Table Of ContentA Note on the Possible Existence of an Instanton-like Self-Dual Solution to Lattice
Euclidean Gravity
S.N. Vergeles∗
Landau Institute for Theoretical Physics,
Russian Academy of Sciences, Chernogolovka,
Moscow region, 142432 Russia
and
Moscow Institute of Physics and Technology,
Department of Theoretical Physics,
Dolgoprudnyj, Moskow region, 141707 Russia
The self-dual solution to lattice Euclidean gravity is constructed. In contrast to the well known
7
Eguchi-HansonsolutiontocontinuousEuclideanGravity,thelatticesolutionisasymptoticallyglob-
1
allyEuclidean, i.e., the boundaryof thespace as r−→∞ is S3 =SU(2).
0
2
PACSnumbers: 11.15.-q,11.15.Ha
n
a
J I. INTRODUCTION II. EGUCHI-HANSON SELF-DUAL SOLUTION
9
First of all it is necessary to describe shortly the
]
t Eguchi-Hanson self-dual solution to continuous Eu-
a
l Soon after the discovery of the self-dual (instanton) clidean Gravity. Let γa, γaγb + γbγa = 2δab, a =
- solution to the 4D Euclidean Yang-Mills theory [1], the 1,2,3,4 be the Hermitian Dirac matrices 4×4 in spinor
p
e self-dual solution to 4D Euclidean Gravity has been ob- representation (σα, α=1,2,3 are Pauli matrices):
h tained [2], [3] (see also [4], [5], [6]). But the space-time
1 [ obfouEngduacrhyi-Hasanrso−n→gr∞aviitsatSio3n/aZl2inasntdanntootn,Sin3 icnontthreasctatsoe γα = −i0σα iσ0α , γ4 = 01 10 ,
v the Yang-Millsone. Otherwise,a ”cone-tipe”singularity (cid:18) (cid:19) (cid:18) (cid:19)
−1 0 1
3 (effective delta-function in the curvature at the instan- γ5 ≡γ1γ2γ3γ4 = , σab = [γa,γb],
0 1 4
7 ton centre for r = a) would be necessary. This means (cid:18) (cid:19)
21 tthonatstohluetsiopnacdei-fftiemrsectroupcoialollgyyforofmthethgeratvoiptoaltoiognyaolfinastraena-l σα4 = 2i σ0α −0σα , σαβ = iεα2βγ σ0γ σ0γ .
0 (cid:18) (cid:19) (cid:18) (cid:19)
space-time. Therefore, though the action of the Eguchi-
. (2.1)
1 Hanson solution is zeroth, the physical meaning of the
0 solution is not clear.
Let’s consider the pure 4D Euclidean Gravity action in
7
1 the Palatini form (the independent variables are tetrad
: and connection):
v I construct here the analogue of the Eguchi-Hanson
i self-dual solution to the 4D Euclidean lattice Gravity
X 1
with zeroth action. The solution transforms locally into A=− trγ5R∧e∧e=
r l2
a the Eguchi-Hanson solution as r −→ ∞. The reason is P Z
1
that the considered lattice theory transforms into Ein- =− Rα ∧Eα −Rα ∧Eα ,
stein theory for a long-wavelength limit. The remark- l2 (+) (+) (−) (−)
P Z n o
able fact is the discrete gravity self-dual solution wipes iσα Rα 0
out a ”cone-tipe” singularity at the center of instanton R ≡2(dω+ω∧ω)= 2 (0+) Rα ,
for the case the space-time boundary as r −→ ∞ is S3. (cid:18) (−) (cid:19)
Thus,thenthegravitationalinstantonswouldexistifthe 1 iσα ωα 0
ω ≡ σabωabdxµ = (+)µ dxµ,
realspace-timeexhibitsthegranularitypropertyatsuper 2 µ 2 0 ωα
(cid:18) (−)µ (cid:19)
small scales.
1
ω(α±) ≡ ±ωµα4+ 2εαβγωµβγ dxµ,
(cid:26) (cid:27)
A preliminary version of the work has been published R(α±) =2dω(α±)−εαβγω(β±)∧ω(γ±),
in [7].
e≡γaeadxµ,
µ
E(α±) ≡ ± eαλe4ρ−e4λeαρ +εαβγeβλeγρ dxλ∧dxρ.
∗e-mail:[email protected] n (cid:0) (cid:1) o (2.2)
2
One can take six 1-forms ωα as independent variables Ψα −→ Ψα +∂ φα . As a result, Eqs. (2.9) fix
(±) (−)λ (−)λ λ (−)
insteadof six 1-formsωab. Obviously,the representation no more than additional 12−3×3 = 3 functions. This
(2.2) is consistent with the representation of the group means that the gauge subgroup Spin(4)(−) is broken by
Spin(4)≈Spin(4) ⊗Spin(4) ≈SU(2) ⊗SU(2) . Eq. (2.8). So we see that the system of equations (2.7)-
(+) (−) (+) (−)
The following equations are equivalent (2.10) is consistent, though it fixes the gauge subgroup
Spin(4) . The Eguchi-Hanson solution is the simplest
(−)
ω =±γ5ω ←→ωab =∓1ε ωcd ←→ωα =0. (2.3) solution of the system. Let’s write out this solution [2],
2 abcd (±) [3].
