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November 3, 2006

del Pezzo

There are several reasons why a lot of recent work on string compactifications has focussed on the case of Type-II orientifold backgrounds. The most obvious is that turning on fluxes gives a mechanism for moduli stabilization.

Equally important, however, is that, in contrast to heterotic backgrounds, many of the phenomena of interest (nonabelian gauge theories coupled to chiral matter) are localized in the target space. This localization (along with the warping due to large fluxes) provides a nice mechanism for generating a large hierarchy of scales. It also emboldens one to take a “tinker-toy” approach, putting together pieces of the desired physics, localized at different locations in some (unspecified) Calabi-Yau orientifold (the Standard Model on a stack of branes at this singularity over here, supersymmetry-breaking over there, …). Whether, in fact, these pieces can be assembled together in arbitrary ways, or whether there are important constraints that one misses in the purely local analysis, is a crucial question.

Another question, to which there is still not a clear-cut answer, is to what extent one can get the Standard Model (without extra junk) out of these local constructions. We know that, with suitable bundles, on the heterotic side, one can get the Standard Model field content, on the nose. (I am thinking, here, of the work by the Penn group in the context of heterotic M-Theory.) Moreover, on the simply-connected covering space (i.e., before Wilson-line breaking), there is a known dictionary between the heterotic compactification, with bundles constructed via the Freedman-Morgan-Witten construction and the F-theory dual. Unfortunately, the same is not true after Wilson-line breaking. So, whereas one can get GUT groups, one cannot, currently, get the Standard Model on the F-theory side.

The “state of the art,” in terms of local constructions, seems to be the model of Verlinde and Wijnholt. The complex codimension-3 singularities of a Calabi-Yau look like a complex cone over a del Pezzo surface, and Verlinde and Wijnholt manage to fit a Standard Model-like theory into the biggest and best of the del Pezzo singularities, dP8. Recently, Buican et al showed how the above construction can be embedded in a compact Calabi-Yau geometry.

S=dP 8S=\text{dP}_8 is 2\mathbb{P}^2, blown up at 8 generic points. The homology, H 2(S)\mathrm{H}_2(S) is generated by the hyperplane class, HH, and the eight exceptional divisors, E iE_i, HH=1,E iE j=δ ij,HE i=0 H\cdot H =1,\quad E_i\cdot E_j = -\delta_{i j}, \quad H\cdot E_i = 0

E_8 Dynkin diagram

The canonical class, K=3H+E iK=-3H+\sum E_i, and the sublattice of H 2(S)\mathrm{H}_2(S), orthogonal to KK, spanned by α i =E iE i+1,i=1,,7 α 8 =HE 1E 2E 3 \begin{aligned} \alpha_i &= E_i - E_{i+1},\quad i=1,\dots , 7\\ \alpha_8 &= H- E_1-E_2-E_3 \end{aligned} is isomorphic to the E 8E_8 root lattice.

At some general point in the Kähler moduli space of S=dP nS=\text{dP}_n, the gauge theory on a D3-brane transverse to this singularity is a quiver gauge theory with n+3n+3 nodes. (The number of nodes is equal to the rank of K 0(S)K^0(S).). As one moves about in the Kähler moduli space, one crosses various surfaces of marginal stability, where the identities of the stable fractional branes changes. Correspondingly, the quiver gauge theories undergo transitions, related to Seiberg dualities.

dPnn, for n5n\geq 5, also has a moduli space of complex structures, of dimension 2n82n-8, corresponding to the freedom to move the blowup points, after fixing the locations of 4 of them via the PGL(3)PGL(3) symmetry. On a suitable locus, S=dP 8S=\text{dP}_8 can develop an A 2A_2 singularity. In the associated CY geometry, the dP88 singular point lies on a curve of A 2A_2 singularities. In this situation, there is a branch of the moduli space, where some of the fractional branes (the ones corresponding to α 1,α 2\alpha_1, \alpha_2) can be moved off the del Pezzo singularity, onto the curve of A 2A_2 singularities, and can be moved far away. This decouples two of the nodes of the quiver, and the bifundamental matter charged under those nodes.

