Five dimensional gauge theories NCTSneo.phys.wits.ac.za/public/workshop_10/hanany.pdf · 3d N=4 Coulomb branch Global symmetry • An imbalance of a node is the number of ﬂavors

Five dimensional gauge theories and Higgs

branch at infinite coupling Amihay Hanany

Wits Rural Facility 2017

Five dimensional gauge theory

• Existence of SCFTs in 4 dimensions and higher

• Brane systems

• D4 - D8

• Webs of five branes

• M theory of CY singularity

• F theory with 8 supercharges

Gauge theory in 5d SQCD

• Today focus on gauge theories with a single gauge group

• Relatively few theories which have a UV fixed point in 5d

• Number of flavors should be small enough

• Conditions similar to the condition for asymptotic freedom in 4d, but not the same

CS coupling

• The CS level k can be turned on, but fewer number of flavors are needed to have a 5d UV SCFT

SU(n) gauge theory

• Further restrict to SU(n) with N flavors and CS level k

• The level k is 1/2 + integer for N odd

• Integer for N even

Coulomb branch parameters

• Inverse gauge coupling 1/g^2

• Masses for flavors

• Both have dimension 1

• Contribute to the central charge formula for BPS states

SU(n) with 2n+4 flavors and 0 CS level

• This theory has a 6d UV fixed point

• Remove 1 flavor and add 1/2 to CS level by integrating out a massive flavor

• Get a condition for the existence of a 5d UV fixed point

• N + 2k <= 2n + 4

• Excluding the case N=2n+4, k=0 which has 6d UV SCFT

Classical flavor symmetry

• U(N) x U(1)

• Two U(1) global symmetries

• Baryon number and the topological U(1) symmetry which counts instanton number

• *tr(F^F) is a conserved current due to Bianchi identity

• For n=2 the global symmetry changes to SO(2N) x U(1)

Gauge instanton in 5d

• This is a particle with a contribution to the mass that scales like |I|/g^2, with instanton charge I

• At infinite coupling, at the origin of the coulomb branch, becomes massless

• Increases the global symmetry

• Generates many new flat directions along the Higgs branch

Get a quantitative description of these physical phenomena

Higgs branch at infinite coupling

• Given a theory with n, N, k set all parameters to 0 and ask what is the Higgs branch at infinite coupling

This problem is at least 20 years old

We present the solution

Higgs branch hyper Kähler cone

• Recall all parameters are set to zero - have a cone

• SU(2) R symmetry

• Chiral (coordinate) ring, finitely generated

• Operators (generators) characterized by representation under SU(2) R and representations under the global symmetry

HyperKähler cones Global symmetry

• Theorem - Kostant Brylinski ?

• All operators with spin 1 under SU(2) R transform in the adjoint representation of the global symmetry F

• Namikawa: if all generators of the chiral ring have spin 1 under SU(2) R, then the moduli space is a clusure of a nilpotent orbit of F

Classical Higgs branch

• Mesons — spin 1 under SU(2) R

• Adjoint representation of U(N)

• Baryons — spin n/2 under SU(2) R

• n-th antisymmetric of SU(N)

• Gaugino bilinear — spin 1 under SU(2) R, adj of U(1)_I

Instanton operators

• Assume there exists chiral operators in the chiral ring which transform under the U(1) instanton, SU(2) R, and classical global symmetries

• Non perturbative operators which do not show in the classical theory and become crucially relevant to the UV SCFT

• Analogs of ‘t Hooft monopole operators in 3d

Instantons and global symmetry

• Instanton operators which carry spin 1 under SU(2) R enhance the global symmetry

• All such instanton operators are highly restricted as they must complete the existing classical global symmetry C to form the adjoint representation of a bigger global symmetry F which contains the classical symmetry as a subgroup

• In all cases we know, C is a Levy subgroup of F - the ranks are equal

Instantons and new flat directions

• Instanton operators which carry spin n/2 under SU(2) R combine with baryons to form very large representations of the global symmetry (spinor or high antisymmetric)

Global symmetry at infinite coupling

• E8: N = 2n + 3, k =1/2: SO(4n + 8)