Let xi = (r, θ, ϕ, ψ), where (θ, ϕ, ψ) be the Euler
Eqs. (2.3) imply the following one: angles. Thecartesiancoordinatesxµ inR4 areconnected
with the coordinates xi as follows:
1
R =±γ5R ←→Rab =∓ ε Rcd ←→Rα =0.
2 abcd (±) θ i
z ≡x1+ix2 =rcos exp (ψ+ϕ) ,
(2.4) 1
2 2
(cid:20) (cid:21)
θ i
The action stationarity condition relative to the connec- z ≡x3+ix4 =rsin exp (ψ−ϕ) . (2.11)
2
tion gives the equation 2 2
(cid:20) (cid:21)
δA/δωab =0−→dea+ωab∧eb =0, (2.5) There is a one-to-one correspondence between these two
µ coordinatesystems ifthe Euleranglesvary inthe ranges
whichdeterminesuniquelytheconnectionformsforfixed
0≤θ ≤π, 0≤ϕ≤2π, 0≤ψ ≤4π. (2.12)
forms ea. Additionally, Eq. (2.5) imply the algebraic
Bianchiidentity for Riemanniantensor,the combination The Eguchi-Hanson solution for the metrics ds2 ≡
of which with Eq. (2.4) leads to the Einstein’s equation eadxµ 2 has the form
µ
Rab ≡Raccb =0. (2.6) (cid:0) (cid:1) 2r (sinψdθ−sinθcosψdϕ)
r (cosψdθ+sinθsinψdϕ)
On the other hand, Einstein’s equation is equivalent to eaµdxµ = 2 −rg(r)(cosθdϕ+dψ) ,
theactionstationarityconditionrelativetotheformseaµ: 2 g(r)−1dr
R =0←→δA/δea =0. (2.7) a4
ab µ g(r)= 1− , r≥a. (2.13)
r4
r
The question arises: why the additional Eq. (2.3) does
For a = 0, the metrics (2.13) transforms to the 4D Eu-
not come into conflict with Eqs. (2.5) and (2.7)? To
clidean metrics in Euler angles on the S3 with ranges
answerthisquestionlet’sconsiderthecasewiththelower
(2.12). But 0 ≤ ψ ≤ 2π for the Eguchi-Hanson solution
sign in Eq. (2.3) when
(when a 6= 0) since the points with coordinates ψ and
ωα =0. (2.8) (ψ +2π) and the same (r, θ, ϕ) are identified. Other-
(−)µ
wise, in order for the Chern-Gauss-Bonnet theorem to
ThecombinationofthepartofEqs. (2.5)δA/δωα =0 be satisfied in the case (2.12), a ”cone-tipe” singularity
(−)µ
(effectivedelta-functioninthecurvatureattheinstanton
and Eq. (2.8) gives the following 12 equations:
centre for r =a) would be necessary (see [3]).
εµνλρ∂ Eα =0. (2.9) The connection 1-forms are given by Eqs. (2.8) and:
ν (−)λρ
2g
Now we must solve the system of equations (2.7), (2.8), ω1 = e1 =g(sinψdθ−sinθcosψdϕ),
(+) r
(2.9) and
2g
ω2 = e2 =g(cosψdθ+sinθsinψdϕ),
δA/δωα =0. (2.10) (+) r
(+)µ 2 g a4
ω3 =2 − e3 =− 1+ (cosθdϕ+dψ).
Note that Eqs. (2.5) and (2.7) do not fix the vari- (+) rg r r4
(cid:18) (cid:19) (cid:18) (cid:19)
ables ωab, ea completely but up to the gauge (orthog- (2.14)
µ µ
onal) transformations. Here the gauge group leaves 6
We have
unfixed functions.
Eqs. (2.9) do not fix the quantities Eα completely 1 1 1
(−)λρ ω1 ∧ ω2 ∧ ω3 =−π2
but up to summands of the forme ∂λΨα(−)ρ−∂ρΨα(−)λ Zr=Const→∞(cid:18)2 (+)(cid:19) (cid:18)2 (+)(cid:19) (cid:18)2 (+)(cid:19) (2.15)
whereΨα are3arbitraryvector(cid:16)fields(12functionsal(cid:17)-
(−)λ
together). But eachof three vector fields Ψα contains and
(−)λ
only 3 independent functions due to invariance of the 1 1 1
ω1 ∧ ω2 ∧ ω3 =0 (2.16)
expression ∂ Ψα −∂ Ψα relative to the changes 2 (+) 2 (+) 2 (+)
λ (−)ρ ρ (−)λ Zr−→a+0(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
(cid:16) (cid:17)
3
for the range 0 ≤ ψ ≤ 2π and orientation θ, ϕ, ψ, r. V = 1,2,..., N → ∞ and W enumerate the vertices
The integral (2.15) would be equal to (−π2) for any and 4-simplices, correspondingly. It is necessary to use
0 < r = Const < ∞ in the case a = 0 (the Euclidean the local enumeration of the vertices a attached to a
V
metrics in Euler angles). Thus, the boundary conditions given 4-simplex: the all five vertices of a 4-simplex with
(2.15)-(2.16) determine the instantonEguchi-Hansonso- index W are enumerated as a , i = 1,2,3,4,5. The
Wi
lution with the same integration constant a as in the laternotationswithextraindexW indicatethatthe cor-
relation (2.16). This means that the system of equa- respondingquantitiesbelongtothe4-simplexwithindex
