nLab RR-field tadpole cancellation

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Contents

Idea

In the presence of D-branes, plain type II string theory in fact has a quantum anomaly reflected on the worldsheet by tadpole Feynman diagrams in the string perturbation series for RR-fields

graphics grabbed from Blumenhagen-Lüst-Theisen 13

and reflected in target spacetime by non-trivial total RR-field flux on compact spaces

graphics grabbed from Ibanez-Uranga 12

This anomaly cancels if the D-branes are accompanied by a suitable collection of O-planes, hence if one considers orientifold backgrounds (Sagnotti 88, pp. 5, Gimon-Polchinski 96, section 3). (For space-filling O-planes this means to consider type I string theory instead.)

Accordingly, tadpole cancellation via orientifolding is a key consistency condition in the construction of intersecting D-brane models for string phenomenology.

Traditionally RR-tadpole cancellation is discussed in ordinary cohomology, the common arguments notwithstanding that D-brane charge should be in K-theory.

Discussion of tadpole cancellation with D-brane charge regarded in K-theory was initated in Uranga 00, Section 5, see also Garcia-Uranga 05, Marchesano 03, Section 4, Marchesano-Shiu 04, CKMNW 05, Section 2.2, Maiden-Shiu-Stefanski 06, Section 5, DFM 09, p. 6-7.

But the situation remains somewhat inconclusive (see also Moore 14, p. 21-22).


In plane and toroidal orientifolds

More details are understood in the special case of plane orbifolds V cptGV^{cpt} \!\sslash\! G and toroidal orientifolds 𝕋 VG\mathbb{T}^V \!\sslash\! G where fractional D-branes may be stuck at orbifold/orientifold singularities, whose D-brane charge is supposed to be in the equivariant K-theory of the point, hence the representation ring of the given isotropy group.

In terms of equivariant K-theory / the representation ring

In this case tadpole cancellation conditions are given by representation theoretic equations, constraining the characters of the linear representations corresponding to the fractional D-branes.

Let GG be a finite group. Let

[1][H 1][H 2][G] [1] \subset [H_1] \subset [H_2] \subset \cdots \subset [G]

be a linear extension of its partially ordered lattice of conjugacy classes of subgroups, with sub- linear order of cyclic subgroups

[1][g 1][g 2][g |ConjCl(G)|]. [1] \subset \left[ \left\langle g_1 \right\rangle \right] \subset \left[ \left\langle g_2 \right\rangle \right] \subset \cdots \subset \left[ \left\langle g_{\vert ConjCl(G)\vert} \right\rangle \right] \,.

This way every virtual representation [V]RU(G)=KU G(*)[V] \in RU(G) = KU_G(\ast) (the D-brane charge of a bound state of fractional D-branes/anti-branes) has a character which is a list of complex numbers of the form

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right]\cdots[g |ConjCl(G)|]\left[\langle g_{\vert ConjCl(G)\vert}\rangle\right]
χ V=\chi_V =dim(V)dim(V)tr V(g 1)tr_V\left( g_1\right)tr V(g 2)tr_V\left(g_2\right)\cdotstr V(g |ConjCl(G)|)tr_V\left(g_{\vert ConjCl(G)\vert}\right)
fractionalD-brane/anti-branebound state{{\text{fractional} \atop \text{D-brane/anti-brane}} \atop \text{bound state}}mass=net numberof branes{ {\text{mass} =} \atop {{\text{net number} \atop \text{of branes}}}}RR-charge ing 1-twisted sector{\text{RR-charge in} \atop {g_1\text{-twisted sector}}}RR-charge ing 2-twisted sector{\text{RR-charge in} \atop {g_2\text{-twisted sector}}}\cdots\cdots

Here dim(V)dim(V) \in \mathbb{Z} is the mass, hence the net number of fractional D-branes/anti-branes in the bound state, while tr V(g k)tr_V\left(g_k\right) is (up to a global rational number-factor 1/|G|1/{\vert G \vert}) supposed to be its charge as seen by the RR-fields in the g kg_k-twisted sector.

In fact, since we are dealing with fractional D-branes, both the charge and mass in the above table are in factional units 1/|G|1/{\vert G\vert} of the order of the isotropy group GG (by this formula), so that normalized mass and charge is

(1)M=1|G|dim(V),AAAQ V(g)=1|G|χ V(g)1|G|tr V(g). M \;=\; \tfrac{1}{{\vert G\vert}} dim(V) \,, \phantom{AAA} Q_V(g) \;=\; \tfrac{1}{\vert G\vert} \chi_V(g) \coloneqq \tfrac{1}{\vert G\vert} tr_V\left( g\right) \,.

