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--- blog:2025-01-01 [2025/01/02 01:30] – pzhou
+++ blog:2025-01-01 [2025/01/02 07:23] (current) – pzhou
@@ Line 16: / Line 16: @@
 In this paper, CK uses $D_n = D(Y_n)$. They also construct, from a tangle $T$, a functor $\Psi(T)$, by composing the elementary functors: merging $F_n^i$, splitting $G_n^i$, and braiding $T$. Here merge and split between $Y_n$ and $Y_{n-2}$ are realized by a correspondence $X_n^i$
-What is the space $Y_n$? First, we fix an $(N, N)$ nilpotent element $z \in End(\C^{2N})$.
+What is the space $Y_n$? First, we fix an $(N, N)$ nilpotent element $z \in End(\C^{2N})$. (From this data, we can build an $N$-step flag, by taking kernel of $z^k$. Hold that thought.) Then, we build a 'weed', $L_1 \In L_2 \cdots \In L_n$, where $\dim L_i=i$ inside $\C^N$. Such that $z L_i \In L_{i-1}$.
-Question: Why we don't care about how large $N$ is? \\
+Let's think a bit. Can we take the limit $N$ goes to $\infty$? Yes, say, the polynomial ring $\C[x]$ as a vector space is the limit of $\C[x]/(x^N)$. Here, we can say, rank $2$ vector bundle over an artin disk $Spec \C[x]/(x^N)$. But, what are those $L_i$? We can start by thinking about $L_1$, we need $z L_1=0$, so that means $L_1$ needs to be in $ker(z)$. I would like to say $z = \d_x$, so that $L_1$ is some 'flat' section. OK. What is $L_2$? We can parametrize $L_2$ by saying, choose a generator $e_1(x)$ for $L_1$, such that $\d_x e_1(x)=0$, then choose a section $e_2$, so that $\d_x e_2 = e_1$, then $L_2$ is generated by $e_1, e_2$. $L_3$, we want something $e_3$, such that $\d_x e_3 = e_2$. No, this is not what it should be.
+There are two models for infinite dimensional vector space where an operator acts locally nilpotently, one is $\d_x$ on $\C[x]$, another is $z\cdot$ on $\C[z,1/z] / \C[z]$. One can certainly take bundles over this. The second one seems more amicable.
+Question: if $z$ is an nilpotent endomorphism of $V$, and $W \In V$ is a subspace invariant under $z$, how do I know how large is $ker(z: V/W \to V/W)$? OK, not sure.
+In our case, for a generic element in $Y_n$, an generic element in $L_i$ takes $i$ step to die under action by $z$. I want to believe that, $L_i$ is just a lattice, no better and no worse than any other lattices in $\C[t,t^{-1}]^2$. This should be the best description. If that is the case, then $Y_n$ is an iterated $\P^1$-bundle. Let's see if that is true.
+Yes, that is true, see http://arxiv.org/abs/0710.3216v2 their second paper on $sl(m)$ case.
+So why did Cautis-Kamnitzer only deal with $sl(m)$? What's so hard about general case?