Let $X$ be a Calabi-Yau threefold. Ordinary Donaldson-Thomas theory is formulated as a virtual count of ideal sheaves $\mathcal{I}$ with discrete invariants $\text{ch}(\mathcal{I}) = (1,0, -\beta, -n)$, which is equivalent to counting one-dimensional subschemes $Y \subseteq X$ with $[Y] = \beta \in H_{2}(X, \mathbb{Z})$ and $\chi(Y)=n$. In the derived category, we have the equivalence $\mathcal{I} \cong [\mathcal{O}_{X} \to \mathcal{O}_{Y}]$. For this reason, in the physical literature, they describe the DT theory as enumerating "bound states of D0-D2 branes with a single D6-brane. This is because $\text{ch}(\mathcal{O}_{X})=(1,0,0,0)$ represents a single D6-brane while $\mathcal{O}_{Y}$ represents the D0-D2 branes.

I was thinking it would be nice if divisors (or what a physicist would call a D4-brane) could be placed on the same footing and incorporated into the DT partition function as well. Mathematically, I feel like it would be nice to sort of have all the holomorphic sub-geometry of $X$ in one generating function and physically, it would be preferable to have all the possible D-branes in the Topological B-model (D0,D2,D4,D6) branes sort of on the same footing.

Of course divisors are special as they're given by vanishing of sections of line bundles, so I'm wondering if it's possible to write something like

$$\text{Hilb}_{D, \beta, n}(X) \cong \text{Pic}_{D}(X) \times \text{Hilb}_{\beta' n'}(X)$$

where $\text{Hilb}_{D, \beta, n}(X)$ would be the Hilbert scheme of points, curves, and divisors. Note that $\beta$ and $n$ will be different from $\beta'$ and $n'$.

We have a deformation/obstruction theory for $\text{Hilb}_{\beta' n'}(X)$ with a virtual class. Since $\text{Pic}_{D}(X)$ is smooth, can we compute $[\text{Hilb}_{D, \beta, n}(X)]^{\text{vir}}$ to define enumerative invariants? I was thinking that since the Picard factor is a torus, maybe by some localization argument it wouldn't contribute to an integral over the moduli space?

Perhaps what I'm asking for is nonsense, but even if it's possible, I guess it would probably destroy modularity properties of the partition function? And would almost certainly lose connection to the Gromov-Witten theory via the MNOP conjecture.


1 Answer 1


The original definition of DT invariants ( https://arxiv.org/abs/math/9806111 ) works for any ch such that there is no strictly semistable objects. Later, this was generalized by Joyce and Song to arbitrary ch (giving rational invariants satisfying a conjectural multicovering formula). In general these invariants depend on a choice of Kähler class defining the notion of stability.

The case of D4-D2-D0 branes has been studied by Gholampour and Sheshmani : see https://arxiv.org/abs/1309.0050 . The most recent work along this line is https://arxiv.org/abs/1601.04030 where you can find further references. The product definition given is the question is too naive because it neglects "interactions" between D4 and D2-D0 branes.

I don't understand the remark on modularity properties: there is no modularity properties in the usual 1D6-D2-D0 case. But there are modularity properties in the D4-D2-D0 cases, which are partially predicted by S-duality.

Concerning the relation with Gromov-Witten theory, there is a very surprising proposal connecting count of D4-D2-D0 branes with Gromov-Witten theory (or via MNOP with 1D6-D2-D0): the OSV conjecture https://arxiv.org/abs/hep-th/0405146 . The OSV conjecture has generated a lot of activity in physics but is still poorly understand mathematically: even its precise mathematical statement is not clear, but it is an interesting direction to pursue.

In the D4-D2-D0 cases, a lot of the geometry can be reduced to some surface geometry which is much easier than the 3-fold geometry. The general case D6-D4-D2-D0 should obviously be the most interesting but I think that almost nothing is known for cases with multiple D6 (except very simple universal cases like those with only D6-D0).


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.