Enumerating the operators of an effective field theory

Until recently, little was known about the high-dimensional operators of the standard model effective field theory (SMEFT). However, in the last few years the number of these operators has been counted up to mass dimension 15 using techniques involving the Hilbert series. In this work, I will show how to perform the same counting with a different method, which is perhaps more intuitive. This alternative approach makes it possible to cross-check results (it confirms the SMEFT numbers), but it also provides some more information on the operators beyond just counting their number. The considerations made here apply equally well to any other model besides SMEFT and, with this purpose in mind, they were implemented in a computer code.


Introduction
It is sometimes useful to consider interactions which are allowed by symmetry, even if they are not renormalizable. Seen as effective interactions, they can be used to study the effects of new, higher-energy physics in a model-independent way.
For example, one can take the Standard Model fields and build all those operators which are invariant under gauge and Lorentz transformations, including those with mass dimension larger than four. This construction is often called the standard model effective field theory (SMEFT), and it has been studied for a long time. The unique dimension 5 operator in SMEFT, which violates lepton number and may explain neutrino masses, was mentioned for the first time in [1]. This very same paper, as well as [2,3] also lists the dimension 6 terms which violate baryon number (hence they can induce nucleon decay), while the remaining baryon number conserving operators with this dimension were listed in [4][5][6]. Nevertheless, for most purposes several of these operators can be shown to be redundant, and the authors of [7] were the first to provide a complete list of non-redundant SMEFT Lagrangian terms up to dimension 6. The counting of how many parameters are contained in such terms was given in [8].
It is remarkable that it took so long to fully understand these operators, given that their dimension is not particularly large: most of them have four fields or less. Currently, the operators of dimension 7 also seem to be well understood [9][10][11], but beyond that there are only partial results. For example, some lepton number violating operators were presented in [12,13], and [14] lists field combinations which violate lepton number in four units but without building the operators explicitly nor checking for redundancies.
However, in the past few years, there has been a remarkable progress in the counting of operators, using the Hilbert series [15][16][17][18]. This technique is not only very elegant and simple to use, but it can also be applied to an arbitrary model, with an arbitrary gauge symmetry group. Using it, the authors of [18] computed the number of SMEFT operators up to dimension 15. Despite these advantages, the Hilbert series method is very different from what I will call the intuitive method of simply multiplying all fields in all possible ways, and retaining those combinations which are invariant under the action of all relevant symmetry groupsusually the Lorentz and gauge groups.
There are two complications to this latter approach. One of them is also present in the Hilbert series method, and it is related to operators with derivatives: they are problematic mainly because some combinations of these operators are redundant (under some assumptions) and it would therefore be desirable to remove them. But it turns out that the ideas used to handle this issue in Hilbert series computations [16,18] work equally well when adapted to the intuitive method of taking all invariant fields combinations.
For this reason, the main focus of this document will be the other complication, which arises when there are repeated fields in an operator. The simplest example in SMEFT is LLHH, which generates neutrino masses when H acquires a vacuum expectation value. For n copies of the field L, one might have thought that such a term in parameterized by n 2 complex numbers. However, this is not true: the coupling matrix κ in flavor space appearing in the expression κ ij L i L j HH is symmetric (κ ij = κ ji ), so the term is parameterized by just n (n + 1) /2 complex numbers. This is purely a consequence of the quantum numbers of L and H, as I will discuss in detail later on. It is easy to find more complicated examples, such as N c N c N c N c and QQQL , (1) where N c represents right-handed neutrino fields (they do not exist in SMEFT), which are gauge singlets. While the dimension of these terms (six) is still quite small, it is non-trivial to derive the number of independent couplings associated to each, assuming n X flavors of the field type X = N c , Q, L. 1 Fortunately, these two complications can be handled in a systematic way. This makes it possible to count operators in models such as SMEFT, up to high mass dimensions, using the intuitive approach mentioned earlier. In fact, besides the number of operators, it is also possible to extract the symmetry of coupling tensors (such as κ above) under flavor index permutations; in some cases, this is a piece of extra information which cannot be inferred from the number of operators.
It is also noteworthy that the intuitive method described in this work seems to be computationally competitive with the Hilbert series approach. The authors of [18] provided the number of each type of SMEFT operator up to dimension 12 plus the total number of operators with dimension d = 13, 14 and 15. Up to now, these results had only been cross-checked by other means up to d = 8. However, with the method explained in this work it was possible to count all types of SMEFT operators up to dimension 15 in a couple of hours, using a standard laptop (all numbers given in [18] were reproduced). Existing Mathematica code [19,20] was modified to make such calculations possible for any model, and such code is publicly available at the web address renatofonseca.net/sym2int.php The remainder of the text is structured as follows: • Section 2 describes the notation and the conventions adopted in this work.
