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Hello,

I am sorry if this question is not appropriate for MO.

Suppose $X$ is the affine line, $i:Z\to X$ is the origin, and $j: U \to X$ is the complement to $Z$ in $X$.

I then have a distinguished triangle: $i_! i^! O \to O \to j_* j^* O \to$, where $O$ is any $D$-module on $X$, but I take it to be for this example just the structure sheaf.

I want to see "explicitly" what do I get. If I am not wrong, $j_* j^* O$ is Laurent polynomials, while $i_! i^! O$ is $\Delta = \oplus k \partial^i $, in degree $1$. Thus the distinguished triangle is equivalent to the data of an exact sequence: $0 \to O \to j_* O_U \to \Delta \to 0$.

My question is:

How in principle should I compute the arrow $j_* O_U \to \Delta$ in the s.e.s. above? It is some connecting arrow in the distinguished triangle, which seems abstract to me.

The second question I have is how to "compute" $j_! O_U$. I have two strategies, about both of which I am not sure exactly. The worse one is to compute $D (j_* O_U)$, the dual of $j_* O_U$. I don't know how to do it, a resolution seems complicated. The other method would be applying duality $D$ to the s.e.s. above, getting $0 \to \Delta \to j_! O_U \to O \to 0$. Then:

How can I compute explicitly how $j_! O_U$ is an extension of $\Delta$ and $O$?

Thank you, Sasha

flag	asked Apr 10 2012 at 17:16 Sasha 1,957●3●15

	Dose your first qustion means "how to discrabe it in general" or "how to discrabe it in this case"? If u mean "how to discrabe it in this case" than you just explained that this is $0$ if I am not wrong – Rami Apr 11 2012 at 3:11
	I also of course want to understand in general, but I wanted to see on that case what happens. Why is it zero? Maybe there is some confusion. – Sasha Apr 11 2012 at 17:01

Jan Weidner · Answer 1 · 2012-04-15 19:11:07Z UTC

1

Some thoughts, which I don't want to put in a comment box, to make them easier to read and edit:

The isomorphisms $\mathbb C[x,x^{-1}]\rightarrow O_U$ and $\mathbb C[x]\rightarrow O$ can be completet to a unique (because our triangles are s.e.s) isomorphism of triangles between $$\mathbb C[x]\rightarrow \mathbb C[x,x^{-1}] \rightarrow \mathbb C[x,x^{-1}]/\mathbb C[x] \rightarrow$$ and $$ O \rightarrow j_* j^* O \rightarrow i_! i^! O[{1}] \rightarrow$$

Now there is a further isomorphism of D-modules $$\Delta=\mathbb C[\partial] \rightarrow \mathbb C[x,x^{-1}]/\mathbb C[x]=O_U/O$$ which maps $\partial^n$ to $ (-1)^n n!x^{-n-1}$. This makes the short exact sequence $0\rightarrow O \rightarrow O_U \rightarrow \Delta\rightarrow 0$ explicit.

To make $j_! O_U$ explicit I would try to use the triangle

$$j_!j^! \rightarrow 1 \rightarrow i_* i^* \rightarrow$$

in a similar way.

edited Apr 15 2012 at 19:11

answered Apr 12 2012 at 10:39

Jan Weidner
6,118●21●39

	Hi. Thank you for your comments. Indeed, the morphism that you write is the one that I suspect. So you sharpened my question: How do I verify that your morphism is the one which I get from the general abstract distinguished triangle? For me, the way the triangle is constructed is by verifying that it is a s.e.s. for some injective D-modules or so, and thus getting it for all complexes of injective D-modules, which are all complexes, up to quasi-isom. – Sasha Apr 12 2012 at 13:47
	Okay, I edited my answer, does this help? – Jan Weidner Apr 12 2012 at 21:44
	I don't know if this is a meaningful question, but in my head, you have a concrete s.e.s. $A\to B\to C$, and a s.e.s. $A \to B \to C_1$ which is concrete except you don't know the last arrow (but you do know concretely the last object). Thus I can fix an isom. between $C_1$ and $C$, and thus get the undetermined arrow. But I can fix another isomorphism (say the first one times 2) and get another arrow. Thus, I can not see the uniqnueness you mention, because the arrow $B \to C_1$ is yet to be determined. – Sasha Apr 15 2012 at 12:01
	I think this question makes sense, if you precisely say what you mean by the statement "$i_!i^! O$ \textbf{is} $\Delta= \oplus k \partial^i$ in deg 1". (By the way, all endomorphisms of $\mathbb C[\partial] are scalars, so in any case the isomorphism is well defined up to scalar.) – Jan Weidner Apr 15 2012 at 19:28
	I just did some concrete calcultaion from definitions of functors. Also, you can think that to give a map from $\Delta$ is the same as to give a section which is killed by $x$, and so $\Delta$ has universal property identifying it with $i_! ...$. So I think I still do not know how to solve the problem, since it seems that everything is well defined to have the right to ask to identify the last arrow concretely. – Sasha Apr 17 2012 at 14:13

Distinguished triangle of closed - open partition, for D-modules

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