# Standardquadrik/Differentialoperatoren/Textabschnitt

Wir betrachten die Fermat-Quadrik
    $X_{1}^{2}+X_{2}^{2}+\cdots +X_{d}^{2}+X_{d+1}^{2}=0$
der Dimension $d\geq 2$ in $d+1$ Variablen. Wir nehmen Charakteristik $0$ an. Der Operator $E_{1}$
    $(d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{j\neq 1}X_{1}\partial _{j}^{2}+2\sum _{j\neq 1}X_{j}\partial _{1}\partial _{j}$
schickt $X_{1}$ auf eine Einheit und die anderen Variablen auf $0$. Sie formen also für die Erzeuger von Grad $1$ ein komplettes unitäres System von Operatoren der Ordnung $\leq 2$.
Es wird $X_{1}^{2}$ auf $2(d-1)X_{1}+2X_{1}$ und $X_{1}X_{i}$ auf $(d-1)X_{i}+2X_{i}$ abgebildet und $X_{i}^{2}$ auf $-2X_{1}$ abgebildet.
Ein Monom $X^{\lambda }$ wird auf
    $\lambda _{1}{\left((d-1)+(\lambda _{1}-1)+2\sum _{j=2}^{d+1}\lambda _{j}\right)}X^{\lambda -e_{1}}-\sum _{j\neq 1}\lambda _{j}(\lambda _{j}-1)X^{\lambda +e_{1}-2e_{j}},$
abgebildet. Es ist ja
    ${\begin{aligned}E_{1}{\left(X^{\lambda }\right)}&={\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{j\neq 1}X_{1}\partial _{j}^{2}+2\sum _{j\neq 1}X_{j}\partial _{1}\partial _{j}\right)}{\left(X^{\lambda }\right)}\\&=(d-1)\lambda _{1}X^{\lambda -e_{1}}+\lambda _{1}(\lambda _{1}-1)X^{\lambda -e_{1}}-\sum _{j\neq 1}\lambda _{j}(\lambda _{j}-1)\partial ^{\lambda +e_{1}-2e_{j}}+2\sum _{j\neq 1}\lambda _{1}\lambda _{j}X^{\lambda -e_{1}}\\&={\left((d-1)\lambda _{1}+\lambda _{1}(\lambda _{1}-1)+2\sum _{j\neq 1}\lambda _{1}\lambda _{j}\right)}X^{\lambda -e_{1}}-\sum _{j\neq 1}\lambda _{j}(\lambda _{j}-1)X^{\lambda +e_{1}-2e_{j}}\\&=\lambda _{1}{\left((d-1)+(\lambda _{1}-1)+2\sum _{j=2}^{d+1}\lambda _{j}\right)}X^{\lambda -e_{1}}-\sum _{j\neq 1}\lambda _{j}(\lambda _{j}-1)X^{\lambda +e_{1}-2e_{j}}.\end{aligned}}$
    ${\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{j=2}^{d+1}X_{1}\partial _{j}^{2}+2\sum _{j=2}^{d+1}X_{j}\partial _{1}\partial _{j}\right)}{\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{k=2}^{d+1}X_{1}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}\partial _{k}\right)}$
Es ist
    $\partial _{1}\circ (\partial _{1})=\partial _{1}^{2}\,,$
    $\partial _{1}\circ (X_{1}\partial _{1}^{2})=\partial _{1}^{2}+X_{1}\partial _{1}^{3}\,,$
    $\partial _{1}\circ (X_{1}\partial _{k}^{2})=\partial _{k}^{2}+X_{1}\partial _{1}\partial _{k}^{2}\,,$
    $\partial _{1}\circ (X_{k}\partial _{1}\partial _{k})=X_{k}\partial _{1}^{2}\partial _{k}\,$
und somit
    $\partial _{1}\circ E_{1}=d\partial _{1}^{2}-\sum _{k=2}^{d+1}\partial _{k}^{2}+F\,.$
Mit Koeffizienten ergibt sich
    $E_{1}^{2}=(d-1){\left(d\partial _{1}^{2}-\sum _{k=2}^{d+1}\partial _{k}^{2}\right)}+F\,,$
wobei $F$ ein Operator mit einem Vorfaktor aus ${\mathfrak {m}}$ ist.
