JHEP10(2019)193 Hz 6 4 − Grand E 6 LISA. ologically Springer E theory and April 18, 2019 nd October 7, 2019 : and October 18, 2019 : : September 17, 2019 : Received Accepted ghter intermediate mass Published Revised ly peaked around 10 t of the symmetry break- o the Standard Model via SO(10) monopoles connected by flux ng gravity waves and we pro- al defects including magnetic rrying a single unit of Dirac a of Dirac charge can appear ble strings, a monopole-string s joined together by flux tubes R ry epoch. We also show the ap- ngineering, nomy, University of Delaware, Published for SISSA by https://doi.org/10.1007/JHEP10(2019)193 and SU(2) c b [email protected] , . It consists of SU(4) R . 3 . This spectrum should be within the detection capability of and Qaisar Shafi models respectively. These lighter monopoles as well as top SU(2) 12 a 1904.06880 6 × − The Authors. E L 10 Solitons Monopoles and Instantons, Gauge Symmetry, String c

We employ a variety of symmetry breaking patterns in SO(10) a ≃ , 2 [email protected] SU(2) h × gw c Newark, DE 19716, U.S.A. E-mail: School of Electrical and ComputerAristotle Engineering, University Faculty of of Thessaloniki, E Thessaloniki 54124, Greece Bartol Research Institute, Department of Physics and Astro b a cosmic strings ArXiv ePrint: Keywords: with Ω system can temporarily appear. Thisvide system an decays example by in emitti which the spectrum of these waves is strong tubes. Necklaces consisting ofare monopoles and also antimonopole identified. Even in the absence of topologically sta stable intermediate scale stringspearance can survive of an a inflationa novelSU(4) necklace configuration in SO(10) broken t Unified Theories to demonstrate themonopoles, appearance strings, of and topologic ing necklaces. pattern, We a show topologicallycharge that stable as independen superheavy well monopoletopologically as ca stable color monopoles magnetic carryingin charge two SO(10) or is and three always quant present. Li Open Access Article funded by SCOAP Abstract: George Lazarides Monopoles, strings, and necklaces in JHEP10(2019)193 r ]. ]. ], ], 1 3 7 an 1 10 14 17 18 17 20 – , 13 19 -plet of ], as we epends on 54 17 e with two – a 13 entations [ GeV. Another breaking via the 16 center of SO(10). 6 4 10 (422, for short) [ E nd some well-known Z ∼ R e example is the appear- stance, in breaking SU(5) e carries one unit of Dirac auge group yields a stable GUT s from SU(2) s depends on the Higgs fields rac magnetic charge [ M namely the lightest sparticle. × y a string (dumbbells) [ -parity, which interchanges the L C R ], and it weighs about ten times the 11 , R SU(2) SU(2) 10 × (333, for short). This breaking produces c × R L SU(3) – 1 – × L SU(3) ψ SU(2) × ] predict the existence of topologically stable magnetic × is precisely equivalent to matter parity which, among L 7 c 2 – U(1) SU(3) 5 Z [ × × 6 SU(3) c E ], and necklaces with monopoles acting as beads kept togethe × 21 c ], and its mass depends on the scale of the 422 breaking which c ], and 12 4 , 3 . In contrast, a 422 breaking to the SM yields a stable monopol monopole that carries one unit of Dirac magnetic charge [ ]. The mass and the magnetic charge carried by the monopoles d 3 ]. Consider, for instance, the breaking of SO(10) to 422 with symmetry in this case happens to be subgroup of the 9 Z GUT , 2 22 8 strings if SO(10) is broken to the SM using only tensor repres Z M 2 ], SO(10) [ Z breaking via SU(3) breaking via SO(10) 2 6 6 Composite topological defects can also appear in many GUTs a The presence of topologically stable strings in these model E E examples include monopole-antimonopole pairs connected b units of Dirac charge [ other things, provides a stable cold dark matter candidate, trinification group SU(3) walls bounded by strings [ on a string [ The gauge interesting example of GUT scale and lighter monopolesa come GUT scale shall verify later. Theintermediate subsequent mass breaking monopole of which carries 333 three to quanta the of SM Di g to the Standard Modelmagnetic (SM) charge gauge (and group, also color the magnetic lightestGUT charge) scale monopol [ be lower, even significantly so, than the standard GUT scale SU(5) [ monopoles [ the underlying GUT and its symmetry breaking pattern. For in In supersymmetric SO(10) this Grand Unified Theories (GUTs) such as SU(4) 6 Conclusions 1 Introduction 3 4 5 Primordial monopoles, strings, and gravity waves Contents 1 Introduction 2 SO(10) breaking via SU(4) Higgs. This leaves unbroken a discrete symmetry, called that are employed to implementance the of symmetry breaking. A prim JHEP10(2019)193 ]. ]. R 21 21 tion. ]. An string SU(2) 2 24 Z × where the klaces can L 2 we show the breaking via Z le monopoles, 2 6 is the mass per × based on pulsar E s SU(2) µ s ry to the SM group ss monopole carry- transition is around × Gµ c y of these primordial SU(5) nt bound from Cosmic mphasis on SO(10) and f necklace configuration ionless string tension is (4) connected by a es from a coalescence of , the Hubble scale during gnetic charge and possi- → c charge in SO(10) models, H nts on pear either in the same way ogous to the SU(5) case, this . This can come about if the in the universe. Another well we show the presence of the breaking via 333, which leads s. It is interesting to note that ed by charge conjugation [ a necklace made of monopoles U(1) 3 ] for some time that monopoles 6 ordial inflation. In addition we they are unstable and disappear we analyze the E × 26 4 ]. This breaking of SO(10) to 422 , d inflation or making an appearance 23 25 , SU(5) ]. 