• Section 3 is devoted to the problem of repeated fields in interactions, suggesting a way of systematically dealing with these cases.
• Section 4 discusses the problem with derivatives, and how they can be handled by using and adapting the solutions proposed in [16,18].
• Section 5 contains a discussion of several topics related to the counting of operators of an effective field theory. The ideas mentioned in sections 3 and 4 were implemented in a computer code, and some of the results obtained with it are presented here and in an appendix.
• A summary is available at the end.

Notation and conventions
The method discussed in this work can be applied to any effective field theory, nevertheless SMEFT will often serve as an example. Its Lagrangian is invariant under the SU (3) C × SU (2) L × U (1) Y gauge group and the restricted Lorentz group 2 . For all practical purposes the algebra of this latter group can be taken to be the same as the one of SU (2) l × SU (2) r so it's representations are given by pairs of non-negative half-integers (j l , j r ). The only caveat to keep in mind that that complex conjugation flips these numbers: (j l , j r ) * = (j r , j l ).
In particular, the representations (0, 0), 1 2 , 0 , 0, 1 2 and 1 2 , 1 2 correspond to a scalar, a left-handed Weyl spinor, a right-handed Weyl spinor and a 4-vector, in this order. A field strength tensor F µν transforms under the Lorentz group in the same way as the anti-symmetric part of the product of two 4-vectors, i.e.
Note that F µν is a real field so the three components transforming as (0, 1) are the complex conjugate of those transforming as (1, 0). I will refer to the part transforming as (1, 0) simply with the letter F and the other part as F * (so F µν = F + F * ). A subscript might be added to F to identify to which of the gauge factor groups it is associated. All SMEFT operators are combinations of the fields shown in table 1, their conjugates and their derivatives (which need to be covariant). There are 3 copies (or flavors) of each fermion field, but we may just treat this number as a variable n. Table 1: SMEFT field content. All operators are built from them, their conjugates and their derivatives. The number of fermion flavors n, which is 3, will often be left unspecified.
Finally, concerning conventions, it is worth noting that sometimes the word operator is used in reference to different things: 1. It might refer to the different gauge and Lorentz invariant contraction of fields, with the flavour indices expanded. In that case the lepton Yukawa interactions L i e c j H * correspond to 9 complex operators, or 18 real ones. That is the meaning I will use for the word operator (which might be real or complex). With this understanding, there are 3045 dimension 6 real operators in SMEFT (546 of which violate baryon number) [8,18].
2. An alternative view is to see some groups of operators such as the lepton Yukawa interactions L i e c j H * as a single structure which I will call a (Lagrangian) term; they can be thought of as operators with the flavor indices unexpanded. With this terminology, SMEFT can be written with just 84 real terms of mass dimension 6 [7].
3. The most general interaction of some combinations of fields cannot be written in the Lagrangian as a single term. In other words, a single coupling tensor κ ij··· with indices ij · · · contracted with the field flavors is not enough. That is what happens, for example, with the fields Q * L * QL: a minimum of two term κ (α) ijkl Q * i L * j Q k L l (α) , α = 1, 2 are needed. We may refer to each collection of terms associated to a common product of fields as a type of operator. 3 There are 72 real types of operators with dimension 6 in SMEFT.

Operators with repeated fields The problem
If there are no derivatives nor repeated fields, the counting of operators of a certain type χ (1) χ (2) · · · χ (m) is straightforward: • Based on the quantum numbers of the fields, we may calculate the number t of independent gauge and Lorentz invariant contractions of their components. Each yields one term.
• Each term is associated to m i=1 n i operators, where n i is the number of flavors of the field χ (i) .
• The number quoted above corresponds to real operators if conjugating the set of fields χ (i) yields back this very same set, i.e. χ (i) * = χ (i) , otherwise there are m i=1 n i complex operators in each term (which is the same as saying that there are 2 m i=1 n i real operators).
So, in the absence of derivatives and repeated fields, there are c t n i real operators of type χ (1) χ (2) · · · χ (n) , where c = 1 if χ (i) * = χ (i) and = 2 otherwise. For example, there are 2n 4 = 162 (for n = 3) real operators of type Q * L * QL. 4 The above is straightforward to compute. Nevertheless, repeated fields complicate this analysis significantly (the problem with derivatives will be discussed in the next section).
