1 An Analytical Comparison of Different Rotation Methods of Anomalous U(1) Charges Kara Merfeld, Caleb Smith

Abstract—Heterotic theories offer the pos- a string. As the quarks are pulled apart, the string sibility of additional gauge groups arising from the connecting them is stretched and the attractive force symmetry breaking of SO(32) or × E8. These increases. At some point, the string would break arise from compactifying the ten space-time dimen- sions of to arrive at the four large and two more quarks would come into existence, space-time dimensions. This study focuses on produc- one at each of the newly created ends[12]. This ing and analyzing string models using the weakly- would account for the absence of individual quarks. coupled free fermionic heterotic string (WCFFHS) When quantum chromodynamics (QCD) came into formalism. Anomalous U(1) gauge groups naturally existence it overshadowed through its arise in models generated by the WCFFHS formalism. Phenomenological analysis and calcu- experimentally confirmed success in modeling the lations are greatly simplified by transforming the U(1) strong force. However, when it was discovered that charges such that only one U(1) is anomalous. Multiple string theory inherently contains gravity, the theory algorithms are derived and presented performing this began to gain support as a grand unifying theory[6]. operation. One of these methods was selected and It would include all of the fundamental particles and implemented into the WCFFHS model building frame- work. Preliminary results of applying this calculation interactions as the result of a single phenomenon: to three generated models are included. different vibrational modes of strings[8]. By the 1980’s there were five distinct string theories, and Index Terms—, Rotation Matrices, Normal- 500 ization, Heterotic, WCFFHS formalism there now exist over 10 string models[14]. It is not a question of which of these models is correct, but rather which models most readily reproduce the I.INTRODUCTION SM. These string models include gravity, allow for A. String Theory dark energy, and depend on only one free parameter: The standard model (SM) of particle physics the length of the strings. These attributes make it an is a collection of the fundamental particles that appealing theory, despite its current inability to be combine together to describe all matter and forces. experimentally verified. The SM contains six quarks and six anti-quarks (not including color), six leptons and six anti-leptons, B. Compactification and twelve gauge bosons. There is also a Higgs bo- Of the five string theories, type I, type IIA, type son, which introduces particle masses in the theory IIB, heterotic E8 × E8 and heterotic SO(32), the through electroweak symmetry breaking. However, focus of this study is on heterotic strings. Het- the SM is incomplete, as it does not account for erotic strings include only closed-string models, in dark matter/energy, and offers no description of which the left-moving modes are supersymmetric, gravity[13]. It also gives no explanation for the and the right-moving modes are bosonic[9]. The wide range of particle masses between different right-movers exist in 26 dimensions, while the left- generations (the hierarchy problem)[7]. Finally, it movers exist in 10. The extra 16 dimensions are possesses about 20 free parameters which must be toroidally compactified, making them internal de- experimentally, rather than theoretically, determined. grees of freedom[11]. In order to make contact with An underlying theory is desired. realistic physics, the remaining dimensions of the String theory first came into existence in the string model are also toroidally compactified down 1960’s, as an attempt to explain the strong force to the four dimensions with which the macroscopic within hadrons. It was hypothesized that the strong world is familiar. The compactified dimensions then force between quarks was due to a “strings” con- exist on scales smaller than that which current necting quarks, with one quark at each endpoint of technology can probe. 2