21 [ 28 → Q monopoles connected by suitable flux tubes. He has been reported recently in ref. [ R 3 – 2 – that is comparable to → − denotes Newton’s constant and string. In section I Q G 2 M Z -plet of SO(10). We will demonstrate the appearance models and discuss the and SU(2) 126 6 -foldings experienced by the GUT scale phase transition. c e E ], where 27 [ -parity, and the strings form boundaries of domain walls [ 7 C − ] for a recent discussion on related topics). In section 10 30 monopoles bound together by flux tubes in a dumbbell configura , × presents a quantitative discussion of how intermediate sca R 2 29 . and show how unstable strings as well as stable strings or nec 5 3 ψ . s -foldings experienced during the intermediate scale phase e U(1) the electric charge operator strings. However, the subsequent breaking of the 422 symmet Gµ 2 × and SU(2) C Z c Of great interest, of course, is the question as to whether an In this paper, we discuss topological defects in GUTs, with e (see also refs. [ 6 topological defects exist in nature, havingafter either the survive inflationary epoch. It has been recognized [ of a new type of necklace if SO(10) breaking occurs via 422. last step is achieved by a Higgs given by Intermediate scale cosmic strings,as on the the monopoles, other hand,Microwave or can Background even ap Radiation after measurements the on end the of dimens inflation. The curre 25-30, rather than the 50-60 number of inflation, may be present in our galaxy at an observable level necessarily breaks this timing observations have been reported in ref. [ is provided by the symmetry breaking SO(10) SO(10) strings, and necklaces in realistic models can survive prim GUT Dirac monopole also in presence of a GUT monopolewhich carrying is one independent unit of of Dirac themonopole magneti symmetry breaking carries pattern. some Anal color magnetic charge. We break the SU E associated with an intermediate scale known option, if available, is toobservation inflate of away walls the bounded domain wall byexample strings of in a necklace made up of monopoles and antimonopoles Under yields Such walls can be toleratedbefore in their realistic energy scenarios density becomes provided the dominant component This symmetry breaking patternconsisting of of 422 alternating also SU(4) yields a new type o left and right components of any representation, accompani A variety of otherand configurations is antimonopoles also connected possible by including a unit length of the string. Somewhat more stringent constrai to intermediate scalebly monopoles to with non-superconducting three stable units strings. of In Dirac section ma SU(4) symmetry to the SMing in two two units steps of and the show Dirac how an charge intermediate (Schwinger ma monopole) emerg appear. Section JHEP10(2019)193 ] π = ≡ 1) 3) he its 12 and − X × , X iπ/ ) O(10) L -parity 2 , D 1 ]. . Thus it , s ¯ 4 21 [ = diag(1 54 s to a rotation are the baryon along its C 3 L + ) = exp(4 c L T ]. It is known [ respectively. This a (4) 3 21 R iπQ onal combination is 45 and T along + B will generate monopoles s comes from ( irac magnetic monopole, π her hand, the 422 singlet and R 1 ves to the Dirac monopole , which belongs to SO(10), . representation of SO(10) is ssible with the space based ], was later called 6 and C X R d loop, which corresponds to = 210 c n disagreement with table III 21 le corresponding to a rotation n gnetic field. U(1) and exp(2 t of magnetic charge as well as 210 rtainly equivalent to taking the d as in ref. [ × ty, and therefore three lines, one 10 c ment. Indeed, the group element is the electric charge operator). It L along × − ), which are both symmetric under Q SU(2) B 1 R , 10 ]. The 3 ( 1 × / , 2) in SU(3) 8 31 -parity which interchanges SU(2) c L U(1) 6 − T ( C × , + L 1 × , Q ) . The latter is the antisymmetric combination , we take rotations by and SU(2) 1 -plet for the SO(10) breaking to 422 so that no – 3 – , R SU(2) c 1 2 = SU(2) 210 , / 210 × 3) acting on down-type quarks and so the combined 6 × 3 R c c T = diag(1 , and so the 422 singlet in iπ/ since color is unbroken. It is easy to see that 3 is exactly the Dirac magnetic monopole as previously 8 c 2 -parity, first found in ref. [ / 2+ ) or ( in SU(4) T L 8 − / c C 2 and SU(2) 210 ( 3 π L T , -plet does not break the discrete symmetry − L + c T 2 + , 54 and the curve between 1 and -1 corresponds to a rotation by B 1 45 3+ 3) lies in the center of SU(3) Q c ( / + its conjugate. One combination of these singlets gives the S 8 c × 1 iπ/ T ) × ) 2 = ]. Note that, in this paper, the SM was embedded in SU(5), but t , 1 2+ 1) respectively, and the overall loop therefore correspond , 2 11 / 16 , ) 2 , − , 1 , L × , and so the 4 10 1) in SU(4) − 3 generator, where − / 54 B 16 , 8 c 3) = exp(2 1 T / -plet of SO(10) comes from the product ] where it is stated that the monopole is unstable.) − 8 ]). This is clear since the SO(10) singlet cannot break c , and conjugates SU(4) = diag(1 30 1 54 32 along the generator )+2 πT , 3 along ( R R 2 L i T π To make the analysis more transparent, we take the curve in SU The next breaking of 422 to SU(3) We choose here to employ a Higgs π − . One combination of them is the SO(10) singlet and the orthog B SU(2) contained in conjugate and ( of these singlets and thus breaks the discrete singlet, and the other the 422 singlet in We will first study the breaking of SO(10) via 422 [ corresponding to rotations by 2 originates from ( argument holds for any compactalong with group the containing ordinary SU(5). magnetic The field, D also carries color ma after the electroweak symmetry breaking.some It color magnetic carries charge. one uni (Thisin conclusion ref. appears [ to be i that the (-1,-1,-1) element ofin 422 each coincides of with the the threea groups identi magnetic between monopole. 1 and We -1 will constitute now show a that close this monopole evol 2 SO(10) breaking via SU(4) contained in and thus it is boundin to the be the symmetric combination. On the ot C observatory LISA. Our conclusions are summarized in sectio discuss how gravity waves emitted by some defects may be acce and along this generator. In SU(2) in ref. [ diag(1 is clear that this rotationexp( brings us back to the identity ele ( by 2 and lepton number operators respectively.curve This along choice is the ce generator strings or subsequent walls bounded by strings are generate acting on up-type quarks or exp( rotation leads to the identity element.by 2 The magnetic monopo shown in refs. [ JHEP10(2019)193 , , is , X X π  (2.2) (2.1) (2.3) (2.4) ) -type c 3 R ν -plet of T 45 + monopole. 3 along . c } X π/ ( 1 monopole have { π 3 2 R i ) =  Y ole corresponds to a 3 and 2 ) VEV could also be 1 its periodicity is 2 enerated between them flux from it. The same , π/ ), combined together to . V) of a Higgs ft, which means that no nuous subgroup, i.e. the 1 U(1) 3 = exp ) R , are, respectively, ether to give the Coulomb , we must define the GUT 3 . Note that the 1/3 of the R T flux. This latter monopole,