Ausgeschrieben ist
    ${\begin{aligned}\partial _{1}\circ E_{1}&=\partial _{1}\circ {\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{k=2}^{d+1}X_{1}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}\partial _{k}\right)}\\&=(d-1)\partial _{1}^{2}+\partial _{1}^{2}+X_{1}\partial _{1}^{3}-\sum _{k=2}^{d+1}\partial _{k}^{2}-X_{1}\sum _{k=2}^{d+1}\partial _{1}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}^{2}\partial _{k}\\&=d\partial _{1}^{2}+X_{1}\partial _{1}^{3}-\sum _{k=2}^{d+1}\partial _{k}^{2}-X_{1}\sum _{k=2}^{d+1}\partial _{1}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}^{2}\partial _{k}.\end{aligned}}$
Somit ist
    ${\begin{aligned}\partial _{1}^{2}\circ E_{1}&=\partial _{1}{\left(d\partial _{1}^{2}+X_{1}\partial _{1}^{3}-\sum _{k=2}^{d+1}\partial _{k}^{2}-X_{1}\sum _{k=2}^{d+1}\partial _{1}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}^{2}\partial _{k}\right)}\\&=d\partial _{1}^{3}+\partial _{1}^{3}+X_{1}\partial _{1}^{4}-\sum _{k=2}^{d+1}\partial _{1}\partial _{k}^{2}-\sum _{k=2}^{d+1}\partial _{1}\partial _{k}^{2}-X_{1}\sum _{k=2}^{d+1}\partial _{1}^{2}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}^{3}\partial _{k}\\&=(d+1)\partial _{1}^{3}+X_{1}\partial _{1}^{4}-2\sum _{k=2}^{d+1}\partial _{1}\partial _{k}^{2}-X_{1}\sum _{k=2}^{d+1}\partial _{1}^{2}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}^{3}\partial _{k}.\end{aligned}}$
Es ist ($j\neq 1$)
    $\partial _{j}\circ (\partial _{1})=\partial _{1}\partial _{j}\,,$
    $\partial _{j}\circ (X_{1}\partial _{1}^{2})=X_{1}\partial _{1}^{2}\partial _{j}\,,$
    $\partial _{j}\circ (X_{1}\partial _{k}^{2})=X_{1}\partial _{j}\partial _{k}^{2}\,,$
    $\partial _{j}\circ (X_{k}\partial _{1}\partial _{k})=X_{k}\partial _{1}\partial _{j}\partial _{k}\,$
(für $j\neq k$) und
    $\partial _{j}\circ (X_{j}\partial _{1}\partial _{j})=\partial _{1}\partial _{j}+X_{j}\partial _{1}\partial _{j}^{2}\,.$
Ausgeschrieben ist
    ${\begin{aligned}\partial _{j}\circ E_{1}&=\partial _{j}\circ {\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{k=2}^{d+1}X_{1}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}\partial _{k}\right)}\\&=(d-1)\partial _{1}\partial _{j}+X_{1}\partial _{1}^{2}\partial _{j}-X_{1}\sum _{k=2}^{d+1}\partial _{j}\partial _{k}^{2}+2\sum _{k=2,k\neq j}^{d+1}X_{k}\partial _{1}\partial _{j}\partial _{k}+2\partial _{1}\partial _{j}+2X_{j}\partial _{1}\partial _{j}^{2}.\end{aligned}}$
Insgesamt ist $E_{1}^{2}$ gleich
    $(d-1){\left(d\partial _{1}^{2}+X_{1}\partial _{1}^{3}-\sum _{k=2}^{d+1}\partial _{k}^{2}-X_{1}\sum _{k=2}^{d+1}\partial _{1}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{k}\partial _{1}^{2}\partial _{k}\right)}$
    $+(d+1)X_{1}\partial _{1}^{3}+X_{1}^{2}\partial _{1}^{4}-2\sum _{k=2}^{d+1}X_{1}\partial _{1}\partial _{k}^{2}-X_{1}^{2}\sum _{k=2}^{d+1}\partial _{1}^{2}\partial _{k}^{2}+2\sum _{k=2}^{d+1}X_{1}X_{k}\partial _{1}^{3}\partial _{k}$
