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October 21, 2005

Shih on OSV

David Shih visited with us this week, which gives me an excuse to talk about the checks he and collaborators have been able to carry out on the Ooguri-Strominger-Vafa conjecture. I’ve written about the work of Dabholkar et al, which focusses on “small” BPS blackholes which, in their examples, have heterotic duals, the Dabholhar-Harvey states (whose degeneracies are, therefore, exactly-known). In the end, that wasn’t a very satisfactory test, because — even in the large-charge limit — the volumes of some cycles on XX turned out to be small. There wasn’t an a-priori reason to expect agreement with OSV.

Shih, Strominger and Yin computed the degeneracy of 1/4 BPS blackholes in N=4N=4 string theory (IIA compactified on K3×T 2K3\times T^2 with nonzero D0, D2, D4 and D6-brane charge, or one of its duals), and 1/8 BPS blackholes in N=8N=8 string theory. These blackholes are “large,” already at the classical level.

Shih and Yin then went on to compare these degeneracy formulæ with OSV.

A single wrapped D6-brane, when lifted to M-theory is replaced by a Taub-NUT space. Taking the radius of the Taub-NUT space to infinity, one obtains a 5D blackhole preserving the same amount of supersymmetry. The degeneracies of those blackholes were computed in the classic paper by Strominger and Vafa and by Breckenridge et al. But the degeneracies can’t depend on the radius, a continuous parameter. So the 4D blackhole degeneracies, with Q 6=1Q_6=1, are given by the same formula. The result is invariant under only a subgroup of the 4D UU-duality group (which preserves Q 6=1Q_6=1), but can be completed in a unique fashion to a formula invariant under the full SL(2,)×SO(6,22,)SL(2,\mathbb{Z})\times SO(6,22,\mathbb{Z}) UU-duality group.

The result, conjectured long ago by Dijkgraaf, Verlinde and Verlinde, can be written as Ω(p,q)=dρdσdνe iπ(ρq m 2+σq e 2+(2ν1)q eq m)Φ(ρ,σ,ν) \Omega(p,q) = \oint d\rho d\sigma d\nu \frac{e^{i\pi(\rho q_m^2 +\sigma q_e^2 +(2\nu-1)q_e\cdot q_m)}}{\Phi(\rho,\sigma,\nu)} where Φ\Phi is the unique automorphic form1 of weight 10 for Sp(2,)Sp(2,\mathbb{Z}). The following combinations of charges are invariant under SO(6,22,)SO(6,22,\mathbb{Z}) q e 2 =2q 0p 1+C abq aq b+c ijq iq j q m 2 =2p 0q 1+C abp ap b+c ijp ip j q eq m =p 0q 0+p 1q 1p aq ap iq i\array{\arrayopts{\colalign{right left}} q_e^2&=2q_0 p^1 +C^{a b} q_a q_b + c^{i j}q_i q_j\\ q_m^2&=2p^0 q_1 + C_{a b}p^a p^b + c_{i j}p^i p^j\\ q_e\cdot q_m&= p^0 q_0 +p^1 q_1 -p^a q_a -p^i q_i } where C abC_{a b} is the intersection form on H 2(K3)H^2(K3) (which has signature (3,19) ) and c ij=(0 σ 1 σ 1 0)c_{i j}=\textstyle{\left(\array{0&\sigma_1\\ \sigma_1&0}\right)} (which has signature (2,2)).

In the IIA language,

  • q 0q_0 is the D0-brane charge
  • q 1q_1 is the number of D2’s wrapped on T 2T^2
  • q aq_a are the D2’s wrapped on 2-cycles of the K3K3.
  • q iq_i are the momentum and winding modes of NS5’s wrapped on K3×S 1K3\times S^1
  • p 0p^0 is the number of wrapped D6-branes
  • p 1p^1 is the number of D4-branes wrapped on K3K3
  • p ap^a are D4-branes wrapped on T 2×T^2\times a 2-cycle of K3K3
  • p ip^i are momentum and winding modes of fundamental strings wrapped on T 2T^2.

