Peng Zhou

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blog:2023-01-22-xy-2 [2023/01/23 08:25] – created pzhoublog:2023-01-22-xy-2 [2023/06/25 15:53] (current) – external edit 127.0.0.1
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 Suppose I build a U brane on the base, and compute its endomorphism.  Suppose I build a U brane on the base, and compute its endomorphism. 
  
-Let me try something else. $xy(y-1)$. How about this function? Is it better in terms of being more generic? partial $x$ gives $y(y-1)=0$, and partial y gives $x (2y-1)=0$, so we get either $(0,0)$ or $(0,1)$ as singular point, all with singular value $0$. OK, so we have two critical points over the same critical value, it seems they don't talk to each other. +Let me try something else. $xy(y-a)$. How about this function? Is it better in terms of being more generic? partial $x$ gives $y(y-a)=0$, and partial y gives $x (2y-1)=0$, so we get either $(0,0)$ or $(0,a)$ as singular point, all with singular value $0$. OK, so we have two critical points over the same critical value, it seems they don't talk to each other. 
  
-Is that what happens with $f=xy^2$? +Is that what happens with $f=xy^2$? Not quite, here is the trouble, the generic fiber was $\C_y \RM \{y=0, y=a\}$, And it is quite different from $a=0$. Maybe you can kill something, so that it is ok.  
 + 
 +What's the story for $f=xy$? Say, we are on the boundary $|x|^2+|y|^2=1$, and we have something stupid, like $xy > 0$ intersecting with it. This says, we have $|x|, |y|$ free, but $\theta_x = -\theta_y$. So, the stop on the boundary is like, an open cylinder in $S^3$, which contracts to $S^1$, and which gives the stop (or the thimble as cone over the stop). 
 + 
 +Now, we do $2 \theta_y + \theta_x = 0$. Good, suppose we now cone over this Legendrian knot, what can I say
  
  
  
  
blog/2023-01-22-xy-2.1674462326.txt.gz · Last modified: 2023/06/25 15:53 (external edit)