Hurwitz' Theorem.

Cyclotomic Quadratic Extensions

Let $K$ be a field, think $\mathbb{R}$ or $\mathbb{Q}$. Choose $\alpha\in K$.

• Define $\left(\frac{\alpha}{K}\right)=K\times K$ with vector addition, $\hat{1}:=(1,0)$ and $i=(0,1)$; and
• Multiplication: $$\begin{array}{c|cc} * & c\hat{1} & di \\ \hline a\hat{1} & ac\hat{1} & adi \\ bi & bci & \alpha bd\hat{1}\end{array}$$ 
• $\overline{a+bi}=a-bi$ "conjugation".

Loose Order: $i\not\leq 0$ and $0\not \leq i$

Cyclotomic Quadratic extensions.

• $$\mathbb{C}\cong \left(\frac{-1}{\mathbb{R}}\right)\cong \left(\frac{-13}{\mathbb{R}}\right)\cong \mathbb{R}[x]/(x^2+1)\cong \cdots$$
• $$\mathbb{R}\times \mathbb{R}\cong \left(\frac{1}{\mathbb{R}}\right)\cong \left(\frac{13}{\mathbb{R}}\right)\cong \mathbb{R}[x]/(x^2-1)\cong \cdots$$
• $$\mathbb{Q}[i]\cong \left(\frac{-1}{\mathbb{Q}}\right)\not\cong \left(\frac{13}{\mathbb{Q}}\right)\cong \mathbb{Q}[\sqrt{13}]\not \cong \left(\frac{4}{\mathbb{Q}}\right)\cong \mathbb{Q}\times \mathbb{Q}$$

not always a field, but always quadratic.

Quatnerions

Fix a  $\left(\frac{\alpha}{K}\right)$ choose $\beta\in K$.

• Define $\left(\frac{\alpha,\beta}{K}\right)=\left(\frac{\alpha}{K}\right)\times \left(\frac{\alpha}{K}\right)$ with vector addition, $\hat{1}:=(\hat{1},0)$ and $j=(0,\hat{1})$; and
• Multiplication: $$\begin{array}{c|cc} * & c\hat{1} & dj \\ \hline a\hat{1} & ac\hat{1} & adj \\ bj & b\bar{c}j & \beta b\bar{d} \hat{1} \end{array}$$
• $\overline{a+bj}=\bar{a}-\bar{b}j$ "conjugation".

Loose commutative: $ij=-ji$

i

j

ij

Octonions

Fix a  $\left(\frac{\alpha,\beta}{K}\right)$ choose $\gamma\in K$.

• Define $\left(\frac{\alpha,\beta,\gamma}{K}\right)=\left(\frac{\alpha,\beta}{K}\right)\times \left(\frac{\alpha,\beta}{K}\right)$ with vector addition, $\hat{1}:=(\hat{1},0)$ and $\ell=(0,\hat{1})$; and
• Multiplication: $$\begin{array}{c|cc} * & c\hat{1} & dj \\ \hline a\hat{1} & ac\hat{1} & da\ell \\ b\ell & b\bar{c}\ell & \beta \bar{d}b \hat{1} \end{array}$$
• $\overline{a+bj}=\bar{a}-\bar{b}\ell$ "conjugation".

Loose Associativity: $i(j\ell)=-(ij)\ell$

$i$

$j$

$ij$

$\ell$

$\ell$

$i\ell$

$j\ell$

$(ij)\ell$

So 4 essential geometries!

• Orthogonal A field $K$
• Unitary a quadratic field extension $$\left(\frac{\alpha}{K}\right)=K[\sqrt{\alpha}],\qquad K\times K$$
• Symplectic a quaternion extension $$\left(\frac{\alpha,\beta}{K}\right)=\mathbb{H}, \quad \mathbb{M}_2(K)$$
• Exceptional an octonion extension $$\left(\frac{\alpha,\beta,\gamma}{K}\right)=\mathbb{O}\qquad ...$$

Fix a ring $K$, e.g. $\mathbb{Z}$ or $\mathbb{Q}, \mathbb{R},\mathbb{Q}[x],\cdots$

$$\mathbb{M}_n(K)=\left\{\begin{bmatrix}A_{11}&\cdots & A_{1n}\\ \vdots & & \vdots\\ A_{n1} & \cdots & A_{nn}\end{bmatrix}\middle| A_{ij}\in K\right\}$$

• Addition: $[A+B]_{ij} = A_{ij}+ B_{ij}$
• Minus: $[-A]_{ij}=-A_{ij}$
• Zero: $[0]_{ij}=0$
• Product: $[A\cdot B]_{ij}=\sum_{k=1}^n A_{ik}B_{kj}$.
• Generators $E_{ij}=e_i^t e_j$
• One: identity matrix $I_n=E_{11}+\cdots+E_{nn}$