• E7: N = 2n + 2, k = 1: SO(4n + 4) x SU(2)

• E7: N = 2n + 2, k = 0: SU(2n + 4)

• E6: N = 2n + 1, k = 3/2: SO(4n + 2) x U(1)

• E6: N = 2n + 1, k = 1/2: SO(2n + 2) x SU(2)

• E5: N = 2n, k = 2: SO(4n) x U(1)

More global symmetry

• E5: N = 2n, k = 1: SU(2n + 1) x U(1)

• E5: N = 2n, k = 0: SU(2n) x SU(2) x SU(2)

• E4: N = 2n - 1, k =5/2: SO(4n - 2) x U(1)

• E4: N = 2n - 1, k =3/2: SU(2n) x U(1)

• E4: N = 2n - 1, k =1/2: SU(2n - 1) x SU(2) x U(1)

n=2 Maximal subgroups

• These are maximal subgroups of E_{N+1}

• SO(16) in E8, etc.

• This is an important clue in the construction below

Instantons and Higgs branch

• If the instanton has a spin under SU(2) R which is higher than 1, the Higgs branch becomes larger, by a significant amount

• Roughly double the dimension

Instantons transform under SU(2) R

• To compute this quantity, we need to evaluate the energy of a vacuum state in the topological sector of 1 instanton

• For a gauge group G this is h/2 where h is the dual Coxeter number of G

• Closest analog to the 4d beta function for a SYM theory

Instanton zero modes• Each flavor of quarks contributes a fermionic zero mode

• Easy to see in the D4-D8 system by looking at D0-D8

• Or D1-D7 for a five brane web

• As a result, the instanton transforms under the global symmetry as

• spinor representation if it is a rotation group

• Antisymmetric representation if it is a unitary group

Higgs branch at infinite coupling

• Given this data, how to proceed?

• Two ways of constructing hyperKahler cones

• HyperKahler quotient (Higgs branch) F & D terms

• 3d N=4 Coulomb branch

• Any attempt to solve using the first method failed

Construction using 3d techniques

• Assume

• The answer is given by a Coulomb branch of 3d N=4 theory

• Use the global symmetry as an input

• The instantons transform in spin n/2 of SU(2) R and spinor or maximally antisymmetric representation of the global symmetry - computed from n=2

3d N=4 Coulomb branch Global symmetry

• An imbalance of a node is the number of flavors minus twice the number of colors

• A node is balanced if the imbalance is 0

• The subset of balanced nodes forms the Dynkin diagram of the global symmetry. Number of U(1) is number of unbalanced nodes minus 1

• If the imbalance of a node is n-2 then there is an operator in the chiral ring with spin n/2 under SU(2) R and representation under the global symmetry given by the node is attached to

E8 sequence

2n 2 SO(4n)⇥ U(1) (7.1)

1 SU(2n+ 1)⇥ U(1) (8.1)

0 SU(2n)⇥ SU(2)⇥ SU(2) (9.2)

2n� 1 5/2 SO(4n� 2)⇥ U(1) (10.1)

3/2 SU(2n)⇥ U(1) (11.1)

1/2 SU(2n� 1)⇥ SU(2)⇥ U(1) (12.1)

Note that the flavour symmetries at infinite coupling for Nf = 2n + 3, 2n +

2, 2n+ 1, 2n were discussed in Table 1 on Page 15 of [1].

2 SU(n)1/2 with Nf = 2n+ 3 flavours

This is the E8 sequence. We choose the CS level k = 12 . The Chern Simons term

vanishes for n = 2.

2.1 n = 2

The theory is SU(2) with Nf = 7 flavours. The global symmetry on the Higgs branch

at infinite coupling is E8. More precisely the Higgs branch at infinite coupling is given

by the Coulomb branch of the 3d quiver:

H1

✓�

SU(2)� ⌅

SO(14)

◆= C3d

0

@�1� �

2� �

3� �

4� �

5�

� 3|�6� �

4� �

2

1

A (2.1)

i.e. the minimal nilpotent orbit of E8.