tions (2.7)-(2.10) together with the boundary conditions W. The Levi-Civita symbol with in pairs different in-
(2.15)-(2.16) possess unique solution (2.13) for the cen- dexes ε = ±1 depending on whether the order of
Wijklm
trallysymmetricalmetrics anzatswith the sameintegra- vertices s4 = a a a a a defines the positive
W Wi Wj Wk Wl Wm
tion constant a as in the relation (2.16). or negative orientation of 4-simplex s4 . An element of
W
Note that the groupSpin(4)and anelement of the Cliffordalgebra
iσα 1
ωα =U−1dU as r −→∞, (2.17) Ω =Ω−1 =exp(ω ), ω ≡ σabωab ,
2 (+) Wij Wji Wij Wij 2 Wij
eˆ =eˆ† ≡ea γa ≡−Ω eˆ Ω−1 . (3.1)
where Wij Wij Wij Wij Wji Wij
iσ3 iσ2 iσ3 are assigned for each oriented 1-simplex aWiaWj. The
U =exp − ϕ exp θ exp − ψ . (2.18) lattice analog of the action (2.2) has the form
2 2 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
1
According to the Eq. (2.17) A= ε trγ5×
Wijklm
5×24
W i,j,k,l,m
1 X X
tr U−1dU ∧ U−1dU ∧ U−1dU = 1
12 × −2l2 ΩWmiΩWijΩWjmeˆWmkeˆWml . (3.2)
1 (cid:0) 1 (cid:1) (cid:0) 1 (cid:1) (cid:0) (cid:1) (cid:26) P (cid:27)
= ω1 ∧ ω2 ∧ ω3 as r −→∞.
2 (+) 2 (+) 2 (+) Thisactionisinvariantrelativeto thegaugetransforma-
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
(2.19) tions
Taking into account Eqs. (2.15) and (2.19), we obtain Ω˜Wij =SWiΩWijSW−1j, e˜Wij =SWieWijSW−1i,
for the angle ranges (2.12):
S ∈Spin(4). (3.3)
Wi
1 tr U−1dU ∧ U−1dU ∧ U−1dU =−2π2. It is natural to interpret the quantity
12
Z (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (2.20) 1 4
l2 ≡ tr(eˆ )2 = (ea )2 ∼l2 (3.4)
Wij 4 Wij Wij P
This equality means that Eq. (2.18) gives the smooth a=1
X
mappingofthespace-timehypersurfaceS3 prescribedby
as the square of the length of the edge a a , and the
Wi Wj
the Euler angles (see Eqs. (2.11) with a fixed parameter
parameterl is ofthe orderofthe lattice spacing. Thus,
P
r) into the SU(2) group space, and the degree of the
the geometricpropertiesofasimplicialcomplexproveto
mapping is equal to (−1).
be completely defined.
Write out also the Riemannian curvature 2-form:
Now, let us show in the limit of slowly varying fields,
that the action (3.2) reduces to the continuous gravity
Rα =0,
(−) action (2.2).
8a4 Considera certain4D sub-complex ofcomplex K with
R1 =− e1∧e4+e2∧e3 ,
(+) r6 the trivialtopologyof four-dimensionaldisk. Realize ge-
8a4 (cid:0) (cid:1) ometricallythissub-complexinR4. Supposethatthege-
R(2+) = r6 e1∧e3−e2∧e4 , ometricrealizationisanalmostsmoothfour-dimensional
surface [11]. Thus each vertex of the sub-complex ac-
16a4(cid:0) (cid:1)
R3 = e1∧e2+e3∧e4 . (2.21) quires the coordinates xµ which are the coordinates of
(+) r6 the vertex image in R4:
(cid:0) (cid:1)
xµ =xµ ≡xµ(a )≡xµ(a ), µ=1, 2, 3, 4
III. THE LATTICE GRAVITY MODEL Wi V Wi V
(3.5)
The next step is to adumbrate the model of lattice Westressthatthesecoordinatesaredefinedonlybytheir
gravity which is used here. A detailed description of the vertices rather than by the higher dimension simplices
model is given in [8]-[10]. to which these vertices belong; moreover,the correspon-
The orientable 4-dimensional simplicial complex and dence between the vertices from the considered subset
its vertices are designated as K and a , the indices and the coordinates (3.5) is one-to-one.
V
4
We have eachfour-dimensionalsimplex. Inthiscase,thefollowing
formula is valid up to O (dx)2 inclusive
|xµ −xµ |∼l . (3.6)
Wi Wj P (cid:0) 1 (cid:1)
It is evident that the four vectors ΩWmiΩWijΩWjm =exp 2 Rµν(xWm)dxµWmi dxνWmj ,
(cid:20) (cid:21)
(3.13)
dxµ ≡xµ −xµ , i=1, 2, 3, 4 (3.7)
Wji Wi Wj where
are linearly independent and R =∂ ω −∂ ω +[ω , ω ]. (3.14)
µν µ ν ν µ µ ν
dx1 dx2 ... dx4 Whenderivingformula(3.13),weusedtheHausdorfffor-
Wm1 Wm1 Wm1
... ... ... ... ≷0, (3.8) mula.