Let us start by taking a close look at the type of operator LLHH, which is not associated to n 2 complex operators, but rather to n (n + 1) /2 as pointed out earlier. Given that L and H are both doublets of SU (2) L , there are two possible contractions: (LL) 1 (HH) 1 and (LL) 3 (HH) 3 . In one of them, the two L's are contracted together as a singlet of SU (2) L , and the same happens with the Higgs fields. The other possibility is for the L's and H's 3 So each operator type is defined by the number of occurrences m i of the various kinds of field χ i in the model (= Q, u c , d c . . . in SMEFT). For these purposes, a derivative may be considered as a field, so we should just count them. 4 The list of fields is self-conjugate, {Q * , L * , Q, L} * = {Q * , L * , Q, L}, so c = 1. On the other hand, there are 2 ways of making the SU (2) L contractions of the fields, so t = 2.
to contract as triplets. Crucially, the singlet contraction of two doublets is anti-symmetric, while the triplet contraction is symmetric. Since there is only one copy of the Higgs field, we retain only the latter. If we now introduce flavor indices and a parameter tensor κ we obtain the term This is the same as κ ij (L j L i ) 3 (HH) 3 so, renaming the dummy indices i and j, we can also write it as κ ji (L i L j ) 3 (HH) 3 . We then conclude that only the symmetric part of κ is relevant: Such a matrix has n (n + 1) /2 complex degrees of freedom. Now let us go beyond SMEFT for a moment in order to consider what would happen to the type of operator LLHH if there were multiple copies of the Higgs field. For that purpose we may add flavor indices to H, noting that now the contraction (LL) 1 (HH) 1 is not null so we can have two terms: It should be clear from the discussion so far that the tensors κ (1) and κ (2) have symmetries: As a consequence, κ (1) and κ (3) contain and [n (n + 1) complex parameters, respectively (n H being the number of Higgs copies). But do we need two terms as in expression (5)? The answer is no; there is no such need. Take a single term O formed from a linear combination of O 1 ≡ (LL) 1 (HH) 1 and O 3 ≡ (LL) 3 (HH) 3 : We can recover instead of expression (5). Nevertheless, the tensor κ is not fully generic, as it contains only nn H (nn H + 1) /2 independent complex parameters (the sum of the number of parameters in κ (1) and κ (3) ). In fact, we may write or, alternatively, κ can be described as the most general tensor with the symmetry κ ijkl = κ jilk . Returning to SMEFT, an identical situation happens with the type of operator L * L * LL: four SU (2) L doublets can be contracted in two different ways [6], but only one term is required [7], and it is associated to a total of n 2 (n 2 + 1) /2 = 45 complex operators [8] (this last expression is the same as the one presented before in relation to LLHH, for the special case where n H = n).
These examples highlight the following points: • The symmetry of the coupling tensors κ is determined by the quantum numbers of the fields in each term.
• From the symmetry of the coupling tensors we may derive the number of independent operators.
• Unlike the number of independent operators, the number of terms is an ambiguous quantity since it is possible to merge multiple terms into a single one. To get around this ambiguity, we may always focus on writing a Lagrangian with a minimal amount of terms.
• Merging terms might seem convenient, but there is a drawback to doing so: it may become more complicated to understand the structure of the associated coupling tensor κ.
In order to derive the symmetry of the coupling tensors under permutations of indices, we may consult tables such as [21] which list the permutation symmetry of products of representations. For example, in the case of SU (2) it is well known that with the subscripts indicating that each component in the product is either symmetric (S) or anti-symmetric (A) under a switch of the two doublets. 6 However, such tables might not be enough: consider for example the product of four scalar SU (2) doublets. There are two invariant contractions because the product of four doublets contains two singlets: Are these singlets symmetric (S) or antisymmetric (A)? Mathematically, the answer is clear but it might be somewhat confusing if one is unaware of this type of complications: the singlets are neither (completely) symmetric nor anti-symmetric. Let us go back to equation (13); as far as subscripts are concerned, all we need to know is what happens to each summand in the right-hand side under the permutation of the two doublets. The 6 To be explicit, it is well known that if (2 1 , 2 2 ) T and (2 1 , 2 2 ) T are two doublets, then the combination transforms as a triplet. Switching the two doublets flips the sign of the first expression, while the triplet remains unchanged. permutation of m objects forms a discrete group of size m! usually denoted by S m , and in the particular case of S 2 everything is very simple. There are two elements in the group, the identity e (do nothing) and a (transpose the two objects), with a 2 = e. There are also only two irreducible representations, which are both 1-dimensional: the trivial/symmetric one (S) under which x → x, and the alternating/anti-symmetric representation (A) under which x → (−1) π x, where (−1) π = +1 if π = e and = −1 if π = a. In other words, the action of the group S 2 can always be translated into signs ±1: in equation (13) the singlet 1 transforms under S 2 as the alternating representation, while the components of the triplet 3 transform trivially.