Verlinde-Wijnholt quiver
The quiver of Verlinde and Wijnholt

What remain, according to Verlinde and Wijnholt, is a quiver with 9 nodes, depicted at left. The gauge group is (U(3)×U(2)×U(1) 7)/U(1) diag(U(3)\times U(2) \times {U(1)}^7)/{U(1)}_{\text{diag}}. There’s a bifundamental chiral multiplet corresponding to each arrow in the quiver, and I’ve suggestively named them by the corresponding supersymmetric Standard Model field. Note that there are 3 generations of quarks, leptons, and right-handed neutrinos as well as 6 generations (six!) of Higgses.

The quiver gauge theory has superpotential terms corresponding to directed loops in the quiver. So we can see, e.g., the usual Yukawa couplings with the Higgses which give masses to the quarks and leptons, as well as a W=+λNH uLW=\dots + \lambda N H_u L, which is part of the mechanism for generating neutrino masses.

I’m not quite sure why this quiver could not arise somewhere in the Kähler moduli space of a dP6 singularity, without the somewhat baroque contrivance of demanding that we have a dP8 singularity sitting on top of a curve of A 2A_2 singularities, and moving off onto a mixed branch of the moduli space.

This quiver gauge theory doesn’t yet look much like the Standard Model. There’s a plethora of extra U(1)U(1)s. One linear combination of the U(1)U(1)s, the one corresponding to the node α 4\alpha_4 of the E 8E_8 Dynkin diagram, is U(1) Y{U(1)}_Y. No problem, say Buican et al., two of the U(1)U(1)s are anomalous, and the corresponding gauge bosons pick up a mass due to the Green-Schwarz mechanism1. The remaining U(1)U(1)s are massless in the noncompact geometry, but are given mass via a Stückelberg mechanism, when this geometry is embedded in a compact Calabi-Yau.

To ensure that U(1) Y{U(1)}_Y remains massless, it suffices to ensure that the homology class, α 4\alpha_4, which is nontrivial on SS, becomes trivial upon embedding in the compact Calabi-Yau geometry. This, they go to some lengths to show can be done, while at the same time ensuring that the dP8 singularity lies on a curve of A 2A_2 singularities in the CY.

Despite the beautiful algebraic geometry at play, there are still some formidable obstacles to interpreting the above theory as the supersymmetric Standard Model. First of all, the μ\mu-term is forbidden by the quiver gauge symmetry. This might actually be a good thing, because we know the μ\mu-term is suppressed. The authors discuss mechanisms for generating the μ\mu-term, either as a nonperturbative correction to the superpotential, or via the Giudice-Masiero mechanism. Much more serious, is that a Majorana mass term for NN is forbidden by the quiver gauge symmetry. Rather than being suppressed (like the μ\mu-term), we need this mass to be large (10 15\sim 10^{15} GeV), in order to generate the observed neutrino masses via the seesaw mechanism.

That seems to me to be a rather serious obstacle to trying to use this quiver to embed the Standard model in a Type IIB orientifold.


1 The Green-Schwarz mechanism, in 4 dimensions, works as follows. Consider some (anomalous) U(1)U(1) gauge symmetry. If the coefficient of Tr(F 3)Tr(F^3) in the anomaly polynomial vanishes (i.e., if we’ve managed to cancel the U(1) 3{U(1)}^3 anomaly, as well as all the pure nonabelian anomalies), then we can cancel the mixed gravitational-U(1)U(1) anomaly (the Tr(F)Tr(R 2)Tr(F)Tr(R^2) term in the anomaly polynomial) and/or the mixed U(1)U(1)-SU(n) 2{SU(n)}^2 anomaly (the Tr(F)Tr(G 2)Tr(F) Tr(G^2) terms in the anomaly polynomial) by picking some pseudoscalar field, aa, and giving it a nontrivial transformation under U(1)U(1) gauge transformations, AA+df,aa+f A\to A +d f, \quad a \to a +f and adding a term a(cTr(RR)+c iTr(G iG i))\int a\, \left(c Tr(R\wedge R) + c_i Tr(G_i\wedge G_i)\right) to the action. In the case at hand, there’s no gravitational anomaly, but there are mixed U(1)U(1)-(U(1)) 2{(U(1)')}^2, U(1)U(1)-SU(3) 2{SU(3)}^2 and U(1)U(1)-SU(2) 2{SU(2)}^2 anomalies, which affect two of the 8 U(1)U(1)s.

