Composition Algebras

Arithmetic+Measurement

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 \[\left(\frac{\alpha,\beta,\gamma}{K}\right)=\mathbb{O}\qquad ...\]

Matrix rings

Boost the dimensions of composition algebras

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.

Adding geometry?

Respect the dot-product!

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!)

Classical Jordan Algebras

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. \((A\bullet A)\bullet (B\bullet A)=((A\bullet A)\bullet B)\bullet A\)

Classical Lie Algebras

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

General Hermitian Geometry

• \(K\) is a field with \(\overline{a+bi}=a-bi\), for some \(i\).
• Hermitian dot-product: \(u*v=\bar{u}^t D v\), \(D=\bar{D}^t\)

Unitary Lie algebras

\[\begin{aligned}\mathfrak{L}_n(*) & = \{ A\in\mathbb{M}_n(K)\mid (u\bar{A}^t)*v+u*(Av)=0\}\\ & = \{ A\in\mathbb{M}_n(K)\mid \bar{A}^tD=-DA\}\end{aligned}\]

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

Hermitian Jordan algebras

\[\begin{aligned}\mathfrak{H}_n(*) & = \{ A\in\mathbb{M}_n(K)\mid (u\bar{A}^t)*v=u*(Av)\}\\ & = \{ A\in\mathbb{M}_n(K)\mid \bar{A}^tD=DA\}\end{aligned}\]

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

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

Important Groups & Loops

Derived form composition 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.