While several formalisms exist for constructing C. U(1) Rotation heterotic string models, the focus of this research The U(1) charges qij are defined by the following is on the weakly-coupled free fermionic heterotic two equations. string (WCFFHS) formalism. In the WCFFHS for- malism, the focus is on fermions: 8 from Qi |ψji = qij |ψji (6) the left-moving SUSY, and 32 from fermionizing hQi | ψji = qij (7) the 16 compact directions[5]. Notationally, the left- movers and right-movers are represented together In equation (6), Qi is the operator for the th th as: i U(1), while |ψji is the j matter state vector. Here qij (the ij element of our U(1) charge matrix) (RL||RR) (1) is an eigenvalue of the operator Qi and eigenvector where R is twice the rank of the non-abelian Lie |ψji. In equation (7), |Qii is the basis vector for th gauge generated, and the subscript indicates the i U(1) (not to be confused with the matrix the direction. For example, in ten large space-time Qi, which is a different basis of matrices). Equation dimensions this is represented by: (7) defines the U(1) charges qij as the inner product of the charge basis vector |Qii and the matter state (8||32). (2) |ψji. A matrix of U(1) charges with elements qij is constructed in accordance with equations (6) and The number of compactified directions is the num- (7). It is this matrix of U(1) charges that shall be ber of bosons, so with each compactified dimension, considered in the following sections. the number of large dimensions decreases by 1, The trace of a U(1) charge is the sum of the and the rank of each the left- and right-movers charges of the physical states under that U(1) factor, increases by two, because there are two coupled and is computed as: fermions created for every fermionized boson[1]. m m When compactification occurs repeatedly, it can be X X T r(Q ) = hψ | Q i = q (8) seen that when D is the number of dimensions[10]: i j i ij j=1 j=1 D = 9 : (10||34) where equation (7) has been applied and m has been D = 8 : (12||36) defined as the total number of matter states. . When the traces of these U(1) charges are com- . puted, it is easily seen that many of the charges D = 4 : (20||44). have non-zero traces, making them anomalous. In the heterotic theories, it is possible to modify the In general, the total rank, RT , can be found using: U(1) basis without changing the physics so that all n but one of the U(1) traces are equal to zero[3]. That RT = 26 − D − (3) is, after the manipulation, only one U(1) factor will 2 remain anomalous. Transforming the U(1) charges where n is the total number of left-right pairings. into this form greatly simplifies phenomenological The total rank is also the sum of the ranks of the computations performed with the U(1) charges, in- abelian and non-abelian charges: cluding supersymmetry calculations[2]. The focus of this study is to develop an algorithm to efficiently R = R + R . (4) T A NA perform such a transformation. Note that in the model building process, only the Specifically, the goal is to transform the U(1) non- factors are constructed on the abelian charges so that: first pass, after which RNA can be determined. A 0 T r(Q1 ) 6= 0 (9) realistic string model must have D = 4 large space- T r(Q 0) = 0 (10) time dimensions[11]. Combining equations (3) and i 0 (4) with D = 4 yields the number of abelian U(1) where Qi is the transformed charge at index i, charges RA: and 1 < i ≤ RA. In order for the physics to be n invariant, it is required that the orthogonal basis R = 22 − − R . (5) A 2 NA formed by the U(1) states be normalized before and after calculation. Therefore, imposing an arbitrary Equation (5) gives the number of abelian U(1) normalization of length squared 2 yields: charges that are explicitly constructed within the model. hQi | Qji = 2δij (11) 3 before the transformation, and: will yield a new basis in which only one U(1) is anomalous. The second requirement (16) follows Q 0 Q 0 = 2δ (12) i j ij from equations (11) and (12) and ensures that we after the transformation, where δij is the normal will preserve the U(1) normalization and orthogo- Kronecker delta. nality. To accomplish this, an orthogonal RA × RA R is defined that, when acting on A. Algorithm I the traces, gives: From the properties of R, if the rotated charges    0  are defined as: T r(Q1) T r(Q1 ) 0 0  T r(Q2)   T r(Q2 )  Qi = ai |Qii + aj |Qji (17) R   =   . (13)  .   .  0  .   .  Qj = aj |Qii − ai |Qji (18) 0 T r(QR ) T r(QR ) A A where ai, aj ∈ ~a, and i and j are the indices of The objective is to find R that will satisfy equations the charges being mixed. The orthogonality between 0 0 (9) and (10) when it is applied to the U(1) traces Qi and Qj is guaranteed[4]. It will be shown and equations (11) and (12) when applied to the that the elimination of all but one anomalous charge U(1) charges. Note that preserving the normalization is also guaranteed. from equations (11) and (12) requires that R must This algorithm explicitly uses this property re- be an (RT = R−1). Once R is peatedly to eliminate half of the traces. The rotation determined, we will apply it to the abelian charge is done in iterations, with every other non-zero trace values in the following manner: being eliminated with each iteration. Each iteration 0 defines a U(1) charge in terms of the traces of qij = h~rj | ~qii (14) the previous U(1) charges, and the abelian values th th where ~rj is the j row of R, ~qi is the i row themselves as described in equations (17) and (18). 0 of the abelian charge matrix, and qij is the ij The matrix associated with the first iteration for even element of the rotated charge matrix. Therefore, n is then:   the rotated charges may be expressed as linear a1 a2 0 0 ··· 0 combinations of the original charges. Three unique, a2 −a1 0 0 ··· 0    yet related algorithms are developed for performing  0 0 a3 a4 ··· 0  R =   . (19) the operation denoted by R. These were written and 1  0 0 a4 −a3 ··· 0  tested in C++ (with additional testing performed in  ......   ......  Mathematica) for the purpose of using one in the   0 0 0 0 ... a FF Framework (the C++ model building program n−1 used Baylor University). When this matrix acts on the trace vector, it can be seen that half of the traces become zero:    2 2 II.METHODS a1 a2 0 0 ··· 0 a1  a1 + a2 a2 −a1 0 0 ··· 0  a2  0         0 0 a3 a4 ··· 0  a3  a 2 + a 2      3 4  A simplified notation is introduced for the de-  0 0 a4 −a3 ··· 0  a4   0      =   .        ......   .   .  scription of our three algorithms. Let the n U(1)  ......   .   .   ......   .   .  traces from equation (8) be represented by the 0 0 0 0 ... an−1 an 0 (20) n-dimension vector ~a = (a1, . . . , an). The U(1) rotation may be completed by applying a rotation This becomes the new trace vector for the next matrix R to the U(1) charges. As discussed earlier, iteration. To ensure that the matrix is orthogonal, the rotation matrix must meet the following two it is necessary to find a normalization constant for requirements: each row. By applying the Pythagorean theorem, it can be seen that the normalization constant of rows    0 a1 a1 i and j are: a2   0  1 R   =   , (15) Ni = Nj = (21)  .   .  p 2 2  .   .  ai + aj an 0 where i and j are the indices of the rows being