Laws

• Addition by Direct Product:  $\mathbb{M}_n(K)\cong \prod_{i=1}^n \prod_{j=1}^n K$ as a $[+,-,0]$-algebraic structure (abelian group).  By variety rules $\mathbb{M}_{n^2}(K)$ is abelian group.
• Multiplication Laws must be proved directly
  • distributive,
  • associative,
  • identity

$$\begin{aligned} [A\cdot(B+C)]_{ij} & = \sum_{k=1}^n A_{ik}(B+C)_{kj}\\ & = \sum_{k=1}^n A_{ik}(B_{kj}+C_{kj})\\ & = \sum_{k=1}^n (A_{ik}B_{kj}+A_{ik}C_{kj})\\ &= \sum_{k=1}^n A_{ik}B_{kj}+\sum_{k=1}^n A_{ik}C_{kj}\\ & = [A\cdot B]_{ij}+[A\cdot C]_{ij} \end{aligned}$$

Congruences/Quotients

• If $K$ has a quotient $K/_{\sim}$ then $\mathbb{M}_{n}(K)/_{\sim}$ exists where $$A\sim B\Longleftrightarrow \forall i\forall j.(A_{ij}\sim B_{ij})$$
• E.g. $\mathbb{M}_n(\mathbb{Z})$ has quotients $\mathbb{M}_n(\mathbb{Z})/_{\sim}\cong \mathbb{M}_n(\mathbb{Z}/n)$, but what if $K$ has not quotients, e.g. $K$ a field?
• Theorem (Wedderburn).  If $K$ is a field or division ring then $\mathbb{M}_n(K)$ is simple (has only the two trivial congruences.)

Proof.  Suppose there is a square matrix $M$ that is nonzero but $M\sim 0$.  There are matrices $X,Y\in \mathbb{M}_n(K)$ where $$0\sim X0Y\sim XMY=\begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix}=:E$$ where r is the rank of $M$ (i.e. row and col. reduce this is where we assume inverses of nonzero coefficients). And so in fact $$0\sim E_{11}\cdot E=E_{11}$$

Chose permutation matrices $\Sigma$ so that $$0\sim \Sigma E_{11}\Sigma^{-1}=E_{ii}.$$

Thus $$0\sim E_{11}+\cdots+E_{nn}=I_n$$

Finally $$\forall X.(0\sim (X\cdot I_n)=X)$$

So if $\exists M.(M\neq 0)\wedge(0\sim M)\Rightarrow \forall M.(0\sim M).$

Remark for the future...

• There is a notion of a one-sided congruence: $$x\equiv y \Rightarrow rx\equiv ry$$
• If you shift to one side $x-y\equiv 0$ you get a subalgebra $$I=\{z\mid z\equiv 0\}$$
• $I$ is a left ideal $$\begin{aligned} x,y\in I & \Rightarrow x+y\in I\\ r\in R, x\in I & \Rightarrow rx\in I \end{aligned}$$
• You get equivalence classes that have "half" the algebra, it is called a "module".

When do you get Linear Algebra?

• Be careful!  There are $K$ where $\mathbb{M}_2(K)\cong \mathbb{M}_3(K)$. Not common but still exist.
• Be careful! Determinants only defined if $K$ is a commutative.
• Be careful! Minimum polynomial, characteristic polynomial unique if $K$ is a field.

Dot product:

$$u*v=u_1v_1+\cdots+u_n v_n=[u_1,\ldots,u_n]\begin{bmatrix} v_1\\ \vdots \\ v_n\end{bmatrix}=u^t v$$

More generally:

$$u*v=[u_1,\ldots,u_n]D\begin{bmatrix} v_1\\ \vdots \\ v_n\end{bmatrix}=u^tD v$$  where $D^t=D$

Even more generally:

$$u*v=[u_1,\ldots,u_n]D\begin{bmatrix} \bar{v}_1\\ \vdots \\ \bar{v}_n\end{bmatrix}=u^tD \bar{v}$$  where $D^t=\pm\bar{D}$

Why $(Au)*v=\pm u*(Av)$?

Why $(Au)*v=\pm u*(Av)$?

Geometry$\to$ Algebra 

• Break into Symmetric & Skew-symmetric
• Symmetric =  $\frac{1}{2}(A+\bar{A}^t)$
• Skew-symmetric =$\frac{1}{2}(A-\bar{A}^t)$

$$A=\frac{1}{2}(A+\bar{A}^t)+\frac{1}{2}(A-\bar{A}^t)$$

So no information lost (except when 2=0!)