(cid:12) (cid:12)
(cid:12) dx1 dx2 ... dx4 (cid:12) In exact analogy with (3.9), let us write out the fol-
(cid:12) Wm4 Wm4 Wm4 (cid:12)
(cid:12) (cid:12) lowing relations for a tetrad field without explanations
(cid:12) (cid:12)
pdeopsietnivd(cid:12)eilnygoornnweghaettihveelythoerifernatmede. XHmWe(cid:12)r1e,,..t.he, XdmWiff4erenis- eˆWmµ dxµWmi =eˆWmi. (3.15)
(cid:0) (cid:1)
tials of coordinates (3.7) correspond to one-dimensional Applying formulas (3.13)–(3.15) to the discrete action
simplices aWjaWi, so that, if the vertex aWj has coor- (3.2)andchangingthe summationto integrationwe find
dinates xµWj, then the vertex aWi has the coordinates that in the long-wavelengthlimit the lattice action (3.2)
xµ +dxµ . transformsintocontinuousaction(2.2)andanyinforma-
Wj Wji
In the continuous limit, the holonomy group elements tionaboutlatticeisforgotteninthemainapproximation.
(3.1) are close to the identity element, so that the quan-
tities ωab tend to zero being of the order of O(dxµ).
ij
Thus one can consider the following system of equation IV. THE SELF-DUAL SOLUTIONS TO
LATTICE GRAVITY
for ω
Wmµ
ωWmµ dxµWmi =ωWmi, i=1, 2, 3, 4. (3.9) Now let us consider the self-dual solution to lattice
gravity. We have the lattice analogue of Eqs. (2.5) and
In this system of linear equation, the indices W and m (2.7):
are fixed, the summation is carried out over the index
δA/δωα =0, (4.1)
µ, and index runs over all its values. Since the deter- (±)Wmi
minant (3.8) is nonzero, the quantities ω are de-
Wmµ
fined uniquely. Suppose that a one-dimensional simplex δA/δea =0. (4.2)
Wmi
XW belongs to four-dimensional simplices with indices
mi
W , W , ... , W . Introduce the quantity Because of the action (3.2) is a homogeneous quadratic
1 2 r
function of the variables {e}, so
1 1
ωµ 2(xWm+xWi) ≡ r ωW1mµ+ ... +ωWrmµ , 2A= eaWmi(δA/δeaWmi)=0 (4.3)
(cid:18) (cid:19) (cid:26) (cid:27)
(3.10) X{e}
on the mass shell according to Euler theorem. Let’s im-
which is assumed to be related to the midpoint of the posetheadditionalconditions(comparewithEqs. (2.8))
segment[xµ , xµ ]. Recallthatthecoordinatesxµ as
Wm Wi Wi
wellasthe differentials(3.7)dependonlyonverticesbut ωα =0. (4.4)
(−)Wmi
not on the higher dimensional simplices to which these
CombiningEqs. (4.1)withindex(−)and(4.4)weobtain:
verticesbelong. Accordingto the definition, we havethe
following chain of equalities δA 2
δωα =−l2 εW′mijkl E(α−)W′[jkl]+
ωW1mi =ωW2mi = ... =ωWrmi. (3.11) (−)Wmi(cid:12)(cid:12)ω(α−)=0 P XW′ jX,k,l (cid:26)
1 (cid:12)
+ E(cid:12)α +Eα +Eα +
It follows from (3.7) and (3.9)–(3.11) that 6 (−)W′m[kl] (−)W′m[lj] (−)W′m[jk]
h(cid:0) (cid:1)
1 + m↔i =0,
ωµ(cid:18)xWm+ 2 dxWmi(cid:19) dxµWmi =ωWmi. (3.12) 1 (cid:0) (cid:1)i(cid:27)
Eα ≡ Eα +Eα +Eα ,
The value of the field element ω in (3.12) is uniquely ∗[mkl] 3 ∗m[kl] ∗k[lm] ∗l[mk]
µ (cid:18) (cid:19)
defined by the corresponding one-dimensional simplex. Eα ≡∓ eα e4 −e4 eα +
(±)∗m[kl] ∗mk ∗ml ∗mk ∗ml
Next, we assume that the fields ω smoothly depend
on the points belonging to the geomµetric realization of +εα(cid:0)βγeβ∗mkeγ∗ml. (cid:1) (4.5)
5
The index W′ in (4.5) enumerates all 4-simplices which We see that the quantities Ψα belonging to the
(−)Wmk
contain a marked 1-simplex aWmaWi. As in continuous common 3-simplices of the adjacent 4-simplices can-
case, the system of equations (4.1), (4.2) and (4.4) is cel on in (4.11) as a consequence of Eqs. (4.10) and
equivalent to the system of equations (4.1) with index
Ψα +Ψα = 0. Note that each quantity
(+), (4.2), (4.4) and (4.5). It will be shown that the (−)Wjk (−)W′kj
square brackets give null equation under the sum (4.5). (cid:16)Ψα(−)W′jk inparenthes(cid:17)esin(4.8)belongstothe3-simplex
Therefore only the first term in the parentheses in Eq. whichis commonto twoadjacent4-simplices. If not, the
(4.5) is significant. cavities and boundaries would be in the simplicial com-
Equations (4.5) do not fix the quantity Eα plex, but such complexes are not considered here. Thus,
(−)W′[jkl]
thesum(4.8)isequaltozeroidentically[12]. Thismeans
completely but up to summands of the kind
that the system of equations (4.1) with index (+), (4.2),
Eα −→ (4.4) and (4.5) is self-consistent and it fixes in part the
(−)W′[jkl]
gauge as well as in continuous case. According to the
−→E(α−)W′[jkl]+ Ψα(−)Wjk+Ψα(−)Wkl+Ψα(−)Wlj , Eqs. (4.2) and (4.3) the action of any solution of this
system of equations is equal to zero.
