going to revise the VGIT paper.

Here, the base is one dimensional, fiber is easy. We can take the easy way out, just prove enough for this case. Or, we can do full proof.

I don't want to be super general, but just for this case. I want to consider skeleton on the total space, and skeleton on a sub-level set.

let's just glue category, that is pretty cool, and useful.

So, we have a constructible sheaf of categories. I think, we can define a sheaf of stable categories. We don't have the notion a microlocal stalk, but we do have the notion of equivalences. In that sense, it is not very satisfactory.

I need to say, non-characteristic deformation lemma. what is 2.7.2?

what is Mittag-Leffler condition? if I give you a projective system of abelian group, (no, not assuming surjectivity at each level), we can ask for projective limit.

It is reasonable to assume ML condition. For example, if each level is a finite dimensional vector space, then it is automatically ML.

It is also easy to get a system that is not ML, for example, $$ k[x] \gets k[x] \gets k[x] \gets \cdots $$ where the map is $x$, well, the limit is zero. I really wonder what is not a ML system.

Well, consider the following examples $$ 0 \to (x^n) \to k[x] \to k[x] / (x^n) \to 0 $$ if we take projective limit on these things, we get $$ 0 \to 0 \to k[x] \to k[[x ] ] $$ note, the right side is not exact, and it should'nt. Is there some derived projective limit? hmm? It should be $k[[x ] ] / k[x]$. This should be derived inverse limit is zero.

So, ML condition implies the acyclic for derived inverse limit functor. This is the poit of 1.12.3.

Then we consider a projective system of chain complex, or equivalently, a chain complex of projective system (since the category of projective system in an abelian category also forms an abelian category). It is natural to compare two things: one is, apply limit functor, two is compute the cohomology.

$$ \phi^k H^k \varprojlim \to \varprojlim H^k $$ Why? Because we have $$ H^k (X_\infty) = \frac{Z_\infty^k}{B_\infty^k} \to \frac{Z_n^k}{B_n^k} = H^k_n $$

Now, is this map $H^k (\varprojlim_n X_n) \to \varprojlim H^k(X_n)$ surjective? We consider taking two exact seq $$ 0 \to B^k(X_\infty) \to Z^k(X_\infty) \to H^k(X_\infty) \to 0 $$ $$ 0 \to \lim B^k(X_n) \to \lim Z^k(X_n) \to \lim H_n^k \to 0 $$ the last part is because $B_n^k$ are ML in $n$, hence the proj lim SES is also right exact.

• we have snake lemma
• middle column isom implies right column surjective
• however, $B^k(X_\infty) =Im(X^{k-1}_\infty) \to \lim_n B^k(X_n)$ may not be surjective without further assumption. This is because $B^k$ is like taking cokernel of a map, and $\lim$ is taking projective lim, they don't commute. If we assume $H^{k-1}(X_n)$ is ML, then we know $Z^{k-1}(X_n)$ is ML. Then, we have

$$ 0 \to Z^{k-1}(X_n) \to X^{k-1}_n \to B^k(X_n) \to 0 $$ then we use the proposition that the first term is ML, then the projective limit preserve the SES, and we get $ B^k(X_\infty) \cong \lim B^k(X_n) $

But why do we care about these ML stuff? Why do we care about the non-characteristic deformation lemma?