saluu/Serres_Tor-Formel_und_Reduktion_auf_die_Diagonale
0
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity

Bereachnung der Randabbildung im Koszukomplex hinzugefügt

Browse source
This commit is contained in:
Johannes Loher 2017-08-08 20:09:19 +02:00
parent 9538a24469
commit e4d915b718
3 changed files with 48 additions and 16 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

1
Ausarbeitung.tex
View file

@ -19,7 +19,6 @@
\input{theorem_environments}
\input{custom_commands}
\allowdisplaybreaks{}
\begin{document}
\include{title}

60
chapters/chapter3.tex
View file

@ -62,29 +62,59 @@ Im Folgenden sei $A$ ein kommutativer Ring.
\begin{equation*}
K_p(\bmx) \cong \bigwedge\nolimits^p(A^r).
\end{equation*}
Die Randabbildung $d\colon K_p(\bmx) \to K_{p - 1}(\bmx)$ ist durch folgende Formel gegeben:
\begin{equation}
\label{eq:koszul-komplex-randabbildung}
d(e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p)\text{-mal}}) = \sum_{k=1}^p {(-1)}^{k+1} x_{i_k}e_{x_{i_1}}\otimes \cdots \otimes \widehat{e_{x_{i_k}}} \otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p + 1)\text{-mal}}
\end{equation}
Dabei verwenden wir die Konvention, dass das Symbol unter \enquote{$\widehat{\;}$} weggelassen wird.
\begin{proof}
Für $p\in \Z$ sei $I_p^r = \{\bmi \subset \{1,\ldots,r\} \mid \#\bmi = p\}$. Wir haben folgendes zu zeigen:
Für $p\in \Z$ sei $I_p^r = \lbrace\bmi \subset \lbrace1,\ldots,r\rbrace \mid \#\bmi = p\rbrace$. Wir haben folgendes zu zeigen:
\begin{equation*}
K_p(\bmx) = \bigoplus_{\bmi \in I_p^r}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right)
K_p(\bmx) = \bigoplus_{\bmi \in I_p^r}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right)
\end{equation*}
Dies beweisen wir durch Induktion über $r$. Im Fall $r = 1$ gilt $I_0^1 = \{\emptyset\}$, $I_1^1 = \{\{1\}\}$ und für $p\notin \{0, 1\}$ gilt $I_p^1 = \emptyset$. Damit folgt:
Dies beweisen wir durch Induktion über $r$. Im Fall $r = 1$ gilt $I_0^1 = \lbrace\emptyset\rbrace$, $I_1^1 = \lbrace\lbrace1\rbrace\rbrace$ und für $p \notin \lbrace0, 1\rbrace$ gilt $I_p^1 = \emptyset$. Damit folgt:
\begin{align*}
K_0(\bmx) &= K_0(x_1) = \bigoplus_{\bmi \in I_0^1}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right)\\
K_1(\bmx) &= K_1(x_1) = \bigoplus_{\bmi \in I_1^1}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right)\\
K_p(\bmx) &= K_p(x_1) = 0 = \bigoplus_{\bmi \in I_p^1}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right) \qquad \text{für }p \notin \{0, 1\}
K_0(\bmx) &= K_0(x_1) = \bigoplus_{\bmi \in I_0^1}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right)\\
K_1(\bmx) &= K_1(x_1) = \bigoplus_{\bmi \in I_1^1}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right)\\
K_p(\bmx) &= K_p(x_1) = 0 = \bigoplus_{\bmi \in I_p^1}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right) && \text{für }p \notin \lbrace0, 1\rbrace
\end{align*}
Für die Randabbildung $d\colon K_1(x_1) \to K_0(x_1)$ im Grad $1$ gilt
\begin{equation*}
d(e_{x_{i_1}}) = x_1 \cdot 1 = {(-1)}^{1 + 1} x_1 \widehat{e_{x_{i_1}}}\otimes 1 = \sum_{k=1}^1 {(-1)}^{k+1} x_{i_k}e_{x_{i_1}}\otimes \cdots \otimes \widehat{e_{x_{i_k}}} \otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(1 - 1 + 1)\text{-mal}}.
\end{equation*}
In den anderen Graden ist die Randabbildung nach \cref{defn:koszul-komplex-einfach} immer $0$ und auch die Formel aus \cref{eq:koszul-komplex-randabbildung} ergibt $0$.
Sei also nun $r > 1$ und die Behauptung für alle $s\in \N$ mit $0 \le s < r$ bereits bewiesen. Sei außerdem $\bmx' = (x_1,\ldots,x_{r-1})$. Dann gilt:
\begin{align*}
K_p(\bmx) &= \bigoplus_{i + j = p} K_i(\bmx') \otimes_A K_j(x_r)\\
&= \bigoplus_{j \in \{0,1\}} K_{p - j}(\bmx') \otimes_A K_j(x_r)\\
&= \quad\phantom{\oplus}\bigoplus_{\bmi \in I_p^{r - 1}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r - 1\} \setminus \bmi}K_0(x_i)\right)\right) \otimes_A K_0(x_r)\\
&\phantom{=}\quad \oplus \bigoplus_{\bmi \in I_{p - 1}^{r - 1}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r - 1\} \setminus \bmi}K_0(x_i)\right)\right) \otimes_A K_1(x_r)\\
&= \quad\phantom{\oplus}\bigoplus_{\bmi \in I_p^{r - 1}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right)\\
&\phantom{=}\quad \oplus \bigoplus_{\bmi \in I_{p - 1}^{r - 1}}\left( \left(\bigotimes_{i \in \bmi \cup \{r\}}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus (\bmi \cup \{r\})}K_0(x_i)\right)\right)\\
&= \quad \phantom{\oplus} \bigoplus_{\substack{\bmi \in I_{p}^{r}\\r \notin \bmi}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right)\\