2.2 n > 2

The theory is SU(n) with Nf = 2n+3 flavours and k = 12 . There is a classical global

symmetry U(2n+ 3)⇥ U(1)I ⇠= SU(2n+ 3)⇥ U(1)B ⇥ U(1)I with rank r = 2n+ 4.

The Higgs branch at infinite coupling is given by the Coulomb branch of the 3d quiver:

H1

�

SU(n) 12

� ⌅2n+3

!= C3d

0

@�1� · · ·� �

2n+1�

� n+1|�

2n+2� �

n+2� •

2

1

A . (2.2)

All the nodes are balanced except for the last one on the right. The global symmetry

can be read o↵ from the quiver, after removal on the unbalanced node: it results on

a flavour symmetry SO(4n+ 8) at infinite coupling, which has the same rank as the

finite coupling global symmetry.

Note that for n = 2, we get a larger global symmetry than this prescription,

since E8 � SO(16).

– 2 –

2n 2 SO(4n)⇥ U(1) (7.1)

1 SU(2n+ 1)⇥ U(1) (8.1)

0 SU(2n)⇥ SU(2)⇥ SU(2) (9.2)

2n� 1 5/2 SO(4n� 2)⇥ U(1) (10.1)

3/2 SU(2n)⇥ U(1) (11.1)

1/2 SU(2n� 1)⇥ SU(2)⇥ U(1) (12.1)

Note that the flavour symmetries at infinite coupling for Nf = 2n + 3, 2n +

2, 2n+ 1, 2n were discussed in Table 1 on Page 15 of [1].


This is the E8 sequence. We choose the CS level k = 12 . The Chern Simons term

vanishes for n = 2.

2.1 n = 2

The theory is SU(2) with Nf = 7 flavours. The global symmetry on the Higgs branch

at infinite coupling is E8. More precisely the Higgs branch at infinite coupling is given


H1

✓�

SU(2)� ⌅

SO(14)

◆= C3d

0

@�1� �

2� �

3� �

4� �

5�

� 3|�6� �

4� �

2

1

A (2.1)

i.e. the minimal nilpotent orbit of E8.

2.2 n > 2

The theory is SU(n) with Nf = 2n+3 flavours and k = 12 . There is a classical global

symmetry U(2n+ 3)⇥ U(1)I ⇠= SU(2n+ 3)⇥ U(1)B ⇥ U(1)I with rank r = 2n+ 4.


H1

�

SU(n) 12

� ⌅2n+3

!= C3d

0

@�1� · · ·� �

2n+1�

� n+1|�

2n+2� �

n+2� •

2

1

A . (2.2)

All the nodes are balanced except for the last one on the right. The global symmetry


a flavour symmetry SO(4n+ 8) at infinite coupling, which has the same rank as the


Note that for n = 2, we get a larger global symmetry than this prescription,

since E8 � SO(16).

– 2 –

flavors of the quiver to form a vector f and the ranks to form a vector r, Set the Car-

tan matrix to be A and the balance condition is given by Ar = f . Given two roots

↵, � with monopole operators Vm1 and Vm2 , respectively, with spin 1 under SU(2)R,

it is straightforward to show that if ↵ + � is a root, then Vm1+m2 also has spin 1

under SU(2)R. There are exceptions to this conjecture with low rank non simply

laced factors in the global symmetry, where this criterion gives rise to a simply laced

maximal sub algebra of the flavor symmetry, but these cases do not show up in the

study of this paper.

We note that for the present case there are no U(1) factors in the global sym-

metry, leading to 1 unbalanced node, and this node is connected to the spinor node,

in order to match to the n = 2 case. This fixes the family uniquely and is presented

below.