× 3 -type SM singlet component R 1 by 4 T 3 B c R 15 3 T L R 2 ν monopole to contribute to the and an SU(2) T ux corresponding to a rotation πT + . c motopy (fundamental) group of − R 2 3 and R i X X T − SU(2) represents right-handed neutrinos). ( U 3 along monopole to the SU(4) 1 2 6 c × π/ ν 1 R √ = c 2 exp monopoles can be connected by a string R 3 and Coulomb fluxes along the unbroken × R 3 is left unbroken by the VEV of the , / ) ) = )  3 monopole, which carries a full flux along R R ) π/ R (SU(3) 3 and 4 3 T R 0 c Q Q – 4 – X,Q 2 T 2 π π/ unbroken ( 2 + √ − + ) components. The ( 1 = 3 √ R X or SU(2) 1 2 − T , X X  c Q ( 1 2 X + = Y , π 3 √ Q 2 X ( ( X and becomes a Schwinger monopole after the electroweak i 15 ] which can transform into Schwinger monopoles. 3 3 Q 3  U(1) 1 1 R 20 √ √ monopole which carries a full monopole towards the SU(2) T × , c R + = = flux sent from the SU(2) L 19 ) and ( 3 = exp B X R U -plet. ). The magnetic fluxes of an SU(4) 3 T 3 , R  1 ) T SU(2) SO(10) , 210 3 2 R 1 . To this end, an SU(4) × T − 3 2 R c T X − -plet, which leaves + X ( 16 SU(3) X π 3  2 . A rotation along this generator by 2 1 ) rotation along i 3 3 along ( R π π  T Note that The smallest broken generator with integral charges so that We can further break these two U(1)’s by the VEV of the π/ 2 flux sent from the SU(4) − exp which pulls them together.flux The of rest of a the doubly fluxes charged are (Schwinger) added monopole tog — see figure in turn, sends 2/3 of its flux to the other one and thus a tube is g sends 1/3 of it to an SU(2) generator to be rearranged in tubes with flux ( to their respective antimonopoles and annihilate. breaking. Needless to say the SU(4) emerge as Coulomb fluxfull from (2 the combined monopole. This monop and two corresponding to rotations by 2 normalized generators by 2 Higgs. Therefore, the generated string contains magnetic fl X Finally, we have four fluxes (two corresponding to rotations To find the broken generator which is perpendicular to Then the normalized unbroken and broken generators X SM group. Consequently,topologically no stable unbroken strings discrete arethe produced symmetry vacuum since is manifold the le first ho and thus this unbroken element belongs to the unbroken conti We only have dumbbells [ SO(10) along its ( breaking can be achieved by the vacuum expectation value (VE tube in between cannotis terminate true on for it but the emerges 2/3 as of Coulomb in a Higgs taken from a Higgs JHEP10(2019)193 u 3 π 1, → h ( − π/ R (red) , d c h 3, 4 , = 1 U(1) and +2 u π/ 3 R h × π 4 T 2 ]. We obtain L − which are the -type VEV of − c the string. For d 33 ν ogether to form a h -type Higgs field c , hase SU(2) ν u of Dirac charge from h = 0 and × L X − change by B d

h oulomb magnetic fluxes of the ) tribute to superconductivity. e that the U(1) 3 ergy dominates over the Higgs perconducting [ e string via × and es not affect the so that the string corresponds to c + + as this one may survive inflation. u ing, the Higgs doublets

3 L 8 h ( T 1 3 2 3 SU(3) +

→ ) 3 monopoles, we can have a network of topo- 2 R

R 2 = – 5 –

3 SU(2) , the phases of ( × 3 L to the down-type ones) with 1 3 L + + T d

h + ) and SU(2) 3 3 R c SU(2) T 2 × + + − c ( X 2 3 . This monopole also carries color magnetic charge. An SU(4) Y U(1) 3 along × π/ L (blue) monopole are connected by a flux tube which pulls them t -plet and also adds the minimal necessary magnetic energy on R 16 1 respectively develop VEVs. As we circle a string they get a p . Emergence of (Schwinger) magnetic monopole with two units SU(2) , 1 × − c If we inflate away the SU(4) = 3 L couples only to right-handed neutrinos and thus does not con only Higgs fields coupling to quarks and charged leptons. Not left-moving and right-moving fermionic zero modes along th monopoles are indicated. Intermediate mass monopoles such Schwinger monopole. The magnetic flux along the tube and the C Figure 1 logically non-stable strings. After the electroweak break and an SU(2) couples to the up-type quarks and the symmetry breaking SU(4) T SU(3) the Higgs respectively around the string. Of course, this addition do a rotation by 2 respectively. If we add to the string 1/3 of flux along definiteness, we will assume throughout thatcontribution the to magnetic the en string energy and so these strings are su JHEP10(2019)193 6 R × c π/ and c , where . Note and an 2 2 Z c monopole × c monopole or Y c ining Coulomb monopoles). We U(1) R × L . We display explicitly . An SU(4) 1 3 to do the breaking of ] — see figure R T

e string and two “loose” 22 ) SU(2) agnetic flux along one of the ing to a rotation by 2 these latter strings to other  3 126 × arrangements of SU(4)  c leads to an unbroken element. laces [ 3 3 along R + + T SU(3)  π/ 2 ( → −

1 3 R X  antimonopoles (SU(2) 3. Now if one imagines opening up one / c  ) U(1) 3 R × T antimonopole connected either to an SU(2) monopoles from the symmetry breaking SU(4) L − – 6 – + 6 along R R -type component of  c X π/ ν

c SU(2) 1 ) ν ×  3 L  antimonopoles connected by a string can also participate − antimonopole connected either to an SU(4) and SU(2) B -plet of SO(10). Notation as in figure R c 3 and 2( c + + / ) 126  3 U(1) R ( T × 2 3 c + antimonopole. Also both tubes emerging from an SU(4) X and SU(2) c c SU(3) → R antimonopole, and the SU(2) minus flux corresponding to a rotation by 2 R antimonopole) can terminate on SU(4) . Necklace with SU(4) . In this case, a rotation by 2 SU(2) monopole are then connected by two such strings with the rema monopoles and antimonopoles. 3 R × R X T R R L 2 Now suppose that we use the − of the two strings,strings one emerging finds from the the two two monopole.similar monopoles One monopole-string connected can structures then by connect in on series and form neck flux in them being ( (SU(2) This yields a string which contains magneticSU(2) flux correspond along SU(2) monopole or SU(4) thus see that a variety of necklaces can appear with different X in the necklace with the SU(4) the last step is achieved by a an SU(2) that pairs of SU(4) only the Coulomb magnetictubes. flux This of necklace two may of survive the inflation. monopoles and the m SU(2) Figure 2 JHEP10(2019)193 c 2 e L − 1) → Z 3 , R 1 T (2.5) com- com- R -type 2 − c , -parity, ν − 1 c C nd an ex- ν U(1) − X monopole, strings too. string, con- and SU(2) × c 3. Since the 2 c (1,1,3) 2 / L (15,1,1) ) Z  does not affect Z 3 3 R 1) = ( R T T 6 along and − enerator of the , π + 2 1 π/ . The first breaking, SU(2) terminate on SU(4) 2 reaking. Recall that ith a n SU(4) , 2 i , X (333, for short) by the × Z − monopole which carries 1) c 1 respectively. As we go R ×  , of Dirac magnetic charge, − c (1,1,1) 1 Y ed monopoles may also be , 1 − ng to the SM group since its exp , d monopole coincides with the SU(4) ings are not superconducting. SU(3) = × (1 U(1) y the VEV of the 3 L × es above, correspond to a rotation   × → R T ) ) L L 3 3 R R 1, T T 3 respectively. If we add to the string changes phase around the string. But − string. Namely, an SU(4) + + , π/ monopoles can form, together with the SU(3) SU(2) 2 SU(3) X X c 1. Moreover, since (1 Z 126 ( ( × = 1 × strings without monopoles on them are also -plet along their × − c c π π 6 6 3 2 R 3, +2 2 2 45 L Z i i T – 7 – π/ , rearranges its magnetic field to form two tubes   ) = 2 X SU(3) iπ − → SU(3) = exp = exp R = 0 and ×  -plet and a c X ) 3 R U(1) . T yields exp( 3 210 × 2 magnetic fluxes. The last breaking is done by the -plet and the SU(4) -type component of c and also adds the minimal necessary magnetic energy along th have L c ν − L ν d ]. Its square is then obviously equivalent to the identity, a c − and are not oriented. The necklaces are themselves 126 h X ν B -plet, yields an intermediate scale SU(4) 126 18 ( 3 , 6 and a Coulomb field around it with flux 2( R [ u -plet, which contains two 333 singlets. One of them breaks / such that the string corresponds to a rotation by 2 π SU(2) T 6 ) L 45 h 2 2 3 L 3 R × i − T − T B 650 L  2 − X B − and U(1) remain constant around the string. Of course, this addition exp is unbroken at this stage, this element is equivalent to the g c X d L h U(1) breaking via SU(3) -type VEV of monopoles are inflated away in this case, these tubes can only monopole we mentioned above since the corresponding loops i , c × 6 along symmetry remains unbroken. Stable u 6 ν R R c h c 2 The group element For an example of a monopole-antimonopole necklace formed w Next let us see what happens after the electroweak symmetry b π/ E ν , Z can break to the trinification group SU(3) 3 L 6 T subgroup of U(1) and SU(2) the 2 antimonopoles — see figure VEV of a Higgs SU(2) which we obtain by circlingaction one on of the these SM strings singlet does not belo -1/3 of flux along 3 are homotopically trivial. The second breaking, achieved b produced and inflated away. InSU(2) particular, the doubly charge ponents respectively, produces awhich GUT presumably monopole with is one inflated unit away. Of course, multiply charg E with flux ( ponent of a Higgs which carries a full magnetic flux along achieved by the VEVs of a string. Thus, only the antimonopoles, a necklace tied together by a both SU(3) sider the following SO(10) breaking pattern: SO(10) SU(3) around the string they acquire a phase component of a Higgs this couples only to right-handed neutrinos and so these str the Higgs doublets by 2 present. These strings, exactly like the ones in the necklac tra JHEP10(2019)193 y L × c (3.3) (3.2) (3.1) SU(3) ct → R are generated. ). The sum of 3 C , U(1) 1 , will be broken, and , × ) 3 3 L , ) goes to itself, while C ( c , 3 1 , Q × ,

¯ 3 ) 3 ) , + SU(2) ¯ 3 1  3 )( , e and the magnetic flux along Q × ¯ 3 1  , , L + 1 ¯ 3 − , d away. We display explicitly only λ B + + , in this case, exchanges SU(3) . ¯ 3 -plet of SO(10). We assume that the ] associated with ( C ≡  ( 21 ) − U(1) 126 . 650 ¯ 3

) ) 2 3 , × ), and ( ) 3 1 + 1 c , , 1  3 , , 1 ¯ 3 ¯ 3  78 , , ¯ 3 , ¯ 3 3 + ¯ 3 2 SU(3) )( )( ( 1 ) + ( is 1 ¯ 3 − – 8 –

, , → × 1 6 = , 3 1 )  R , , E 3 1 ( , 3 ¯ 3 ,

27 ( ( 3 1 6 3 ) , U(1) ×  3 − − 3 singlet. The other two orthogonal combinations are in ×  ) ) ) + ( 27 L 6 ¯ 3 1 ), ( 3 monopoles (red) and antimonopoles (green) from the symmetr , , E , ¯ 3 + + ¯ 3 3 c , ¯ 3 , , ,  3 1 ¯ 3 SU(2) , 1 ( )( )( 1 × ( 2 3 1 3 c = ( , , × 3 ¯ 3 , , ) . Both these singlets can acquire VEVs and thus 27 ) are interchanged. There are three 333 singlets in the produ 3 1 3 , where the last step is achieved by a C ( , 2 SU(4) ¯ 3 2( , ¯ 3 Z , 1 → × , 1 has no 333 singlet. They could be ¯ 3 Y and conjugates the representation, in which case ( R 78 U(1) . Necklace with SU(4) and ( × L since The fundamental representation of but the other one does not. Note that the symmetry monopoles from the firstthe step Coulomb of magnetic symmetry flux breakingone of of are the one inflate tubes. monopole This and necklace one may antimonopol survive inflation. and SU(3) (3,3,1) SU(2) breaking SO(10) They are the ( Figure 3 650 these three singlets gives the we expect that no strings or walls bounded by strings [ The latter violates JHEP10(2019)193 , , c H Q 2), 3), 333 − (3.5) (3.6) (3.4) (3.7) of π/ 3+ , of the 2 → / 1 3 i 8 , c 6 3 Z − T E Z is the weak ) = . In SU(3) . As shown in R Y ) = c H 3) respectively. along ( ( along the gener- 1 H , similarly to the π = diag(1 netic charge. π/ ( 3), exp( 6 π 1 2 π 27 i E 8 π R π/ along this generator − T 2 i π , i.e. three curves in the , and c ) = R Q, 3) in SU(3) netic flux or charge as well and , exp( tates in /H =   π/ L enter of SU(3) . 6 c 2 3 3 L R i E T ( T coincides with the identity ele- , g vial, and the breaking monopoles, we define the gen- − = 2 monopole in c 1 2 2 1 π 3 3 ) is a rotation by 2 H c , d + + Z Z 3), or 1 to exp( c ’s of SU(3) H 8 u R ( Y 3 2), which leads from 1 to the element π/ 1  T 2 − π        = 6 1 i 0) of SU(3) c , c = 3 , 1 e ν R − c + 1 , T d u 3 Q L l N − 2 1 h h antitriplet respectively. , T . Similarly, a rotation by 2 3), and a rotation by 2 2 1 L – 9 –        + R / -plet and, consequently, on all the representations 8 R 1 + = and T − 8 L 27 λ = diag(1 1 6 , along these generators closes a circle. We see that T 3 8 c 1 6 / + and the columns    = diag(1 T π 1 3 L + L . Note that we use integer elements in these defini- q g 3), or 1 to exp( − 3 T L 8 , R c 3) in SU(3)    2 1 T 3 π/ T / 2 π/ 1 3 = i + rotation along the generator 2 8 2), i 3 along L Q π interpolates between 1 and exp( T − − is then a loop that connects (1,1,1) with π/ , 1 6 magnetic monopoles. 3 R 1 ’s of SU(3) , triplet and an SU(3) T 3 ¯ 3 /H = diag(2 Z 6 L 2) 0) of SU(3) / 3 , E L 1 T of the unbroken trinification subgroup − 2) , + (1 /  = diag(1 8 R ] this is the ordinary Dirac monopole also carrying color mag 3) T 8 L 11 3). The generator of the first homotopy (fundamental) group + (1 6) T π/ , / 2 8 L π/ i . The generator of the second homotopy group 10 T 2 6 i − = diag(1 In order to understand the structure of these One can very easily verify that the element 6) monopole in SO(10), carries one (Dirac) unit of ordinary mag E / 3 2 R can be represented by a 2 exp( we take a rotation by 2 therefore generates ator (1 brings us from 1 to exp( erators (1 tions so that a full rotation by 2 T Obviously, the third power of this loop is homotopically tri three SU(3)’s from 1 to exp( of exp( ment as it acts like the identity on the vacuum manifold which are an SU(3) as color magnetic flux corresponding to the generator of the c refs. [ It is easy to check that Z with the rows being exactly as in the SO(10) case. As a consequence, the the electric charge operator, by applying it on the various s where hypercharge). Finally, we see that the generator of JHEP10(2019)193 ]. Y 4, / 18 ) (3.8) (4.1) (3.9) ψ ) = L − + 5 to U(1) B , for short) χ L ial loops in L − , we get an singlet in the − 8 B R 3 = 2 and has = ( 3221 unbroken. So, B / T L charges for the ′ / ) ]. This breaking L superconducting ψ 8 L + ψ − U(1) 34 T . B 8 (333 L × L (3221 + -plet. The generator 2 T strings as in ref. [ − R π 8 L R B 2 27 − T Z , where B ′ U(2) e ψ = ( ) component. In particular, does not generate any new 6 rotation along the generator L , y 3221 Y U(1) ct, see ref. [ ectroweak Higgs doublets have π ¯ 6 , U(1) − × Higgs doublet, analogous to the , 1 . R B × = 2 and the SU(2) R ) (1) . This can be achieved by a Higgs 8 R -type component of it. This belongs L ) in a Higgs subgroup of U(1) c ψ 16 T to U(1) invariant and thus remains unbroken. 