Posted by distler at November 3, 2006 11:32 PM

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Re: del Pezzo

Hi Jacques,

I just noticed your post on Del Pezzo’s, and thought I would comment on a few things.

“Another question, to which there is still not a clear-cut answer, is to what extent one can get the Standard Model (without extra junk) out of these local constructions. We know that, with suitable bundles, on the heterotic side, one can get the Standard Model field content, on the nose.”

It seems you are asking for two things here. First with D-branes you can only get the MSSM with some extra massive junk. For instance the SU(3) comes with an extra gauged U(1)_B, which will eat a Stueckelberg field and become massive. Secondly, there is the question of interactions. Here it was Herman who first emphasized that in the local context, getting the MSSM spectrum right ought to be sufficient; by open/closed intuition, any deformation of the MSSM should then map to a deformation of the local geometry.

” (I am thinking, here, of the work by the Penn group in the context of heterotic M-Theory.)”

I don’t want this to be seen as belittling the work of the Penn group, since they did a lot of impressive work, but if I remember correctly they did have a massless U(1)_B-L that was hard to get rid off.

“I’m not quite sure why this quiver could not arise somewhere in the Kähler moduli space of a dP6 singularity, without the somewhat baroque contrivance of demanding that we have a dP8 singularity sitting on top of a curve of A 2 singularities, and moving off onto a mixed branch of the moduli space.”

Well, DP6 and DP8 are related by blowing up -1 curves, not -2 curves.

I do agree that it seems somewhat arbitrary from the geometry. I don’t see anything that would make the Standard Model seem “special.”

Finally, the phenomenological problems have to do with the fact that we were restricting ourselves to oriented quivers. I have been looking at unoriented quivers (including orientifolds) recently; then it appears that one can indeed get exactly the MSSM, though I still have to finish checking that indeed the surplus non-chiral matter can be lifted.

Posted by: Martijn on November 5, 2006 3:59 PM | Permalink | Reply to this

Re: del Pezzo

I don’t want this to be seen as belittling the work of the Penn group, since they did a lot of impressive work, but if I remember correctly they did have a massless U(1) BL{U(1)}_{B-L} that was hard to get rid off.

That depends. They study models with rank-4 visible gauge groups (descended from SU(5)SU(5) GUTs) and models with rank-5 gauge group (descended from SO(10)SO(10) GUTs). The latter, indeed, have an extra U(1) BLU(1)_{B-L} and, indeed, it is hard to see how to break it at a suitably high scale, so as to give a large Majorana mass to the right-handed neutrinos (that also appear in your model).

They can also get models with minimal (supersymmetric) Higgs content, rather than the embarrassingly large Higgs content that you find.

On the downside, they don’t have suitable candidates for the stable bundle in the hidden sector. This difficulty is the result of trying to satisfy a constraint that I don’t see the analogue of in your case: one needs that the sum of the second Chern classes of the bundles in the visible and hidden E 8E_8s, minus the second Chern class of the tangent bundle must be an effective curve class, so that it can be cancelled by wrapping heterotic 5-branes.

Well, DP6 and DP8 are related by blowing up -1 curves, not -2 curves.

Yes, I’m just asking whether there’s an intuition for why this 9-node quiver doesn’t arise in somewhere in the moduli space of dP 6\text{dP}_6.