In the case of equation (14) we must find out what happens to its right-hand side under arbitrary permutations of the four doublets, so we must deal with the irreducible representations of the S 4 group. It turns out that the singlets in equation (14) transform as an irreducible 2-dimensional representation of S 4 . To better grasp the significance of this statement, let us see these two SU (2)-invariant contractions of four doublets explicitly. We can write them as follows: where the subscripts take the value 1 or 2 (these are doublet indices unrelated to flavor) and is the Levi-Civita tensor. Note than even though there are m! permutations which can be performed on m objects x i , all of them can be obtained through successive applications of only two of them: So, let us consider in the following what happens to c (1) and c (2) under the changes 2 ↔ 2 and 2 → 2 → 2 → 2 → 2. It is rather easy to see that No change of basis c (α) new = B αβ c (β) will simultaneously diagonalize the two matrices above, so c (1), (2) form an irreducible 2-dimensional representation of the permutation group S 4 . As a consequence, in this case the effect of permuting the doublets cannot be reduced to a simple matter of signs ±1.
We now introduce flavor, so there will be two sets of indices: the g i will stand for group indices, while flavor indices will be called f i . For scalar doublets φ with multiple flavors we should write down two terms: where c (1),(2) g 1 g 2 g 3 g 4 represent the tensors in front of the doublets in equations (15) and (16). As for κ (1) and κ (2) , they are tensors containing free parameters, and there is some symmetry associated to them, which is yet to be determined. (19) explicitly, we may use the short-hand notation κ

Instead of writing down all the indices in equation
where repeated indices are summed over. We know two things: {g} , where the two-by-two matrices P (π −1 ) can be obtained, for any π, from the product of the two matrices in equations (17) and (18). 7 It follows that To be explicit, These are the only constraints on the parameter tensors κ (1) and κ (2) , so these objects can be seen as the most general tensors which satisfy the above equations. It is not very useful for our purposes to find their exact form, which is basis dependent in any case (κ (1) and κ (2) would change under a redefinition c (α) new = B αβ c (β) ). However, just like with the LLHH type of operator, it would be interesting to know how many operators are there in expression (19) or, in other words, how many independent entries are there in the κ (α) tensors. The answer, which is n 2 φ n 2 φ − 1 /12 for n φ flavors of φ, depends only on the fact that the c (α) (and consequently the κ (α) as well) transform under the 2-dimensional representation of S 4 . We shall discuss below how to compute these numbers.
There is one last remark to be made about this example. It might have seemed that we need to write down two terms as shown in expression (19), but it is easy to see that they 7 The inverse permutation of π, π −1 (π −1 • π = id), appears here rather than π itself for the following reason, which is not important for the present discussion. Strictly speaking, for each α the numbers c (α) {g} are not a tensor but rather the components of a tensor in some basis |e g1 e g2 · · · e gm ≡ e {g} . In other words, the tensors are Under a permutation π, the basis of the tensors changes, e {g} → π e {g} = e π{g} , so π C (α) = c (α) However, we know that this results must be a linear combination of the C (α) , P (π) αβ C (β) , therefore we conclude that can be merged into a single one. That is because we may write one of the c (α) as a function of the other, with the indices permuted. In particular, from equations (17) and (18) it can be inferred that c (2) g 1 g 2 g 3 g 4 = c (1) g 1 g 3 g 2 g 4 or, in a short-hand notation, c Note that κ merged is not a completely generic 4-index tensor since it has only n 2 φ n 2 φ − 1 /12 independent entries, rather than n 4 φ .