T mixed. The normalization constants will have the RR = I. (16) general form: The first requirement (15) follows from equations 1 Ni = Nj = (22) p 2 2 (9), (10), and (13) and ensures that the rotation si + sj 4 where i and j are the indices of the rows being applying equations (17) and (18) directly. For this mixed and si and sj are the traces being mixed. reason, care must be taken regarding maintaining an Thus in equation (21) we mixed traces si = ai orthogonal rotation by recording normalization con- and sj = aj. Because charges are ideal stants for the Qi charges throughout. Once mixing for analysis, the normalization constants are stored occurs, we have found that if the final matrix R were in a separated vector of doubles in the program to be computed, the normalization constants are not throughout the course of the algorithm, and then uniform for each row (for instance, the simplest are passed along to the next phase of analysis. case of such an R contains a row in which half During the following iterations, many of the traces the row has one normalization constant, while half are already zero. The charges that corresponds to the row has another). This is due to the constraint these zero traces are not altered. Instead the charges that an integer rotation be applied to the charges, that remain anomalous are selected out and mixed with normalization constants included separately. with each other. The matrix corresponding to the second iteration for n divisible by 4 is then: B. Algorithm II  2 2 2 2  a1 + a2 0 a3 + a4 0 ··· 0 This algorithm applies repeated Gauss-Jordan  0 1 0 0 ··· 0 elimination to a working basis B to successively  2 2 2 2  a3 + a4 0 −a1 − a2 0 ··· 0 determine an additional row vector orthogonal to all R =   . 2  0 0 0 1 ··· 0 other row vectors in the basis. This additional vector    ......  is then appended to the basis, and the process is  ......  0 0 0 0 ··· 1 repeated until a complete n × n orthogonal basis is obtained. The rows are then normalized to create an Note that the ones along the diagonals of the rows orthonormal basis, which is used as our orthogonal that are not being mixed do not alter the correspond- rotation matrix R. ing charge values, so the normalization constant for We will develop the final form of R by consid- these rows is one. ering the case of three anomalous U(1)’s with three Applying equation (22) to R2 yields normaliza- non-zero traces such that ~a = (a1, a2, a3). The first tion constants N1 and N3: row of our basis B will be the trace vector: 1   N1 = N3 = , B = a1 a2 a3 . (25) p 2 2 2 2 2 2 (a1 + a2 ) + (a3 + a4 ) Applying Gauss-Jordan elimination yields: 2 2 where the traces mixed are s1 = a1 + a2 and 2 2 1 a2 a3 ~b = 0. (26) s2 = a3 + a4 . Applying this second matrix R2 a1 a1 2 to the trace rotated by R (without normalization 1 Equation (26) represents the following equation with constants included) yields: ~ which we wish to determine the values for b2, the  2 2 2 2   2 2 a1 + a2 0 a3 + a4 0 ··· 0 a1 + a2 m second row of B0.  0 1 0 0 ··· 0  0  0       a 2 + a 2 0 −a 2 − a 2 0 ··· 0 a 2 + a 2  0   3 4 1 2   3 4         0 0 0 1 ··· 0  0   0  a a     =   . 2 3        ......   .   .  (1)b21 + b22 + b23 = 0 (27)  ......   .   .   ......   .   .  a1 a1 0 0 0 0 ··· 1 an 0 (23) Note that we have two free variables. We will always where our anomalous non-zero trace m is: set the first (left-most) free variable to 1 and any