Local/twisted tadpole cancellation

The twisted (local) tadpole cancellation condition for fractional D-branes at orbifold singularities is that the RR-charges in all non-trivially twisted sectors vanish:

(2)Q V(g)=0AAhence equivalentlyAAχ V(g)=0,AAAge Q_V(g) = 0 \phantom{AA}\text{hence equivalently} \phantom{AA} \chi_{V}\left(g\right) \;=\; 0 \,, \phantom{AAA} g \neq e
Example

(regular representation solves tadpole cancellation for fractional D-branes)

For every finite group GG, the homogeous tadpole cancellation condition (2) is satisfied by all multiples nk[G/1]n \cdot k[G/1] of the regular representation k[G/1]k[G/1] (since no non-trivial element gGg \in G has fixed points when acting on GG, and using this Prop.). Hence the mass and charge (1) of the fractional D-brane corresponding to the regular representation is

M k[G/1]=1,AAQ k[G/1](g)=0. M_{{}_{k[G/1]}} \;=\; 1 \,, \phantom{AA} Q_{{}_{k[G/1]}}(g) \;=\; 0 \,.

These multiples of the regular representation are regarded as trivial solutions to (2).

Proposition

In fact, the multiples of the regular representation (Example ) are the only solutions to the local/twisted tadpole cancellation condition (2) for fractional D-branes.

Proof

Consider the truncated character morphism

Q|G|:Rep k(G)χk |ConjCl(G)|forget dimension/massk |ConjCl(G)|1. Q \cdot {\vert G \vert} \;\colon\; Rep_k(G) \overset{\chi}{\longrightarrow} k^{\left\vert ConjCl(G) \right\vert} \overset{ \text{forget dimension/mass} }{\longrightarrow} k^{\left\vert ConjCl(G)\right\vert -1 } \,.

We have to show that the kernel of this map is the free abelian group generated by the regular representation:

ker(Q|G|)k[G/1]. ker\big( Q \cdot {\vert G \vert} \big) \;\simeq\; \mathbb{Z} \cdot k[G/1] \,.

Now over a ground field kk of characteristic zero (such as the real numbers or complex numbers, in the case at hand) we have (from this Example) that

  1. for ρ1\rho \neq \mathbf{1} a non-trivial irreducible representation we have

    gG{e}Q ρ(g)|G|gG{e}χ ρ(g)=dim(ρ) \underset{g \in G \setminus \{e\}}{\sum} Q_{\rho}(g) \cdot {\vert G \vert} \;\coloneqq\; \underset{g \in G \setminus \{e\}}{\sum} \chi_\rho(g) \;=\; - dim(\rho)
  2. for ρ=1\rho = \mathbf{1} the trivial irreducible representation we have

    gG{e}Q ρ(g)|G|gG{e}χ ρ(g)=|G|1=dim(1)mod|G| \underset{g \in G \setminus \{e\}}{\sum} Q_{\rho}(g) \cdot {\vert G \vert} \;\coloneqq\; \underset{g \in G \setminus \{e\}}{\sum} \chi_\rho(g) \;=\; {\left\vert G\right \vert} - 1 \;=\; - dim(\mathbf{1}) \;mod\; {\vert G\vert}

Since every VR k(G)V \in R_{k}(G) is a \mathbb{Z}-linear combination of these irreps, it follows generally that the fractional part of the mass of a fractional D-brane is recovered from its charges:

dim(V)mod|G|=gG{e}Q V(g)|G|gG{e}χ V(g). dim(V) \;mod\; {\vert G \vert} \;=\; - \underset{g \in G \setminus \{e\}}{\sum} Q_{V}(g) \cdot {\vert G \vert} \;\coloneqq\; - \underset{g \in G \setminus \{e\}}{\sum} \chi_V(g) \,.

But this means that all VV in the kernel of Q|G|Q \cdot {\vert G \vert} must have

dim(V)=0mod|G|. dim(V) \;=\; 0 \;mod\; {\vert G \vert} \,.

This is indeed the case for the multiples V=nk[G/1]V = n\cdot k[G/1] of the regular representation (Example ). Conversely, the injectivity of the full character morphism χ\chi (this Prop.) says that every VV with dim(V)=n|G|dim(V) = n \cdot {\vert G\vert } and Q V(g)=0Q_V(g) = 0 must be the nnth multiple of the regular representation.

Global/untwisted tadpole cancellation

On the other hand, at an orientifold singularity, the O-plane itself carries such charge – O-plane charge (see there):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right]\cdots[g |ConjCl(G)|]\left[\langle g_{\vert ConjCl(G)\vert}\rangle\right]
χ O=\chi_O =dim(O)dim(O)tr O(g 1)tr_O\left( g_1\right)tr O(g 2)tr_O\left(g_2\right)\cdotstr O(g |ConjCl(G)|)tr_O\left(g_{\vert ConjCl(G)\vert}\right)
O-plane\text{O-plane}O-plane charge ing 1-twisted sector{\text{O-plane charge in} \atop {g_1\text{-twisted sector}}}O-plane-charge ing 2-twisted sector{\text{O-plane-charge in} \atop {g_2\text{-twisted sector}}}\cdots\cdots

(These are O O^--plane charges. There may also be O +O^+-plane charges. Alternatively, these are O O^--branes with a fractional D-brane stuck on them.)

Now the untwisted (global) tadpole cancellation condition is that (all representations are real and) this O-plane charge is cancelled against the D-brane charge:

(3)χ V(g k1)=0AAandAAdim(V)=dim(O). \chi_{V}\left(g_{k \geq 1}\right) \;=\; 0 \phantom{AA} \text{and} \phantom{AA} dim(V) = dim(O) \,.

