6d SCFTs and F-theory: a bottom-up perspective · Iknow it’s not quite the Dave day yet,but let me say one thing. Dave was originally a pure mathematician. When he was about my

6dSCFTsandF-theory:

abottom-upperspective

YujiTachikawa

Feb. 2016, F-theory@20, Caltech

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I knowit’snotquitetheDavedayyet, butletmesayonething.

Davewasoriginallyapuremathematician.

Whenhewasaboutmyagerightnow, hestartedtolearnstringtheory,andbecameoneoftheleadingfiguresinthefield.

Isn’titamazing?

MaybeI shouldtrysomethingotherthanstringtheory,followinghisexample!

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Isn’titamazing?


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Isn’titamazing?


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6dSCFTsandF-theory:

a bottom-up perspective

3/81

Asyouknow, I’mnotreallyanF-theoryperson.

I don’tevenremembertheordersof (f, g,∆) inKodaira’stablebyheart!

I wasaskedtogiveareviewtalkfromanoutsiderpointofview,soit’sprobablyOK.

I guesstheideawaslikeinvitinganLQG persontotheStringsconference...

I alsoapologizeinadvancethatmytalkwillbeverysubjective,andwillnotcite/mentionmanyrelevantpapers,althoughI knowI should.

ButatleastI wouldliketoillustrate...

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The UnreasonableEffectiveness ofF-theory

inthestudyof6dSCFTs

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Q. Why6dSCFT?

A. When [Heckman-Morrison-Vafa 1312.5746] cameout, mystudentHiroyukiShimizugotinterested, soasanadviserI neededtostudyitandcomeupwithaprojecthecouldworkon.

A. I’vebeenstudying4d N=2 SUSY forquitesometime,and6dSCFTsfeltfamiliar

A. 6disthe maximal dimensionwhere superconformalgroupsareavailable. So, somethinginterestingmightbegoingon.

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http://arxiv.org/abs/1312.5746

Q. Why6dSCFT?




6/81


Q. Why6dSCFT?




6/81


Q. Why6dSCFT?




6/81


Sketchoftheproofthat 6disthemaximaldimension:

1 Superconformalalgebrasare simple.

2 Fermionicpartsofthesuperconformalalgebrasare spinors.

3 Simple superalgebrasareclassified.

4 Fermionicpartsofalmostallofthemareinthe fundamental.

5 Needanaccidentalisomorphism spinors ≃ fundamental.

6 Themaximalcaseistherefore so(8), or so(6, 2) forourpurpose.

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6d N=(n, 0) SCFT correspondsto osp(6, 2|2n).

Openquestion

Showthat N=(n > 2, 0) SCFTsdon’texist.

cf. 5d N=1 SCFT correspondsto F (4), whosebosonicpartisso(7) ⊕ su(2) andthefermionicpartis spinor ⊗ doublet.

Theresimplyisno5d N>1 superconformalalgebra,sothere’snocorrespondingopenquestion.

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6d N=(n, 0) SCFT correspondsto osp(6, 2|2n).

Openquestion

Showthat N=(n > 2, 0) SCFTsdon’texist.

cf. 5d N=1 SCFT correspondsto F (4), whosebosonicpartisso(7) ⊕ su(2) andthefermionicpartis spinor ⊗ doublet.

Theresimplyisno5d N>1 superconformalalgebra,sothere’snocorrespondingopenquestion.

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Q. Howdoyoustudy6dSCFTs?

• Conformalbootstrap.• AnalysisoftheLagrangianonthetensorbranch.• AnalysisoftheLagrangianofthe S1 compactification.• BraneconstructionsinM,(massive)typeIIA,ortypeI• F-theory!

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M-theoryconstructionsandF-theory

↓

structureonthetensorbranchandF-theory

↓

massivetypeIIA constructionsandF-theory

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↓


↓


11/81

A largeclassof6d N=(1, 0) SCFTscanbeobtainedbyputtingM5-branesontheALE singularities:

R6 ×

R1

C2/

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When Γ = Zk, wehave SU(k) gaugefieldsatthesingularity,andanM5justgivesabifundamentalof SU(k) × SU(k):

R6 ×

SU(k)

bifundamental

SU(k)

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Butsurprisingthingshappenwhen Γ isoftype Dk or Ek.[delZotto-Heckman-Tomasiello-Vafa, 1407.6359]

Forexample, take Γ oftype Dk andput1M5:

R6 ×

SO(2k)

nontrivialSCFT

SO(2k)

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TheM5becomestwofractionalM5s:

R6 ×

USp(2k−8)

bifund. bifund.

