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RR-field tadpole cancellation

Contents

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, Simon-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.

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

For fractional D-branes at orbifold singularities

More details are understood in the special case of fractional D-branes 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.

Detailed review of this is in Marchesano 03, Section 4, based on ABIU 99, Honecker 02.

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 G DEG_{DE} (by this formula), so that 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 1). 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_1\right) \,.

Now in terms of this, the tadpole cancellation condition for fractional D-branes 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

(Marchesano 03 (4.9))

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 homogeneous 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.


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

Now the tadpole cancellation condition is that (all representations are real and) this O-plane charge is cancelled against the D-brane charge, an affine version of the previous condition:

(3)χ V(g k1)χ O(g k1)=0. \chi_{V}\left(g_{k \geq 1}\right) - \chi_{O}\left( g_{k \geq 1} \right) \;=\; 0 \,.

(Marchesano 03 (4.15), following Honecker 02 (28))

The general solution to (3) is of course the sum OO with any solution to the homogeneous equation (2). By Prop. the latter are exactly the multiples nk[G/1]n \cdot k[G/1] of the regular representation, so that the general solution to the tadpole cancellation condition (3) for fractional D-branes is

V=O+nk[G/1]. V \;=\; O + 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 the general solution to the tadpole cancellation condition then is the affine version of (2), being +2 p41+ 2^{p-4} \cdot \mathbf{1} plus, by Prop. , any numbers of the regular representation:

V=2 p4n O1+nk[G/1]. V \;=\; 2^{p-4} n_O \cdot \mathbf{1} \;+\; n \cdot k[G/1] \,.

Examples

We discuss more explicitly the solutions to the homogeneous 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 homogenous 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 (homogenous part of the) 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 3\mathbb{Z}_3-orientifold singularity

For G= 3G = \mathbb{Z}_3 the cyclic group of order | 3|=3{\vert \mathbb{Z}_3\vert} = 3, 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 1]\left[\langle g_1\rangle\right]
χ V 1=\chi_{V_1} =1\phantom{-}11\phantom{-}11\phantom{-}1
χ V 2=\chi_{V_2} =1\phantom{-}1exp(132πi)\exp\left( \tfrac{1}{3} 2 \pi i \right)exp(232πi)\exp\left( \tfrac{2}{3} 2 \pi i \right)
χ V 3=\chi_{V_3} =1\phantom{-}1exp(232πi)\exp\left( \tfrac{2}{3} 2 \pi i \right)exp(132πi)\exp\left( \tfrac{1}{3} 2 \pi i \right)

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

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

whose minimal positive mass (net brane number) is

M 3 =dim(V)/| 3| =χ V([e])/| 3| =(11+11+11)/3 =3/3 =1 \begin{aligned} M_{\mathbb{Z}_3} & = dim(V) / {\vert \mathbb{Z}_3 \vert } \\ & = \chi_V([\langle e\rangle]) / {\vert \mathbb{Z}_3 \vert} \\ & = \big( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 \big) / {3} & \\ & = 3 / 3 \\ & = 1 \end{aligned}

This pattern immediately continues for all cyclic groups n\mathbb{Z}_n.

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 (homogenous part of the) 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 (homogenous part of the) 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 (homogenous part of the) 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 (homogenous part of the) 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 (homogenous part of the) 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 (homogenous part of the) 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 (homogenous part of the) 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 (homogenous part of the) 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 (homogenous part of the) 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 (homogenous part of the) 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

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

See also

  • G. Aldazabal, D. Badagnani, Luis Ibáñez, Angel Uranga, Tadpole versus anomaly cancellation in D=4,6D=4,6 compact IIB orientifolds, JHEP 9906:031, 1999 (arXiv:hep-th/9904071)

  • Angel Uranga, D-brane probes, RR tadpole cancellation and K-theory charge, Nucl.Phys.B598:225-246, 2001 (arXiv:hep-th/0011048)

  • Gabriele Honecker, Intersecting brane world models from D8-branes on (T 2×T 4 3)/Ω 1(T^2 \times T^4\mathbb{Z}_3)/\Omega\mathcal{R}_1 type IIA orientifolds, JHEP 0201 (2002) 025 (arXiv:hep-th/0201037)

  • Maria E. Angulo, David Bailin, Huan-Xiong Yang, Tadpole and Anomaly Cancellation Conditions in D-brane Orbifold Models, Int.J.Mod.Phys.A18:3637-3694, 2003 (arXiv:hep-th/0210150)

  • Fernando Marchesano, section 4 of Intersecting D-brane Models (arXiv:hep-th/0307252)

  • Fernando Marchesano, Gary Shiu, Building MSSM Flux Vacua, JHEP0411:041, 2004 (arXiv:hep-th/0409132)

  • C.-M. Chen, G. V. Kraniotis, V. E. Mayes, D. V. Nanopoulos, J. W. Walker, A K-theory Anomaly Free Supersymmetric Flipped SU(5) Model from Intersecting Branes, Phys.Lett. B625 (2005) 96-105 (arXiv:hep-th/0507232)

  • Inaki Garcia-Etxebarria, Angel Uranga, From F/M-theory to K-theory and back, JHEP 0602:008, 2006 (arXiv:hep-th/0510073)

  • John Maiden, Gary Shiu, Bogdan Stefanski, D-brane Spectrum and K-theory Constraints of D=4, N=1 Orientifolds, JHEP0604:052,2006 (arXiv:hep-th/0602038)

  • Tetsuji Kimura, Mitsuhisa Ohta, Kei-Jiro Takahashi, Type IIA orientifolds and orbifolds on non-factorizable tori, Nucl.Phys.B798:89-123, 2008 (arXiv:0712.2281)

For the topological string:

  • Johannes Walcher, Evidence for Tadpole Cancellation in the Topological String (arXiv:0712.2775)

The character tables for virtual permutation representations above are taken from

Last revised on March 19, 2019 at 01:07:17. See the history of this page for a list of all contributions to it.