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

Posted by Dennis in algebraic geometry.
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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 \mathbb{P}^1_k over a field. The basic question is that, since we know that there exist nontrivial {\rm Ext} groups between line bundles on \mathbb{P}^1, what explicit extensions do they correspond to?

For example, given that we we know {\rm Ext}^1(\mathscr{O},\mathscr{O}(-2))=H^1(\mathscr{O}(-2))=k, there should be a nontrivial extension

0\rightarrow \mathscr{O}(-2)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0.

Recall that all vector bundles over \mathbb{P}^1_k splits into a direct sum of line bundles. So E=\mathscr{O}(a)\oplus \mathscr{O}(b). By taking determinant bundles, we know \mathscr{O}(a+b)=\mathscr{O}\otimes \mathscr{O}(-2)=\mathscr{O}(-2), so a+b=-2. Furthermore, since it is a nontrivial extensions, we need both a,b<0, or else we would have a section \mathscr{O}\rightarrow E. Therefore, E=\mathscr{O}(-1)\oplus \mathscr{O}(-1).

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

0\rightarrow k[x,y]\xrightarrow{\begin{pmatrix} -y\\x \end{pmatrix}}k[x,y]\oplus k[x,y]\xrightarrow{\begin{pmatrix} x & y\end{pmatrix}}k[x,y]\rightarrow k\rightarrow 0

is exact. Turning this into sheaves over \mathbb{P}^1_k, we get exactly a short exact sequence

0\rightarrow \mathscr{O}(-2)\rightarrow \mathscr{O}(-1)\oplus \mathscr{O}(-1)\rightarrow \mathscr{O}\rightarrow 0.

Notice that this only works because k is identified with the zero sheaf when we try to turn it from a k[x,y]-module to a sheaf over \mathbb{P}^1_k.

Repeating the argument above, we see that any extension

0\rightarrow \mathscr{O}(-n)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0.

must have E\cong \mathscr{O}(a)\oplus \mathscr{O}(b) for a+b=-n and a,b<0. We might ask if all such values of a,b 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

0\rightarrow k[x,y]\xrightarrow{\begin{pmatrix} -y^j\\x^i \end{pmatrix}}k[x,y]\oplus k[x,y]
\xrightarrow{\begin{pmatrix} x^i & y^j\end{pmatrix}}k[x,y] \rightarrow k[x,y]/(x^i,y^j)\rightarrow 0

and turn it into sheaves over \mathbb{P}^1_k. This yields

0\rightarrow \mathscr{O}(-i-j)\rightarrow   \mathscr{O}(-i)\oplus \mathscr{O}(-j)\rightarrow \mathscr{O}\rightarrow 0.

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 {\rm Ext}^1 and extensions

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

0\rightarrow A\rightarrow E\rightarrow B\rightarrow 0?

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

To do so, we can find a surjection F\rightarrow B\rightarrow 0 onto B from a free module, and lift it to a map F\rightarrow E. Then, we have a surjection F\oplus A\rightarrow E. This gives us the diagram:


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



Conversely, if we have such a map K\rightarrow A, 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 A and B 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 K\rightarrow A give equivalent extensions precisely when it extends to a map F\rightarrow A 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 I containing \mathscr{O}(-n). Let the cokernel of 0\rightarrow \mathscr{O}(-n)\rightarrow I be K. Then, each E in the middle of a short exact sequence

0\rightarrow \mathscr{O}(-n)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0.

is a pullback of the diagram


for some map \mathscr{O}\rightarrow K and conversely, every map \mathscr{O}\rightarrow K gives a pullback diagram, and the pullback E fits into a short exact sequence 0\rightarrow \mathscr{O}(-n)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0.

Unfortunately, I don’t know of a good way of writing such an I 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 \mathscr{O}(-n)\rightarrow I extended to a map E\rightarrow I.

The long exact sequence applied to 0\rightarrow \mathscr{O}(-n)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0 in {\rm Ext} yields {\rm Hom}(E, I)\rightarrow {\rm Hom}(\mathscr{O}(-n), I)\rightarrow {\rm Ext}^1(\mathscr{O},I). Therefore, it suffices for H^1(I)=0.