SO(2k)SO(2k)

Somehowthemiddleregionthegaugegroupis USp(2k − 8),andeachhalf-M5givesabifundamental.

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Similarly, when Γ isoftype E6, afullM5-branefractionates...

R6 ×

E6

nontrivialSCFT

E6

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Similarly, when Γ isoftype E6, afullM5-branefractionates...

R6 ×

E6E6SU(3)

∅∅

into4fractionalM5s, andthegaugegroupsoccurinthesequence

E6, ∅, SU(3), ∅, E6.

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Ingeneral, wehave

A : doesn’tfractionate.

Dk : SO(2k), USp(2k − 8), SO(2k)

E6 : E6,∅, SU(3),∅, E6

E7 : E7,∅, SU(2), SO(7), SU(2),∅, E7

E8 : E8,∅,∅, SU(2), G2,∅, F4,∅, G2, SU(2),∅,∅, E8.

Mynaturalreactionwasthis:

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What the hell are thesesequences of groups?

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M-theoretically, youcangoasfollows[Ohmori-Shimizu-YT-Yonekura, 1503.06217, Sec. 3.1]:

Tostudythe tensorbranch ofthissystem,

R6 ×

G

M5

G

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Wecaninsteadstudythe Coulombbranch ofits T 3 compactification:

R3 × ×

G

M5

G

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ReduceittoIIA:

R3 × ×

G

D4

G

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TakethedoubleT-dual:

R3 × ×

G

D2

G

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LiftitbacktoM-theory:

R3 × ×

G

M2

G

We’renowinterestedinits Higgsbranch,sincewe’veeffectivelytakenthe3dmirror.

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AnM2candissolveintothe G gaugefieldasaninstantonon T 3 × R:

G

CS

0

1

TheplotbelowshowstheevolutionoftheChern-Simonsinvarianton T 3 ateachslice.

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When G = SO(2k), theinstantoncanfractionate:

SO(2k)

CS

0

1

½

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Inanextremesituation, wehavethis:

SO(2k)

CS

0

1

½

Thebundleisflatbutnontrivial.

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SO(2k)

Threeholonomiesareknowntobegivenby

diag(+,+,+,−,−,−,−,+2k−7)

diag(+,−,−,+,+,−,−,+2k−7)

diag(−,+,−,+,−,+,−,+2k−7)

Originallynoticedby [Witten, hep-th/9712028].

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http://arxiv.org/abs/hep-th/9712028

Sotheunbrokengaugegroupis

SO(2k) → SO(2k−7)

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Sowehave

R3 × ×

½M2

SO(2k−7)SO(2k)SO(2k)

½M2

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Goingbackthedualitychain, wehave

R3 × ×

½M5

USp(2k−8)SO(2k)SO(2k)

½M5

sinceweneedtotake4dS-duality/3dmirrorsymmetry:

SO(2k−7) ↔ USp(2k−8)

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Theanalysiscanbecarriedoutinasimilarmannerforany G,usingtheresultsin [Borel,Friedman,Morgan math.GR/9907007].

Whatneedstobedoneistheclassificationofflat G bundleson T 3

andthecomputationoftheirChern-Simonsinvariants.

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http://arxiv.org/abs/math.GR/9907007

Example: G = E7.

Youcanconstructabundlewith CS = 1/2 asfollows. Take

diag(+,+,+,−,−,−,−)

diag(+,−,−,+,+,−,−)

diag(−,+,−,+,−,+,−)

in SO(7). Infacttheyarein G2.

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E7 hasamaximalsubgroup G2 × USp(6). Therefore

CS

0

1

½

E7 USp(6) E7

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Youcanfractionatefurther, sincetheallowedCS invariantsare

CS = 0,1

4,1

3,1

2,2

3,3

4.