2 2 2 2 2 2 following free variables to 0 in order to obtain a m = (a1 + a2 ) + (a3 + a4 ) . (24) simple matrix with a consistent form. However, note The algorithm also simplifies rows by determining that this choice is arbitrary, and there exist a myriad the great common divisor (gcd) of its elements of other valid choices. In equation (27), we set b22 = ~ and then dividing each element by the gcd. This 1 and b23 = 0 and append b2 to B from equation 0 is allowed before normalization, as a row may be (25) to obtain B : arbitrarily scaled before being normalized. Applying a2 b21 = − , (28) normalization afterwards gives the row unit length. a1 This general process of mixing Qi’s will be repeated  a a a  until only one Q has non-zero trace. B0 = 1 2 3 . (29) i − a2 1 0 Additionally, note that at no point in the algorithm a1 is the matrix R ever explicitly computed by the Note that the two row vectors in B0 are orthogonal program, but an equivalent operation is done by as they have a dot product of zero, as required. Now 5

0 applying Gauss-Jordan elimination to B yields: where we define αi and βi as:

 a1a3  1 1 0 2 2 − a1 +a2 ~b = ~0. (30) i ! 2 a2a3 3 X 0 1 2 2 2 a1 +a2 αi = ak i > 0, (39) Equation (30) represents the following two equa- k=1 tions with which we wish to determine the values −1 ~ 00 βi = αi . (40) for b3, the third row of B .   th a1a3 Note that each i row of n × n matrix R contains (1)b31 + (0)b32 + 2 2 b33 = 0 (31) a1 + a2 its own normalization constant Ni of the form:   a2a3 ( (0)b31 + (1)b32 + 2 2 b33 = 0 (32) αn : i = 1 a1 + a2 Ni = . (41) α α : 1 < i ≤ n Note that we have one free variable. In equations i−1 i ~ 0 (31) and (32), we set b33 = 1 and append b3 to B The final n × n rotation matrix R for some n- 00 from equation (29) to obtain B : dimensional trace ~a becomes: a1a3  N1a1 N1a2 N1a3 ··· N1an  b = − , (33) 2 31 2 2  −N a a N β 0 ··· 0  a + a  2 1 2 2 1  1 2  −N a a −N a a N β 2 ··· 0   3 1 3 3 2 3 3 2    , a a  . . . . .  2 3  . . . . .   .  b32 = − , (34)  . . . .  2 2 2 a1 + a2 −Nna1an −Nna2an Nna3an ··· Nnβn−1 (42)  a a a  1 2 3 where we have assumed a 6= 0. B00 = − a2 1 0 . 1  a1  (35) a1a3 a2a3 1) Proof by Induction: We shall prove the form − 2 2 − 2 2 1 a1 +a2 a1 +a2 given in equation (42) by induction. For the base At this point we are allowed to arbitrarily scale case, consider the original trace (a1, a2). Applying each row, as we have not yet normalized B00. It is Gaussian elimination gives: advantageous to do so in order to obtain a matrix a a  → 1 a2  (43) with integer components rather than rational compo- 1 2 a1 nents. After obtaining an integer matrix B00, we will normalize the length of each row to 1 to obtain the which yields a final rotation matrix (before normal- orthogonal matrix R. Multiplying the second row of ization) of: 00 2 00 2 2 B by a1 and the third row of B by a1 + a2     a1 a2 a1 a2 from equation (35) gives: a2 → 2 . (44) − 1 −a1a2 a1   a1 a1 a2 a3 00 2 Normalizing gives the form from equation (42): B = −a1a2 a1 0  . (36) −a a −a a (a 2 + a 2)   1 3 2 3 1 2 N1a1 N1a2 2 . (45) Finally, note that though the row vectors are orthog- −N2a1a2 N2a1 onal, the matrix is not orthogonal unless the rows We now suppose that this method will work for the are normalized. If the rows are not normalized, then kth case. Consider the k + 1 case, with traces of the matrix multiplied by its transpose with not give (a , . . . , a ). As we have the correct form (before the (the definition of an orthogonal 1 k+1 normalization) for the kth case, for the k+1 case we matrix). Each row vector is normalized to a length may perform Gaussain elimination on the following of 1 by dividing by its magnitude.√ For instance, the 2 2 2 matrix to obtain: first row has a length of a1 + a2 + a3 and is   normalized by multiplying by α3, which we define a1 a2 ··· ak ak+1 2 as: −a1a2 a1 ··· 0 0  1   , α3 = √ . (37)  . . . . .  2 2 2 ...... a1 + a2 + a3  . . . .  −a a −a a ··· a 2 + ... + a 2 0 Applying normalization constants to B00 produces 1 k 2 k 1 k−1 R the orthogonal rotation matrix ,  −a1ak+1  1 0 ··· 0 2 2   a1 +...+ak α3a1 α3a2 α3a3 −a2ak+1 0 1 ··· 0 2 2  2  a1 +...+ak  R = −α1α2a1a2 α1α2β1 0  , . . . . .  . (46) 2 ......  −α2α3a1a3 −α2α3a2a3 α2α3β2 . . . .  −akak+1 (38) 0 0 ··· 1 2 2 a1 +...+ak 6