2.2 E8 Sequence, n > 2

The theory is SU(n) with Nf = 2n + 3 flavors and k = 12 . The Higgs branch at

infinite coupling is given by the Coulomb branch of the 3d quiver:

H1

�

SU(n)± 12

� ⌅U(2n+3)

!= C3d

0

@�1� · · ·� �

2n+1�

� n+1|�

2n+2� �

n+2� •

2

1

A . (2.2)

From which the dimension can be derived simply,

dimH1 = 2n2 + 7n+ 7 (2.3)

This increases the finite coupling Higgs branch which has dimension (2n + 3)n �(n2 � 1) = n2 + 3n + 1 by adding n2 + 4n + 6 new flat directions. All the nodes

are balanced except for the last one on the right. Henceforth unbalanced nodes

are depicted in blue. There is a classical global symmetry U(2n + 3) ⇥ U(1)I ⇠=SU(2n+3)⇥U(1)B ⇥U(1)I with rank r = 2n+4. The global symmetry conjecture

implies that the flavor symmetry at infinite coupling is SO(4n + 8), which has the

same rank as the finite coupling global symmetry.

Indeed, for n = 2, by our construction, we verify that there is a larger set of

balanced nodes, resulting in a global symmetry E8 � SO(16).

Let us now concentrate on the last node on the right hand side. it has an

imbalance of Nf � 2Nc = n � 2. Consequently, the lowest SU(2)R spin for a 3d

monopole operator with non-zero fluxes associated to this last gauge node is n/2.

Such a monopole transforms in the spinor representation of SO(4n + 8). Thus the

chiral ring at infinite coupling is generated by an SO(4n+ 8) adjoint rep at SU(2)Rspin-1 and an SO(4n+ 8) spinor rep at SU(2)R spin-(n/2).

The resulting highest weight generating function (HWG) is

PE

"n+1X

i=1

µ2it2i + t4 + µ2n+4(t

n + tn+2)

#. (2.4)

– 9 –

Generators of the chiral ring

• The global symmetry of this Coulomb branch SO(4n + 8)

• An adjoint valued operator at spin 1 of SU(2) R

• A spinor valued operator at spin n/2 of SU(2) R

• Mesons, baryons, gaugino bilinear, instantons

• satisfying relations that are dictated by the quiver and the symmetries

E7 sequences

4.1 n = 2

The gauge theory is SU(2) with Nf = 6 flavours and it flows from a SCFT at infinite

coupling which displays symmetry enhancement. The Higgs branch is indeed given


H1

✓�

SU(2)� ⌅

SO(12)

◆= C3d

0

@�1� �

2� �

3�•2|�4� �

3� �

2� �

1

1

A (4.1)

which has global symmetry E7. Such a global symmetry can be evinced by recalling

that the above is precisely the a�ne Dynkin diagram for E7.

4.2 n > 2

The theory is SU(n)0 with Nf = 2n+2 flavours. There is a classical global symmetry

U(2n+ 2)⇥U(1)I ⇠= SU(2n+ 2)⇥U(1)B ⇥U(1)I with rank r = 2n+ 3. The Higgs

branch at infinite coupling is given by the Coulomb branch of the 3d quiver:

H1

✓�

SU(n)0� ⌅

2n+2

◆= C3d

0

@�1� · · ·� �

n+1�

•2|�

n+2� �

n+1� · · ·� �

1

1

A . (4.2)

All the nodes are balanced except for the node at the top. The global symmetry


a flavour symmetry SU(2n+4) at infinite coupling, which has rank 2n+3, the same

as the finite coupling global symmetry.

For n = 2 there is again a further enhancement, since the infinite coupling global

symmetry is E7 � SU(8).

The top node has again an imbalance of Nf � 2Nc = n� 2. Analogously to the

previous case, there is a monopole operator at SU(2)R spin-n/2. Here the unbalanced

node is connected to the (n+2)th node, hence the monopole transforms in the (n+2)th

antisymmetric representation of SU(2n+ 4).

Thus the chiral ring at infinite coupling is generated by two SU(2n+4) reps: the

adjoint at SU(2)R spin-1 and the (n+2)th antisymmetric rep at SU(2)R spin-(n/2).



H1

�

SU(n) 32

� ⌅2n+1

!= C3d

0

BBBB@�1� · · ·� �

2n�2�

•1|� n|�

2n�1� �

n� •

1

1

CCCCA. (5.1)

– 4 –

Let us now concentrate on the last node on the right hand side. it is imbalanced

by an amount Nf � 2Nc = n� 2. The lowest SU(2)R spin for a monopole with non-

zero fluxes associated to this last gauge node is n/2. Such a monopole transforms in

the spinor representation of SO(4n+ 8).