Wehave

CS

E7 USp(6) E7SU(2) SU(2) ∅∅

1

0

1/41/31/22/33/4

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Intheoriginaldualityframewehave

E7

E7SU(2)

∅

∅SU(2)

SO(7)1/4M5

1/12M5

1/6M5 1/6

M51/12M5

1/4M5

NotethattheM5chargesare notequallydistributed.

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InF-theory, theanalysisisdoneasfollows[Aspinwall-Morrison, hep-th/9705104]:

RecallthattheM-theoryconfiguration

R6 ×

G

M5

G

isdualto...

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ThisF-theoryconfiguration:

R6 ×

G

G

C2

wheretwoF-theory7-branesintersecttransversallyatapoint.

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I knowI don’thavetoreviewthefollowinginthisworkshop, butanyway.

Let’ssayweputtheellipticfibrationtotheWeierstrassform

y2 = x3 + fx + g

where f , g arefunctionsonthebase.

Let ∆ = 4f3 + 27g2 beitsdiscriminant.

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g G ord(f) ord(g) ord(∆)

Ik

(1 k0 1

)SU(k) 0 0 k

II

(1 1−1 0

)∅ ≥ 1 1 2

III

(0 1−1 0

)SU(2) 1 ≥ 2 3

IV

(0 1−1 −1

)SU(3) ≥ 2 2 4

I∗k

(−1 −k0 −1

)SO(2k + 8) 2 3 k + 6

IV ∗(−1 −11 0

)E6 ≥ 3 4 8

III∗(0 −11 0

)E7 3 ≥ 5 9

II∗(0 −11 1

)E8 ≥ 4 5 10

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So, supposetwo E7 7-branesintersect.

E7

E7

(3, 5, 9)

(3, 5, 9)

Here (3, 5, 9) meansthat (f, g,∆) vanishtotheseordersthere.

Attheintersection,

(3, 5, 9) + (3, 5, 9) = (6, 10, 18) ≥ (4, 6, 12).

A smoothellipticfibrationcan’texceed (4, 6, 12).

Soweblow-uptheintersectionpoint.

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Wenowgetthisconfiguration

E7

E7

(3, 5, 9)

(3, 5, 9)

(2, 4, 6)

where(2, 4, 6) = (3, 5, 9) + (3, 5, 9) − (4, 6, 12).

Lookingupthetable, thiscorrespondsto I∗0 with SO(8).

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A moredetailedanalysisshowsthatthereisan outer-automorphism actionof SO(8) aroundthis S2 of I∗

0 curve

E7

E7

(3, 5, 9)

(3, 5, 9)

(2, 4, 6)

SO(7)

giving SO(7).

Theintersectionof (2, 4, 6) and (3, 5, 9) isstillsingularsince

(2, 4, 6) + (3, 5, 9) ≥ (4, 6, 12).

Weneedtoblowup, repeat...

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Weendupwiththisfinalconfiguration:

E7

E7SO(7)SU(2) SU(2)∅ ∅

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RecallthatinM-theorythiswas

E7

E7SU(2)

∅

∅SU(2)

SO(7)1/4M5

1/12M5

1/6M5 1/6

M51/12M5

1/4M5

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whichreflectedpossiblechoicesofflat E7 connectionson T 3

CS

E7 USp(6) E7SU(2) SU(2) ∅∅

1

0

1/41/31/22/33/4

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Thecorrespondenceworksforany G = Ak, Dk and E6,7,8.

E7

E7SO(7)SU(2) SU(2)∅ ∅

CS

E7 USp(6) E7SU(2) SU(2) ∅∅

1

0

1/41/31/22/33/4

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HowonearthdoestheF-theoryknowtheflatconnectionson T 3?

Notethat [Aspinwall-Morrison, hep-th/9705104] appearedbefore [Borel,Friedman,Morgan math.GR/9907007].

F-theoryworksinmysteriousways.

Openquestion

TheF-theoryshouldknowtheCS invariantsoftheflatconnections,buthow?

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HowonearthdoestheF-theoryknowtheflatconnectionson T 3?

Notethat [Aspinwall-Morrison, hep-th/9705104] appearedbefore [Borel,Friedman,Morgan math.GR/9907007].


Openquestion

TheF-theoryshouldknowtheCS invariantsoftheflatconnections,buthow?

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↓


↓


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We’reactivatingscalarsinthetensormultiplet.