By Prop. the only possible solution of this is the nnth multiple of the regular representation, if dim(O)dim(O) is nn times the dimension of the regular representation:

(4)V=Nk[G/1]. V = N \cdot k[G/1] \,.

In basic examples the O-plane-charge

O=2 p4n1 O = 2^{p-4} n \cdot \mathbf{1}

is for n On_O coincident O-planes is the corresponding multiple by the O-plane charge μ Op=2 84\mu_{Op} = -2^{8-4} (here) of the trivial irrep, whence a solution to the tadpole cancellation exists if 2 p4|G| \frac{2^{p-4}}{\vert G\vert } \in \mathbb{N} \subset \mathbb{Q} and is then given by

V=2 p4|G|k[G/1]. V \;=\; \frac{2^{p-4}}{\vert G\vert } \cdot k[G/1] \,.

Sometimes the condition (5) is found with an offset by a trivial representation

(5)V=Nk[G/1]+p triv. V = N \cdot k[G/1] + \mathbf{p}_{triv} \,.

This corresponds to single fractional D-branes sitting on top of the O-planes, turning O O^--planes into O +O^+-planes.

Examples for toroidal orientifolds

RR-field tadpole cancellation conditions for D-branes wrapped on toroidal orientifolds
in terms of their D-brane charge VK G(*)=R GV \in K_G(\ast) = R_G in equivariant K-theory = representation ring
(here n reg=k[G/1]\mathbf{n}_{reg} = k[G/1] denotes the regular representation of dimension n=n = ord(G))

single D-brane species
on toroidal orientifold
local/twisted
tadpole cancellation condition
global/untwisted
tadpole cancellation condition
comment
D5-branes
transv. to 𝕋 4 2\mathbb{T}^{\mathbf{4}_{\mathbb{H}}}\!\sslash\! \mathbb{Z}_{2}
V=N2 regV = N \cdot \mathbf{2}_{reg}
(Buchel-Shiu-Tye 99 (19))
V=162 regV = 16 \cdot \mathbf{2}_{reg}
(Buchel-Shiu-Tye 99 (18))
following
Gimon-Polchinski 96,
Gimon-Johnson 96
D5-branes
transv. to 𝕋 4 4\mathbb{T}^{\mathbf{4}_{\mathbb{H}}} \!\sslash\! \mathbb{Z}_{4}
V=N4 regV = N \cdot \mathbf{4}_{reg}
(Buchel-Shiu-Tye 99 (19))
V=84 regV = 8 \cdot \mathbf{4}_{reg}
(Buchel-Shiu-Tye 99 (18))
following
Gimon-Polchinski 96,
Gimon-Johnson 96
D4-branes
transv. to 𝕋 1 triv+4 k\mathbb{T}^{\mathbf{1}_{\mathrm{triv}} + \mathbf{4}_{\mathbb{H}}} \!\sslash\! \mathbb{Z}_k
V=Nk regV = N \cdot \mathbf{k}_{reg}
(AFIRU 00a, 4.2.1)
D4-branes
transv. to 𝕋 1 triv+4 3\mathbb{T}^{\mathbf{1}_{\mathrm{triv}} + \mathbf{4}_{\mathbb{H}}} \!\sslash\! \mathbb{Z}_3
V=N3 regV = N \cdot \mathbf{3}_{reg}
(AFIRU 00b, (7.2))
V=43 reg+41 trivV = 4 \cdot \mathbf{3}_{reg} + 4 \cdot \mathbf{1}_{triv}
(Kataoka-Shimojo 01, (14)-(17)
D8-branes
on 𝕋 1 triv+4 3\mathbb{T}^{\mathbf{1}_{\mathrm{triv}} + \mathbf{4}_{\mathbb{H}}} \!\sslash\! \mathbb{Z}_3
V=N3 regV = N \cdot \mathbf{3}_{reg}
V=43 reg+41 trivV = 4 \cdot \mathbf{3}_{reg} + 4 \cdot \mathbf{1}_{triv}
(Honecker 01, 4,
Honecker 02a, (25) ∧ (28) ⇔ (29),
Honecker 02b, (3.19)-(3.27))
equivalent to D4 case by T-duality:
Honecker 01, p. 2,
Honecker 02a, 6,
Honecker 02b, p. 15
review in:
Marchesano 03, Sec. 4
D6-branes
on 𝕋 6 4\mathbb{T}^6 \!\sslash\! \mathbb{Z}_4
V=84 regV = 8 \cdot \mathbf{4}_{reg}
(Ishihara-Kataoka-Sato 99, (4.16))
D3-branes
on 𝕋 4 k\mathbb{T}^{\mathbf{4}_{\mathbb{H}}} \!\sslash\! \mathbb{Z}_k
V=Nk regV = N \cdot \mathbf{k}_{reg}
(Feng-He-Karch-Uranga 01, (25))
D7-branes
on 𝕋 4 k\mathbb{T}^{\mathbf{4}_{\mathbb{H}}} \!\sslash\! \mathbb{Z}_k
V=Nk regV = N \cdot \mathbf{k}_{reg}
(Feng-He-Karch-Uranga 01, (5), (6))


graphics grabbed from Sati-Schreiber 19

See also at equivariant Hopf degree theorem.

graphics grabbed from Sati-Schreiber 19

Examples of non-compact singularities

We discuss more explicitly the solutions to the local/twisted tadpole cancellation condition (2) for fractional D-branes at orbifold singularities for isotropy group one of the non-abelian finite subgroups of SU(2),

G DESU(2) G_{DE} \;\subset\; SU(2)

hence those in the D- and E-series, hence the binary dihedral groups 2D 2n2 D_{2n} and the three exceptional cases: 2T, 2O and 2I.