&\phantom{=}\quad \oplus \bigoplus_{\substack{\bmi \in I_{p}^{r}\\r \in \bmi}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right)\\
&= \bigoplus_{\substack{\bmi \in I_{p}^{r}\\r \in \bmi}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right)\\
&= \bigoplus_{\bmi \in I_p}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \{1, \ldots, r\} \setminus \bmi}K_0(x_i)\right)\right)
&= \bigoplus_{j \in \lbrace0,1\rbrace} K_{p - j}(\bmx') \otimes_A K_j(x_r)\\
&= \poplus\bigoplus_{\bmi \in I_p^{r - 1}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r - 1\rbrace \setminus \bmi}K_0(x_i)\right)\right) \otimes_A K_0(x_r)\\
&\peq \oplus \bigoplus_{\bmi \in I_{p - 1}^{r - 1}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r - 1\rbrace \setminus \bmi}K_0(x_i)\right)\right) \otimes_A K_1(x_r)\\
&= \poplus\bigoplus_{\bmi \in I_p^{r - 1}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right)\\
&\peq \oplus \bigoplus_{\bmi \in I_{p - 1}^{r - 1}}\left( \left(\bigotimes_{i \in \bmi \cup \lbrace r \rbrace}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus (\bmi \cup \lbrace r \rbrace)}K_0(x_i)\right)\right)\\
&= \poplus \bigoplus_{\substack{\bmi \in I_{p}^{r}\\r \notin \bmi}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right)\\
&\peq \oplus \bigoplus_{\substack{\bmi \in I_{p}^{r}\\r \in \bmi}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right)\\
&= \bigoplus_{\substack{\bmi \in I_{p}^{r}\\r \in \bmi}}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right)\\
&= \bigoplus_{\bmi \in I_p}\left( \left(\bigotimes_{i \in \bmi}K_1(x_i)\right) \otimes_A \left(\bigotimes_{i \in \lbrace1, \ldots, r\rbrace \setminus \bmi}K_0(x_i)\right)\right)
\end{align*}
Nun betrachten wir die Randabbildung $d\colon K_p(\bmx) \to K_{p - 1}(\bmx)$. Dabei unterscheiden wir die zwei Fälle $r \in \lbrace i_1, \ldots, i_p\rbrace$ und $r \notin \lbrace i_1, \ldots, i_p\rbrace$. Im ersten Fall gilt:
\begin{align*}
& d(e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - 1 - p)\text{-mal}} \otimes \underbrace{1}_{\in K_0(x_r)}) \\
=& d(e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - 1 - p)\text{-mal}}) \otimes 1 + {(-1)}^p e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - 1 - p)\text{-mal}} \otimes \underbrace{d(1)}_{= 0} \\
=& \left(\sum_{k=1}^p {(-1)}^{k+1} x_{i_k}e_{x_{i_1}}\otimes \cdots \otimes \widehat{e_{x_{i_k}}} \otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p)\text{-mal}}\right) \otimes 1 \\
=& \sum_{k=1}^p {(-1)}^{k+1} x_{i_k}e_{x_{i_1}}\otimes \cdots \otimes \widehat{e_{x_{i_k}}} \otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p + 1)\text{-mal}}
\end{align*}
Im zweiten Fall können wir ohne Einschränkung $r = i_p$ annehmen. Dann gilt:
\begin{align*}
& d(e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_{p - 1}}} \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p)\text{-mal}}) \\
=&\pplus d(e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_{p - 1}}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{((r - 1) - (p - 1)\text{-mal}}) \otimes e_{x_{i_p}} \\
&+ {(-1)}^{p + 1} e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_{p - 1}}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p)\text{-mal}} \otimes d(e_{x_{i_p}}) \\
=& \pplus \left( \sum_{k=1}^{p - 1} {(-1)}^{k+1} x_{i_k}e_{x_{i_1}}\otimes \cdots \otimes \widehat{e_{x_{i_k}}} \otimes \cdots \otimes e_{x_{i_{p - 1}}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p + 1)\text{-mal}}\right) \otimes e_{x_{i_p}} \\
& + {(-1)}^{p+1} e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_{p - 1}}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p)\text{-mal}} \otimes (x_{i_p} \cdot 1) \\
=& \pplus \sum_{\substack{k=1\\k\ne p}}^p {(-1)}^{k+1} x_{i_k}e_{x_{i_1}}\otimes \cdots \otimes \widehat{e_{x_{i_k}}} \otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p + 1)\text{-mal}} \\
& + {(-1)}^{p + 1} x_{i_p} e_{x_{i_1}}\otimes \cdots \otimes e_{x_{i_{p - 1}}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p + 1)\text{-mal}} \\
=& \sum_{k=1}^p {(-1)}^{k+1} x_{i_k}e_{x_{i_1}}\otimes \cdots \otimes \widehat{e_{x_{i_k}}} \otimes \cdots \otimes e_{x_{i_p}} \otimes \underbrace{1 \otimes \cdots \otimes 1}_{(r - p + 1)\text{-mal}}
\end{align*}
\end{proof}
\end{lem}

3
custom_commands.tex
View file

@ -18,3 +18,6 @@
\DeclareMathOperator{\gr}{gr}
\DeclareMathOperator{\map}{map}
\DeclareMathOperator{\Supp}{Supp}
\DeclareMathOperator{\pplus}{\phantom{+}}
\DeclareMathOperator{\poplus}{\phantom{\oplus}}
\DeclareMathOperator{\peq}{\phantom{=}}