E7

E7

E7

E7SO(7)SU(2) SU(2)∅ ∅

Ongenericpointsonthetensorbranch, wejusthave

• tensormultiplets• vectormultiplets• hypermultiplets

soonecanapplyamoretraditionalfield-theoreticalanalysis.

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Caveat: weareassumingthatallnontrivialSCFTshavetensorbranch.

Openquestion

Showthatifa6dSCFT doesnothaveanytensorbranch,itisatheoryoffreehypermultiplets.

Openquestion

Showthatifa4d N=2 SCFT doesnothaveanyCoulombbranch,itisatheoryoffreehypermultiplets.

Comments:

• Ifyoubelieveevery6dSCFT comesfromF-theoreticsingularities,thentheanswerisyes:3-dimCY singularitiescanalwaysberesolved.

• Bootstrap?

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E7

E7SO(7)SU(2) SU(2)∅ ∅

InF-theory, each P1 isassociatedto

• a tensor multiplet• a string comingfromD3wrappedthere,whosetensionisgivenbythescalarinthetensormultiplet

• possiblya gaugemultiplet, whoseinversecouplingsquaredisalsogivenbythescalarinthetensormultiplet

• When ∃ gaugemultiplet, the string isthe instanton-string.

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E7

E7SO(7)SU(2) SU(2)∅ ∅

InF-theory, eachintersectionoftwo P1 supporting G1 and G2 givesaparticularhypermultipletchargedunder G1 × G2.

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Soyoucantrytowork bottom-up: whatisthepossiblestructureof

• tensors Hi, i = 1, . . . , nT

• vectorsfor Ga, a = 1, . . . , nV

• hyperschargedunder∏

Ga

ongenericpointsofthetensorbranchofa6dSCFT?

Strongconstraintscomefromthe anomalycancellation.

Animportantroleisplayedbythe Diracpairing ⟨·, ·⟩onthechargelattice Λ ≃ ZnT ofthestrings.

Notethatin6d, thepairingis symmetric.

SCFT requiresittobe positivedefinite.

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Beforegettingfurther, I shouldsaytheanalysisthatfollowswouldbe ratherdefective.

• There’s noguarantee thatagivenanomaly-freecombinationcomesfroma6dSCFT thatreallyexists.

• Theanalysisdoesnottellusanythingaboutthe modelswheretherearenovectors.

I willcomebacktothefirstpointlater.

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Onthesecondpoint:

Thereshouldbeawaytounderstandbetterthe modelsthatdo nothavevectormultipletsonthetensorbranch.

Sofar, knownexamplesare

• ADE N=(2, 0) theories, and• E-string theoryanditshigher-rankanalogues, whichare N=(1, 0).

Whydoes N=(2, 0) theoriesclassifiedby ADE?

Wheredoesthe E8 symmetry comefrom, for N=(1, 0) examples?

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Whydoes N=(2, 0) theoriesclassifiedbyADE?

Thereisaniceargument [Henningson, hep-th/0405056] showingthatthe anomalycancellationonthestringworldsheet requiresthatthechargelatticeofany N=(2, 0) theoryshouldbea simply-lacedrootlattice.

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Essentialpointsofhisideaareasfollows.

• Takethestringofchargevector q ∈ Λ.• ThestringbreakshalftheSUSY.Nambu-Goldstonemodesbecomeworldsheetfields.Theyarechiral, andthereforeanomalous.

• Italsocouplestotheself-dualfieldsinthebulk.Thisgivestheanomalyinflowproportionalto ⟨q, q⟩.

• Thecancellationrequires ⟨q, q⟩ = 2.• So Λ isanintegrallatticegeneratedbyelementsof (length)2 = 2.

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Howabout N=(1, 0) examples?

Openquestion

Giveanargumentthatifagenuine N=(1, 0) SCFT doesnothaveanyvectormultipletongenericpointsonthetensorbranch,itistheE-stringoritshigher-rankanalogue.

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Thefirststepwouldbetostudytherank-1case.

AssumethattheDiracpairingofthechargelattice Λ ≃ Z isminimal.

Thenitseemsthatthecancellationoftheworldsheetanomalyrequiresthat ∃ aleft-moving c = 8 modular-invariantsector,showingthatitautomaticallyhas E8 symmetry.