[NM: Talk about the generators of the infinite coupling Higgs branch

first. After that, we do the representation decomposition.]

Thus the chiral ring at infinite coupling is generated by an SO(4n+8) adjoint rep

at SU(2)R spin-1 and an SO(4n+8) spinor rep at SU(2)R spin-(n/2). Decomposing

these into representations of U(2n + 3) ⇥ U(1), we obtain: mesons, the gaugino

bilinear, instantons, baryons and baryonic instantons

[NM: To be edited:] with transformations properties as tabulated in Table 2.

The instantons carrying spin-1 lead to the flavour symmetry enhancement whilst the

mesons baryons baryonic instantons gaugino bilinear instantons

SU(2)R spin-1 n/2 n/2 1 1, n/2

SO(4n+ 8) adjoint antisymm rep spinor trivial spinor rep

Table 2.

instantons carrying spin-n/2 constitute some exotic directions of the Higgs branch.

3 SU(n)1 with Nf = 2n+ 2 flavours

There is a classical global symmetry U(2n+2)⇥U(1)I ⇠= SU(2n+2)⇥U(1)B⇥U(1)Iwith rank r = 2n+3. The Higgs branch at infinite coupling is given by the Coulomb

branch of the 3d quiver:

H1

✓�

SU(n)1� ⌅

2n+2

◆= C3d

0

@�1� · · ·� �

2n�1�

� n|�2n

� �n+1

� •2� �

1

1

A . (3.1)

All the nodes are balanced except for •2. The global symmetry can be read o↵ from

the quiver, after removal on the unbalanced node: it results on a flavour symmetry

SO(4n + 4) ⇥ SU(2) at infinite coupling, which has rank 2n + 3, the same as the


4 SU(n)0 with Nf = 2n+ 2 flavours

This corresponds to the E7 sequence.

– 3 –

4.1 n = 2




H1

✓�

SU(2)� ⌅

SO(12)

◆= C3d

0

@�1� �

2� �

3�•2|�4� �

3� �

2� �

1

1

A (4.1)



4.2 n > 2




H1

✓�

SU(n)0� ⌅

2n+2

◆= C3d

0

@�1� · · ·� �

n+1�

•2|�

n+2� �

n+1� · · ·� �

1

1

A . (4.2)















H1

�

SU(n) 32

� ⌅2n+1

!= C3d

0

BBBB@�1� · · ·� �

2n�2�

•1|� n|�

2n�1� �

n� •

1

1

CCCCA. (5.1)

– 4 –

E6 sequences

The global symmetry at infinite coupling can be read o↵ from the quiver, after

removing the blue nodes: this operation results on a flavour symmetry SO(4n+2)⇥U(1), which has rank 2n+2, preserving the rank of the symmetry at finite coupling.


This is the E6 sequence.

6.1 n = 2

The gauge theory is SU(2) with Nf = 5 flavours. The Higgs branch at infinite cou-

pling is given by the Coulomb branch of the 3d quiver:

H1

✓�

SU(2)� ⌅

SO(10)

◆= C3d

0

BBBB@�1� �

2�

� 1|� 2|�3� �

2� �

1

1

CCCCA(6.1)

which is the a�ne Dynkin diagram for E6. This is precisely the global symmetry of

the Higgs branch at infinite coupling.

6.2 n > 2

The theory is SU(n)1/2 with Nf = 2n + 1 flavours. There is a classical global sym-

metry U(2n+ 1)⇥U(1)I ⇠= SU(2n+ 1)⇥U(1)B ⇥U(1)I with rank r = 2n+ 2. The

Higgs branch at infinite coupling is given by the Coulomb branch of the 3d quiver:

H1

�

SU(n) 12

� ⌅2n+1

!= C3d

0

BBBB@�1� · · ·�

� 1|•2|�

n+1� · · ·� �

1

1

CCCCA. (6.2)


cutting the vertical leg: this operation results on a flavour symmetry SU(2)⇥SU(2n+

2), which has rank 2n+ 2, preserving the rank of the symmetry at finite coupling.