    $-(d-1)X_{1}\sum _{j=2}^{d+1}\partial _{1}\partial _{j}^{2}-X_{1}^{2}\sum _{j=2}^{d+1}\partial _{1}^{2}\partial _{j}^{2}+\sum _{2\leq j,k\leq d+1}X_{1}^{2}\partial _{j}^{2}\partial _{k}^{2}-2\sum _{2\leq j,k\leq d+1,j\neq k}X_{1}X_{k}\partial _{1}\partial _{j}^{2}\partial _{k}-4\sum _{2\leq j\leq d+1}X_{1}\partial _{1}\partial _{j}^{2}$
    $+2(d-1)\sum _{j=2}^{d+1}X_{j}\partial _{1}^{2}\partial _{j}+2\sum _{j=2}^{d+1}X_{1}X_{j}\partial _{1}^{3}\partial _{j}+2\sum _{j=2}^{d+1}X_{j}\partial _{1}^{2}\partial _{j}$
    $-2\sum _{1\leq j,k\leq d+1}X_{j}\partial _{j}\partial _{k}^{2}-2\sum _{1\leq j,k\leq d+1}X_{j}X_{1}\partial _{1}\partial _{j}\partial _{k}^{2}+4\sum _{2\leq j,k\leq d+1}X_{j}X_{k}\partial _{1}^{2}\partial _{j}\partial _{k}+4\sum _{2\leq j\leq d+1}X_{j}\partial _{1}^{2}\partial _{j}$
Es ist
    ${\begin{aligned}\partial _{1}^{2}\circ E_{1}&=(d-1)\partial _{1}^{3}+2\partial _{1}^{3}-2\sum _{j=2}^{d+1}\partial _{1}\partial _{j}^{2}+G\\&=(d+1)\partial _{1}^{3}-2\sum _{j=2}^{d+1}\partial _{1}\partial _{j}^{2}+G\end{aligned}}$
und
    ${\begin{aligned}\partial _{j}^{2}\circ E_{1}&=(d-1)\partial _{1}\partial _{j}^{2}+4\partial _{1}\partial _{j}^{2}+H\\&=(d+3)\partial _{1}\partial _{j}^{2}+H.\end{aligned}}$
Somit ist bis auf die Anteile mit einem Vorfaktor aus ${\mathfrak {m}}$
    ${\begin{aligned}E_{1}^{3}&=E_{1}^{2}\circ E_{1}\\&=(d-1){\left(d\partial _{1}^{2}-\sum _{j=2}^{d+1}\partial _{j}+F\right)}\circ E_{1}\\&=(d-1){\left(d(d+1)\partial _{1}^{3}-2d\sum _{j=2}^{d+1}\partial _{1}\partial _{j}^{2}-\sum _{j=2}^{d+1}(d+3)\partial _{1}\partial _{j}^{2}+F\right)}\\&=(d-1){\left(d(d+1)\partial _{1}^{3}-3(d+1)\sum _{j=2}^{d+1}\partial _{1}\partial _{j}^{2}+F\right)}\\&=(d-1)(d+1){\left(d\partial _{1}^{3}-3\sum _{j=2}^{d+1}\partial _{1}\partial _{j}^{2}+F\right)}\end{aligned}}$
Es ist
    ${\begin{aligned}\partial _{1}^{m}(E_{1})&=\partial _{1}^{m}{\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{j=2}^{d+1}X_{1}\partial _{j}^{2}+2\sum _{j=2}^{d+1}X_{j}\partial _{1}\partial _{j}\right)}\\&=(d-1)\partial _{1}^{m+1}+m\partial _{1}^{m-1+2}-m\sum _{j=2}^{d+1}\partial _{1}^{m-1}\partial _{j}^{2}+H\\&=(d+m-1)\partial _{1}^{m+1}-m\sum _{j=2}^{d+1}\partial _{1}^{m-1}\partial _{j}^{2}+H\end{aligned}}$
und
    ${\begin{aligned}\partial _{1}^{k}\partial _{j}^{2}(E_{1})&=\partial _{1}^{k}\partial _{j}^{2}{\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{j=2}^{d+1}X_{1}\partial _{j}^{2}+2\sum _{j=2}^{d+1}X_{j}\partial _{1}\partial _{j}\right)}\\&=(d-1)\partial _{1}^{k+1}\partial _{j}^{2}+k\partial _{1}^{k-1+2}\partial _{j}^{2}-k\sum _{j=2}^{d+1}\partial _{1}^{k-1}\partial _{j}^{4}+4\partial _{1}^{k+1}\partial _{j}^{2}+H.\end{aligned}}$
Es ist