Posted by: Jacques Distler on November 5, 2006 6:36 PM | Permalink | PGP Sig | Reply to this

Re: del Pezzo

“They can also get models with minimal (supersymmetric) Higgs content, rather than the embarrassingly large Higgs content that you find.”

That’s another consequence of restricting to oriented quivers. The SU(2)_weak comes with an extra U(1), which disallows the use of the epsilon tensor. This, plus the requirement of reproducing all the classical Yukawa couplings, forces the extra Higgses on you. In unoriented models however this problem evaporates, because you can take the gauge group to be USp(2) = SU(2).

“On the downside, they don’t have suitable candidates for the stable bundle in the hidden sector. This difficulty is the result of trying to satisfy a constraint that I don’t see the analogue of in your case: one needs that the sum of the second Chern classes of the bundles in the visible and hidden E 8 s, minus the second Chern class of the tangent bundle must be an effective curve class, so that it can be cancelled by wrapping heterotic 5-branes.”

Similar issues show up when you consider compactification of the local geometry.

“Yes, I’m just asking whether there’s an intuition for why this 9-node quiver doesn’t arise in somewhere in the moduli space of dP 6 .”

Not really. The `basic’ quivers that you can get by placing D-branes at a Del Pezzo singularity are reasonably well understood, but all the stuff you can get after Higgsing these quivers is very hard to characterize.

Posted by: Martijn on November 6, 2006 6:30 AM | Permalink | Reply to this

Re: Heterotic models

Just for the record, in order to not mix up different models:

  • Ron Donagi’s model does not have a U(1) BLU(1)_{B-L}, slope-stable visible and hidden bundle, and supersymmetric (wrapping an effective curve) five-branes.
  • My et al’s model has a U(1) BLU(1)_{B-L}, slope-stable visible and hidden bundle, and non-supersymmetric five-branes.

Our proposed supersymmetry breaking mechanism (over there) is precisely of the kind that you describe above, only localized in the 11-d interval instead of a Calabi-Yau singularity.

As for the F-theory dual, I’d love to see that as well. But even before the Wilson-line breaking I don’t know how to apply the dictionary: What is the dual of an extension of bundles?

Posted by: Volker Braun on November 6, 2006 10:00 AM | Permalink | Reply to this

Re: Heterotic models

Volker said:

As for the F-theory dual, I’d love to see that as well. But even before the Wilson-line breaking I don’t know how to apply the dictionary: What is the dual of an extension of bundles?

Oh. Yes, perhaps you’re right. I thought one knew the F-theory dual of bundles constructed via the Friedman-Morgan-Witten construction. But I guess that doesn’t apply to extensions …

Martijn said:

Similar issues show up when you consider compactification of the local geometry.

Since one of your demands is that most of the H 2(S)\mathrm{H}^2(S) lift to nontrivial homology classes in the Calabi-Yau, and since we have fractional branes wrapped around these cycles, do we have to worry about cancelling the net D5-brane charge, when the CY is compact?

Posted by: Jacques Distler on November 6, 2006 11:25 AM | Permalink | PGP Sig | Reply to this

Re: Heterotic models

Hi Jacques:

I think I can respond to that last question. So long as the D-brane of the standard model is pure D3-brane plus torsion in homology (or K-theory), I think you are in business.


There are many examples where the blow-up mode of various conifold singularities really correspond to a torsion cycle globally, so even wrapping fractional branes at each such conifold locus does not introduce real five-branes globally. Locally, of course you can’t tell the difference in the effective field theory.

Posted by: David Berenstein on November 6, 2006 12:26 PM | Permalink | Reply to this

Re: Heterotic models

Hi Volker,

Thanks for the clarification.

Jacques,

“Since one of your demands is that most of the H 2 (S) lift to nontrivial homology classes in the Calabi-Yau, and since we have fractional branes wrapped around these cycles, do we have to worry about cancelling the net D5-brane charge, when the CY is compact? “

I think David has answered it: we really started out with a D3-brane, so that’s the only net charge you have to cancel. Under heterotic/F-theory duality, M5-branes wrapped on the T^2 fibre on the heterotic side get mapped to D3-branes transverse to the CY four-fold, so the heterotic constraint that you mentioned is morally very similar to cancelling net D3 charge.