(cid:16) ((cid:17)4.6)
and the lattice 1-form Ψα = −Ψα can be
(−)Wmk (−)Wkm
varied according to V. LATTICE ANALOGUE OF THE
EGUCHI-HANSON SOLUTION
Ψα −→Ψα + φα −φα . (4.7)
(−)Wmk (−)Wmk (−)Wk (−)Wm
NowproceedtostudythelatticeanalogoftheEguchi-
(cid:16) (cid:17)
Hansonsolution. Butitisimpossibletogivetheirregular
It follows from here that Eqs. (4.5) fix no more than
3 additional real-valued parameters at each vertex of latticesolutioninanexplicitformincontrasttothecon-
tinuous case. Thus, the problem reduces to the solution
complex, leading to the fixation of gauge subgroup
Spin(4) . existence proof and finding of its asymptotics.
(−)
Let’s prove the given statements. For this end one Suppose that the complex K can be considered as a
must check that the equation triangulation of R4.
Let’sintroducethefollowingnotations: k⊂Kmeansa
finitesub-complexcontainingthecentreofinstantonwith
εW′mijkl Ψα(−)W′jk+Ψα(−)W′kl+ the boundary ∂k = S ≈ S3; S means the boundary
∞
XW′ (cid:26)jX,k,l (cid:0) of extra-large sub-complex of complex K containing the
centre of instanton, so that in a wide vicinity of S the
Ψα =0 (4.8) ∞
(−)W′lj long-wavelength limit is valid and the continuous solu-
(cid:27)
(cid:1) tion (2.13), (2.14) approximates correctly the exact lat-
issatisfiedidentically. ItisevidentthatthebracesinEq. tice solution and the hypersurface S is given by the
∞
(4.8) vanishes identically at eachfixed value ofindex W′ Eq. r = R = Const −→ ∞. Evidently, the Euler char-
if Ψα = φα −φα . acteristics χ(k) = χ(K) = 1. We have the exact lattice
(−)W′mk (−)W′k (−)W′m
equivalents of the instanton boundary conditions (2.15)
Consider two(cid:16)adjacent positively o(cid:17)riented 4-simplices
and (2.16):
s4 =a a a a a ,
W Wm Wi Wj Wk Wl 1
s4W′ =aW′maW′iaW′kaW′jaW′l′, 12·4! εSijkmtr Ω−(+1)SjmΩ(+)Sji ×
aWm =aW′m, aWi =aW′i, aWj =aW′j, S(XS∞)iXjkm (cid:0) (cid:1)
× Ω−1 Ω Ω−1 Ω =−2π2, (5.1)
aWk =aW′k, aWl 6=aW′l′, (4.9) (+)Skm (+)Skj (+)Sim (+)Sik
(cid:0) (cid:1)(cid:0) (cid:1)
so that
ε tr Ω−1 Ω ×
Sijkm (+)Sjm (+)Sji
εWmijkl =εW′mikjl′ =1. (4.10) SX(S)iXjkm (cid:0) (cid:1)
× Ω−1 Ω Ω−1 Ω =0. (5.2)
Select from the sum (4.8) two summands corresponding (+)Skm (+)Skj (+)Sim (+)Sik
to the 4-simplices (4.9):
(cid:0) (cid:1)(cid:0) (cid:1)
HeretheindicesS(S )andS(S)enumerate3-simplices
∞
εWmijkl Ψα(−)Wjk+Ψα(−)Wkl+Ψα(−)Wlj + on the boundaries S∞ and S, correspondingly, and the
Levi-Civita symbol ε = ±1 depending on whether
j,k,l Sijkm
X (cid:0) (cid:1) theorderofverticess3 =a a a a definesthepos-
+ εW′mikjl′ Ψα(−)W′kj +Ψα(−)W′jl′ +Ψα(−)W′l′k . itive or negative orienStationSiofStjhiSsk3-Ssimmplex.
jX,k,l′ (cid:0) (cid:1) Sincethelong-wavelengthlimitisvalidasr−→∞,one
(4.11)
can use the instanton solution for the dynamic variables
6
(2.13)-(2.14) in this region: Ψab is determined up to the lattice gradient (φab −
Wmk Wk
φab ). So it follows from Eqs. (5.3) (just like as in the
iσα Wm
Ω ≈1+ ωα dxµ . case of (4.4)) that now not only the gauge sub-group
(+)Wmi 2 (+)µ Wmi Spin(4) is broken but also the gauge group Spin(4) is
(−)
brokentoonk\∂kalmostwhollyexceptforthe centerof
Thereforethesumin(5.1)transformsintointegral(2.15)
the sub-group SU(2) (see Eqs. (4.6)-(4.11)).
with the only difference that now the angle ψ varies in (+)
From here and throughout the following discus-
the interval (2.12), and so the boundary condition (5.1)
sions the pairs of indices (V V ) enumerate 1-simplexes
is just. 1 2
a a ∈K.