Anybodyinterestedinfillinginthegaps?

Orisitalreadygivenintheliterature?

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Letuscomebacktothesetup:

• tensors Hi, i = 1, . . . , nT

• vectorsfor Ga, a = 1, . . . , nV

• hyperschargedunder∏

Ga

Letusfurtherassume nT = nV .Instantonstringsarechargedunderthetensors, sowehave

dHa = c2(Ga).

Thiscontributestotheanomalyby

I tensor8 =1

2Ωabc2(Ga)c2(Gb)

where Ωab istheintegralDiracpairingofthechargelattice.

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A sidecomment:

TheIIB F5 alsosatisfies dF5 = (something), andthereforethereis

IGS12 =1

2(something)2

whichisnotusuallydiscussed. Isitzero?

Ifitisn’t, itruinstheanomalycancellationofIIB supergravity.

Yes, since (something) = H3 ∧ G3.

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A sidecomment:

TheIIB F5 alsosatisfies dF5 = (something), andthereforethereis

IGS12 =1

2(something)2

whichisnotusuallydiscussed. Isitzero?

Ifitisn’t, itruinstheanomalycancellationofIIB supergravity.

Yes, since (something) = H3 ∧ G3.

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Comingbackto6d, thetotalanomalyshouldvanish:

0 =1

2Ωabc2(Ga)c2(Gb) +

∑a

Ivector8 (Ga) +∑

Ihyper8

FirstanalyzedbySeiberg [hep-th/9609161] for nV = nT = 1.

For SU(2) with 2Nf half-hypersinthedoublet, wehave

0 =1

2Ω c2

2 −32 − 2Nf

24c2

2

Weneed 32 > 2Nf , because Ω > 0.

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For SU(2) with Nf half-hypersinthedoublet, wehave

0 =1

2Ωc2

2 −32 − 2Nf

24c2

2

Weneed 32 > 2Nf , because Ω > 0.

BershadskyandVafain [hep-th/9703167] pointedoutthattherearemoreconstraints.

• F-theoryconstructionsonlygave Nf = 4 and = 10. Why?• In6d, there’saglobalanomalyassociatedto π6(SU(2)) = Z12,anditrequires 32 − 2Nf = 0 mod 12.

• Also, Ω needstobeaninteger, whichgivesthesamecondition.

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HowonearthdoestheF-theoryknowtheglobalanomalyassociatedtothesubtlehomotopygroup π6(SU(2)) = Z12?


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Wecanextendthefield-theoreticalanalysistogeneral nT = nV ,witharbitrary

∏Ga andarbitraryhypers.

[Heckman-Morrison-Rudelius-Vafa 1502.05405][Bhardwaj, 1502.06594]

Oneexampleis nT = nV = 2, G = su(2) × so(8),withahalf-hyperin 2 ⊗ 8V andfullspinorsin 8S ⊕ 8C .

Thisisfreeofbothlocalandglobalanomalies, withthechargepairing(2 −1−1 3

)

But thisisneverrealizedinF-theory!Isthereafield-theoreticalwaytoseethis?

Yes. [Ohmori-Shimizu-YT-Yonekura, 1508.00915, AppendixA]

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Thismodeltriestogaugethe so(8) flavorsymmetryofthe6dmodel

su(2) and 4 doubletsonthetensorbranch.

ThepointisthatthisSCFT onlyhas so(7) ⊂ so(8)where8doubletstransformasaspinorof so(7).

Soyoucan’tgauge so(8).

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Toseethatthereisonly so(7), compactifyon T 2 the6dmodel


Youget

4d N=2 su(2)G with 4 doublets

+ anadditional su(2)T comingfromthe6dtensor.

TheWeylgroupof su(2)T istheS-dualityofthis4d N=2 su(2)G with 4 doublets, andmaps 8V to 8S .

So, onlythe so(7) ⊂ so(8) s.t. 8C → 7 + 1 iscompatible.

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Anotherwayistocompactifyon S1 the6dmodel


Youget

5d N=1 su(2)G with 4 doublets

+ anadditional su(2)T comingfromthe6dtensor.

5d su(2)G with 4 doublets hasanenhanced so(10) flavorsymmetry,andthe su(2)T gaugesthe so(3) subgroup,whichcomesfromtheenhancedinstantonnumbersymmetry.