Posted by: Martijn on November 6, 2006 1:03 PM | Permalink | Reply to this

Re: Heterotic models

Under heterotic/F-theory duality, M5-branes wrapped on the T^2 fibre on the heterotic side get mapped to D3-branes transverse to the CY four-fold, so the heterotic constraint that you mentioned is morally very similar to cancelling net D3 charge.

Except that the effective curve class, that they end up wrapping M5-branes around, is typically not the class of the T 2T^2 fiber.

we really started out with a D3-brane, so that’s the only net charge you have to cancel.

That’s maybe a little less than obvious from the description. The Stückelberg mechanism certainly requires that (most of) the classes in H 2(S)\mathrm{H}^2(S) correspond to nontrivial 2-cycles in XX. However, you say, the net wrapping number of the fractional branes around those cycles is zero.

I guess that must be true …

Posted by: Jacques Distler on November 6, 2006 1:40 PM | Permalink | PGP Sig | Reply to this

Re: Tadpole cancellation

There are many examples where the blow-up mode of various conifold singularities really correspond to a torsion cycle globally, so even wrapping fractional branes at each such conifold locus does not introduce real five-branes globally.

Due to the BPS inequality, supersymmetric cycles such as holomorphic submanifolds or special Lagrangians come in pointy-tipped cones: any non-zero linear combination of supersymmetric cycles is a free generator in the Homology/K-theory lattice.

The only way around this is to either look at cycles that are not supersymmetric at all, or at least consider linear combinations of cycles with mutually incompatible supersymmetries. So you can cancel the charge of a brane wrapped on, say, a resolved or deformed conifold, but only if you are willing to introduce anti-branes somewhere (which is probably a good idea anyhow).

That still does not guarantee that you can put the anti-branes over there. The Mayer-Vietoris sequence tells you if there is a topological obstruction, but in the end you’ll have to analyze the brane–anti-brane potential (as we had to do in the heterotic context).

Posted by: Volker Braun on November 6, 2006 3:57 PM | Permalink | Reply to this

Re: Tadpole cancellation

Volker said:

Due to the BPS inequality, supersymmetric cycles such as holomorphic submanifolds or special Lagrangians come in pointy-tipped cones: any non-zero linear combination of supersymmetric cycles is a free generator in the Homology/K-theory lattice.

It turns out that torsion cycles in homology with branes wrapped on them can become non-torsion generators in K-theory,
with integer D3-brane charge plus a fraction. This is because the K-theory lattice and the homology lattice are different.

There is no contrivance necessary, It just happens that you can not locally blow up the corresponding cycles without breaking SUSY and thus the singularities are essentially unavoidable. The typical example involves defroming T^6/Z_2xZ_2 with discrete torsion.

Posted by: David Berenstein on November 6, 2006 6:43 PM | Permalink | Reply to this

Torsion stays torsion

It turns out that torsion cycles in homology with branes wrapped on them can become non-torsion generators in K-theory… This is because the K-theory lattice and the homology lattice are different.

That doesn’t sound right.

Rationally, K-theory and cohomology are isomorphic. The torsion subgroups are different, but modulo torsion, they are isomorphic as rings (not just as abelian groups), and the Chern character provides the ring isomorphism.

Moreover, even if they were different, rationally, it makes no sense to say that something torsion can become non-torsion, when passing from cohomology to K-theory. If it’s torsion, then it’s a cyclic element of order NN (for some NN). A cyclic element of finite order cannot become infinite order by waving a magic wand (the Atiyah-Hirzebruch Spectral Sequence).

Posted by: Jacques Distler on November 6, 2006 8:53 PM | Permalink | PGP Sig | Reply to this

Re: Torsion stays torsion

It’s kind of funny, a naive torsion class can acquire anomalous D3-brane charge because of B-field.