Anyway, one wishes to Laplace-transform, Z(p,ϕ)= qΩ(p,q)e ϕq Z(p,\phi)= \sum_q \Omega(p,q) e^{- \phi\cdot q} and compare the result with the square of the topological string partition function, evaluated at the attractor values for the moduli, |Z top(p+iϕ/π)| 2|Z_{\text{top}}(p+i\phi/\pi)|^2.

For T 6T^6, Z top=e F top,F top=(2πi) 3D ABCt At Bt Cg top 2 Z_{\text{top}} = e^{F_{\text{top}}}, \qquad F_{\text{top}}=\frac{(2\pi i)^3 D_{A B C} t^A t^B t^C}{g_{\text{top}}^2} and for K3×T 2K3\times T^2 case, F top=(2πi) 3C MNt Mt Nt 1g top 224logη(t 1) F_{\text{top}}= \frac{(2\pi i)^3 C_{M N} t^M t^N t^1}{g_{\text{top}}^2} -24\log \eta(t^1)

where t M=X MX 0=p M+iϕ M/πp 0+iϕ 0/π t^M = \frac{X^M}{X^0}=\frac{p^M+i\phi^M/\pi}{p^0+i\phi^0/\pi} are the Kähler moduli and g top=4πiX 0=4πip 0+iϕ 0/π g_{\text{top}}= \frac{4\pi i}{X^0}= \frac{4\pi i}{p^0 +i\phi^0/\pi} is the topological string coupling.

Shih and Yin carry out the computation for p 0=0p^0=0. They find disagreement, even at the perturbative level, but the discrepancy can be summarized by a modified formula, Z(p,ϕ)= ϕϕ+2πik|Z top(p+iϕ/π)| 2g top 2(b 12)V(X)+𝒪(e V(X)/g top 2) Z(p,\phi) = \sum_{\phi\to \phi +2\pi i k}|Z_{\text{top}}(p+i\phi/\pi)|^2 g_{\text{top}}^{2(b_1-2)}V(X) + \mathcal{O}\left(e^{-V(X)/g_{\text{top}}^2}\right) where V(X)V(X) is the attractor value of the volume of X=K3×T 2X=K3\times T^2 (or X=T 6X=T^6 in the N=8N=8 case) and b 1b_1 is the first Betti number3 of XX.

It would be very interesting to see whether this persists for other N=4N=4 models (orbifolds of the above), and it would be very nice to understand the origin of these corrections to OSV.


1 The 4×44\times 4 matrix, (A B C D)Sp(2,)\left(\array{A&B\\ C&D}\right)\in Sp(2,\mathbb{Z}) acts on τ=(ρ ν ν σ)\tau= \left(\array{\rho&\nu\\ \nu&\sigma}\right) as τ(Aτ+B)(Cτ+D) 1\tau\to (A\tau +B)(C\tau+D)^{-1}. The weight-10 modular form,

Φ(τ)=2 12 αevenΘ[α](τ)\Phi(\tau) = 2^{-12}\prod_{\alpha \text{even}}\Theta[\alpha](\tau)

is the product of all the genus-2 theta functions for even spin structures.

2 Later, we’ll take the direct sum, and denote by C MNC_{M N}, the quadratic form of signature (5,21), C abc ijC_{a b} \oplus c_{i j}.

3 Note that extrapolating that the power of g topg_{\text{top}} in the correction depends linearly on b 1(X)b_1(X), based on only two data point, X=K3×T 2,T 6X=K3\times T^2,\, T^6, is a bit suspect.

Posted by distler at October 21, 2005 3:21 AM

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MathML

I think a while ago you asked to report problems with the display of mathematical characters.

On my Explorer+Mathplayer on WinXp setup the above text produces a red question mark for the first symbol right after the equality sign in your very first equation, as well as two red question marks right in front of the bracket term in your very last non-inline equation.

Posted by: Urs Schreiber on October 21, 2005 7:31 AM | Permalink | Reply to this

Re: MathML

…a red question mark for the first symbol right after the equality sign in your very first equation

You mean the \oint in ...=dρ... ...= \oint d\rho...

two red question marks right in front of the bracket term in your very last non-inline equation.