Toimplementtheboundarycondition(5.2)wesuggest V1 V2
Theaction(3.2)isequaltozeroidenticallyforthecon-
the following solutions on the sub-complex k.
figuration (5.4) for any values of ea , where 1-simplex
Consider the following solutions of Eqs. (4.1), (4.2) a a ∈ k, a a ∈/ ∂k. ThereforVe1,VE2 qs. (4.2) are sat-
and (4.4) on k: V1 V2 V1 V2
isfied automatically in this case, they do not give any
Ω(+)Wij =−1, Ω(−)Wij =1, s4W ∈k. (5.3) ctioonnsttroaiEnqtso.n(t5o.8t)h:e corresponding1-forms eaV1V2 in addi-
From here it follows that
δA/δea ≡0, a a ∈k, a a ∈/ ∂k. (5.9)
V1V2 V1 V2 V1 V2
Ω Ω Ω =−1 on k. (5.4)
(+)Wmi (+)Wij (+)Wjm Now let us prove that the lattice configuration of the
variables ωab , ea satisfyingthesystem
Theequalities(5.4)holdtrueifweperformagaugetrans- V1V2(inst) V1V2(inst)
formation (see (3.3)) with S = ±1, S = 1. of equationns and constraints (4.1),o(4.2), (4.4), (5.1) and
(+)Wi (−)Wi
The boundary condition (5.2) is true for each configura- (5.2)doesexist. Theconfigurationisthelatticeanalogue
tion obtained in such a way. of the Eguchi-Hanson continuous instanton.
Obviously, 1-form eˆ can be considered as a 1- Recall that the sets of equations (4.2) with the sign
Wij
cochain on complex, and the quantity (−) and (4.4) are equivalent to the set of equations (see
Eqs. (4.5))
Eab =(Eα , (Eα )=
Wm[kl] (+)Wm[kl] (−)Wm[kl] l2 δA
=ε ec ed , (5.5) Φα ≡− P =
abcd Wmk Wml (V1V2) 2 δωα
(−)V1V2(cid:12)ω(−)=0
(cid:12)
ipsroad2u-cctoscohfai1n-cwohchicahinissetahe s.uperposition of the exterior = εW′mijklE(α−)W′[jk(cid:12)(cid:12)l] =0 (5.10)
Wij W′ j,k,l
It may be verified (compare with (4.5)) that the left XX
hand side of Eq. (4.1) is nothing but the exterior lattice and Eqs. (4.4). Here 1-simplexes a a are renamed
Wm Wi
derivativeofa2-cochain(5.5),andEq. (4.1)impliesthat as a a . Further the set of equations (5.10) is consid-
V1 V2
thederivativeisequaltozerointhecaseof(5.4),i.e. the ered as the set of constraints.
quantity (5.5) is a cocycle [13]: At the first stage we shall solve the problem for fi-
nite complex K with extra-large (though finite) number
εW′mijklEWab′[jkl] =0, of vertices N and the boundary ∂K=S∞. There is the
W′ j,k,l estimation
XX
aWmaWi ∈s4W′, aWmaWi ∈k, aWmaWi ∈/ ∂k. lP2N∼R4. (5.11)
(5.6)
Theconstraint(5.2)isrealizedevidentlybytakingthe
Evidently, Eqs. (5.6) are satisfied for variables Ω on the sub-complex k as in (5.3). We
(+)V1V2
believe also that the relations (5.7) are valid. The con-
eaV1V2 =φaV2 −φaV1, aV1aV2 ∈k, aV1aV2 ∈/ ∂k. (5.7) straint (5.1) also is realized evidently by taking
where φa is any scalar field on k. This statement is iσα
V Ω =exp ωα dxµ ≈
checkedeasily by a directcalculationand it follows from (+)Smi 2 (+)µ Smi
(cid:26) (cid:27)
thefactthatthe cochain(5.5)isthesuperpositionofthe iσα
exterior products of exact 1-forms in the case (5.7). ≈1+ ωα dxµ (5.12)
2 (+)µ Smi
Since the second cohomology group H2(k)=0, so the
cocycle (5.5) is a lattice coboundary: on the boundary S∞, where the field ω(α+)µ is given by
Eq. (2.17).
Eab = Ψab −Ψab , So, the problem is equivalent to the searching for the
Wm[kl] Wmk Wml
Ψab (cid:0)=−Ψab . (cid:1) (5.8) stationary point of the Lagrange function
Wmk Wkm
L=A− λα Φα (5.13)
Eqs. (5.6)fixnomorethan6realnumbersateachvertex (V1V2) (V1V2)
ofthesub-complexkforthereasonthatthelattice1-form α,(V1V2),XaV1aV2∈/k\∂k
7
relative to the variables so one may put
ω(α+)(V1V2), ea(V1V2), λα(V1V2) , λαV1V2 p =0 (5.17)
n o atthestationarypoints. The(cid:12)(cid:12) lastequationsfixthegauge
assuming that the conditions (4.4), (5.3) and (5.12) take
subgroup Spin(4) . Due to Eqs. (5.17) we have
(−)
place.
