Talk on BGG resolution - 편집 역사

2020년 11월 14일 (토) 09:14에 Pythagoras0님의 편집

2020-11-14T09:14:51Z

2020-11-13T07:13:32Z

2017-12-28T03:26:24Z

exactness

2017-12-28T03:25:32Z

characters

2016-05-05T09:02:28Z

exactness

2016-05-05T09:01:43Z

exactness

2016-05-05T09:01:10Z

exactness

2016-05-05T08:59:01Z

exactness

2016-05-05T08:58:15Z

exactness

2016-05-05T08:48:26Z

standard resolution of trivial module

@@ 144번째 줄: / 144번째 줄: @@
 ;proof
-Suppose $u\in E_{j,k}$ satisfies $\partial(u)=0$ in $E_{j+1,k-1}$.
+Suppose $u\in E_{j,k},\, j,k\geq 1$ satisfies $\partial(u)=0$ in $E_{j+1,k-1}$.
 show : there exists $v\in E_{j-1,k+1}$ such that $\partial(v)=u$ in

@@ 25번째 줄: / 25번째 줄: @@
 * if we have a long exact sequence, we still get a similar alternating sum = 0
 ** why? [[Euler-Poincare principle|Euler-Poincare mapping]] : a long exact sequence can be decomposed into short exact sequences.
-** then the Euler characteristic of a resolution makes sense
+** then the Euler characteristic of a finite resolution makes sense
 * goal : realize the alternating sum \ref{WCF} as an Euler characteristic of a suitable resolution of $L(\lambda)$
 * The BGG resolution resolves a finite-dimensional simple $\mathfrak{g}$-module $L(\lambda)$ by direct sums of Verma modules indexed by weights "of the same length" in the orbit $W\cdot \lambda$

@@ 120번째 줄: / 120번째 줄: @@
 * see Lang, Algebra ' Koszul complex'
-* example
+* examples
 $$
 \to \wedge^2 \to S^1\otimes \wedge^1 \to S^2\to 0
@@ 130번째 줄: / 130번째 줄: @@
 $$
-\to \wedge^2 \to S^1\otimes \wedge^1 \to S^2\to 0
+\to \wedge^1 \to S^1 \to 0
 $$
 or

@@ 122번째 줄: / 122번째 줄: @@
 * example
 $$
-\to \wedge^2 \to S^1\otimes \wedg^1 \to S^2\to 0
+\to \wedge^2 \to S^1\otimes \wedge^1 \to S^2\to 0
 $$
 or
@@ 130번째 줄: / 130번째 줄: @@
 $$
-\to \wedge^2 \to S^1\otimes \wedg^1 \to S^2\to 0
+\to \wedge^2 \to S^1\otimes \wedge^1 \to S^2\to 0
 $$
 or

@@ 105번째 줄: / 105번째 줄: @@
 ;prop
-For each $j\ge 1$, $\{(E_{j,k}/E_{j-1,k},\overline{\partial_k})\}_{-1\le k\le m+1}$ is an exact sequence. (here the first and $m+1$-th terms are zero)
+For each $j\ge 1$, $\{(E_{j,k}/E_{j-1,k},\overline{\partial_k})\}_{-1\le k\le m+1}$ is an exact sequence. (here $-1$ and $m+1$ terms are zero)
 ;proof
@@ 121번째 줄: / 121번째 줄: @@
 ;prop
-For each $j\ge 1$, $\{(E_{j,k},\partial_k)\}_{-1\le k \le m+1}$ is exact. (here the first and $m+1$-th terms are zero)
+For each $j\ge 1$, $\{(E_{j,k},\partial_k)\}_{-1\le k \le m+1}$ is exact. (here $-1$ and $m+1$ terms are zero)
 ;proof

@@ 105번째 줄: / 105번째 줄: @@
 ;prop
-For each $j\ge 1$, $\{(E_{j,k}/E_{j-1,k},\overline{\partial_k})\}_{0\le k\le m+1}$ is an exact sequence. ($\overline{\partial_0}=\overline{\partial_{m+1}}=0$)
+For each $j\ge 1$, $\{(E_{j,k}/E_{j-1,k},\overline{\partial_k})\}_{-1\le k\le m+1}$ is an exact sequence. (here the first and $m+1$-th terms are zero)
 ;proof
@@ 120번째 줄: / 120번째 줄: @@
 * see Lang, Algebra ' Koszul complex'
 ;proof
 Suppose $u\in E_{j,k}$ satisfies $\partial(u)=0$ in $E_{j+1,k-1}$.
@@ 153번째 줄: / 151번째 줄: @@
 u=\partial(v_1)+\cdots +\partial(v_j).
 $$
-Thus if we set $v:=v_1+\cdots +v_j\in E_{j-1,k+1}$ and $\partial(v)=u$.
+Thus if we set $v:=v_1+\cdots +v_j\in E_{j-1,k+1}$ and $\partial(v)=u$. ■
-So far we have proved the sequence is exact at $D_k,\, k\ge 1$.
+If $\partial(u)=0$, then it implies $u_0 = 0$ as only degree zero term survives under $\partial$.
-We need to show the exactness at $D_0$.
+Therefore $j\ge 1$ and the above proposition still applies to conclude that there exists $v\in E_{j-1,1}$ such that $\partial(v)=u$.
-■
+This proves the exactness at $D_0$. ■
 ==weak BGG resolution of $L(0)$==

@@ 124번째 줄: / 124번째 줄: @@
 $0\to D_m \to \cdots \to D_k\to \cdots \to D_1 \to D_0 \to L(0)\to 0$ is exact.
 ;proof
-Assume $j\ge 1$ and $k\ge 0$.
+We show that for each $j\ge 1$, $\{(E_{j,k},\partial_k)\}_{0\le k \le m}$ is exact.
 Suppose $u\in E_{j,k}$ satisfies $\partial(u)=0$ in $E_{j+1,k-1}$.