarrow zw\in U(A)\]

Boring start \(U(\mathbb{R})=\{z\mid |z|=1\}=\{\pm 1\}\cong \mathbb{Z}/2\) but with complex gets interesting....

• \[\begin{aligned} U(\mathbb{C})& =\{z\mid |z|=1\}\\ & =\{a+bi\mid 1=|a+bi|=(a+bi)(a-bi)=a^2+b^2\}\\ & = S^1\end{aligned}\]
• which you know because unit circle is \(e^{i\theta}=\cos\theta+i\sin\theta\).
• I.e. group is \[e^{i\theta}e^{i\tau}=e^{i(\theta+\tau)}\]

Dihedral Groups

Try with quaternions.

• \[\begin{aligned} U(\mathbb{H})& =\{z\mid |z|=1\}\\ & =\{a+bi+cj+dk\mid 1=a^2+b^2+c^2+d^2\}\\ & = S^3\end{aligned}\] which you may know from i,j,k rotation in graphics
• Super important subgroup (right-hand rule group) is \[Q_8=\{\pm 1, \pm i,\pm j,\pm k\}\] "The Quaternion group"

Quaternion Groups

Try with octonions...

• \[\begin{aligned} U(\mathbb{Q})& =\{z\mid |z|=1\}\\ & =\{a+bi+cj+dk+e\ell+f(i\ell)+g(j\ell)+h(k\ell) \\ & \quad \mid 1=a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2\}\\ & = S^7\end{aligned}\] you don't know this one,
• neither to do I;
• Physics pretends to know it.
• It is not even a group (nonassociative), it is what is known as a Moufang Loop.
• If you learn this one you might become a powerful wizzard.

Important Groups

Derived form matrices over composition algebras

General Linear Group

• \(GL_d(K)=\{A\in \mathbb{M}_d(K)\mid \exists A^{-1}\}\)
• If \(K\) commutative it is enough to have \(det(A)\) invertible.
• \(det:GL_d(K)\to K^{\times}\) is a group homomorphism.  Its kernel is \(SL_d(K)=\{A\in \mathbb{M}_d(K)\mid \det(A)=1\}\)
• \(A\sim B\Leftrightarrow A^{-1}B=sI_d\) is a congruence, the quotient is \(PGL_d(K)\) "projective general linear group"
• Most important \(PSL_d(K)\).  This is simple unless d=2 and \(|K|=2,3\).

Simplicity in Groups?

• We prove simplicity using many ad hoc methods.
• Iwasawa's lemma is a big help.
• A hugely important and controversial theorem is the Classification of Finite Simple Groups.
• \(PSL_d(K)\) is by far the most important one to know on this list.
• \(PSL_2(K)\) is even more important.

\(PSL_2(K)\)

\[PSL_2(K)= \left\{\begin{bmatrix} a & b\\ c & d\end{bmatrix} \middle| ad-bc=1\right\}\mod{\pm I_1}\]

Best explored as "fractional linear maps":

\[z\mapsto \frac{az+b}{cz+d}\]

Notice scaling by negatives top and bottom will not change it.

\[PSL_2(K)= \left\{\begin{bmatrix} a & b\\ c & d\end{bmatrix} \middle| ad-bc=1\right\}\mod{\pm I_1}\]

If \(|K|=q\) then 

• \(|GL_2(K)=(q^2-1)(q^2-q)=(q-1)q(q^2-1)\)
• \(|SL_2(K)=\frac{(q^2-1)(q^2-q)}{q-1}=q(q^2-1)\)
• \(|PSL_2(K)|=\frac{q(q^2-1)}{2}\)

\[PSL_2(K)= \left\{\begin{bmatrix} a & b\\ c & d\end{bmatrix} \middle| ad-bc=1\right\}\mod{\pm I_1}\]

Generators:

\[\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix},\begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix}\]

I.e. it can do row elimination except for the scalars and permuations.