Epistemology
Introduction E9F5FC Questions FFFFC0 Software |
Bott periodicity, Bott periodicity models divisions Linear complex structures The inverse of an orthogonal matrix is given by its transpose. {$J_1=\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}\equiv i$} {$J_2=\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix}\equiv \begin{pmatrix} 0 & \varphi \\ -\varphi & 0 \\ \end{pmatrix}$} {$J_1J_2=\begin{pmatrix} i & 0 \\ 0 & i \\ \end{pmatrix}\begin{pmatrix} 0 & \varphi \\ -\varphi & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & i\varphi \\ -i\varphi & 0 \\ \end{pmatrix}=\alpha$} {$I_1J_1J_2=\begin{pmatrix} I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & -I & 0 \\ 0 & 0 & 0 & -I \\ \end{pmatrix}\begin{pmatrix} 0 & i\varphi & 0 & 0 \\ -i\varphi & 0 & 0 & 0 \\ 0 & 0 & 0 & i\varphi \\ 0 & 0 & -i\varphi & 0 \\ \end{pmatrix}=\begin{pmatrix} 0 & i\varphi & 0 & 0 \\ -i\varphi & 0 & 0 & 0 \\ 0 & 0 & 0 & -i\varphi \\ 0 & 0 & i\varphi & 0 \\ \end{pmatrix}$} {$J_3=(I_1J_1J_2)^{-1}=\begin{pmatrix} 0 & -i\varphi & 0 & 0 \\ i\varphi & 0 & 0 & 0 \\ 0 & 0 & 0 & i\varphi \\ 0 & 0 & -i\varphi & 0 \\ \end{pmatrix}\Rightarrow \begin{pmatrix} 0 & -i\varphi \\ i\varphi & 0 \\ \end{pmatrix}$} {$LJ_3=\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} 0 & -i\varphi & 0 & 0 \\ i\varphi & 0 & 0 & 0 \\ 0 & 0 & 0 & i\varphi \\ 0 & 0 & -i\varphi & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & i\varphi \\ 0 & 0 & -i\varphi & 0 \\ 0 & -i\varphi & 0 & 0 \\ i\varphi & 0 & 0 & 0 \\ \end{pmatrix}$} {$J_4=(LJ_3)^{-1}=\begin{pmatrix} 0 & 0 & 0 & -i\varphi \\ 0 & 0 & i\varphi & 0 \\ 0 & i\varphi & 0 & 0 \\ -i\varphi & 0 & 0 & 0 \\ \end{pmatrix}$} {$J_\gamma=\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}=i$} {$\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\begin{pmatrix} v_+ \\ v_- \\ \end{pmatrix} = \begin{pmatrix} v_- \\ v_+ \\ \end{pmatrix}$} is the form of an isometry {$L_\beta$} between two vector subspaces. {$\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}\begin{pmatrix} v_+ \\ v_- \\ \end{pmatrix} = \begin{pmatrix} v_+ \\ -v_- \\ \end{pmatrix}$} is a nontrivial antilinear operator {$I_\alpha$} that squares to {$+1$}. {$\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$} Suppose we have a linear operator that squares to {$+1$}. What does that look like? |