2020 James B. Wilson
Colorado State University
Composition Algebra: \(A\) has length & \([*,1,\div,+,-,0]\)-arithmetic
Theorem( Hurwitz) There 4 families like this:
Let \(K\) be a field, think \(\mathbb{R}\) or \(\mathbb{Q}\). Choose \(\alpha\in K\).
Loose Order: \(i\not\leq 0\) and \(0\not \leq i\)
not always a field, but always quadratic.
Fix a \(\left(\frac{\alpha}{K}\right)\) choose \(\beta\in K\).
Loose commutative: \(ij=-ji\)
i
j
ij
Fix a \(\left(\frac{\alpha,\beta}{K}\right)\) choose \(\gamma\in K\).
Loose Associativity: \(i(j\ell)=-(ij)\ell\)
\(i\)
\(j\)
\(ij\)
\(\ell\)
\(\ell\)
\(i\ell\)
\(j\ell\)
\((ij)\ell\)
Boost the dimensions of composition algebras
Fix a ring \(K\), e.g. \(\mathbb{Z}\) or \(\mathbb{Q}, \mathbb{R},\mathbb{Q}[x],\cdots\)
\[\mathbb{M}_n(K)=\left\{\begin{bmatrix}A_{11}&\cdots & A_{1n}\\ \vdots & & \vdots\\ A_{n1} & \cdots & A_{nn}\end{bmatrix}\middle| A_{ij}\in K\right\}\]
\[\begin{aligned} [A\cdot(B+C)]_{ij} & = \sum_{k=1}^n A_{ik}(B+C)_{kj}\\ & = \sum_{k=1}^n A_{ik}(B_{kj}+C_{kj})\\ & = \sum_{k=1}^n (A_{ik}B_{kj}+A_{ik}C_{kj})\\ &= \sum_{k=1}^n A_{ik}B_{kj}+\sum_{k=1}^n A_{ik}C_{kj}\\ & = [A\cdot B]_{ij}+[A\cdot C]_{ij} \end{aligned}\]
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 \(\exists M.(M\neq 0)\wedge(0\sim M)\Rightarrow \forall M.(0\sim M).\)
\(u*v=u^t D v\) so
\[(Au)*v=(Au)^tD v=u^t (A^t D)v\]
\[u*(Av)=u^t D (Av)=u^t (DA) v\]
\((Au)*v=u*(Av)\Longleftrightarrow A^tD=DA\)
\((Au)*v=u*(Av)\Longleftrightarrow A^tD=D\bar{A}\)
Defn. \(u\perp v\) if \(u*v=0\)
Lemma. \((Au)*v=u*(Av)\) implies \(Null(A)\perp Im(A)\)
Proof. \(Au=0\Rightarrow 0=(Au)*v=u*(Av).\)
So these matrices turn geometry into algebra!
Lemma. \((Au)*(Av)=u*v\) (length does not change "isometry") if, and only if, \((Au)*v=u*(\bar{A}^tv)\) and \(A\) is invertible. I.e. \(A\bar{A}^t=I\)
\[A=\frac{1}{2}(A+\bar{A}^t)+\frac{1}{2}(A-\bar{A}^t)\]
So no information lost (except when 2=0!)
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?
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?
\[\begin{aligned} [A, B]^t & = (i(AB-BA))^t\\ & = \bar{i}((AB)^t-(iBA)^t)\\ & = -i(\bar{B}\bar{A}-\bar{A}\bar{B})\\ & = i (\bar{A}\bar{B}-\bar{B}\bar{A}) \\ &= i\overline{(AB-BA)}\\ & = - \overline{i(AB-BA)}\\ & =-\overline{[A,B]}\end{aligned}\]
Hence: \(i\) sprinkled everywhere in quantum,e.g. Schrodinger
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 \(A\bullet B=\frac{1}{2}(AB+BA)\).
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.
These algebras are simple! They are therefore key building blocks of geometry.
If you go into science, get to know these algebras
Boring start \(U(\mathbb{R})=\{z\mid |z|=1\}=\{\pm 1\}\cong \mathbb{Z}/2\) but with complex gets interesting....
Try with quaternions.
Try with octonions...