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Let $X$ be the total space of $\mathscr{O}(-1)^{\oplus{n}}\rightarrow\mathbb{P}^m$. I think there should be some general way to compute its quantum cohomology $QH^\ast(X)$.

However, since I'm not familiar with these things, I found it hard to carry out the computations directly by hand without assuming that $m\gg n$, or both of $m$ and $n$ are very small, and of course it's trivial when $n=1$. When $m\gg n$, by checking dimensions one sees easily that the degrees of the holomorphic curves must be 0 or 1 in order to obtain non-trivial Gromov-Witten invariants, but this simplification does not seem to happen in general.

Is there any coherent way to compute $QH^\ast(X)$ for all the possible $m,n\geq1$? I think the problem should be trivial to experts and it's quite possible that it can be found in the literature.

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    $\begingroup$ Naive question: since these bundles have no sections, won't any holomorphic curves all land in the base $\mathbb P^m$? $\endgroup$ Oct 17, 2015 at 20:24
  • $\begingroup$ @AllenKnutson Sure. $\endgroup$
    – YHBKJ
    Oct 18, 2015 at 3:06
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    $\begingroup$ the question made me think of this paper, but I have no idea whether it's relevant (thought I'd share anyway) projecteuclid.org/euclid.nmj/1415383898 $\endgroup$
    – pro
    Oct 18, 2015 at 7:51
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    $\begingroup$ Regarding Allen's question: are you asking about Gromov-Witten invariants taking values in cohomology / Chow theory, or are you asking about equivariant Gromov-Witten invariants for equivariant cohomology / equivariant Gromov-Witten theory. When $n\gg m$, of course the "usual" Gromov-Witten invariants are zero, while the equivariant theory can be nonzero. If you are asking about equivariant invariants, what is your choice of group action? $\endgroup$ Oct 18, 2015 at 13:31
  • $\begingroup$ @JasonStarr I'm talking about GW invariants taking values in cohomology. There may still be some non-trivial GW invariants, and when the rank of the bundle is not small compared to the dimension of the case, I think the computaton of $QH$ will still be involved as there are potentially many non-trivial GW invariants. $\endgroup$
    – YHBKJ
    Oct 18, 2015 at 13:37

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I'm not sure whether what I'm going to describe is good enough for your purpose. But the following will be the first thing that I would think about if I'm asked to compute your quantum rings. I apologize if you have been already aware of this stuff.

The keyword is Quantum Lefschetz hyperplane theorem (for concave bundle). Although computing the quantum cohomology might be hard, there is a beautiful formula for the small $I$-function associated to these (non-equivariant limit of) local Gromov-Witten invariants. Once the $I$-function is given, there is a systematic algorithm to reconstruct all genus-$0$ Gromov-Witten invariants if the cohomology of the variety is generated by divisors. I don't think I know a reference for the whole algorithm written down in one particular paper. But I think it should be done in the following order:

Birkhoff factorization -> Mirror transformation -> "Divisor relation" + Topological recursion relation

(by "divisor relation" I mean Theorem 1 of this paper)

Come back to your question. Here is how to write the (small) $I$-function in your case. Let $h$ be the hyperplane class in $H^2(\mathbb P^m)$. Let $t_0, t_1$ be the formal parameters parametrizing the fundamental class $1$ and the hyperplane class $h$ (just like what you would do in the small quantum cohomology). Your $I$-function is the following series:

$$I(z)=e^{\frac{1t_0+ht_1}{z}}\left( 1+ \sum\limits_{d\geq 1}\frac{\prod\limits_{i=1}^{d-1}(-h-iz)^n}{\prod\limits_{i=1}^d(h+iz)^{m+1}}q^de^{dt_1}\right) $$

where the denominators should be expanded into $z^{-1}$ series. (Depending on the author, $I$ functions might differ by a factor $z$ and/or by a sign on the variable $z$)

As far as I can recall, I don't think this $I$-function has any exception for $m,n\geq 1$. And I believe the rest of the procedures will be the same for different $m,n$. Therefore I would say it's a coherent way, though extremely complicated.

As an example, when $n=m+1$, the $I$-function and a few other invariants are described in section 3.3 of this paper.

But I still want to make a few further remarks about the computation. As you may know, there is the small $J$-function as a generating series of one point descendant invariants in the form of $\langle \tau_i h^j\rangle_{0,n,d}$. It has a variable $z$ as well (sometimes denoted by $\hbar$). The above $I$-function agrees with the $J$-function when the factor in the bracket (after $e^{\frac{1t_0+ht_1}{z}}$) doesn't have $z^{-1}$ term, where all the denominators in the fractions are already expanded into $z^{-1}$ series. In your case, I believe it happens when $n\leq m+1$ and $m$ is not too small. In this case, you can read off enough amount of one-point invariants (i.e., skip the Birkhoff factorization and Mirror transformation). One can also read off some of the relations in the small quantum ring by looking at the differential operators in $t_1$ that annihilates the $J$-function. I'm not sure how nice and explicit your quantum rings can be described. But these might be a good start.

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  • $\begingroup$ I've found a solution using toric geometry and Stanley-Reisner theory, which should be similar to your description. $\endgroup$
    – YHBKJ
    Feb 29, 2016 at 13:38
  • $\begingroup$ That's great. Toric geometry should suffice to compute the small quantum ring. And the involved I-function is indeed the same as what I was saying. $\endgroup$
    – Honglu
    Mar 1, 2016 at 19:06

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