For these groups, by BSS 18, Theorem 4.1 the virtual permutation representations span precisely the sub charge lattice of integral (non-irrational) characters/RR-charges in the orientifold charge lattice of the corresponding ADE-singularity, namely of the equivariant KO-theory=real representation ring

KO G DE 0(*)=RO(G DE). KO^0_{G_{DE}}(\ast) \;=\; RO\left( G_{DE} \right) \,.

Since the tadpole cancellation condition (2) in particular requires the characters/charges to be integral (specifically: zero) the general solution to the tadpole cancellation condition is indeed in this sub-lattice, and so that is where we may and do solve it, below.

In accord with the general Prop. we find that in each case there is precisely a 1-dimensional (i.e. \simeq \mathbb{Z}) sublattice of the charge lattice (the representation ring) which solves the twisted tadpole cancellation condition (2), hence a sublattice given by the integer-multiples NV 0N \cdot V_0 of one single fractional D-brane bound state V 0KO G 0(*)V_0 \in KO^0_G(\ast). There are then necessarily two of these generators ±V 0\pm V_0. We check below that in all cases the normalized mass of these is ±\pm unity, as it must be for the regular representation, by Prop. .

At a 2\mathbb{Z}_2-orientifold singularity

For G= 2G = \mathbb{Z}_2 the cyclic group of order | 2|=2{\vert \mathbb{Z}_2\vert} = 2, the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (e.g. here)

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11-1

One sees immediately that the general solution to the local/twisted tadpole cancellation condition (2) for G= 2G = \mathbb{Z}_2 is

V=n(1V 1+1V 2),AAAn. V \;=\; n \cdot \Big( 1 \cdot V_1 + 1 \cdot V_2 \Big) \,, \phantom{AAA} n \in \mathbb{Z} \,.

whose minimal positive mass (net brane number) is

M 2 =dim(V)/| 2| =χ V([e])/| 2| =(11+11)/2 =2/2 =1 \begin{aligned} M_{\mathbb{Z}_2} & = dim(V) / {\vert \mathbb{Z}_2 \vert } \\ & = \chi_V([\langle e\rangle]) / {\vert \mathbb{Z}_2 \vert} \\ & = \big( 1 \cdot 1 + 1 \cdot 1 \big) / {2} & \\ & = 2 / 2 \\ & = 1 \end{aligned}

At a 4\mathbb{Z}_4-orientifold singularity

For G= 4G = \mathbb{Z}_4 the cyclic group of order 4, the characters/D-brane charges of the complex irreducible representations/fractional D-branes are (e.g. here)

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11-11-1
χ V 3=\chi_{V_3} =1\phantom{-}11-1i-ii\phantom{-}i
χ V 4=\chi_{V_4} =1\phantom{-}11-1i\phantom{-}ii-i

One sees immediately that the general solution to the local/twisted tadpole cancellation condition (2) for G= 3G = \mathbb{Z}_3 is

V=n(1V 1+1V 2+1V 3+1V 4),AAAn. V \;=\; n \cdot \Big( 1 \cdot V_1 + 1 \cdot V_2 + 1 \cdot V_3 + 1 \cdot V_4 \Big) \,, \phantom{AAA} n \in \mathbb{Z} \,.

whose minimal positive mass (net brane number) is

M 4 =dim(V)/| 4| =χ V([e])/| 4| =(11+11+11+11)/4 =4/4 =1 \begin{aligned} M_{\mathbb{Z}_4} & = dim(V) / {\vert \mathbb{Z}_4 \vert } \\ & = \chi_V([\langle e\rangle]) / {\vert \mathbb{Z}_4 \vert} \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 \big) / {4} & \\ & = 4 / 4 \\ & = 1 \end{aligned}

At a 2D 42 D_4-orientifold singularity

For G=2D 4=Q 8G = 2 D_4 = Q_8 the binary dihedral group of order |2D 4|{\vert 2 D_4\vert} (equivalently: the quaternion group), the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.1):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11-11\phantom{-}11-1
χ V 3=\chi_{V_3} =1\phantom{-}11\phantom{-}11-11-11\phantom{-}1
χ V 4=\chi_{V_4} =1\phantom{-}11\phantom{-}11\phantom{-}11-11-1
χ V 5=\chi_{V_5} =4\phantom{-}44-40\phantom{-}00\phantom{-}00\phantom{-}0