3 in the previous section. ν − 2 , SU(2) L L ¯ 3 SU(5) Z N B − , − × ( U(1) , 2) + 1 B B 8 2 1 L R → − × T ( subgroup of SO(10). The ψ triplet with − U(1) , which belongs to the continuous part of the ) = χ 10 -plet with ψ along the R 8 8 L – 10 – L R SU(2) 6 × U(1) T T × R 27 R SO(10) U(1) . This rotation corresponds to the group element × c + becomes homotopically trivial and can be removed. 8 (4) + L U(1) 8 L c × 1 T → 4 along the unbroken generator T = 1 and generates no topological defects. ( × 6 = ) with a VEV along its ( remains, which corresponds to a monopole with triple the 1 6 π/ L E 27 π subgroup of U(1) 27 1) in SU(3) . It thus leaves the − -plet are given in parentheses , c × . This means that no additional discrete symmetries are left 2 1 ν B L 2 along Z 27 c -type component of ( by 6 27 − − ν , π/ N B 1 Q − -plet in the SU(3) 4 along this generator leaves = 4. This is an SU(2) 3 8 L π/ R T , which means that it consists of the homotopically non-triv doublet with (contained in 333 with R ) ′ 2) = diag( and thus remain constant around the string. The string is not L L / ) remains in the unbroken subalgebra since , and so the loop in SU(3) which are trivial in 333. The minimal loop is a 6 8 − L L breaking via SO(10) Q 351 B corresponding to the SU(5) L 3.6 − πT 6 − χ 2 B B i We could alternatively use for the spontaneous breaking of S The second homotopy group of the vacuum manifold The subsequent breaking of SU(2) We can further break 333 to SU(3) E 3 + / -plet. We can further break SO(10) (3221 of SU(3) 8 c 1 to an SU(2) a Higgs 78 T electroweak breaking — for a detailed explanation of this fa with SO(10) components of the 3221 Only the rotation along topological objects provided that it is done by an SU(2) ordinary magnetic charge and no color magnetic flux at all. π Let us now turn to the case of zero However, there are no necklaces in this case. Note that the el can be achieved by the VEV of a Higgs by giving a VEV to the just as the string from the we take the SU(2) ¯ 6 unbroken subgroup 3221 exp( in this case, in addition to the two types of monopoles, we hav unbroken. In other words, the unbroken subgroup is precisel the quantum numbers of 4 in eq. ( The orthogonal broken generator is but a rotation by 2 Adding to thisequivalent a rotation rotation by by 2 2 JHEP10(2019)193 . , l ], ψ ψ → 10 35 Γ (4.4) (4.3) (4.2) (4.5) (4.6) i 16 − U(1) together × ) rotation . ψ (-5,1), and π . 1  . 1) ) , to only one of π 69 1 ) is of negative ¯ 5 σ SO(10) 3) in SU(5). It is − is 2 ( 4 along coincides with the 10 + − ′ → ′ 4 corresponds to the . + 3 π/ ψ 10 ψ 69 78 charges given as 6 / . This corresponds to direction has elements ) − σ ′ ¯ 5 σ E of the center of SO(10) ′ ψ (1) , ψ ψ 78 charges are the minimal 2 ψ . Now + + 1) + , 10 σ + ′ 10 1 , 2 1 Γ 45 ψ χ 45 , U(1) i king ponds to a full (2 (3 σ σ → − ¯ 5 × 12 again is the identity as one can charge since it gives -5 for the 5) in U(1) we use the notation of ref. [ + 1 (2) + σ . ’s being -1, the ¯ 5 ψ χ 12 π/ -plet ( 6 σ 03 along ( 2 σ 1) + 4 with the i and ψ σ √ , 16 / π = diag(2 2 5 + ) . A rotation by 2 4 along it. = ¯ is achieved by the VEV of (0) + − 10 ψ Y 10 along 03 = 9 ( ′ 1 π/ Γ σ 1 Γ ψ π 2 i ψ ( 6 center. This center is generated by + 5 − − Γ → − 2 4 iπ χ 8 3) + to three of them being -1 and the rest +1. So Z 2) + U(1) Γ , it is more transparent to use rotation along  ,Q − 10 7 = ( χ ( − × – 11 – Γ . So the generator , 10 ′ ¯ 5 10 5 2 ¯ 5 χ ψ Γ √ − 4 and ]. The SO(10) 2 acts on the SO(10) representations as follows: is homotopically trivial in SO(10). ( = exp corresponds to all Γ . A rotation by 2 and also an SO(10) flux corresponding to the inverse 2) + ¯ 5 ψ 2 SU(5) 36 χ ψ = ψ ’s coincides with the 69 1 − Γ ψ σ ( we divided by 4 so that the 1 χ is a closed loop in SO(10) iσ → 5 ′ Γ 4+ Q 2) + χ 3 78 direction has no common elements with the center of SU(5) ψ / ψ ) − Γ charges), so that the periodicity of U(1) , and 3 for iσ , 0 ψ ψ 4 along Γ 45 ψ (2 (5) + 10 . A fourfold monopole, i.e. a monopole with magnetic flux equa 5 U(1) + correspond to the following GUT normalized generators i 5 1 π/ iσ 10 χ and, consequently, × and thus coincides with ψ Γ = 4 along 12 = and i 1 = ( 5) of the center of SU(5) with iσ 16 4) + 10 π/ χ i ′ 27 , Γ 03 and → ψ ¯ Y/ (0 → iσ χ 1 π 1 intersects with SO(10) in its 2 , , -1 for the charges. The ) charges of its SU(5) parts are i = 1 16 = ψ and corresponds to a rotation by ψ − 10 , 10 χ charges of the SU(5) singlets are 4 and 1. However, the χ, ψ 10 27 Γ ] that i ψ Γ i → − 37 , since a full rotation along − . Note that ψ ′ 10 ψ Also, a rotation by 2 Instead of using rotations along The U(1) , 16 with a rotation by 2 is the chirality operator in ten Euclidean dimensions. Here lies in U(1) to four times the flux of the minimal charged monopole, corres the smallest charge magnetic monopole generated by the brea i generator of its center It has 1/4 of magnetic flux along along where SU(5) singlet thus the unbroken U(1) corresponds to Note that in the definition of under and integer ones (as the them being -1 and all others +1, and the identity as we have just seen, and a rotation by 2 The breaking of SO(10) It is easy to see that the sum of which follows the notation of ref. [ while the ( known [ chirality and so the SU(5) singlet element exp( see from the which coincide with the center of SU(5). Namely, exp( since the JHEP10(2019)193 ′ × ψ s of (4.9) (4.8) (4.7) s, no (4.10) 4? As lead to / ) ′ coincide. (SU(5) charge of ′ 0 ′ ′ χ 5 along χ π χ χ π/ + 4 flux of the ′ 4 fluxes of the and also carry / cuum manifold ′ / ψ and ′ ′ 4, which are not χ ′ ψ / ψ ′ 4 along ψ , χ − -type component of , 1) c π/ 4 = ( } ν symmetry remains, of / les with magnetic flux 1 − . ) ( { ′ by the SO(10) singlet in le. We therefore obtain belongs to SU(5), as we χ ψ ψ ′ , ′ 5 along rotation by 2 4 and 10 ψ ψ and thus the unbroken sub- too), the ψ / + ) = ′ π/ ′ ′ , ′ χ ψ , ψ by the ψ wever, new monopoles appear ψ subgroup since the 6 ′ he story proceeds as usual with ′ 10 ′ (2) + 4 ψ √ χ ψ ¯ 5 √ subgroup contained in the center Z 2 U(1) 2 subgroups from 5 × charges. It is, as we will see, more 4 = Z U(1) = 4) + ′ Z   5) of the center of SU(5). We are left at × rather than χ − ψ ψ ( ′ , -plet? The U(1) Q ′ ¯ 1 χ Q Y/ ψ , π (SU(5) 16 subgroup of U(1) − ′ 15 belongs to U(1) 0 2 i ψ χ 5 π 4 breaks to its √ 1) + rotation. Z Q ′ Z – 12 – = − χ + π ( 15 χ  ¯ 5 together with a rotation by 2 √ ′ is Q ψ ′  ′  is a 2 ψ 1 4 ψ 4 1 ′ (2) + χ = U(1) 5 = 6 ′ ′ charges are × χ E ψ ′ Q monopoles. Next we can break U(1) Q χ 1) + , which is connected, i.e. its zeroth homotopy group 4 along ′ ] which disappear. The Coulomb ′ − ψ ψ subgroup (which belongs to U(1) ( π/ 20 are -1 and -4. However, the SU(5) 1 4 , monopoles to antimonopoles leading them to annihilate. Thu  direction has no common elements with SU(5) since the charge Z = 27 U(1) ′ 19 1 ′ 4. The flux. Indeed, since the π ψ χ / × ′ ) 27 ψ ψ − . Therefore, the first homotopy (fundamental) group of the va } χ 1 -plet, i.e. the singlet in its SO(10) -plet which, of course, leaves unbroken its { 4 and obtain a network of cosmic strings with magnetic flux / 27 27 = (3 ) breaks to its ′ ′ ) = -type component is -4. However, this ′ χ ′ χ c ψ χ What happens in the next breaking to SU(5) It is interesting to note that we could inflate away the monopo Note that the How about the previous monopole with magnetic flux ( ν + ′ ψ the course, unbroken, but the orthogonal U(1) U(1) group is just SU(5) the identity element as one can see from the convenient to use the orthogonal generators a Higgs topological defects survive at thethe end. breaking From of this point SU(5). on, t which connect the this stage only with these a Higgs ( showed above, these monopoles correspond to a rotation by 2 of SU(5). Then strings are formed with flux corresponding to a SU(5) flux corresponding to the element exp( Namely, a rotation by 2 monopole is confined tounstable a dumbbells tube [ and connects it to an antimonopo which carry U(1) monopole and antimonopole, of course, cancel each other. Ho such that the full rotation along U(1) the SU(5) singlets in with and the orthogonal generator is Then the normalized generator for and no stable strings will appear at this stage. JHEP10(2019)193 , , c e Y 27 , urth c -plet must . g 2 which which (4.11) i in the E change c ij 16 c u , ν ). Also, ). Then, ¯ 5 α d ν e h h to remain h E , , changes by ( , i around such 3) which are u E 4.11 d N c , N h h changes phase π 2 h D , c ν ) by e down-quark and = 1 at hase of 4.11 3 respectively, change portional to also remains constant, in the SO(10) i, j ( − mber of transverse zero d , very similar analysis can h ij 2 10 . The coefficients ). Then the masses of the , ′ ak breaking, the phases of changes by 2 α − move from the mass matrix c ψ gh to the minimal necessary in- in eq. (    E o the one in eq. ( N te from the VEV of ero modes are generated along = ( onducting. d d ′ D , D ψ -type component of the Higgs    5 and SU(5) flux to get a network D / N ′ -plet, which we call    , and U(1) c d ψ ′ d 10 h ] which says that, if a particular mass d χ 1 and , ij 3 blocks. Three of them are of the order of 38 c − by the , × D , h ′ c around the string. However, this string is not ψ N, α ν violating SM singlet VEVs, i.e. = 2 π    ′ – 13 – ′ 2 ψ  χ − c remain constant around it. Only the VEV of the -type VEV contributes to their masses. c d , d 5 respectively. If we minimally add to the string -2/5 ν c h in the SO(10) -plet, which we call -type charged leptons) exist not only in the , D π/ e ¯ 5 u but multiplied by coefficients 10 4  5. Recall that the phase of h and U(1) / d − ′ = ′ h remain unaltered around the string, so all rows and columns ψ χ d , the Planck mass. This is due to fact that a direct trilinear d P , which have h as indicated with constant unsuppressed coefficients. The fo M d m 5 and d h respectively. Adding minimally on the string 2/5 of flux alon h π/ and π remains constant around the string, while the phase of and N 5)2 -type quarks ( u in the SO(10) , and c u remains constant. The phases of the electroweak doublets -type charged leptons) exist not only in the / c h c d , the VEVs of 3 h c 5 ν e ν Y − , N ,( π -plet, but also in the 16 5)2 / 2 20 magnetic flux is added along these strings in order for the p . Again, we have no zero modes from the up-quark sector. For th Recall that We could also inflate away the monopoles with We can now apply a theorem given in ref. [ / − ′ π -type Higgs field changes its phase by ψ 2 c -type quarks ( ν around the string by 4 could generate zero modes since the a string, while around the string doesbe not done generate for zero the modes in charged this leptons case. by replacing A of flux along 2 of strings with magnetic flux respectively. We conclude that these strings are not superc remains constant around the string,the and thus string. no The transverse z down-type quarks and charged lepton, althou by ( constant around them. Thiscrease addition of certainly the corresponds magneticthe energy Higgs of fields the string. At the electrowe can be removed and no zero modes appear. We see that the fact th we can remove the rows and columns which contain elements pro matrix element remains constant aroundthe the row string, and we the canmodes. column re In that our contain case it when calculating the nu we then see that SO(10) a superconducting. Indeed, the up-type quark masses origina − topologically stable. At the breaking of U(1) charged lepton sectors, we can write mass matrices similar t Yukawa coupling is forbidden, in this case, by U(1) then necessarily contain U(1) suppressed by powers of is proportional to the VEV of down-type quarks can be schematically written as where the mass matrix is given in terms of four 3 d but also in the the VEVs of JHEP10(2019)193 , , 2 2 ′ Y Z Z nge , so ′ have 351 ψ (4.12) klaces d strings . Thus, h ong 2 -foldings d 2 . Again, , c e ′ Z u ψ at inflation D h 25 . − monopoles and 2 ′ Z ψ 8, the phase of the / ′ monopoles we obtain ) = χ uring the breaking of ′ 2 ), but merely rotates it. ψ Z ak doublets -type component of opoles from 422 