fixAedcc1o-rsdiminpgletxo a(3.1a) eaicshaofsmthoeotehlemmaentrtisxΩfu(+n)cVt1ioVn2 foonr ∂e1VN−1VN =0. (5.18)
tohneSu3nit 3DwshpihcherVae1rSe(V3cV2o1Vn2v)en(ωie(αn+t)Vfo1Vr2thareelothneg-cwoaovredliennagttehs ∂λαV1V2 (cid:12)(cid:12)(cid:12)(cid:12)p
(V1V2) Let’sscrutinize equationsfo(cid:12)llowingfromthe condition
transition). Denote alsoby SH the hyperspherewhich is dL|p =0.
determined by the equation Notefirstlythatthesummandwiththesign(−)inthe
action (3.2) is equal to zero identically for Ω =1 and
(−)
ea 2+ λα 2 =l2N. any values of the variables ea:
V1V2 (V1V2) P
(V1V2),aXV1aV2∈/k\∂k(cid:26)(cid:0) (cid:1) (cid:16) (cid:17) (cid:27) (5.14) A(−)|Ω(−)=1 ≡0, ∂∂Ae(a−) ≡0.
(cid:12)Ω(−)=1
(cid:12)
Let’s search for a stationary point for the Lagrange (cid:12)
Itisseenfromhere thatEqs. (5.17)(cid:12)donotcontradictto
function(5.13)onthecompactsmoothmanifoldwithout
the equations cited below.
boundary:
The conditions
∂L
=0
C =SH S3 , ∂ωα
(V1V2) (+)(V1V2)
[ (V1V2),aV[1aV2∈/k,S∞
give the set of Eqs. (4.1) with the sign ”+”.
∂C =∅. (5.15)
We have also the set of Eqs. (4.4) supplemented by
the conditions
Since the Lagrange function (5.13) is a smooth function
defined on the compact metric space (5.15), so it is a ∂L ∂L ∂e1
bounded function and it has the local maximum(s) and =−Φα + · VN−1VN =
minimum(s) at some points pξ ∈ C. Moreover, since the ∂λαV1V2 (V1V2) ∂e1VN−1VN ∂λαV1V2 (cid:12)(cid:12)p
space C is without boundary, so the total differential of =−Φα =0. (cid:12)(cid:12) (5.19)
the Lagrange function at the points p is equal to zero. (V1V2) (cid:12)
ξ
Itwillbe appreciatedthatthe Lagrangefunctionvari- Here Eqs. (5.18) are taken into account. Eqs. (4.4) and
ables are constrained due to Eq. (5.14). Let’s express (5.19)areequivalenttoEqs. (4.4)and(4.1)withthesign
the onlyone variablee1VN−1VN for some aVN−1aVN ∈S∞ ”−”.
in terms of the rest variables involved in the constraint The last set of equations is (due to Eqs. (5.17)) as
(5.14): follows:
e1VN−1VN =±fN(...), ∂e∂aL = ∂∂eAa(+) ± ∂e∂1A(+) · ∂∂eafN =0. (5.20)
fN(...)= lP2N− eaV1V2 2+... . It folloVw1Vs2from hVe1rVe2that (VseNe−1(V4N.3), (5V.113V)2and (5.19))
s (V1V2),aXV1aV2∈/k\∂kn(cid:0) (cid:1) o
(5.16) 2A=2L| =∓ ea ∂A(+) · ∂fN 6=0.
p V1V2∂e1 ∂ea
As stated before (see Sections II and IV), the set of (VX1V2) VN−1VN V1V2
constraints (5.10) fixes at most three real number at (5.21)
each vertex which is equivalent to the gauge fixing. Ev-
idently, in the limit ω(α−)V1V2 −→ 0 the Lagrange func- TvahteurqeuaRntity∂A(s(e+e)/(∂3e.11V3N)−)1iVnNtihseprnoepigohrtbioounrahlotoodthoefctuhre-
tion(5.13)is obtainedfromthe action(3.2) by replacing (+)µν
ωα −→(l2/2)λα . Therefore,thesetofLagrange boundary S∞. Due to Eqs. (2.21) and (5.11) there is
(−)V1V2 P V1V2 the chain of estimations:
multipliers {λα } contains only tree significant real-
valued parame(Vte1rVs2)per each vertex. Since the infinites- ∂A ∼R−6 ∼N−3/2,
imal gauge transformations act on the set of Lagrange ∂e1
VN−1VN
multipliers according to the rule ea ∂A · ∂fN ∼N−1/2 ∼R−2. (5.22)
δλαV1V2 =ψV2 −ψV1, ψV −→0, (VX1V2) V1V2∂e1VN−1VN ∂eaV1V2
8
According to the Eqs. (5.20), (5.21) and estimations Ω Ω Ω ∼±1. This is the reason why one can
Wmi Wij Wjm
(5.22) we obtaine the final result takethe rangeofangles(2.12)inalattice gravity. More-
over,all lattice equations are satisfied and the action for
∂A −→0, A−→0 as R−→∞. (5.23) the instanton solution is equal to zero.
∂ea Thus,the setting ofthe problemforfinding a self-dual
V1V2
solution to lattice Euclidean gravity is as follows: one
The result can be explained once again as follows. must solve (in anti-instanton case) the difference lattice
Let the set of variables {ΩV1V2(inst)(e)} be such that system of equations (4.1) with indices (+), (4.2), (4.4),
the boundary conditions (5.4) and (5.12) as well as Eqs. theconstraints(5.10),andtheboundaryconditions(5.4),
(4.1) are fulfilled for any values of variables {e}. This is (5.12).
possible since the action (3.2) is a smooth real function Here some questions are not enough clear or remain
of the variables {Ω } (for fixed set of values {e}), i.e. unclear.