You mean ... αevenΘ[α]... ... \prod_{\alpha \text{even}} \Theta [\alpha] ... or a different equation?

It’s hard to see what’s wrong with either of those.

Posted by: Jacques Distler on October 21, 2005 9:27 AM | Permalink | PGP Sig | Reply to this

Re: Shih on OSV

There is nothing wrong with the MathML.

MathPlayer generally displays a red question mark when it cannot find a font on the local system containing the desired character. In this case, the first red question mark is a contour integral, which on my system displays fine. However, using a debug version of MathPlayer, I determined it is coming from Mathematical Pi 3, which is not a common font. Code 2000 is a shareware font with good coverage of math characters, so if you install that, the contour integral should display.

The pair of red question marks in the formula for Z(p, phi) is a different problem. That is a plane 1 Unicode character, and MathPlayer cannot currently display these characters. It is interpreting it as two BMP Unicode characters, neither of which it can render, which accounts for it displaying two question marks. I note, however, that Firefox cannot display this character on my system either. It apparently recognizes it as a plane 1 character, but I don’t have a font that contains it, so I get a black question mark.

Plane 1 support is on the list of feature requests for MathPlayer, but there isn’t a timeline for it yet.

Robert Miner
Director, New Product Development
Design Science, Inc.

Posted by: Robert Miner on October 21, 2005 4:25 PM | Permalink | Reply to this

Fonts

Ah, font problems …

The Code2001 font contains all the relevant Plane-1 characters. I recommend everyone install it. Waiting around for the STIX fonts won’t do you any good.

The thing is, I thought Urs had installed Code2001, which is why I did not expect him to see such errors. But, perhaps that was on his other machine …

Posted by: Jacques Distler on October 21, 2005 4:42 PM | Permalink | PGP Sig | Reply to this

Re: Fonts

Right, my fault. I had these fonts installed on my other machine, but was under the wrong impression that MathPlayer came with its own font libraries.

Posted by: Urs on October 22, 2005 6:43 AM | Permalink | Reply to this

Re: Fonts

MathPlayer does come with a small set of fonts that we have a license to distribute. But they only cover a core set of math characters, so one generally needs other fonts as well. I have Code 2000 installed as my “backup” font, but as per Jacques’ advice, I am going to switch to Code 2001. Sorry for the confusion.

Posted by: Robert Miner on October 22, 2005 10:26 AM | Permalink | Reply to this

Re: Fonts

Unfortunately, on my system at least, Firefox + Code2001 still displays two characters incorrectly (as little boxes) in the last term of the formula for Z(p,ϕ)Z(p, \phi).

Posted by: Dan on November 1, 2005 4:42 PM | Permalink | Reply to this

Re: Fonts

It’s actually just a single character, $\mathcal{O}$ (𝒪).

Works OK here on Mozilla/Mac. Don’t know what the problem might be on Firefox/Windows.

Sorry.

Posted by: Jacques Distler on November 1, 2005 4:56 PM | Permalink | PGP Sig | Reply to this

Re: Shih on OSV

MathPlayer does as much as it needs to ensure well-formedness, and how to render MathML, but that’s about it. Even with MathPlayer, XML stylesheet declarations are ignored, there’s an implicit tbody element, and other features such as xml:base are ignored. Obviously, there’s nothing stopping anyone writing a plugin to make IE handle application/xhtml+xml according to the specification, but as far as I’m aware, no one has.

Posted by: Stephanie Rose on October 28, 2005 1:46 PM | Permalink | Reply to this

Not a Replacement for a Real Browser

Obviously, there’s nothing stopping anyone writing a plugin to make IE handle application/xhtml+xml according to the specification, but as far as I’m aware, no one has.

Some would consider that the responsibility of the IE Development Team.

We’re just grateful MathPlayer enables IE to render MathML. Turning it into a decent modern browser would be considerably more work.

Posted by: Jacques Distler on October 28, 2005 2:52 PM | Permalink | PGP Sig | Reply to this

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