breaking c × during observable inflation to the SM can also survive ν ′ s of the product 4.12 ese c e SM via the 422 subgroup, superconducting. man-Weinberg potential and superconducting. ψ ν er density of these intermedi- H an survive inflation in realistic subgroup of it is in U(1) -plet to break U(1) 8. The monopoles can then be 4 . As a consequence, / 27 2 U(1) ′ Z 4 flux of the monopole with total Z in eq. ( χ 5 as we have seen above. The VEV . This matrix also does not change / × ′ / d 2 ′ strings, i.e. the string and antistring × χ h Z ψ ′ 2 ij ψ Z α (SU(5) 0 U(1) center of SO(10) is broken, we do not have on the string with flux π ′ 2 × strings themselves. We can also have simple is achieved by the = ψ – 14 – Z ]. This model predicts that the tensor-to-scalar ′ GeV, which has important implications for pri- 2 ψ . In practice, one needs about 23  Z 39 2 14 H , Z 10 U(1) 25 ∼ 5 magnetic flux which connect the × / ′ I − × ′ ψ ψ M 13 6 E ] — see below. By the same token the intermediate scale U(1) remains unbroken. But the 3 respectively and thus, around the string, their phases cha ′ 26 × 8. Note that these are χ − , / , ′ monopoles will also appear at this stage as before. 2 χ 25 ′ − remain constant around it. In summary, these strings and nec 8 without monopoles on them since ψ / SU(5) d ′ = ]. In other words, the Hubble parameter h ′ χ  5 respectively. If we minimally add to the string 1/5 of flux al subgroup of the center of SO(10) remains unbroken. Assume th , 1 40 ψ 2 u 4 splits into two tubes, each with flux 3 matrix which is proportional to π/ π / Z h ) × ′ 02 [ . ψ 0 1 and 5, +2 -type component does not break the extra + & subgroup of U(1) − ′ may survive inflation if N , π/ 8 χ r I 2 Now if the breaking to SU(5) We can further use the SO(10) singlet in a Higgs Z − M = 2 -type component remains unchanged around it. The electrowe ′ 5 Primordial monopoles, strings,As and previously mentioned gravity primordial waves monopolesmodels. and strings c Consider, forsuch instance, that the the breaking of SO(10) to th survive even after the electroweak breaking but they are not the VEVs of by strings with flux connected to form necklaces which are is driven by an SO(10) singletwith scalar minimal field coupling with a to Higgs gravity or [ Cole are formed with flux χ N antimonopoles and lead them to annihilation. If we inflate th of the flux ( coincide (they are not oriented). In this case, the a network of non-superconducting strings with flux we obtain flux tubes carrying actually the unbroken subgroup is SU(5) no transverse zero modes exist and these strings are also not the Needless to say the ratio Actually, if we add 1/40 of the flux along cosmic strings which arethe produced during inflationary the epoch. breaking of In 422 the case the phase around the string as one can see from the various charge leaves the 3 for adequate suppression andate still mass leave an monopoles observable [ numb is estimated to be of order 10 at mordial monopoles and strings.SO(10) to 422 The are GUT inflated monopoles away, but produced the intermediate d mass mon JHEP10(2019)193 ] ] 14 39 41 − with (5.3) (5.2) (5.4) (5.1) 10 η e × . In this he subse- being the 2 in ref. [ 43 r by another ce between λ . t v t at an observ- = 1 ] or with A 3 39 / 2 ) volution of monopoles rg potential of ref. [ 333 yields a superheavy /τ r t in ref. [ opologically stable or un- → . ry scenarios we have men- 1 ) should not be larger than 6 ravity waves which may be ologically non-stable strings. − GeV. Also M ble realization of this scenario ly charged intermediate mass E cts (monopoles or strings) at can be readily constructed. 5.1 . For the defects to enter into ediate scale monopoles, strings 13 1 uced during the breaking of 422 factor ( − GeV 10 7 × · −  r 0 ]. In this case, the inflationary scale is 2 2 10 25 T T . 1 GeV · 41 7 3 H M × , and from eq. (12) of the same reference / ≃ 2 2 2 2 2 1 ≃ 13 . λ r π λ  t 1 −  r H 24 64 τ t 10 ∼ ln  – 15 – = 1 × η A − e 9 A . 1 H = − . During inflation it acquires an extra factor 2 0 1 / = 1 λ 1 H − ) 2 . The VEV of the inflaton 0 0 7 π H λ λ − GeV. From the formula ∼ (8 10 19 ∼ × 10 τ 14 . -foldings following the generation of the defects. During t × 6 e in eqs. (5) and (6) of this reference much smaller than 17 ≃ . 3 GeV, which implies that 7 . λ 2 the rollover time, and from reheating until the present time 0 , as we have shown, the symmetry breaking λ is the present temperature). So all together the mean distan t 16 6 τ ≃ 0 E P 10 T GeV and for the SM spectrum, ( m × 9 0 4 . ], we then obtain that 10 75 . /T GeV) monopoles from the breaking of 333 to the SM may be presen r 29 1 39 ≃ T 14 r ≃ ≃ We will now give some details concerning the production and e We consider the inflationary scenario with a Coleman-Weinbe Regarding T 4 10 / M being the number of 1 0 ∼ defects becomes factor η stable strings. Theproduction mean is distance estimated between to topological be defe able level in our galaxy.and composite Other objects realistic examples that of survive interm an inflationary scenario and cosmic strings as well as the gravity waves generated by t case, eq. (6) of this reference reduces to tioned here. However, analogous to( the SO(10) case, the trip GUT monopole which is inflated away, at least in the inflationa the horizon until today, their present mean distance in eq. ( of ref. [ reheat time and V is the present time with the coupling quent inflaton oscillations this distance is multiplied by a To be more specific, we will takewhich as appears an example in a the particular fourth via line of table 4 in ref. [ are connected, in the nextThe stage of monopole-string symmetry system breaking, by eventuallydetectable top decays by future by experiments emitting (see g below). topologically stable strings. However, the monopoles prod For that corresponding to JHEP10(2019)193 . d R is 9 in to − b m R (5.5) (5.6) 10 n We see ]). The × and 7 25 . reduces to a SU(2) 3 R × . 2 L ) U(1) P (red) and SU(2) , c × 2 0 /m ]). For the particular without topologically SU(2) s L h strings, we can employ R ] and for the numerical 39 M -foldings from the time ×  aton oscillations, ( oles or simply annihilate e 2 the couplings GeV, are allowed by the ) ) c 44 0 0 Z – t ≃ t onopoles enter the horizon e monopole pairs with the monopole. forming random walks with SU(2) GeV. In the case of strings ( 12 ( e s ce: c γ 42 · SU(2) on (i.e. they are not inflated × s eventually decay to gravity pologically stable strings with ρ ρ 10 14 2 ], we conclude that the parame- × Gµ L  − × (see ref. [ L 10 − 25 3 4  . During inflation, the monopoles , and the corresponding symmetry B × r 3 45  τ M t . 