One sees (here) that the general solution to the local/twisted tadpole cancellation condition (2) for G=2D 4G =2 D_4 is

V=n(1V 1+1V 2+1V 3+1V 4),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 1 \cdot V_3 + 1 \cdot V_4 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2D 4 =dim(V)/|2D 4| =χ V([e])/|2D 4| =(1+1+1+1+4)/8 =8/8 =1 \begin{aligned} M_{2 D_4} & = dim(V)/ {\vert 2 D_4\vert} \\ & = \chi_V\left( [\langle e\rangle]\right) / {\vert 2 D_4\vert} & = \big( 1 + 1 + 1 + 1 + 4 \big) / 8 \\ & = 8 / 8 \\ & = 1 \end{aligned}

At a 2D 62 D_6-orientifold singularity

For G=2D 6G = 2 D_6 the binary dihedral group of order |2D 6|=12{\vert 2 D_6\vert} = 12, the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.2):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11\phantom{-}11-11-11\phantom{-}1
χ V 3=\chi_{V_3} =2\phantom{-}22\phantom{-}21-10\phantom{-}00\phantom{-}01-1
χ V 4=\chi_{V_4} =2\phantom{-}22-22\phantom{-}20\phantom{-}00\phantom{-}02-2
χ V 5=\chi_{V_5} =4\phantom{-}44-42-20\phantom{-}00\phantom{-}02\phantom{-}2

One finds (here) that the general solution to the local/twisted tadpole cancellation condition (2) for G=2D 6G =2 D_6 is

V=n(1V 1+1V 2+2V 3+1V 4+1V 5),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 2 \cdot V_3 + 1 \cdot V_4 + 1 \cdot V_5 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2D 6 =dim(V)/|2D 6| =χ V([e])/|2D 6| =(11+11+22+12+14)/12 =12/12 =1 \begin{aligned} M_{2 D_6} & = dim(V) / {\vert 2 D_6 \vert } \\ & = \chi_V([\langle e\rangle]) / {\vert 2 D_6\vert} \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 2 + 1 \cdot 2 + 1 \cdot 4 \big) / {12} & \\ & = 12 / 12 \\ & = 1 \end{aligned}

At a 2D 82 D_8-orientifold singularity

For G=2D 8G = 2 D_8 the binary dihedral group of order |2D 8|=16{\vert 2 D_8\vert} = 16, the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.3):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right][g 6]\left[\langle g_6\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11\phantom{-}11-11\phantom{-}11-11-1
χ V 3=\chi_{V_3} =1\phantom{-}11\phantom{-}11\phantom{-}11-11-11\phantom{-}11\phantom{-}1
χ V 4=\chi_{V_4} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11-11-11-1
χ V 5=\chi_{V_5} =2\phantom{-}22\phantom{-}22-20\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}0
χ V 6=\chi_{V_6} =8\phantom{-}88-80\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}0

One finds (here), that the general solution to the local/twisted tadpole cancellation condition (2) for G=2D 8G =2 D_8 is

V=n(1V 1+1V 2+1V 3+1V 4+2V 5+1V 6),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 1 \cdot V_3 + 1 \cdot V_4 + 2 \cdot V_5 + 1 \cdot V_6 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2D 8 =dim(V)/|2D 8| =χ V([e])/|2D 8| =(11+11+11+11+22+18)/16 =16/16 =1 \begin{aligned} M_{2 D_8} & = dim(V) / {\vert 2 D_8\vert} \\ & = \chi_V([\langle e\rangle]) / { \vert 2 D_8\vert } \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 2 + 1 \cdot 8 \big) / 16 & \\ & = 16 / 16 \\ & = 1 \end{aligned}

At a 2D 102 D_{10}-orientifold singularity

For G=2D 10G = 2 D_{10} the binary dihedral group of order |2D 10|=20{\vert 2 D_{10}\vert} = 20, the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.4):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right][g 6]\left[\langle g_6\rangle\right][g 7]\left[\langle g_7\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11-11-11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 3=\chi_{V_3} =2\phantom{-}22-20\phantom{-}00\phantom{-}02\phantom{-}22\phantom{-}22-22-2
χ V 4=\chi_{V_4} =4\phantom{-}44\phantom{-}40\phantom{-}00\phantom{-}01-11-11-11-1
χ V 5=\chi_{V_5} =8\phantom{-}88-80\phantom{-}00\phantom{-}02-22-22\phantom{-}22\phantom{-}2

One finds (here) that the general solution to the local/twisted tadpole cancellation condition (2) for G=2D 10G =2 D_{10} is

V=n(1V 1+1V 2+1V 3+2V 4+1V 5),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 1 \cdot V_3 + 2 \cdot V_4 + 1 \cdot V_5 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2D 10 =dim(V)/|2D 10| =χ V([e])/|2D 10| =(11+11+12+24+18)/20 =20/20 =1 \begin{aligned} M_{2 D_{10}} & = dim(V) / {\vert 2 D_{10}\vert} \\ & = \chi_V([\langle e\rangle]) / {\vert 2 D_{10}\vert} \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 2 + 2 \cdot 4 + 1 \cdot 8 \big) / 20 & \\ & = 20 / 20 \\ & = 1 \end{aligned}