The extension works like

Z\to Z \to Z_2\to 0

where the Z in the center is the one in the K-theory lattice, and the ones on the sides are in homology.

In this situation, the naive torsion class counts as half a D3-brane, and the rank of the lattice does not change, but you have a generator that is `smaller’.

I’m not explaining this too well, am I?

Posted by: David Berenstein on November 6, 2006 10:13 PM | Permalink | Reply to this

Re: Torsion stays torsion

I’m not explaining this too well, am I?

No, but I think I understand what you mean.

This is one of the ways in which the torsion in cohomology can get modified (in this case, wiped out), in passing to K-theory.

The Atiyah-Hirzebruch spectral sequence converges to the associated-graded of the filtration of the K-theory. One then has an extension problem to solve.

In particular, the relevant piece, on a Calabi-Yau 3-fold, looks like 0H 6(X)FH 4(X)0 0 \to \mathrm{H}^6(X) \to F \to \mathrm{H}^4(X) \to 0 The generator of H 6(X)=\mathrm{H}^6(X)=\mathbb{Z} can be taken to be supported at a point on XX — the location of a D3-brane transverse to the Calabi-Yau. (Some of) the torsion in H 4(X)\mathrm{H}^4(X) can be killed by this extension, in the fashion you indicated.

[The thing I called FF, above, then fits into some further extensions involving the other terms in the AHSS, ultimately yielding K 0(X)K^0(X).]

Posted by: Jacques Distler on November 7, 2006 2:37 AM | Permalink | PGP Sig | Reply to this

Re: K-theory torsion

For a (connected) Calabi-Yau threefold XX, we have

(1)K 0(X)H 2(X,)H 4(X,), K 1(X)H 1(X,)H 3(X,)H 5(X,),\array{ K^0(X) \simeq \mathbb{Z}\oplus H^2(X,\mathbb{Z})\oplus H^4(X,\mathbb{Z})\oplus \mathbb{Z}, \\ K^1(X) \simeq H^1(X,\mathbb{Z})\oplus H^3(X,\mathbb{Z})\oplus H^5(X,\mathbb{Z}), }

see Doran and Morgan. In other words, none of the AHSS subtleties arise. The K-theory and Homology lattices are identical (maybe not completely canonical, but who cares).

The only way around this are fluxes (including discrete torsion), in which case the D-brane charges are no longer classified by the topological K-theory but some twisted version thereof. So David’s example is necessarily of this non-geometric kind.

Posted by: Volker Braun on November 7, 2006 10:57 AM | Permalink | Reply to this

Re: K-theory torsion

What I understood from David’s example was that he was considering the case of twisted K-theory, with [H]H 3(X,)[H]\in \mathrm{H}^3(X,\mathbb{Z}) torsion.

Second, I’m not sure I understand the statement in the case where XX has a nontrivial Brauer group, H 3(X,) torsion0\mathrm{H}^3(X,\mathbb{Z})_{\text{torsion}}\neq 0. Can’t d 3:H 0(X,)H 3(X,)d_3: \mathrm{H}^0(X,\mathbb{Z})\to \mathrm{H}^3(X,\mathbb{Z}) be nontrivial?

Also, on a compact Calabi-Yau (whose holonomy is SU(3)SU(3), and not a proper subgroup), H 1(X,)=0\mathrm{H}^1(X,\mathbb{Z})=0.

Posted by: Jacques Distler on November 7, 2006 11:47 AM | Permalink | PGP Sig | Reply to this

Re: K-theory torsion

Yes, I agree that David’s example has a torsion “flux” in

(1)H 2× 2 3(T 6,).H^3_{\mathbb{Z}_2\times \mathbb{Z}_2}(T^6,\mathbb{Z}).

I just wanted to point out that this is necessary if you want different K-theory/Homology lattices.

As for the d 3:H 0H 3d_3: H^0\to H^3 being zero, you proved that in hep-th/0102018. Doran and Morgan have a more high-brow way by identifying classifying spaces.