Matrices with orthogonal geometry

• $K$ is a field (so we get linear algebra),
• Dot-product: $u*v=u^t v$ on $K^n$

Hermitian Jordan algebras

$$\begin{aligned}\mathfrak{jo}_n(*) & = \{ A\in\mathbb{M}_n(K)\mid (Au)*v=u*(Av)\}\\ & = \{ A\in\mathbb{M}_n(K)\mid A^t=A\}\end{aligned}$$

Addition as in matrices, product $A\bullet B=\frac{1}{2}(AB+BA)$.

$$\begin{aligned} (A\bullet B)^t & = \left(\frac{1}{2}(AB+BA)\right)^t\\ & = \frac{1}{2}((AB)^t+(BA)^t)\\ & = \frac{1}{2}(B^t A^t+A^t B^t)\\ & = \frac{1}{2}(BA+AB)\\ &= \frac{1}{2}(AB+BA)=A\bullet B.\end{aligned}$$

Why 1/2? $$(A\bullet I)=\frac{1}{2}(AI+IA)=\frac{1}{2}(2A)=A$$

Laws?

1. $A\bullet B=B\bullet A$
2. $A\bullet I=A$
3. $(A\bullet A)\bullet (B\bullet A)=((A\bullet A)\bullet B)\bullet A$

Matrices with orthogonal geometry

• $K$ is a field (so we get linear algebra),
• Dot-product: $u*v=u^t v$ on $K^n$

Orthogonal Lie algebras

$$\begin{aligned}\mathfrak{0}_n(*) & = \{ A\in\mathbb{M}_n(K)\mid (Au)*v=-u*(Av)\}\\ & = \{ A\in\mathbb{M}_n(K)\mid A^t=-A\}\end{aligned}$$

Addition as in matrices, product $[A, B]=(AB-BA)$.

$$\begin{aligned} [A, B]^t & = (AB-BA)^t\\ & = (AB)^t-(BA)^t\\ & = B^t A^t-A^t B^t\\ & = (-B)(-A)-(-A)(-B)\\ &= -(AB-BA)\\ & =-[A,B]\end{aligned}$$

Why no 1/2? $$[A,I]=(AI-IA)=0$$

Laws?

1. "Altenrating" $[A,A]=0$
2. $[A,B]=-[B,A]$
3. "Jacobi" $[A,[B,C]]=[[A,B],C]+[B,[A,C]]$

Physics trick...

1. Suppose $A^t=\bar{A}$ and $B^t=\bar{B}$, which is a form of "symmetric matrices" so ought to be Jordan.
2. Sly trick: $$[A,B]=i(AB-BA)$$

$$\begin{aligned} [A, B]^t & = (i(AB-BA))^t\\ & = \bar{i}((AB)^t-(iBA)^t)\\ & = -i(\bar{B}\bar{A}-\bar{A}\bar{B})\\ & = i (\bar{A}\bar{B}-\bar{B}\bar{A}) \\ &= i\overline{(AB-BA)}\\ & = - \overline{i(AB-BA)}\\ & =-\overline{[A,B]}\end{aligned}$$

Hence: $i$ sprinkled everywhere in quantum,e.g. Schrodinger

E.g. Symplectic (quaternion) geometry

• $K$ is a field (so we get linear algebra),
• Hermitian dot-product: $u*v=\bar{u}^t J v$ on $K^n$, $J=\begin{bmatrix} 0 & I_m\\ -I_m & 0 \end{bmatrix}$

Symplectic Jordan algebras

$$\begin{aligned}\mathfrak{H}_{2m}(K) & =  \{ A\in\mathbb{M}_{2m}(K)\mid A^t=JAJ^{-1}\}\end{aligned}$$

Symplectic Lie algebras

$$\begin{aligned}\mathfrak{sp}_{2m}(K) & = \{ A\in\mathbb{M}_{2m}(K)\mid A^t=-JAJ^{-1}\}\end{aligned}$$

$$\mathbb{M}_m(\mathbb{H})\cong\mathfrak{H}_{2m}(K)\oplus\mathfrak{sp}_{2m}(K)$$ where $\mathbb{H}$ is Hamilton's quatnerions.

Laws

• Addition, sometimes a direct product, but careful not on the diagonal! I.e. $A_{ii}=\bar{A}_{ii}$.  So in unitary cases needs special purpose study of laws.
• Multiplication laws are far more complicated.  $$A\bullet B=B\bullet A\qquad A\bullet I_n=A$$ $$[A,A]=0\qquad [A,B]=-[B,A]$$$$(A\bullet A)\bullet (B\bullet A)=(((A\bullet A)\bullet B)\bullet A$$ $$[[A,B],C]=[B,[A,C]]+[A,[B,C]]$$

Congruences

These algebras are simple!  They are therefore key building blocks of geometry.

If you go into science, get to know these algebras

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 $$e^{i\theta}e^{i\tau}=e^{i(\theta+\tau)}$$

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"

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.