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.