    ${\begin{aligned}\partial _{1}^{m}\partial _{j}^{k}(E_{1})&=\partial _{1}^{m}\partial _{j}^{k}{\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{j=2}^{d+1}X_{1}\partial _{j}^{2}+2\sum _{j=2}^{d+1}X_{j}\partial _{1}\partial _{j}\right)}\\&=(d-1)\partial _{1}^{m+1}\partial _{j}^{k}+m\partial _{1}^{m-1+2}\partial _{j}^{k}-m\sum _{j=2}^{d+1}\partial _{1}^{m-1}\partial _{j}^{k+2}+2k\partial _{1}^{m+1}\partial _{j}^{k}+H\\&=(d+m+2k-1)\partial _{1}^{m+1}\partial _{j}^{k}-m\sum _{j=2}^{d+1}\partial _{1}^{m-1}\partial _{j}^{k+2}+H\end{aligned}}$
Es ist
    ${\begin{aligned}\partial ^{\nu }(E_{1})&=\partial ^{\nu }{\left((d-1)\partial _{1}+X_{1}\partial _{1}^{2}-\sum _{j=2}^{d+1}X_{1}\partial _{j}^{2}+2\sum _{j=2}^{d+1}X_{j}\partial _{1}\partial _{j}\right)}\\&=(d-1)\partial ^{\nu +e_{1}}+\nu _{1}\partial _{1}^{\nu +e_{1}}-\nu _{1}\sum _{j=2}^{d+1}\partial ^{\nu -e_{1}+2e_{j}}+2\sum _{j=2}^{d+1}\nu _{j}\partial ^{\nu +e_{1}}+H\\&={\left(d-1+\nu _{1}+2\sum _{j=2}^{d+1}\nu _{j}\right)}\partial ^{\nu +e_{1}}-\nu _{1}\sum _{j=2}^{d+1}\partial ^{\nu -e_{1}+2e_{j}}+H.\end{aligned}}$
Beispielsweise ist
    $\partial _{1}^{2}(E_{1})=(d+1)\partial _{1}^{3}-2\sum _{j=2}^{d+1}\partial _{1}\partial _{j}^{2}\,$
und
    $\partial _{2}^{2}(E_{1})=(d+3)\partial _{1}\partial _{2}^{2}\,.$
Damit ist
    $\partial _{2}^{2}(E_{1})-\partial _{3}^{2}(E_{1})=(d+3){\left(\partial _{1}\partial _{2}^{2}-\partial _{1}\partial _{3}^{2}\right)}\,,$
diese Operatordifferenz gehört also dazu. Damit gilt wiederum
    ${\begin{aligned}{\left(\partial _{1}^{2}-\partial _{2}^{2}\right)}(E_{1})&=(d+1)\partial _{1}^{3}-2\sum _{j=2}^{d+1}\partial _{1}\partial _{j}^{2}-(d+3)\partial _{1}\partial _{2}^{2}\\&=(d+1)\partial _{1}^{3}-2d\partial _{1}\partial _{2}^{2}-(d+3)\partial _{1}\partial _{2}^{2}\\&=(d+1)\partial _{1}^{3}-3(d+1)\partial _{1}\partial _{2}^{2}.\end{aligned}}$
Somit sollte $\mu !\partial ^{\nu }-\nu !\partial ^{\nu }$ für benachbarte Monome darin sein (unitär ausdrückbar).
    ${\begin{aligned}{\left(\mu !\partial ^{\nu }-\nu !\partial ^{\mu }\right)}{\left(E_{1}\right)}&=\mu !{\left(d-1+\nu _{1}+2\sum _{j=2}^{d+1}\nu _{j}\right)}\partial ^{\nu +e_{1}}-\mu !\nu _{1}\sum _{j=2}^{d+1}\partial ^{\nu -e_{1}+2e_{j}}-\nu !{\left(d-1+\mu _{1}+2\sum _{j=2}^{d+1}\mu _{j}\right)}\partial ^{\mu +e_{1}}+\nu !\mu _{1}\sum _{j=2}^{d+1}\partial ^{\mu -e_{1}+2e_{j}}\\&=\mu !{\left(d-1+\nu _{1}+2\sum _{j=2}^{d+1}\nu _{j}\right)}\partial ^{\nu +e_{1}}-\nu !{\left(d-1+\mu _{1}+2\sum _{j=2}^{d+1}\mu _{j}\right)}\partial ^{\mu +e_{1}}-\mu !\nu _{1}\sum _{j=2}^{d+1}\partial ^{\nu -e_{1}+2e_{j}}+\nu !\mu _{1}\sum _{j=2}^{d+1}\partial ^{\mu -e_{1}+2e_{j}}.\end{aligned}}$
Wenn $\nu _{1}=\mu _{1}$ ist, so stimmen die Zahlen vorne in den Klammer überein, und die neuen und alten Fakultäten unterscheiden sich wie gehabt an den anderen Stellen. Die hinteren Summen heben sich nach Induktion über $\nu _{1}=\mu _{1}$ weg.
Sei nun $\nu _{1}>\mu _{1}=0$. Dann muss man $\nu _{1}\sum _{j=2}^{d+1}\partial ^{\nu -e_{1}+2e_{j}}$ verstehen.
