Peng Zhou

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blog:2023-07-24 [2023/07/25 06:09] – pzhoublog:2023-07-24 [2023/07/25 06:32] (current) – pzhou
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   * watched a youtube video [[https://www.youtube.com/watch?v=NOl0v54DaXo | 8 traits of successful people]]   * watched a youtube video [[https://www.youtube.com/watch?v=NOl0v54DaXo | 8 traits of successful people]]
   * try to fix the annoying keyboard on a macbook pro, which turns out to be [[https://support.apple.com/keyboard-service-program-for-mac-notebooks| not my (or my wife's) fault.]]    * try to fix the annoying keyboard on a macbook pro, which turns out to be [[https://support.apple.com/keyboard-service-program-for-mac-notebooks| not my (or my wife's) fault.]] 
 +  * found an interesting lecture note of Teleman on rep theory. Never really understood what is character formula. 
  
  
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 For our flag variety, why is it true that, $T$-equivariant homology and $G$-equivariant homology only differ by Weyl covering.  For our flag variety, why is it true that, $T$-equivariant homology and $G$-equivariant homology only differ by Weyl covering. 
  
-(1) First of all, why is that for any Borel $B$ of $G$, the abstract Cartan $B/[B,B]$ for different $B$ are identified? Well, we may consider conjugation action of $G$ on the set of $B$. Suppose $B_1 = g B_2 g^{-1}$ for some $g$, the ambiguity of $g$: is right multiplication by $B_2$, left multiplication by $B_1$. For any $g' \in B_1 g B_2$, maybe consider the induced map on. +(1) (what is T? no, not the diagonal, but a quotient) First of all, why is that for any Borel $B$ of $G$, the abstract Cartan $B/[B,B]$ for different $B$ are identified? Well, we may consider conjugation action of $G$ on the set of $B$. Suppose $B_1 = g B_2 g^{-1}$ for some $g$, the ambiguity of $g$: is right multiplication by $B_2$, left multiplication by $B_1$. For any $g' \in B_1 g B_2$, maybe consider the induced map on. 
  
 What is conjugation action on $B$? Fix a $b_0 \in B$, we can send $b \mapsto b_0 b b_0^{-1} = b b^{-1} b_0 b b_0^{-1}$. So, thing in the same conjugacy class in $B$ are also in the same unipotent orbit.  What is conjugation action on $B$? Fix a $b_0 \in B$, we can send $b \mapsto b_0 b b_0^{-1} = b b^{-1} b_0 b b_0^{-1}$. So, thing in the same conjugacy class in $B$ are also in the same unipotent orbit. 
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 A conjugacy class of $B$ is contained in the unipotent orbit. But not the other way around. For example, identity matrix is its own conjugacy class (even under $G$ conjugation). However, multiplying by $[B,B]$ gives a lot.  A conjugacy class of $B$ is contained in the unipotent orbit. But not the other way around. For example, identity matrix is its own conjugacy class (even under $G$ conjugation). However, multiplying by $[B,B]$ gives a lot. 
  
-(2) On the Lie algebra and Lie group level. Taking $W$ invariant.  +(2) $\C[x_1, \cdots, x_n]$ is a free $\C[x_1, \cdots, x_n]^{S_n}$-mod. So is the multiplicative case.  
 + 
 +Not trivial, Steinberg-Pittie proved this. It turns out not be about permutation representation. 
  
 Ex: can you show that $\C[x_1,x_2]$ is a $\C[x_1+x_2,x_1 x_2]$ free module? Well,let's choose some generator in $ \C[x_1,x_2] / (x_1+x_2, x_1 x_2)$, say $e_1 = 1, e_2 = x_1$. Well, that is a bit arbitrary, how about $e_1 = 1, e_2 =x_1 - x_2$, so under $W=S_2$ action, they are eigenvectors. Then, for any polynomial $f(x_1, x_2)$, we can do  Ex: can you show that $\C[x_1,x_2]$ is a $\C[x_1+x_2,x_1 x_2]$ free module? Well,let's choose some generator in $ \C[x_1,x_2] / (x_1+x_2, x_1 x_2)$, say $e_1 = 1, e_2 = x_1$. Well, that is a bit arbitrary, how about $e_1 = 1, e_2 =x_1 - x_2$, so under $W=S_2$ action, they are eigenvectors. Then, for any polynomial $f(x_1, x_2)$, we can do 
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 You have a coordinate ring, of the torus, or of the plane. And you have the $W$ action. You pass to the $W$ invariant subring. Then, the original ring turns out to be a **free module** of finite rank over the W-invariant ring.  You have a coordinate ring, of the torus, or of the plane. And you have the $W$ action. You pass to the $W$ invariant subring. Then, the original ring turns out to be a **free module** of finite rank over the W-invariant ring. 
  
-You have a bunch of fundamental weight, they are element in the weight lattice, that are dual to the coroots. (ok, what are roots and coroots? Just do $gl_n$. you pick a cartan subalgebra, and let the cartan acts on Lie algebra. I guess I am just not used to having a family of commuting operator acting on something. (how about module over a commutative ring?) ok fine. ok, as h-mod, $\frak g$ lives over a bunch of points on $\frak h^*$. So far, these are canonical, we don't have Killing form. Let's assume 'semi-simple', which says, the root vectors span $\frak h^*$. Take a half-plane, and take some primitive roots, call them simple root. +You have a bunch of fundamental weight, they are element in the weight lattice, that are dual to the coroots. (ok, what are roots and coroots? Just do $gl_n$. you pick a cartan subalgebra, and let the cartan acts on Lie algebra. I guess I am just not used to having a family of commuting operator acting on something. (how about module over a commutative ring?) ok fine. ok, as h-mod, $\frak g$ lives over a bunch of points on $\frak h^*$. So far, these are canonical, we don't have Killing form. Let's assume 'semi-simple', which says, the root vectors span $\frak h^*$. Take a half-plane, and take some primitive roots, call them simple root.  read this note https://math.mit.edu/~dav/roots.pdf for what is coroot) 
 + 
 +So, we have $R(G) = R(T)^W$ and $\C[G]^G = \C[T]^W$. The first one is the 'integral form'. What does a 'coordinate ring over $\Z$' even mean? And, how much information did I lose? 
  
-cut the crap. read this note https://math.mit.edu/~dav/roots.pdf+ 
  
  
  
blog/2023-07-24.1690265353.txt.gz · Last modified: 2023/07/25 06:09 by pzhou