At a 2D 122 D_{12}-orientifold singularity

For G=2D 12G = 2 D_{12} the binary dihedral group of order |2D 12|=24{\vert 2 D_{12}\vert} = 24, the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.5):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right][g 6]\left[\langle g_6\rangle\right][g 7]\left[\langle g_7\rangle\right][g 8]\left[\langle g_8\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11\phantom{-}11-11-11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 3=\chi_{V_3} =1\phantom{-}11\phantom{-}11\phantom{-}11-11\phantom{-}11-11\phantom{-}11-11-1
χ V 4=\chi_{V_4} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11-11-11\phantom{-}11-11-1
χ V 5=\chi_{V_5} =2\phantom{-}22\phantom{-}21-10\phantom{-}00\phantom{-}02\phantom{-}21-11-11-1
χ V 6=\chi_{V_6} =2\phantom{-}22\phantom{-}21-10\phantom{-}00\phantom{-}02-21-11\phantom{-}11\phantom{-}1
χ V 7=\chi_{V_7} =4\phantom{-}44-44\phantom{-}40\phantom{-}00\phantom{-}00\phantom{-}04-40\phantom{-}00\phantom{-}0
χ V 8=\chi_{V_8} =8\phantom{-}88-84-40\phantom{-}00\phantom{-}00\phantom{-}04\phantom{-}40\phantom{-}00\phantom{-}0

One sees (here) that the general solution to the local/twisted tadpole cancellation condition (2) for G=2D 12G =2 D_{12} is

V=n(1V 1+1V 2+1V 3+1V 4+2V 5+2V 6+1V 7+1V 8),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 1 \cdot V_3 + 1 \cdot V_4 + 2 \cdot V_5 + 2 \cdot V_6 + 1 \cdot V_7 + 1 \cdot V_8 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2D 12 =dim(V)/|2D 12| =χ V([e])/|2D 12| =(11+11+11+11+22+22+14+18)/24 =24/24 =1 \begin{aligned} M_{2 D_{12}} & = dim(V) / {\vert 2 D_{12}\vert } \\ & = \chi_V([\langle e\rangle]) / {\vert 2 D_{12}\vert } \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 + 1 \cdot 4 + 1 \cdot 8 \big) / 24 & \\ & = 24 / 24 \\ & = 1 \end{aligned}

At a 2D 142 D_{14}-orientifold singularity

For G=2D 14G = 2 D_{14} the binary dihedral group of order |2D 14|=28{\vert 2 D_{14}\vert} = 28, the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.6):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right][g 6]\left[\langle g_6\rangle\right][g 7]\left[\langle g_7\rangle\right][g 8]\left[\langle g_8\rangle\right][g 9]\left[\langle g_9\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11-11-11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 3=\chi_{V_3} =2\phantom{-}22-20\phantom{-}00\phantom{-}02\phantom{-}22\phantom{-}22\phantom{-}22-22-22-2
χ V 4=\chi_{V_4} =6\phantom{-}66\phantom{-}60\phantom{-}00\phantom{-}01-11-11-11-11-11-1
χ V 5=\chi_{V_5} =112\phantom{-1}\mathllap{12}112\phantom{-1}\mathllap{-12}0\phantom{-}00\phantom{-}02-22-22-22\phantom{-}22\phantom{-}22\phantom{-}2

One sees by immediate inspection, that the general solution to the local/twisted tadpole cancellation condition (2) for G=2D 14G =2 D_{14} is

V=n(1V 1+1V 2+1V 3+2V 4+1V 5),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 1 \cdot V_3 + 2 \cdot V_4 + 1 \cdot V_5 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2D 14 =dim(V)/|2D 14| =χ V([e])/|2D 14| =(11+11+12+26+112)/28 =28/28 =1 \begin{aligned} M_{2 D_{14}} & = dim(V) / {\vert 2 D_{14}\vert } \\ & = \chi_V([\langle e\rangle]) / {\vert 2 D_{14}\vert } \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 2 + 2 \cdot 6 + 1 \cdot 12 \big) / 28 \\ & = 28 /28 \\ & = 1 \end{aligned}

At a 2D 162 D_{16}-orientifold singularity

For G=2D 16G = 2 D_{16} the binary dihedral group of order |2D 16|=32{\vert 2 D_{16}\vert} = 32, the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.7):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right][g 6]\left[\langle g_6\rangle\right][g 7]\left[\langle g_7\rangle\right][g 8]\left[\langle g_8\rangle\right][g 9]\left[\langle g_9\rangle\right][g 9]\left[\langle g_9\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11\phantom{-}11-11\phantom{-}11\phantom{-}11\phantom{-}11-11-11-11-1
χ V 3=\chi_{V_3} =1\phantom{-}11\phantom{-}11-11-11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 4=\chi_{V_4} =1\phantom{-}11\phantom{-}11-11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11-11-11-11-1
χ V 5=\chi_{V_5} =2\phantom{-}22\phantom{-}20\phantom{-}00\phantom{-}02\phantom{-}22-22-20\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}0
χ V 6=\chi_{V_6} =4\phantom{-}44\phantom{-}40\phantom{-}00\phantom{-}04-40\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}0
χ V 7=\chi_{V_7} =116\phantom{-1}\mathllap{16}116\phantom{-1}\mathllap{-16}0\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}00\phantom{-}0