Therefore, instead of using an injective object for I, we can use any object whose first cohomology vanishes. One source of such an object I is the Cech complex. As an abuse of notation, if \mathscr{F} is a sheaf on X, I’ll write \mathscr{F}|_U for an open U\subset X as \iota_{*}\iota^{*}\mathscr{F}, where \iota: U\rightarrow X is the inclusion.

The Cech complex gives us the exact sequence of sheaves

0\rightarrow \mathscr{O}(-n)\rightarrow \mathscr{O}(-n)|_{D^+(x)}\oplus \mathscr{O}(-n)|_{D^+(y)}\rightarrow \mathscr{O}(-n)|_{D^+(xy)}\rightarrow 0,

so we have I=\mathscr{O}(-n)|_{D^+(x)}\oplus \mathscr{O}(-n)|_{D^+(y)} and K=\mathscr{O}(-n)|_{D^+(xy)}.

Recall that a basis of H^1(\mathscr{O}(-n)) is represented by the sections s_i: \mathscr{O}\rightarrow \mathscr{O}(-n)|_{D^+(xy)} that sends 1 to x^{-i}y^{i-n} for 1\leq i< n. So it is natural to ask what extensions these sections s_i 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 E given by s_i has the transition map given by

\begin{pmatrix}  t^n & - t^i\\  0 & 1  \end{pmatrix}

from D^+(x) to D^+(y), where t:=\frac{y}{x}. To get this answer, we need to compute the pullback


Over D^+(x), a section of the pullback E is equivalent to choosing sections x^{-n}g, x_1^{-n} h over \mathscr{O}(-2)(D^+(x)) and \mathscr{O}(-2)(D^+(xy)) respectively and a section f of \mathscr{O}(D^+(x)), such that x^{-n}g-y^{-n}h=x^{-i}y^{i-n}f.

Equivalently, h = (\frac{y}{x})^n g -(\frac{y}{x})^i f. So a basis for E over D^+(x) is given by f,g\in \mathscr{O}(D^+(x)), and h is determined uniquely by f and g.

The situation over D^+(y) is exactly the same. To compute the transition map, we need to take a compatible choice (x^{-n}g, y^{-n}h, f) over D^+(xy), express it in terms of the basis over D^+(x), and then see what it should be in terms of the basis over D^+(y). In the basis over D^+(x), it is (g, (\frac{y}{x})^n g -(\frac{y}{x})^i f, f), and in the basis over D^+(x), it is ((\frac{x}{y})^n h -(\frac{x}{y})^{n-i}, h,  f).

This means the transition matrix is given by \begin{pmatrix} (\frac{y}{x})^n & - (\frac{y}{x})^i \\ 0 & 1\end{pmatrix}, which is what we said above.

Finally, the identify E, we need to diagonalize our transition matrix. Since we are allowed to change bases over D^+(x) and over D^+(y), we are allowed to multiple on the right by an element in GL_2(k[t]) and on the left by an element of GL_2(k[t^{-1}]). Through row and column operations, we get

\begin{pmatrix} (\frac{y}{x})^n & - (\frac{y}{x})^i \\ 0 & 1\end{pmatrix}\rightarrow  \begin{pmatrix} 0 & t^{-i} \\ t^{n-i} & 1\end{pmatrix} \rightarrow \begin{pmatrix} 0 & t^{-i} \\ t^{n-i} & 0\end{pmatrix} \rightarrow \begin{pmatrix} t^i & 0 \\ 0 & t^{n-i}\end{pmatrix}.


This means the section s_i: \mathscr{O}\rightarrow \mathscr{O}(-n)|_{D^+(xy)} that sends 1 to x^{-i}y^{i-n} represents an element in H^1(\mathscr{O}(-n)) that corresponds to an extension

0\rightarrow \mathscr{O}(-n)\rightarrow \mathscr{O}(-i)\oplus \mathscr{O}(i-n)\rightarrow \mathscr{O}\rightarrow 0


Lifting idempotents à la Grothendieck August 29, 2009

Posted by Akhil Mathew in algebra, algebraic geometry, commutative algebra.
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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 1 Let {A} be a ring complete with respect to the {I}-adic filtration. Then if {\bar{e} \in A/I} is idempotent (i.e. {\bar{e}^2=\bar{e}}) then there is an idempotent { e \in A} such that {e} reduces to {\bar{e}}  (more…)