V1V2
the action (3.2) is a smooth real function on a compact 1)Istheofferedsolutionwithχ(K)=1stableoritcan
manifold without boundary. Now let’s impose the con- be contracted smoothly into the trivial one?
straintsΩ(−)V1V2(inst)(e)=0whichareequivalenttothe The answer to this question seems to be as follows.
constraints(5.10), andresolvethe constraintsrelative to Let’sconsiderasmoothpathintheconfigurationspace
the a subset of variables {ea }′. As a result each vari-
ablefromthe subsetbecomeVs1Va2homogeneousfunctionof Ω (t), ea (t), 0≤t≤1,
V1V2 V1V2
degree 1 of the rest set {ea }′′ mutually independent
V1V2
variables. As a result the action (3.2) becomes the func- such that
tion of the variables {ea }′′ such that Eqs. (4.1) are
satisfied and, hence V1V2 ΩV1V2(0)=ΩV1V2(inst), eaV1V2(0)=eaV1V2(inst),
Ω (1)=1, ea (1)=φa −φa . (6.1)
2A= ea δA/δea . (5.24) V1V2 V1V2 V2 V1
V1V2 V1V2
{Xe}′′ (cid:0) (cid:1) Here the lower index (inst) designates the dis-
cussed self-dual solution. The field configuration
Now let’s bound the function A on the hypersphere SH′′
Ω (1), ea (1) is a trivial solution of Eqs. (4.1),
determined by the relation V1V2 V1V2
(4.2). Evidently,theglobal continuousdescriptionofthis
(cid:8) (cid:9)
trivial solution is possible. Nevertheless, the considered
ea 2 =l2N.
V1V2 P instantoncontractionscenarioseemstobeunsatisfactory
{Xe}′′(cid:0) (cid:1) since to do this, one must contract topologically non-
trivialconnectionelements(see(5.1)and(5.12))intounit
The subsequent consideration and conclusion are identi-
in the infinite space-time.
cal to those which has been given with the help of Eqs.
Another instanton contraction scenario which is de-
(5.16), (5.20)-(5.23): there is a stationary point of the
scribedincontinuallimit bylimit processa−→0andin
action (3.2) relative to its variables with the additional
the lattice case by reducing the sub-complex k up to its
equations (4.4) and boundary conditions (5.4), (5.12);
disappearance also seems to be unsatisfactory. Indeed,
the action is equal to zero at the stationary point at the
in this case we would have resulted in the failure of the
limit R−→∞.
Chern-Gauss-Bonnet theorem in R4. So, this scenario is
also impossible, it can be considered as a decreasing of
the instanton scale a up to a lattice scale.
VI. CONCLUSION
Therefore, the considered instanton solution for the
case χ(K)=1 seems to be stable.
Aforesaid means that the lattice analogue of the
2)Thecaseχ(k)=(2k+1), k =1,2,...and∂k=S≈
Eguchi-Hanson self-dual solution to continuous Eu-
clidean Gravity does exist for the complex K≈R4. S3 is interesting, but it is not considered here. So, the
question remains unanswered: for what values of χ(K)
It is crucially important that the problem of possible
the lattice self-dual solution does exist and it would be
singularities for the curvature tensor does not exist on
stable?
the lattice gravity. In continuous gravityfor the ψ range
(2.12), the manifold would have ”cone-tip” singularities
at r = a; this implies the necessity of delta-functions in
Acknowledgments
the curvature at r = a (see [2, 3]). But delta-functions
transforms into Kronecker symbol which is of the order
of unity in discrete mathematics. The same is true with I thank A. Maltsev for many consultations. The work
respect to the lattice analogue of the curvature tensor has been supported by the RScF grant 16-12-10151.
9
[1] A.A. Belavin, A.M. Polyakov, A.S. Schwarz and [8] S.N. Vergeles, Nuclear PhysicsB 735, 172 (2006).
Yu.S.Tyupkin,Physics Letters B 59, 85 (1975). [9] S.N. Vergeles, JETP 106, 46 (2008).
[2] T. Eguchi and A.J. Hanson, Physics Letters B 74, 249 [10] S.N. Vergeles, Physical Review D 92, 025053 (2015).
(1978). [11] Here,byanalmost smoothsurface, wemeanapiecewise
[3] T. Eguchi and A.J. Hanson, Annals of Physics 120, 82 smooth surface consisting of flat four-dimensional sim-
(1979). plices, suchthattheanglesbetweenadjacent 4-simplices
[4] G.W.GibbonsandC.N.Pope,CommunicationsinMath- tendtozeroandthesizesofthesesimplicesarecommen-
ematical Physics 66, 267 (1979). surable.
[5] A.J.HansonandT.Regge,ProceedingsoftheIntegrative [12] Hence the statement that the square brackets give null
ConferenceonGroupTheoryandMathematicalPhysics, equation underthesum (4.5) is proved.
Universityof Texas at Austin (1978). [13] It was provedtobethecase for thequantityE(α−)W[mkl]
[6] R. d’Auria and T. Regge, Nuclear Physics B 195, 308 (see(4.5)).Thecorrespondingprovementforthequantity
(1982). E(α+)W[mkl] is thesame.
[7] S.N.Vergeles, JETP Letters 104, 494 (2016).