One sees by immediate inspection, that the general solution to the local/twisted tadpole cancellation condition (2) for G=2D 16G =2 D_{16} is

V=n(1V 1+1V 2+1V 3+1V 4+2V 5+2V 6+1V 7),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 1 \cdot V_3 + 1 \cdot V_4 + 2 \cdot V_5 + 2 \cdot V_6 + 1 \cdot V_7 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2D 16 =dim(V)/|2D 16| =χ V([e])/|2D 16| =(11+11+11+11+22+24+116)/32 =32/32 =1 \begin{aligned} M_{2 D_{16}} & = dim(V) / {\vert 2 D_{16}\vert} \\ & = \chi_V([\langle e\rangle]) / {\vert 2 D_{16}\vert} \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 4 + 1 \cdot 16 \big) /32 \\ & = 32 / 32 \\ & = 1 \end{aligned}

At a 2T2 T-orientifold singularity

For G=2TG = 2 T the binary tetrahedral group (whose order is |2T|=24{\vert 2T \vert} =24), the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.8):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right][g 6]\left[\langle g_6\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =2\phantom{-}22\phantom{-}21-11-12\phantom{-}21-11-1
χ V 3=\chi_{V_3} =3\phantom{-}33\phantom{-}30\phantom{-}00\phantom{-}01-10\phantom{-}00\phantom{-}0
χ V 4=\chi_{V_4} =4\phantom{-}44-41\phantom{-}11\phantom{-}10\phantom{-}01-11-1
χ V 5=\chi_{V_5} =4\phantom{-}44-42-22-20\phantom{-}02\phantom{-}22\phantom{-}2

One finds (here) that the general solution to the local/twisted tadpole cancellation condition (2) for G=2TG = 2T is

V=n(1V 1+1V 2+3V 3+2V 4+1V 5),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 3 \cdot V_3 + 2 \cdot V_4 + 1 \cdot V_5 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2I =dim(V)/|2I| =χ V([e])/|2I| =(11+12+33+24+14)/24 =24/24 =1 \begin{aligned} M_{2I} & = dim(V) / {\vert 2 I \vert } \\ & = \chi_V([\langle e\rangle]) / {\vert 2 I \vert } \\ & = \big( 1 \cdot 1 + 1 \cdot 2 + 3 \cdot 3 + 2 \cdot 4 + 1 \cdot 4 \big) / 24 & \\ & = 24 / 24 \\ & = 1 \end{aligned}

At a 2O2 O-orientifold singularity

For G=2OG = 2 O the binary octahedral group (whose order is |2O|=48{\vert 2O \vert} = 48), the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.9):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right][g 6]\left[\langle g_6\rangle\right][g 7]\left[\langle g_7\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11-11\phantom{-}11-11-1
χ V 3=\chi_{V_3} =2\phantom{-}22\phantom{-}21-12\phantom{-}20\phantom{-}01-10\phantom{-}00\phantom{-}0
χ V 4=\chi_{V_4} =3\phantom{-}33\phantom{-}30\phantom{-}01-11\phantom{-}10\phantom{-}01-11-1
χ V 5=\chi_{V_5} =3\phantom{-}33\phantom{-}30\phantom{-}01-11-10\phantom{-}01\phantom{-}11\phantom{-}1
χ V 6=\chi_{V_6} =8\phantom{-}88-82\phantom{-}20\phantom{-}00\phantom{-}02-20\phantom{-}00\phantom{-}0
χ V 7=\chi_{V_7} =8\phantom{-}88-84-40\phantom{-}00\phantom{-}04\phantom{-}40\phantom{-}00\phantom{-}0

One finds (here) that the general solution to the local/twisted tadpole cancellation condition (2) for G=2OG = 2O is

V=n(1V 1+1V 2+2V 3+3V 4+3V 5+2V 6+1V 7),AAAn V \;=\; n \Big( 1 \cdot V_1 + 1 \cdot V_2 + 2 \cdot V_3 + 3 \cdot V_4 + 3 \cdot V_5 + 2 \cdot V_6 + 1 \cdot V_7 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2O =dim(V)/|2O| =χ V([e])/|2O| =(11+11+22+33+33+28+18)/48 =48/48 =1 \begin{aligned} M_{2O} & = dim(V) / {\vert 2 O \vert } \\ & = \chi_V([\langle e\rangle]) / {\vert 2 O \vert } \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 3 \cdot 3 + 2 \cdot 8 + 1 \cdot 8 \big) / 48 & \\ & = 48 / 48 \\ & = 1 \end{aligned}

At a 2I2 I-orientifold singularity

For G=2IG = 2 I the binary icosahedral group (whose order is |2I|=120{\vert 2I \vert} = 120), the characters/D-brane charges of the elementary virtual permutation representations/fractional D-branes are (BSS 18, 4.10):

