All varieties appearing below are assumed smooth projective over $\mathbb C$ and all vector bundles, sections etc are assumed to be algebraic/holomorphic. We use the word resolution to mean quasi-isomorphism.

Suppose $X\subset\mathbb P^n$ is a closed subvariety of codimension $r$. Suppose we can a find a locally free sheaf $E$ on $\mathbb P^n$ of rank $r$ and a section $s$ such that $s^{-1}(0)$ is cut out transversely and equals $X$ (for example, if $X$ is a complete intersection). Then, we get the Koszul resolution

$0\to\wedge^rE^\vee\to\cdots\to\wedge^2E^\vee\to E^\vee\to\mathcal O_{\mathbb P^n}\to\mathcal O_X\to 0$

I would like to view this as a resolution of $\mathcal O_X$ by a differential graded commutative $\mathcal O_{\mathbb P^n}$-algebra $K(s)$ with $E^\vee$ placed in degree $-1$ (call $K(s)$ the Koszul algebra of $s$). Notice that as a graded commutative algebra $K(s)$, is simply the (graded commutative) symmetric algebra $\text{Sym}^\bullet_{\mathcal O_{\mathbb P^n}}(E^\vee[1])$. Now, suppose we had an extension $0\to E\to E'\to F\to 0$ of vector bundles, then we could consider the section $s'$ of $E'$ which is the image of $s$. Now, the Koszul algebra $K(s')$ is no more a resolution of $\mathcal O_X$, in fact its cohomology algebra is $\mathcal O_X\otimes_{\mathcal O_{\mathbb P^n}}\wedge^\bullet F^\vee$. Thus, by adding the free generator $F^\vee$ in degree $-2$ to $K(s')$, we get a new commutative differential graded algebra with the differential mapping $F^\vee$ to $E'^\vee$ by the dual of the map $E'\to F$. The evident map from this new algebra to $K(s)$ is a quasi-isomorphism and thus, we again have a resolution of $\mathcal O_X$.

More generally, there is the notion of Koszul-Tate resolution of $\mathcal O_X$ which consists of a commutative differential graded algebra $A$ over $\mathcal O_{\mathbb P^n}$ resolving $\mathcal O_X$ and having the following additional property. Locally on $\mathbb P^n$, $A$ is given (as a graded algebra) by a free (graded commutative) polynomial ring (in possibly infinitely many variables) over $\mathcal O_{\mathbb P^n}$ with finitely many generators in each negative degree. We'll say that $A$ is locally finitely generated if, locally on $\mathbb P^n$, it is a polynomial algebra over $\mathcal O_{\mathbb P^n}$ in finitely many variables (with the variables placed in negative degrees).

We have seen above that $K(s)\to\mathcal O_X$ is a resolution with $K(s)$ finitely generated. Are there any other examples of locally finitely generated resolutions besides the Koszul algebras $K(s)$ (and quasi-isomorphic modifications coming from exact sequences of the form $0\to E\to E'\to \cdots\to F\to 0$)?

In particular, are there examples of smooth $X\subset\mathbb P^n$ which are not complete intersections, for which $\mathcal O_X$ admits a Koszul-Tate resolution by a locally finitely generated differential graded commutative $\mathcal O_{\mathbb P^n}$-algebra? It seems pretty easy to show that Koszul-Tate resolutions exist if we do not require them to be locally finitely generated.

Note that not every $X\subset\mathbb P^n$ admits such a resolution, since a Koszul-Tate resolution of $\mathcal O_X$ would give a resolution (using the theory of the cotangent complex) of $\Omega^1_X$ by vector bundles on $X$ pulled back from $\mathbb P^n$, and therefore, if the algebra resolution is finitely generated, then so is the resolution of the cotangent bundle. Thus, taking determinants, we see that the canonical bundle $K_X$ lies in the image of the restriction map $\text{Pic}(\mathbb P^n)\to\text{Pic}(X)$. For instance, this shows that the twisted cubic in $\mathbb P^3$ doesn't admit a locally finitely generated Koszul-Tate resolution. There are also intrinsic conditions on $X$, for instance: any $X$ admitting a finite resolution must necessarily have $K_X$ to be either trivial, very ample or negative of very ample (since every line bundle on $\mathbb P^n$ is such). This, for example, rules out any $X$ which is a product of a Fano variety with a Calabi-Yau variety.

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    $\begingroup$ When you say "not complete intersections" you mean not zero loci of regular sections of vector bundles, right? Or literally not complete intersections? $\endgroup$ – Sasha Nov 22 '18 at 14:49
  • $\begingroup$ I would prefer examples which are not zero loci of regular sections of vector bundles. Though if you have such examples of regular zero loci involving vector bundles which are not the direct sum of line bundles, that will also be useful to me as I don't know too many concrete examples of those. $\endgroup$ – Mohan Swaminathan Nov 22 '18 at 15:13

Horrocks-Mumford bundle on $\mathbb{P}^4$ gives an abelian surface as the zero locus of its general section, which is not a complete intersection.

  • $\begingroup$ Thanks for the example. I'll wait to see if there are more interesting examples (i.e., something which is not the transversely cut out zero locus of a section of a vector bundle) before accepting an answer. $\endgroup$ – Mohan Swaminathan Nov 22 '18 at 16:26

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