This yields the final matrix (before normalization): This final trace has only one anomalous U(1), as

 a1 a2 ··· ak ak+1 desired. Additionally, as both R1 and R2 are special 2  −a a β ··· 0 0   1 2 1    orthogonal, R = R2R1 is special orthogonal as well,  . . . . .   . . . . .  . (47)  .   . . . .  since the set of special orthogonal matrices is closed  2   −a1ak −a2ak ··· βk−1 0  2 under multiplication. −a1ak+1 −a2ak+1 · · · −akak+1 βk Applying normalization gives:     β2 0 a3 a1 a2 0  N1a1 N1a2 ··· N1ak N1ak+1 2 R = α3α2  0 β3 0  −a2 a1 0   −N a a N β ··· 0 0   2 1 2 2 1     . . . . .  −a3 0 β2 0 0 β2  . . . . .  .  .   . . . .   2     −Nka1ak −Nka2ak ··· Nkβk−1 0  β2a1 β2a2 β2a3 −N a a −N a a · · · −N a a β 2 k+1 1 k+1 k+1 2 k+1 k+1 k k+1 k R = α3α2 −β3a2 β3a1 0  2 By induction, as the k +1 case agrees with equation −a1a3 −a2a3 β2 equation (42), R will have the form of equation (42). Now observe that from equations (39) and (40), This final form holds for any real traces with a1 6= 0. 2 − 1 −1 2 1 It is left to the reader to show that of α1 = (a1 ) 2 = a1 and β1 = (a1 ) 2 = a1. our final rotation matrix is +1, implying that we Distributing α2α3 and multiplying the second row are in fact using special orthogonal matrices in our by α1a1 (note that α1a1 = 1) gives: rotation (this will also touched upon in Algorithm   α3a1 α3a2 α3a3 III). 2 R = −α1α2a1a2 α1α2β1 0  , 2 C. Algorithm III −α2α3a1a3 −α2α3a2a3 α2α3β1 (50) We will develop the final “coupling” method by which matches the form obtained from Algorithm considering the case of three anomalous U(1)’s with II in equation (38). Again, as in equation (38), we three non-zero traces such that ~a = (a1, a2, a3). see that equation (50) has normlization for each We shall determine the form of R by applying th i row of the form Ni as definded in equation intermediate rotations R1 and R2. We will find that (41). Hence the final n × n rotation matrix R with R = R2R1. intermediate rotation matrices R1 ...Rn−1 for some The first orthogonal coupling matrix R1 is given n-dimensional trace ~a becomes: by:   n−1 a1 a2 0 Y R = R × ... × R = R (51) R1 = α2 −a2 a1 0  (48) n−1 1 n−i i=1 0 0 β2 where from equations (39) and (40), α2 =  N1a1 N1a2 N1a3 ··· N1an  √ 2 1  −N2a1a2 N2β1 0 ··· 0  2 2 − 2 2 2   (a + a ) and β = a + a . Note that  −N a a −N a a N β 2 ··· 0  1 2 2 1 2  3 1 3 3 2 3 3 2  T   (52)  . . . . .  R1R1 = I and det(R1) = +1, making R1  . . . . .   .   . . . .  2 a special orthogonal matrix. Applying R1 to the −Nna1an −Nna2an Nna3an ··· Nnβn−1 original trace vector ~a yields the rotated trace ~a0.       Thus R has components rij of the form a1 a2 0 a1 β2 ~a0 = R ~a = α −a a 0 a = 0  1 2  2 1   2   Niaj : i = 1 0 0 β a a  2 3 3 0 : i < j rij = (53) The second orthogonal coupling matrix R2 is given −Niajai : i > j by:  2   Niβi−1 : i = j β2 0 a3 R = α 0 β (49) 2 3  3  Note that each row has a normalization con- −a3 0 β2 stant Ni. In implementing the rotation in the where from equations (39) and√ (40), α3 = FF Framework, it is prefered that the final rotated 2 2 2 − 1 2 2 2 (a1 + a2 + a3 ) 2 and β3 = a1 + a2 + a3 . U(1) charges are . For this reason the ma- T Note that R2R2 = I and det(R2) = +1, making trix R is constructed in integer form without the R2 special orthogonal. Applying R2 to the rotated normalization constants Ni included (as they are trace vector ~a0 yields the final trace ~a00. irrational). The integer rotated charges are then used       β2 0 a3 β2 β3 for phenomenological analysis, while the normaliza- 00 0 tion constants are stored as doubles and included ~a = R2~a = α3  0 β3   0  =  0  −a3 0 β2 a3 0 separately. 7