[H]=[H] = [e]\left[\langle e\rangle\right][g 1]\left[\langle g_1\rangle\right][g 2]\left[\langle g_2\rangle\right][g 3]\left[\langle g_3\rangle\right][g 4]\left[\langle g_4\rangle\right][g 5]\left[\langle g_5\rangle\right][g 6]\left[\langle g_6\rangle\right][g 7]\left[\langle g_7\rangle\right][g 8]\left[\langle g_8\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =4\phantom{-}44\phantom{-}41\phantom{-}10\phantom{-}01-11-11\phantom{-}11-11-1
χ V 3=\chi_{V_3} =5\phantom{-}55\phantom{-}51-11\phantom{-}10\phantom{-}00\phantom{-}01-10\phantom{-}00\phantom{-}0
χ V 4=\chi_{V_4} =6\phantom{-}66\phantom{-}60\phantom{-}02-21\phantom{-}11\phantom{-}10\phantom{-}01\phantom{-}11\phantom{-}1
χ V 5=\chi_{V_5} =112\phantom{-1}\mathllap{12}112\phantom{-1}\mathllap{-12}0\phantom{-}00\phantom{-}02\phantom{-}22\phantom{-}20\phantom{-}02-22-2
χ V 6=\chi_{V_6} =8\phantom{-}88-82\phantom{-}20\phantom{-}02-22-22-22\phantom{-}22\phantom{-}2
χ V 7=\chi_{V_7} =8\phantom{-}88-84-40\phantom{-}02-22-24\phantom{-}42\phantom{-}22\phantom{-}2

One finds (here) that the general solution to the local/twisted tadpole cancellation condition (2) for G=2IG = 2I is

V=n(1V 1+4V 2+5V 3+3V 4+3V 5+2V 6+1V 7),AAAn V \;=\; n \Big( 1 \cdot V_1 + 4 \cdot V_2 + 5 \cdot V_3 + 3 \cdot V_4 + 3 \cdot V_5 + 2 \cdot V_6 + 1 \cdot V_7 \Big) \,, \phantom{AAA} n \in \mathbb{Z}

whose minimal positive mass (net brane number) is

M 2I =dim(V)/|2I| =χ V([e])/|2I| =(11+44+55+36+312+28+18)/120 =120/120 =1 \begin{aligned} M_{2I} & = dim(V) / {\vert 2I \vert } \\ & = \chi_V([\langle e\rangle]) / {\vert 2I \vert } \\ & = \big( 1 \cdot 1 + 4 \cdot 4 + 5 \cdot 5 + 3 \cdot 6 + 3 \cdot 12 + 2 \cdot 8 + 1 \cdot 8 \big) / 120 \\ & = 120 / 120 \\ & = 1 \end{aligned}

References

General

The issue was first highlighted in

  • Augusto Sagnotti, Open strings and their symmetry groups in G. Mack et. al. (eds.) Cargese ’87, “Non-perturbative Quantum Field Theory,” (Pergamon Press, 1988) p. 521 (arXiv:hep-th/0208020)

The argument is recalled in

Details are in

Textbook accounts include

Quick illustrations include:

Critical outlook in

The above discussion follows and character tables for virtual permutation representations above are taken from

See also

In view of consistency of flux compactifications:

  • Philip Betzler, Erik Plauschinn, Type IIB flux vacua and tadpole cancellation (arXiv:1905.08823)

For the topological string:

Examples and Models

Specifically K3 orientifolds (𝕋 4/G ADE\mathbb{T}^4/G_{ADE}) in type IIB string theory, hence for D9-branes and D5-branes:

Specifically K3 orientifolds (𝕋 4/G ADE\mathbb{T}^4/G_{ADE}) in type IIA string theory, hence for D8-branes and D4-branes:

The N\mathbb{Z}_N action with even NN contains an order 2 element [...][ ...] Then there will be D8-branes in the type IIA D4-brane theory. Since the concept of intersecting D-branesinvolves use of the same dimensional D-branes, we restrict ourselves to the case that the order NN of N\mathbb{Z}_N is odd. (p. 4)

The Witten-Sakai-Sugimoto model on D4-D8-brane bound states for QCD with orthogonal gauge groups on O-planes:

Specifically D5 brane models T-dual to D6/D8 models:

Specifically for D6-branes:

  • S. Ishihara, H. Kataoka, Hikaru Sato, D=4D=4, N=1N=1, Type IIA Orientifolds, Phys. Rev. D60 (1999) 126005 (arXiv:hep-th/9908017)

  • Mirjam Cvetic, Paul Langacker, Tianjun Li, Tao Liu, D6-brane Splitting on Type IIA Orientifolds, Nucl. Phys. B709:241-266, 2005 (arXiv:hep-th/0407178)

Specifically for D3-branes/D7-branes:

Various:

  • Dieter Lüst, S. Reffert, E. Scheidegger, S. Stieberger, Resolved Toroidal Orbifolds and their Orientifolds, Adv.Theor.Math.Phys.12:67-183, 2008 (arXiv:hep-th/0609014)

Last revised on January 4, 2024 at 08:57:09. See the history of this page for a list of all contributions to it.