There is an imbalance for the lowest of the two vertical nodes. Again it is given

by Nf � 2Nc = n � 2. The monopole operators carrying the smallest flux under

this node carry spin-(n/2) and transform in the (spin-1/2, (n + 1)-antisymm) of

SU(2)⇥ SU(2n+ 2).

As it is by now clear, the n = 2 case is special. Spin-n/2 is in this case spin-1,

i.e there are extra operators at spin-1 and thus the enhancement of the symmetry is

larger. Indeed E6 � SU(2)⇥ SU(6)

– 5 –


removing the blue nodes: this operation results on a flavour symmetry SO(4n+2)⇥U(1), which has rank 2n+2, preserving the rank of the symmetry at finite coupling.


This is the E6 sequence.

6.1 n = 2



H1

✓�

SU(2)� ⌅

SO(10)

◆= C3d

0

BBBB@�1� �

2�

� 1|� 2|�3� �

2� �

1

1

CCCCA(6.1)

which is the a�ne Dynkin diagram for E6. This is precisely the global symmetry of

the Higgs branch at infinite coupling.

6.2 n > 2

The theory is SU(n)1/2 with Nf = 2n + 1 flavours. There is a classical global sym-

metry U(2n+ 1)⇥U(1)I ⇠= SU(2n+ 1)⇥U(1)B ⇥U(1)I with rank r = 2n+ 2. The

Higgs branch at infinite coupling is given by the Coulomb branch of the 3d quiver:

H1

�

SU(n) 12

� ⌅2n+1

!= C3d

0

BBBB@�1� · · ·�

� 1|•2|�

n+1� · · ·� �

1

1

CCCCA. (6.2)


cutting the vertical leg: this operation results on a flavour symmetry SU(2)⇥SU(2n+

2), which has rank 2n+ 2, preserving the rank of the symmetry at finite coupling.

There is an imbalance for the lowest of the two vertical nodes. Again it is given

by Nf � 2Nc = n � 2. The monopole operators carrying the smallest flux under

this node carry spin-(n/2) and transform in the (spin-1/2, (n + 1)-antisymm) of

SU(2)⇥ SU(2n+ 2).

As it is by now clear, the n = 2 case is special. Spin-n/2 is in this case spin-1,

i.e there are extra operators at spin-1 and thus the enhancement of the symmetry is

larger. Indeed E6 � SU(2)⇥ SU(6)

– 5 –

4.1 n = 2




H1

✓�

SU(2)� ⌅

SO(12)

◆= C3d

0

@�1� �

2� �

3�•2|�4� �

3� �

2� �

1

1

A (4.1)



4.2 n > 2




H1

✓�

SU(n)0� ⌅

2n+2

◆= C3d

0

@�1� · · ·� �

n+1�

•2|�

n+2� �

n+1� · · ·� �

1

1

A . (4.2)















H1

�

SU(n) 32

� ⌅2n+1

!= C3d

0

BBBB@�1� · · ·� �

2n�2�

•1|� n|�

2n�1� �

n� •

1

1

CCCCA. (5.1)

– 4 –

E5 sequences7 SU(n)2 with Nf = 2n flavours


H1

✓�

SU(n)2�⌅

2n

◆= C3d

0

@�1� �

2� · · ·� �

2n�3�n�1 �

|�2n�2

�•1|�n� •

1

1

A . (7.1)

The global symmetry at infinite coupling can be read o↵ from the above quiver, after

removing the blue nodes: this operation results on a flavour symmetry SO(4n)⇥U(1),

which has rank 2n+ 1, as expected from finite coupling.

8 SU(n)1 with Nf = 2n flavours


H1

✓�

SU(n)1�⌅

2n

◆= C3d

0

@�1� �

2� · · ·� �

n�1�1•|�n�•1|�n� �

n�1� · · ·� �

1

1

A . (8.1)

The global symmetry at infinite coupling can be read o↵ from the above quiver,

after removing the blue nodes: this operation results on a flavour symmetry SU(2n+

1)⇥ U(1), which has rank 2n+ 1, as expected from finite coupling.