1) Proof by Induction: We will prove that the each column of the final rotated charges. In the in equation (51) produces a program implementation, no intermediate matrices matrix of the form in equation (52) by induction. are created or multiplied; this has been included to For the base case, an original trace of ~a = (a1, a2) show the derivation of the final of R. Instead, R is gives the final R: simply constructed based on the original trace ~a, as it adheres to a systematic form. It is then applied  N a N a  R = R = 1 1 1 2 (54) (in its integer form) to the abelian U(1) charges as 1 −N a a N a 2 2 1 2 2 1 in equation (14), taking the inner product of the jth th We now suppose that equation (51) produces a row of R and i row of the qij charge matrix to th 0 matrix of the form in equation (52) for the k yield qij element of the rotated charge matrix. case. For the k + 1 case, we first apply the Rk−1 Tables I - IV on page 8 are the results of applying matrix (from our induction hypothesis) to the k +1- Algorithm III to the U(1) charges generated from dimensional trace ~a. four WCFFHS models. Model 1 had an abelian rank      of 6 (and thus 6 U ’s), Model 2 had an abelian rank N1a1 ··· N1ak 0 a1 βk i ...... of 7, and Models 3 and 4 had abelian ranks of 8.  . .. . .  .   .   . . .  .  =  . Tables I - IV show the original and final Ui traces,  2      −Nka1ak ··· Nkβk−1 0  ak   0 as well as the normalization constants for each Ui. 0 ··· 0 1 ak+1 ak+1 Observe that U1 always has a non-zero final trace, Therefore our remaining two anomalous traces that while Ui for i 6= 1 always has a final trace of zero. This is precisely what the rotation meant to we shall couple in our Rk matrix are βk and ak+1.

 0 0  accomplish. N1 βk ··· 0 N1 ak+1    . . . .   . . . .   .   . . .    IV. CONCLUSION  0 ··· 1 0  0 0 −N1 ak+1 ··· 0 N1 βk This study has proved quite profitable for its  N1a1 ··· N1ak 0 constituents, and we hope that it has been of some  . . .   .   . . . .  ×  . . . .  value to the reader as well. All three algorithms  2  −Nka1ak ··· Nkβk−1 0 0 ··· 0 1 have properly rotated the anomalous abelian U(1)

 0 0 0  charges such that only one U(1) trace per remains N1N1 a1βk ··· N1N1 akβk N1 ak+1    . . . .  anomalous. This was a successful endeavor because  . . . .  =  .   . . .   2  in each algorithm, a rotation matrix R was de-  −Nka1ak ··· Nkβk−1 0  m 0 0 0 T −N1N1 a1ak+1 · · · −N1N1 akak+1 N1 βk termined such that RR = I and R~a = [ ~0 ].

