This is a followup to here.

Consider Lemma 2 from Beilinson's paper "Coherent Sheaves on $\mathbb{P}^n$ and Problems of Linear Algebra", as follows.

Lemma 2. For any pair $i$, $j$ such that $0 \le i$, $j \le n$, and $l \ge 1$:$$\text{Hom}(\Omega^i(i), \Omega^j(j)) = \Lambda^{i-j}(V^*),\text{ Ext}^l(\Omega^i(i), \Omega^j(j)) = 0,$$$$\text{Hom}(\mathcal{O}(-i),\mathcal{O}(-j)) = S^{i-j}(V),\text{ Ext}^l(\mathcal{O}(-i),\mathcal{O}(-j)) = 0,$$where composition of morphisms coincides with multiplication in $\Lambda^\bullet(V^*)$ and $S^\bullet(V)$, respectively.

The lemma is proved by induction with the aid of the exact sequence$$0 \to \Omega^i(i) \to \Lambda^i(V) \otimes \mathcal{O} \to \Omega^{i-1}(i) \to 0.$$

My question is, how do others think about it/the (hopefully underlying geometric) intuition for this? What role does the "dual exceptional collection" intuition play here?

  • 3
    $\begingroup$ At least for the case $\textrm{Hom}(O(-i),O(-j))$ I am puzzled by your surprise. $O(i)$'s are line bundles and so invertible and hence $R\textrm{Hom}(O(-i),O(-j)) = R\textrm{Hom}(O,O(i-j)) = R\Gamma(O(i-j)) = S^{i-j}(V)$. This is just a computation of the cohomology of the $O(i)$'s. $\endgroup$ Nov 11, 2015 at 2:47

1 Answer 1


$\newcommand\Ocal{\mathcal{O}}\newcommand\Hom{\mathop{\mathrm{Hom}}}\newcommand\Ext{\mathop{\mathrm{Ext}}\nolimits}$I think the following is a rather clumsy way of proving it. It is an induction on two variables, but probably there are better ways to set up the induction. The idea is to prove it for

  1. $i$ between 0 and $n$, $j=0$
  2. $i=0$, $j$ between 0 and $n$
  3. show that the cases $(i,j)$ and $(i-1,j-1)$ are isomorphic

I will denote the sequence that Beilinson suggests to do the induction with by $(\#_i)$.

First part The case $i=0$ is clear. To see that $\Hom(\Omega^i(i),\Ocal)=\bigwedge^iV^*$ for $i\geq 1$ we apply $\Hom(-,\Ocal)$ to $(\#_i)$. One gets a four-term exact sequence

$$0\to\Hom(\Omega^{i-1}(i),\Ocal)\to\Hom(\bigwedge\nolimits^iV\otimes\Ocal,\Ocal)\to\Hom(\Omega^i(i),\Ocal)\to\Ext^1(\Omega^{i-1}(i),\Ocal)\to 0$$

and isomorphisms $\Ext^l(\Omega^i(i),\Ocal)\cong\Ext^{l+1}(\Omega^{i-1}(i),\Ocal)$ for $l\geq 1$. The second term in the sequence is isomorphic to $\bigwedge^i V^*$, which is what we are after. So we would like to have that $\Ext^l(\Omega^{i-1}(i),\Ocal)=0$ for all $l$.

Lemma $\Ext^l(\Omega^{i-1}(i),\Ocal)=0$ for all $l$, and $i=1,\dotsc,n$.

Proof The case $i=1$ is clear. So assume that $i\geq 2$, and now we apply $\Hom(-,\Ocal(-1))$ to $(\#_{i-1})$. In the long exact sequence we get we see that $\Ext^l(\bigwedge\nolimits^{i-1}V^*\otimes\Ocal,\Ocal(-1))=0$ for all $l$. But then we apply the induction hypothesis, and conclude that everything is zero. $\square$.

So we get the desired isomorphism in this case.

Second part The case $j=0$ is clear. The case $j=n$ is also clear, because $\Omega^n(n)=\Ocal(-1)$. Then we rewrite the short exact sequence $(\#_{j+1})$ as

$$0\to\Omega^{j+1}(j)\to\bigwedge\nolimits^{j+1}\otimes\Ocal(-1)\to\Omega^j(j)\to 0$$

and then applying $\Hom(\Ocal,-)$ to this modified sequence gives a long exact sequence where $\Ext^l(\Ocal,\bigwedge\nolimits^{j+1}V\otimes\Ocal(-1))=0$, and similar to the previous lemma we can sandwich the desired term between zeroes.

Third part By now the ideas should be clear. Assume that $i\geq 1$ and $j\geq 1$. Apply $\Hom(-,\Omega^j(j))$ to $(\#_i)$. By the second part we see that the second (and fifth, etc.) term vanishes. We get another $\Ext^l=\Ext^{l+1}$ isomorphism.

Now apply $\Hom(\Omega^{i-1}(i),-)$ to $(\#_j)$. We get yet another long exact sequence, where by induction the third term $\Hom(\Omega^{i-1}(i),\Omega^{j-1}(j))$ is isomorphic to $\bigwedge\nolimits^{i-j}V^*$ (and higher $\Ext^l$ vanish).

Now we do yet another induction on as in the lemma, to see that the second (and fifth, etc.) term vanishes. Hence tracing back all the isomorphisms we see that

\begin{align} \bigwedge\nolimits^{i-j}V^* &\cong\Hom(\Omega^{i-1}(i-1),\Omega^{j-1}(j-1)) \\ &\cong\Hom(\Omega^{i-1}(i),\Omega^{j-1}(j)) \\ &\cong\Ext^1(\Omega^{i-1}(i),\Omega^j(j)) \\ &\cong\Hom(\Omega^i(i),\Omega^j(j)) \end{align}

as desired. Hurray!

The geometric intuition should correspond to its interpretation as a dual exceptional collection, but I don't see a way to make this into an actual explanation. The above is just a proof by tedious and boring computation. I hope I didn't make mistakes.

  • $\begingroup$ After rereading the question it turns out you didn't ask for the details of a proof. Well, now you have them (if you didn't already). $\endgroup$
    – pbelmans
    Nov 12, 2015 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.