This is the E5 sequence. CS=0 is consistent. The (p, q) webs with D7 branes as

flavours, which apply for the E8, E7, E6 sequences don’t apply here.

9.1 n = 2



H1

✓�

SU(2)� ⌅

SO(8)

◆= C3d

0

BBB@�1�1�� 2� �1�

|�2� �

1

1

CCCA. (9.1)

This is the a�ne Dynkin diagram for SO(10). The Coulomb branch of this quiver

is known to correspond to the moduli space of one SO(10) instanton, namely the

minimal nilpotent orbit of SO(10).

– 6 –



H1

✓�

SU(n)2�⌅

2n

◆= C3d

0

@�1� �

2� · · ·� �

2n�3�n�1 �

|�2n�2

�•1|�n� •

1

1

A . (7.1)






H1

✓�

SU(n)1�⌅

2n

◆= C3d

0

@�1� �

2� · · ·� �

n�1�1•|�n�•1|�n� �

n�1� · · ·� �

1

1

A . (8.1)







9.1 n = 2



H1

✓�

SU(2)� ⌅

SO(8)

◆= C3d

0

BBB@�1�1�� 2� �1�

|�2� �

1

1

CCCA. (9.1)




– 6 –



H1

✓�

SU(n)2�⌅

2n

◆= C3d

0

@�1� �

2� · · ·� �

2n�3�n�1 �

|�2n�2

�•1|�n� •

1

1

A . (7.1)






H1

✓�

SU(n)1�⌅

2n

◆= C3d

0

@�1� �

2� · · ·� �

n�1�1•|�n�•1|�n� �

n�1� · · ·� �

1

1

A . (8.1)







9.1 n = 2



H1

✓�

SU(2)� ⌅

SO(8)

◆= C3d

0

BBB@�1�1�� 2� �1�

|�2� �

1

1

CCCA. (9.1)




– 6 –

9.2 n > 2

The theory is SU(n)0 with Nf = 2n flavours. There is a classical global symmetry

U(2n)⇥ U(1)I ⇠= SU(2n)⇥ U(1)B ⇥ U(1)I with rank r = 2n+ 1. The Higgs branch

at infinite coupling is given by the Coulomb branch of the 3d quiver:

H1

✓�

SU(n)0�⌅

2n

◆= C3d

0

BBB@�1� · · ·�

1��2•�1�|�n� · · ·� �

1

1

CCCA. (9.2)


after cutting o↵ the subquiver connected to the node with label n: this operation

results on a flavour symmetry SU(2)⇥ SU(2)⇥ SU(2n), which has rank 2n+ 1, as

expected from finite coupling.

The node which is not balanced is middle node on the top line of the quiver

and again by an amount Nf � 2Nc = 2(n � 2). Similarly to the previous cases, the

monopole operator with non-zero flux under this gauge group has SU(2)R spin-n/2

and transforms in the (1/2, 1/2, n-antisymm) of SU(2)⇥ SU(2)⇥ SU(2n+ 2).

As expected, for n = 2, there is a further symmetry enhancement: indeed

SO(10) � SU(2)⇥ SU(2)⇥ SU(4).

10 SU(n)5/2 with Nf = 2n� 1 flavours

There is a classical global symmetry U(2n�1)⇥U(1)I ⇠= SU(2n�2)⇥U(1)B⇥U(1)Iwith rank r = 2n. The Higgs branch at infinite coupling is given by the Coulomb


H1

�

SU(n) 52

� ⌅2n�1

!= C3d

0

BB@ �1� �

2� · · ·� �

2n�4�

•1|� n�1|�

2n�3� �

n�1� •

1

1

CCA . (10.1)


removing the blue nodes: this operation results on a flavour symmetry SO(4n�2)⇥U(1), which has rank 2n, as expected from finite coupling.


H1

�

SU(n) 32

� ⌅2n�1

!= C3d

0

@ �1� �

2� · · ·� �

n�2�

1•|�

n�1� �

n�1�

•1|�

n�1� �

n�2� �

2� �

1

1

A .

