##
Extensions of Line Bundles on the Projective Line
*January 26, 2015*

*Posted by Dennis in algebraic geometry.*

add a comment

add a comment

The following is probably obvious to experts, as it is easy to find generalizations using google. However, this humble master’s student didn’t find this most basic case, so it’s probably worth writing down.

We will be working over the projective line over a field. The basic question is that, since we know that there exist nontrivial groups between line bundles on , what explicit extensions do they correspond to?

For example, given that we we know , there should be a nontrivial extension

Recall that all vector bundles over splits into a direct sum of line bundles. So . By taking determinant bundles, we know , so . Furthermore, since it is a nontrivial extensions, we need both , or else we would have a section . Therefore, .

One way to produce an example of such a nontrivial extension is from the Koszul complex. Recall that the complex

is exact. Turning this into sheaves over , we get exactly a short exact sequence

Notice that this only works because is identified with the zero sheaf when we try to turn it from a -module to a sheaf over .

Repeating the argument above, we see that any extension

must have for and . We might ask if all such values of are possible. The answer is yes, and we can construct them, though I didn’t think of something as nice as the Koszul complex.

________________________________________________________________________________________________

Edit: Sometime after posting this, I realized there was a way to get these extensions in exactly the same way as above with the Koszul complex. Namely, we take the exact sequence

and turn it into sheaves over . This yields

However, it’s at least useful for myself to remember how the machinery below works in a basic example.

________________________________________________________________________________________________

To do so, we first recall the general recipe for how to get a extension from an element of the Ext group. Instead of just recalling the construction, we will derive part of it in a way that I think a beginner could come up with, as I don’t see this written down either. Those who aren’t interested can just skip the next section.

**Aside on correspondence between and extensions**

Since projectives are easier for me to think about than injectives, we’ll motivate this by constructing extensions of -modules for some (commutative) ring . A natural question is to ask, if we fix modules and , what modules can fit in the middle of the short exact sequence

I know I have had the experience of trying to ask this on mathstackexchange and getting a bunch of answers telling me to learn homological algebra and Ext, which is more effort than a poor undergraduate wants to spend to get at a concrete problem. However, I think there is a natural direct approach.

Instead of being in the middle of a short exact sequence, we would instead like to have a presentation for . For example, of and and are finitely generated (which is a lot of the time), then this allows us to compute explicitly. To do so, we want to find a surjection onto .

To do so, we can find a surjection onto from a free module, and lift it to a map . Then, we have a surjection . This gives us the diagram:

Here, is defined to be the kernel of . The top row is exact by the 9-lemma. Therefore, we have recovered the fact that there exists some map such that is the pushout of

.

Conversely, if we have such a map , then we can construct the first two rows of the commutative diagram above, and 9-lemma imples the exactness of the last row. Therefore, constructing all extensions if and are finitely generated abelian groups, say, is not a mysterious thing.

The main issue I see with this approach is that it’s not clear to me how to classify that two such maps give equivalent extensions precisely when it extends to a map directly from the large diagram.

**Constructing the extensions**

Now, we want to use the section above to construct our extensions. One technical issue is that the quasicoherent sheaves don’t usually have enough projectives, so we would have to work with injective objects instead. The same argument in the previous section works exactly, so, we can fix an injective object containing . Let the cokernel of be . Then, each in the middle of a short exact sequence

is a pullback of the diagram

for some map and conversely, every map gives a pullback diagram, and the pullback fits into a short exact sequence .

Unfortunately, I don’t know of a good way of writing such an down. To work around this, we note that the argument we used to construct extensions didn’t use the full strength of injectivity. Namely, we only need to know that the injection extended to a map .

The long exact sequence applied to in yields . Therefore, it suffices for .

Therefore, instead of using an injective object for , we can use any object whose first cohomology vanishes. One source of such an object is the Cech complex. As an abuse of notation, if is a sheaf on , I’ll write for an open as , where is the inclusion.

The Cech complex gives us the exact sequence of sheaves

so we have and .

Recall that a basis of is represented by the sections that sends 1 to for . So it is natural to ask what extensions these sections represent.

Reading somebody else's computations isn't usually that enlightening, so I'll state the answer first. I got that the rank 2 bundle given by has the transition map given by

from to , where . To get this answer, we need to compute the pullback

Over , a section of the pullback is equivalent to choosing sections , over and respectively and a section of , such that .

Equivalently, . So a basis for over is given by , and is determined uniquely by and .

The situation over is exactly the same. To compute the transition map, we need to take a compatible choice over , express it in terms of the basis over , and then see what it should be in terms of the basis over . In the basis over , it is , and in the basis over , it is .

This means the transition matrix is given by , which is what we said above.

Finally, the identify , we need to diagonalize our transition matrix. Since we are allowed to change bases over and over , we are allowed to multiple on the right by an element in and on the left by an element of . Through row and column operations, we get

.

This means the section that sends 1 to represents an element in that corresponds to an extension

.

##
Lifting idempotents à la Grothendieck
*August 29, 2009*

*Posted by Akhil Mathew in algebra, algebraic geometry, commutative algebra.*

Tags: completions, connectedness, idempotents, lifting idempotents, schemes

add a comment

Tags: completions, connectedness, idempotents, lifting idempotents, schemes

add a comment

I am going to get back shortly to discussing algebraic number theory and discrete valuation rings. But this tidbit from EGA 1 that I just learned today was too much fun to resist. Besides, it puts the material on completions in more context, so I think the digression is justified.

**Lifting Idempotents **

The theorem says we can lift “approximate idempotents” in complete rings to actual ones. In detail:

Theorem 1Let be a ring complete with respect to the -adic filtration. Then if is idempotent (i.e. ) then there is an idempotent such that reduces to .(more…)