Es ist
    ${\begin{aligned}E_{2}\circ E_{1}&=(d-1)\partial _{2}(E_{1})+H\\&=(d-1)((d-1)\partial _{1}\partial _{2}+2\partial _{1}\partial _{2})+H\\&=(d-1)(d+1)\partial _{1}\partial _{2}+H\end{aligned}}$
und
    ${\begin{aligned}E_{2}\circ E_{1}\circ E_{1}&=(d-1)(d+1)\partial _{1}\partial _{2}(E_{1})+H\\&=(d-1)(d+1){\left((d+2)\partial _{1}^{2}\partial _{2}-\sum _{j=2}^{d+1}\partial _{2}\partial _{j}^{2}\right)}+H\end{aligned}}$
und
    ${\begin{aligned}E_{2}\circ E_{2}\circ E_{1}\circ E_{1}&=(E_{2}\circ E_{2}\circ E_{1})(E_{1})\\&={\left((d+2)\partial _{1}\partial _{2}^{2}-\sum _{j\neq 2}\partial _{1}\partial _{j}^{2}\right)}(E_{1})\\&=(d+2){\left((d+4)\partial _{1}^{2}\partial _{2}^{2}-\sum _{j\neq 1}\partial _{j}^{2}\partial _{2}^{2}\right)}-\sum _{j\neq 2}{\left((d+4)\partial _{1}^{2}\partial _{j}^{2}-\sum _{k\neq 1}\partial _{j}^{2}\partial _{k}^{2}\right)}\end{aligned}}$
    ${\begin{aligned}{\frac {1}{\nu !}}\partial ^{\nu }(E_{1})-{\frac {\nu _{1}}{(\nu _{2}+1)\nu !}}\partial ^{\nu -e_{1}+e_{2}}(E_{2})&={\frac {d-1+\nu _{1}+2\sum _{j\neq 1}\nu _{j}}{\nu !}}\partial ^{\nu +e_{1}}-{\frac {\nu _{1}}{\nu !}}\sum _{j\neq 1}\partial ^{\nu -e_{1}+2e_{j}}-{\frac {\nu _{1}(d-1+\nu _{2}+1+2\sum _{j\neq 2}\nu _{j}-2)}{(\nu _{2}+1)\nu !}}\partial ^{\nu -e_{1}+2e_{2}}+{\frac {\nu _{1}}{(\nu _{2}+1)\nu !}}(\nu _{2}+1)\sum _{j\neq 2}\partial ^{\nu -e_{1}+2e_{j}}\\&={\frac {d-1+\nu _{1}+2\sum _{j\neq 1}\nu _{j}}{\nu !}}\partial ^{\nu +e_{1}}-{\frac {\nu _{1}(d-2+\nu _{2}+2\sum _{j\neq 2}\nu _{j})}{(\nu _{2}+1)\nu !}}\partial ^{\nu -e_{1}+2e_{2}}-{\frac {\nu _{1}}{\nu !}}\partial ^{\nu -e_{1}+2e_{2}}+{\frac {\nu _{1}}{\nu !}}\partial ^{\nu -e_{1}+2e_{1}}\\&={\frac {d-1+2\sum _{j}\nu _{j}}{\nu !}}\partial ^{\nu +e_{1}}-{\frac {\nu _{1}(d-1+2\sum _{j}\nu _{j})}{(\nu _{2}+1)\nu !}}\partial ^{\nu -e_{1}+2e_{2}}\\&={\left(d-1+2\sum _{j}\nu _{j}\right)}{\left({\frac {\nu _{1}+1}{(\nu +e_{1})!}}\partial ^{\nu +e_{1}}-{\frac {\nu _{2}+2}{(\nu -e_{1}+2e_{2})!}}\partial ^{\nu -e_{1}+2e_{2}}\right)}.\end{aligned}}$
Es gibt die (nichtunitären) Derivationen $X_{i}\partial _{j}-X_{j}\partial _{i}$. Es ist beispielsweise
    ${\begin{aligned}{\left(X_{1}\partial _{2}-X_{2}\partial _{1}\right)}\circ {\left(X_{1}\partial _{3}-X_{3}\partial _{1}\right)}&=X_{1}^{2}\partial _{2}\partial _{3}-X_{1}X_{3}\partial _{1}\partial _{2}-X_{1}X_{2}\partial _{1}\partial _{3}-X_{2}\partial _{3}+X_{2}X_{3}\partial _{1}\partial _{1}\\\end{aligned}}$