On a proper (holonomy the full SU(3)SU(3)) Calabi-Yau manifold H 1(X,)H^1(X,\mathbb{Z}) is indeed zero, because H 1H^1 never has torsion and b 1=rankH 1(X,)=0b_1=\mathrm{rank} H^1(X,\mathbb{Z})=0. But the theorem of Doran and Morgan only relies on c 1(X)=0c_1(X)=0, so they cover strictly smaller holonomy groups as well where H 1(X,)= b 1H^1(X,\mathbb{Z})=\mathbb{Z}^{b_1} need not vanish.

Posted by: Volker Braun on November 7, 2006 12:45 PM | Permalink | Reply to this

Re: K-theory torsion

As for the d 3:H 0H 3d_3:\mathrm{H}^0→\mathrm{H}^3 being zero, you proved that in hep-th/0102018.

In the untwisted case, d 3=Sq 3d_3 = {Sq}^3, the Steenrod square, which satisfies Sq 3:H i(X)H i+3(X)=0fori<3 {Sq}^3: \mathrm{H}^i(X)\to \mathrm{H}^{i+3} (X) = 0\; \text{for}\; i\lt 3

I was referring to the twisted case, where d 3=Sq 3+[H]d_3 = {Sq}^3 + [H]. Sorry for not making that clear.

Posted by: Jacques Distler on November 7, 2006 1:36 PM | Permalink | PGP Sig | Reply to this

Re: K-theory torsion

It seems that Doran and Morgan only cover the case where the Calabi-Yau is toric. If one has a non-abelian fundmental group without fixed points (a clearly non-toric case), then I believe it is possible to have non-trivial torsion in the third homology group as well.

Any thoughts on this possibliity?

Posted by: David Berenstein on November 7, 2006 1:47 PM | Permalink | Reply to this

Re: K-theory torsion

At the end of section two, they say that the K-theory results apply to all CY 3-folds, not just hypersurfaces in toric varieties.

Posted by: Aaron Bergman on November 7, 2006 1:52 PM | Permalink | Reply to this

Re: Brauer group

If one has a non-abelian fundmental group without fixed points (a clearly non-toric case), then I believe it is possible to have non-trivial torsion in the third homology group as well.

For the record, there are examples of toric Calabi-Yau hypersurfaces with a torsion subgroup in H 3(X,)H^3(X,\mathbb{Z}). For that the ambient toric variety needs to be singular, but the hypersurface misses these singularities. No need to go to non-toric constructions just to get an example.

But looking at non-toric Calabi-Yau manifolds is certainly interesting. Did anybody compute the integral cohomology of the Beauville manifold?

I agree that the twisted K-theory will differ from the homology groups. Doran and Morgan’s result certainly does not apply to twisted K-theory.

Posted by: Volker Braun on November 7, 2006 2:12 PM | Permalink | Reply to this

Re: Brauer group

Did anybody compute the integral cohomology of the Beauville manifold?

Yes.

Doran and Morgan’s result certainly does not apply to twisted K-theory.

Before looking at their paper, I was not aware that, in the untwisted case, the extension problem (going from the associated graded to the K-theory) was necessarily trivial. That’s interesting.

Posted by: Jacques Distler on November 7, 2006 4:37 PM | Permalink | PGP Sig | Reply to this

Re: Brauer group

Doh! I wanted to say: Did anybody compute the integral cohomology of the mirror of the Beauville manifold? Conjecturally, this has H 3(B^,)= 4 2 2H^3(\hat{B},\mathbb{Z}) = \mathbb{Z}^{4}\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2. This would be a cute non-toric example.

Posted by: Volker Braun on November 7, 2006 5:05 PM | Permalink | Reply to this

Beauville

Try Jae’s paper. Not sure if it has what you want.

Posted by: Jacques Distler on November 7, 2006 5:18 PM | Permalink | PGP Sig | Reply to this

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