 0 0 0  Only Algorithm III was used to incorporate in the N1 a1 ··· N1 ak N1 ak+1    . . . .  FF Framework; it required much less computation  . . . .  =  . . . .     2  time than Algorithm II and allowed for a simpler  −Nka1ak ··· Nkβk−1 0  2 −Nk+1a1ak+1 · · · −Nk+1akak+1 Nk+1βk normalization method than required for Algorithm I. 0 However, the others were successfully used to verify We have introduced N1 where: its validity. N1 = αk 0 N1 = αk+1 0 N1N1 = αkαk+1 = Nk+1 0 2 2 N1 βk = αkαk+1βk = Nk+1βk . Wherewithal, equation (51) produces a matrix of the form in equation (52) for the k+1 case, by induction (51) produces a matrix of the form in equation (52) for any real traces ~a with a1 6= 0.

III.RESULTS AND ANALYSIS The third algorithm was included with the FF Framework C++ model building program. Shown below are three rotated models, including charges and charge traces before and after the ro- tation, with normalization constants included for 8

TABLE I: Model 1 Rotation REFERENCES

Vector U1 U2 U3 [1] Ignatios Antoniadis, C. P. Bachas, and C. Kounnas. Four- Original Trace -96 0 0 Dimensional Superstrings. Nucl. Phys., B289:87, 1987. Final Trace 336 0 0 [2] Gerald B. Cleaver. in free fermionic Normalization 0.756 2.00 2.00 strings. Nucl. Phys., B456:219–256, 1995. [3] Gerald B. Cleaver. Advances in old-fashioned heterotic string model building. Nucl. Phys. Proc. Suppl., 62:161– 170, 1998. Vector U4 U5 U6 [4] Gerald B. Cleaver and Alon E. Faraggi. On the anomalous u(1) in free fermionic superstring models. Int. J. Mod. Original Trace 96 -192 -96 Phys., A14:2335–2356, 1999. Final Trace 0 0 0 [5] Keith R. Dienes, Michael Lennek, David Senechal, and Normalization 1.41 1.15 0.309 Vaibhav Wasnik. Supersymmetry versus gauge symmetry on the heterotic landscape. Phys. Rev., D75:126005, 2007. [6] Alon E. Faraggi. Construction of realistic standard - like TABLE II: Model 2 Rotation models in the free fermionic superstring formulation. Nucl. Phys., B387:239–262, 1992. Vector U U U U [7] Alon E. Faraggi. Generation mass hierarchy in superstring 1 2 3 4 derived models. Nucl. Phys., B407:57–72, 1993. Original Trace -108 -108 108 -20 [8] Maurizio Gasperini. Elements of . Cam- Final Trace 8862 0 0 0 bridge University Press, 2007. Normalization 0.0300 1.41 0.816 0.0123 [9] David J. Gross, Jeffrey A. Harvey, Emil J. Martinec, and Ryan Rohm. The Heterotic String. Phys.Rev.Lett., 54:502– 505, 1985. [10] John H. Schwarz Katrin Becker, Melanie Becker. String Vector U5 U6 U7 Theory and M-Theory: A Modern Introduction. Cambridge Original Trace -20 -52 180 University Press, 2007. Final Trace 0 0 0 [11] Hikaru Kawai, David C. Lewellen, and S. H. Henry Tye. Normalization 0.000450 0.000431 0.000919 Construction of fermionic string models in four- dimen- sions. Nucl. Phys., B288:1, 1987. [12] Elias Kiritsis. String Theory in a Nutshell. Press, 2007. TABLE III: Model 3 Rotation [13] K. Nakamura et al. Review of particle physics. J. Phys., G37:075021, 2010. Vector U1 U2 U3 U4 [14] Barton Zwiebach. A First Course in String Theory. Cam- Original Trace -48 210 -26 70 bridge University Press, 2004. Final Trace 9060 0 0 0 Normalization 0.0297 0.111 0.000513 0.000809

Vector U5 U6 U7 U8 Original Trace 114 68 24 -48 Final Trace 0 0 0 0 Normalization 0.000275 0.000476 0.00275 0.00269

TABLE IV: Model 4 Rotation

Vector U1 U2 U3 U4 Original Trace -128 100 -116 -116 Final Trace 9230 0 0 0 Normalization 0.0294 0.0985 0.000987 0.000694

Vector U5 U6 U7 U8 Original Trace 76 68 -12 100 Final Trace 0 0 0 0 Normalization 0.000570 0.000522 0.00151 0.00233