Thecommutantisclearly so(7), since so(3) ⊕ so(7) ⊂ so(10).

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HowdoestheF-theoryknowthissubtleissue?


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↓


↓


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Wecanalsouse(massive)IIA toengineer6dSCFTs.[Brunner-Karch, hep-th/9712143] [Hanany-Zaffaroni, hep-th/9712145]

Twoexamples:

n D6s

NS5½NS5

O8−+ 16 D8s

n − 8 D6s

gives su(n) withan antisymmetric and n + 8 fundamentals, and

n D6s

NS5½NS5

O8+

n − 8 D6s

gives su(n) witha symmetric and n − 8 fundamentals.

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Theyarefreefromanomalies, bothlocalandglobal.

However, itsohappensthattheF-theoreticatomicclassification[Heckman-Morrison-Rudelius-Vafa, 1502.05405] includes

su(n) withan antisymmetric and n + 8 fundamentals

but doesnotinclude

su(n) witha symmetric and n − 8 fundamentals.

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Thisconundrumwasnoticedbymanypeoplesimultaneously,intheUS,inCanada, inJapan, andmaybeelsewheretoo.

I didn’tnoticeitmyself, butI learnedaboutitfrommultiplesources.

Doesthemodel

su(n) witha symmetric and n − 8 fundamentals.

havesomesecretinconsistency?

No. ClassificationsinF-theorysofarforgottoinclude O7+.[workinprogress, withBhardwaj, MorrisonandTomasiello]

74/81

Youcantakeeither

n D6s

NS5½NS5

O8−+ 16 D8s

n − 8 D6s

or

n D6s

NS5½NS5

O8+

n − 8 D6s

andcompactifyonetransversedirection,thentaketheT-dualtogototheIIB.

75/81

Youget

O7−+ 8 D7s

n D7s

n − 8 D7s

transversetangent

or

O7+

n D7s

n − 8 D7s

transversetangent

76/81

IntheF-theorylanguage, wehave

In

transversetangent

I 4* In−8

or

transversetangent

In

In−8“ I4* ”

What’sthis “I∗4”?

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ThepointisthatinF-theory, thereare two distincttypesof I∗

n singularitieswhen n ≥ 4.

• I∗n with so(2n + 8) symmetry, deformabletobesmooth.

• “I∗n” with usp(2n − 8) symmetry, deformableonlydownto “I∗

4”.

Thelatterisfrozenbyamysteriousdiscreteflux[Witten, hep-th/9712028].

YoumightworrythatotherKodairatypesmighthavefrozenversions,butthatdoesn’thappen[YT, 1508.06679].

F-theorygeometrizesmostofthefieldtheoryphenomena,butitstillneedssomeadditionaldata. cf. T-brane.

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Summary

• 6dSCFTscanbestudiedinmanyways.

• Varioussubtlefield-theoreticalfeaturesarealwaysencodedinF-theory, butinmysteriousways.

–Propertiesof ADE Instantonson R × T 3

–Globalanomaly π6(SU(2)) = Z12

–Reductionofflavorsymmetry so(8) ⊃ so(7)of nT = 1 su(2) with 4 doublets

–Issueson O+-planes

• Therearemanyopenquestions.

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Openquestion

Doesevery6dSCFT comefromF-theory?

Openquestion

Doesevery4d N=2 SCFT comefromstringtheory?

80/81

Openquestion

Doesevery6dSCFT comefromF-theory?

Openquestion

Doesevery4d N=2 SCFT comefromstringtheory?

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cf. AsfarasI know, thereisneitherastringyrealizationof

• 4d N=2 SU(7) or SU(8) with 3-indexantisymmetric• 4d N=2 Sp(3) or Sp(4) with 3-indexantisymmetric• 4d N=2 SO(13) or SO(14) with spinor

northeSeiberg-Wittensolutionstothem. DoesF-theoryhelp?

NotethatalmostexactlythesamelistofgroupsweregivenbyDaveonthe1stdayoftheworkshop, assubtlecasesinTate’salgorithm.Anyrelation?

81/81

Documents

6d SCFTs and F-theory: a bottom-up perspective · Iknow it’s not quite the Dave day yet,but let me say one thing. Dave was originally a pure mathematician. When he was about my