(11.1)

– 7 –

E4 sequences

9.2 n > 2




H1

✓�

SU(n)0�⌅

2n

◆= C3d

0

BBB@�1� · · ·�

1��2•�1�|�n� · · ·� �

1

1

CCCA. (9.2)










SO(10) � SU(2)⇥ SU(2)⇥ SU(4).




H1

�

SU(n) 52

� ⌅2n�1

!= C3d

0

BB@ �1� �

2� · · ·� �

2n�4�

•1|� n�1|�

2n�3� �

n�1� •

1

1

CCA . (10.1)




H1

�

SU(n) 32

� ⌅2n�1

!= C3d

0

@ �1� �

2� · · ·� �

n�2�

1•|�

n�1� �

n�1�

•1|�

n�1� �

n�2� �

2� �

1

1

A .

(11.1)

– 7 –

9.2 n > 2




H1

✓�

SU(n)0�⌅

2n

◆= C3d

0

BBB@�1� · · ·�

1��2•�1�|�n� · · ·� �

1

1

CCCA. (9.2)










SO(10) � SU(2)⇥ SU(2)⇥ SU(4).




H1

�

SU(n) 52

� ⌅2n�1

!= C3d

0

BB@ �1� �

2� · · ·� �

2n�4�

•1|� n�1|�

2n�3� �

n�1� •

1

1

CCA . (10.1)




H1

�

SU(n) 32

� ⌅2n�1

!= C3d

0

@ �1� �

2� · · ·� �

n�2�

1•|�

n�1� �

n�1�

•1|�

n�1� �

n�2� �

2� �

1

1

A .

(11.1)

– 7 –


removing the blue nodes: this operation results on a flavour symmetry SU(2n)⇥U(1),

which has rank 2n, as expected from finite coupling.




H1

�

SU(n) 32

� ⌅2n�1

!= C3d

0

BB@ �1� �

2� · · ·� �

n�2�

1•|�

n�1�

•1|�

n�1� �

n�2� �

2� �

1

�1

1

CCA .

(12.1)


removing the blue nodes: this operation results on a flavour symmetry SU(2n�1)⇥SU(2)⇥ U(1), which has rank 2n, as expected from finite coupling.

References

[1] H. Hayashi, S.-S. Kim, K. Lee, M. Taki, and F. Yagi, “A new 5d description of 6d

D-type minimal conformal matter,” JHEP 08 (2015) 097, arXiv:1505.04439

[hep-th].

– 8 –

Minimally unbalanced quivers

• Quivers with very few unbalanced nodes

• Global symmetry

• 1 unbalanced node gives either a single non Abelian factor or 2 non Abelian factors

• If there are p unbalanced nodes, expect p-1 U(1) factors

• Gradually complicated moduli spaces

Multiplicity free varieties

• Given an operator in the chiral ring it transforms under SU(2) R and F

• Given a representation r, it’s multiplicity m_r is the number of operators transforming under this representation

• A multiplicity free variety has m_r either 0 or 1

• All moduli spaces we saw are multiplicity free

Example H^n Multiplicity free

• The smooth variety H^n has global symmetry Sp(n) and at spin n/2 of SU(2) R precisely 1 operator at Sym^n (fund)

• HWG = PE of μ1 t

HWG

• Generating function of all highest weights in the algebraic variety

• μ are fugacities for highest weights

• t is fugacity for SU(2) R

Example Nilpotent orbits of height 2

• All closures of nilpotent orbits of height 2 are multiplicity free

• The closure of the nilpotent orbit 2^k 1^{m-2k} of SL(m)

• HWG = PE of sum_i μ_i μ_{m-i} t^{2i}

• Where the sum goes from 1 to k

Summary • The Higgs branch at infinite coupling of a 5d theory is

conveniently encoded by a coulomb branch of a 3d N=4 theory

• The chiral ring can be derived from this description

• The generators of the chiral ring are simple and describe the solution to this problem

• Get a window to non perturbative effects made by instantons and the precise way they correct classical relations in the chiral ring

Thank you !

Documents

Five dimensional gauge theories NCTSneo.phys.wits.ac.za/public/workshop_10/hanany.pdf · 3d N=4 Coulomb branch Global symmetry • An imbalance of a node is the number of ﬂavors