JHEP07(2019)083
supplemented by the constraint:
T
√
−γγ
AB
M
MN
∂
B
Y
N
=
AB
q
MN
∂
B
Y
N
. (1.10)
This formulation is equivalent to that where the constraint is implemented by gauging the
shift symmetry in dual directions (a consequence of the section condition), leading to the
introduction of V
M
A
as the gauge field for this symmetry [4, 5, 28, 31]. (Note that one
could view this, when the background metrics are flat, as describing the exceptional sigma
model on the background R
1,n−1
× T
dim R
1
, i.e. on an “exceptional torus”.) Varying (1.9)
gives the following boundary terms:
δS ⊃ −
Z
d
2
σ ∂
A
T
√
−γγ
AB
δX
µ
∂
B
X
ν
g
µν
+
1
2
T
√
−γγ
AB
δY
M
M
MN
∂
B
Y
N
. (1.11)
Let us now work in conformal gauge, γ
00
= −1 = −γ
11
, γ
01
= 0,
01
= −1. Our interest
is in the boundary conditions for the extended coordinates Y
M
. For the time being we
will assume Neumann boundary conditions for the X
µ
, that is ∂
1
X
µ
= 0 at the worldsheet
boundaries at σ = 0, π, and comment on the imposition of Dirichlet boundary conditions
in these directions later (in section 2.3). So we are studying the boundary condition:
δY
M
M
MN
∂
1
Y
N
bry
= 0 . (1.12)
At σ
1
= 0, let (P
Dir
)
M
N
denote the projector onto Dirichlet directions, and let (P
Neu
)
N
M
=
δ
M
N
− (P
t
Dir
)
N
M
denote the Neumann projector. We have P
Dir
δY = 0 = δY P
t
Dir
. This
implies that we have to require
(P
Neu
)
M
N
M
NP
∂
1
Y
P
= 0 (1.13)
at σ
1
= 0. (So note the Neumann projector naturally acts on M
MN
∂
1
Y
N
, hence its index
structure). Compatibility with the constraint
q
MN
∂
0
Y
N
= T M
MN
∂
1
Y
N
, (1.14)
then means
(P
Neu
)
M
N
q
NP
∂
0
Y
P
= 0 , (1.15)
at σ
1
= 0. This can be achieved if
(P
Neu
)
M
N
q
NP
= q
MN
(P
Dir
)
N
P
, (1.16)
which in turn implies
P
t
Dir
qP
Dir
= 0 = P
Neu
qP
t
Neu
. (1.17)
Evidently, at the other endpoint, σ
1
= π, we introduce similarly projectors
g
P
Dir
and
]
P
Neu
,
which need not coincide with the ones at σ
1
= 0. Thus each endpoint of the string can
be attached to a different subspace of the full extended space. (However, for the rest
of this paper, we will assume that both endpoints of the string obey identical boundary
conditions.)
– 5 –
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