12 5 2 9 H . − /  9 75 1 2 1 SU(2) . . U(1) m ∼ 10 λ 3 . of ref. [ × η 10 . GeV. Of course, this is just an order of ]. The present abundance of these gravity × 3 c s × 0 ∼ is the number of d − c  5 45 M . e 16 5. This implies that, for topologically stable 1 2 . M 1 3 m c/λ ) 10 s η  GUT 23 & r – 16 – is the breaking scale corresponding to the defects. × = 3 2 M H T Gµ & (the Parker bound) [ ) d ( 6 η .  (this was not taken into account in ref. [ m 2 1 5 M 30 , we have the formation of SU(4) η , namely /M 2 ∼  − R ∼ − m 11 2 Γ ), where 3 ) r m − GeV . , where SU(3) ] and the GUT gauge coupling constant. M m T  n . These monopoles are subsequently partially diluted by s ( /τ η 10 U(1) r 2 m 13 39 3 t GUT M , where ≡ ∼ of topologically stable magnetic monopoles can be estimate × × − ∼ M 10 12 r M m ) 5 L − m . c 0 × t 1 n 10 ( 2 77 × . . h 3 8 SU(2) s . 4 gw ], which holds for strings surviving until the present time. & 7. From the formula × . Ω Gµ . does not exceed 10 L m 28 67 2 − r ) M B . will be possibly measurable by LISA and BBO in the future. ]. At production it is expected to be η /M 20 strings. In this case, during the breaking of SU(4) d 25 ), this gives for the dimensionless string tension U(1) s − 2 M ( × M 10 Z c ∼ ≡ & For the analysis of this case, we follow ref. [ Note that the GUT scale is given by The number density Next let us consider SO(10) broken via SU(4) For models predicting the existence of topologically stabl c s d M the potential in eq. (5) of ref. [ as in ref. [ multiplied by another factor ( magnitude estimate since we do not know the precise values of ( of monopole production until the end of inflation. During infl are diluted by a factor exp( stable SU(3) figure 1 of ref. [ if they are a red or blue monopole with the corresponding anti the SM gauge group, thesestep monopoles are about connected the by horizon strings connected size in at pairs subsequent by times.string one segment Later, string behave the segment. like m waves and pressureless After the matter. this monopoles time, merge The th to string form either Schwinger monop breaking scale current experimental bounds. It isGµ important to note that to (blue) monopoles at a scale waves is given by combining eqs. (63) and (64) of this referen example discussed here, we find that final relative monopole number density is example we are discussing, that strings with ter Requiring that This scale should then satisfy the inequality monopoles, the parameter away) gives The requirement that the defects eventually enter the horiz inflation. At a breaking scale JHEP10(2019)193 ], f × ≃ 35, L 28 . . Of (5.8) (5.7) , the GeV. sec H 48 27 sec ) with H t . t 11 2 12 ≃ 5.1 SU(2) 10 10 ≃ m × η Hz provided × × H L 3 4 t . − − 6 . After 35 B . c monopole that can be obtained. 14 10 1 ≃ . The frequency ] — is consistent. yrs. Consequently, − emit gravity waves s H ≃ ≃ t 12 . For 4 U(1) d 45 10 M − H t d f lightest one with mass uld be equal to t t × 10 ed × 10 c constant such as SU(5), × . From figure 1 in ref. [ ≃ 71 7 th respect to the electron . n. As explained above, the the monopoles . 12 2 arries color magnetic charge, 6 ing pattern of the underlying − h nd that ) in ref. [ rings at y LISA. The spectrum of these ≃ = 4 , gw s 2 3 GeV, which is consistently lower eq = 10 t  4 2 Gµ 0 − eq h t H t t 10 gw 2  corresponding to 1 × 2 1 after reheating is given by eq. ( Hz. − 10) is the GUT fine structure constant. H )  4 / s T 74 T 1 , which corresponds to − . d eq t 1 t 14 Gµ ∼ GeV. ( −  – 17 – Hz and Ω ≃ (Γ 1 ]— 4 10 14 − H H is the equidensity time at which matter starts dom- − t ∼ T × 45 10 GUT d eq 7 ∼ α t t . × 6 ) = 10 7 is the present value of the Hubble parameter in units 0 . t 26 GeV, the monopoles re-enter the horizon after reheating ≃ 0 f ( . s 1 ) are the present photon and critical energy densities of the f ≃ 0 11 t GeV and the subsequent breaking scale of this group to the ≃ 0 ( Gµ 10 c h m ρ 14 × M . GeV, where 10 1 32 . − × ) should be replaced by 17 6 ) and 0 10 , and 26 5.1 t ≃ . ( Mpc all predict the existence of a topologically stable magneti 12 ∼ γ s 1 − ρ 6 − M E 10 27 sec. The decay time of the string segments is then . GUT 50, yrs, which is prior to matter domination at 3 sec connected by a string segment with 2 in eq. ( × . is the time at which the monopoles enter the horizon, 4 2 /α ∼ ≃ T H 07 . At horizon re-entrance of the monopoles, this distance sho is about 1 10 . t ≃ H 3 m t R GUT × η Summarizing, we see that if the breaking scale of 422 to SU(3) Now the question arises under what circumstances the requir As an example we take H ≃ t M = 28 . m which at present have frequency monopole string structures behave like particles and the st the above calculation, which requires this — see e.g. eq. (65 at U(1) SM group is 4 inating the universe. For the example under discussion, we fi than the reheat temperature. Our requirement then gives for c that carries a singlecharge). unit This of superheavy GUT Dirac scaleand magnetic magnetic monopole this also conclusion charge c holds (quantized independentGUT of wi model. the symmetry In break SU(5),∼ this magnetic monopole happens to be the 6 Conclusions Grand Unified TheoriesSO(10), with and a unified (single) gauge coupling and for the SM spectrum, we obtain is the decay time of the strings, and waves is expected to strongly peak around 10 we see that such gravity waves will be perfectly detectable b η This will be decidedmean by intermonopole the distance monopole at production temperature and evolutio universe respectively, and course, where Γ where of 100 km sec The present abundance of gravity waves, in this case, is Ω of these waves is given by — see ref. [ JHEP10(2019)193 ommons SO(10), , ]. 6 . E (1974) 194 (1974) 275 (1974) 276 unified gauge SPIRE 6 20 , E IN [ D 10 B 79 (1975) 575 ank Dr. Digesh Raut , 23 arge. We have observed le level. We have depicted JETP Lett. ermediate scale. Hence they redited. ry two or three units of the (1978) 301 , Phys. Rev. space based observatory LISA. s as well as composite objects . The gravity waves emitted by Nucl. Phys. , , B 79 ]. ]. AIP Conf. Proc. ]. , SPIRE SPIRE monopoles, magnetic symmetry and IN ]. IN [ Phys. Lett. , SPIRE ]. Consequences of a monopole with Dirac magnetic IN – 18 – [ ]. SPIRE A universal gauge theory model based on ]. Phase transitions and magnetic monopoles in ]. IN ]. 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