The permutation group of m objects
These examples highlight the fact that in order to list operators and terms in the Lagrangian, it is important to have an understanding of the permutation group S m . As such, I will briefly review here some of its properties. The irreducible representations of S m can be labeled by the different ways in which the number m can be partitioned. For example It is common to depict the partition λ = {λ 1 , λ 2 , · · · } of m (symbolically λ m) with λ 1 boxes in a row, followed by λ 2 boxes in a second row, and so on, such that i λ i = m and λ i+1 ≤ λ i . These are called Young diagrams. For example: The product of S m representations can easily and efficiently be decomposed in irreducible parts as in this example: There are two other important properties of the irreducible representations λ of S m which are worth mentioning. First, the dimension of λ can calculated with the famous Hook length formula [22], where u represents each box of the Young diagram of λ and h (u) is equal to the number of boxes to the right of u plus the number of boxes below u plus 1: The second important property of a partition λ is the following. A semi-standard Young tableaux with shape λ is obtained by filling the Young diagram of λ with natural numbers up to some limit n (possibly repeating these numbers) in such a way that the numbers increase along each column, and they do not decrease along rows. For example, in the case of λ = {2, 2} and n = 3 there are six possibilities: The number of semi-standard Young tableaux with a shape λ using numbers up to n can be calculated with the Hook content formula (see [23]), where u represent each box of the diagram, h (u) was mentioned earlier, and c (u) is equal to the column number minus the row number of u: As an example, notice that {2, 2} is the 2-dimensional representation of S 4 , d ({2, 2}) = 2, and furthermore S ({2, 2} , n φ ) = n 2 φ n 2 φ − 1 /12 using formula (32). As we have seen, this last expression corresponds to the number of quartic operators which can be build with n φ scalars φ transforming as doublets of SU (2).

Permuting the indices of tensors
Consider now a tensor T i 1 i 2 ···im where each index goes from 1 to some numbers n. It is well known to mathematicians that the n m components can either be referenced with the m indices (i 1 , i 2 , · · · , i m ) or alternative by just 3 special indices (λ, α, j), Rather than trying to write explicitly the components of T λαj as a function of those of T i 1 i 2 ···im , it is more illuminating to describe the nature of the 3 new labels. The first one (λ) can be any partition of m, the second one (α) takes values from 1 to S (λ, n), and the last one (j) goes from 1 to d (λ). The reader will correctly infer from here that the following identity holds: For a fixed α, T λαj transforms as the irreducible representation λ of S m when the i x indices are permuted: where P λ (π) stands for the matrices of the irreducible representation λ of S m . This means that the tensor T contains S (λ, n) irreducible representations λ of S m (one for each value of α). All the above is true for any tensor. It is interesting now to consider the particular scenario where there is not just one tensor, but rather a list of tensors T for some partition µ of m; the φ 4 interactions we considered earlier are the motivation to study this kind of lists of tensors (see text just below equation (20), where the T 's were called c's). Note that the P matrices can always be made unitary and real, so let us assume for simplicity that they do have these properties. With the (λ, α, j) indices it is easy to see what is the most general form of the above list of tensors; it must obey the constraint 8 and so, according to Shur's lemma from group theory, the components of T (a) λαj are non-zero only if λ = µ and j = a, plus their value is independent of j: The T (α) are free numbers; they are the free parameters in the most general list of tensors T (a) subjected to the constraint in equation (38), and there are S (µ, n) of them since the index α can go from 1 to this number. There is one final aspect of lists of tensor with permutation symmetries which is worth looking at, as it will be relevant for the counting of the minimum number of terms in an effective field theory. Consider again a list of tensors which obeys equation (38)  π(i 1 ,··· ,im) for different π (the same is true if we used T (2) or any other component of the list of tensors). This is a consequence of the orthogonality relations among the entries of the unitary (and real) matrices P µ (π) of an irreducible representation: The precise form of this last expression is not particularly relevant; rather, it suffices to keep in mind that all the entries of any of the tensors T (a) can be obtained from the entries of a single one of them (T (1) for example), given that there is the relation (38). However, we have started with a list of tensors T (a) which, under permutations, transforms as an irreducible representation µ of S m . It is helpful to consider what happens if instead of µ we had a reducible representation which is a direct sum (µ 1 ) r 1 + (µ 2 ) r 2 + · · · of distinct irreducible parts µ i repeated r i times. It turns out that if each µ i appears at most once in this sum (r i = 0 or 1), then it is possible to pick a single combination of the T (a) and generate all T (a) from it as we did before with T (1) . On the other hand, if there are µ i which appear more than once, max (r i ) linear combinations of the T (a) are needed in order to generate all of them via permutations of the i 1 , · · · , i m indices. 9 In summary, two important facts stand out from this discussion about tensors: 1. A list of tensors T (a) which transforms under some irreducible representation µ when its indices are permuted (see equation (38)) has S (µ, n) independent entries, where n in the number of values each of the tensor indices can take. For multiple irreducible representations µ i , the number of independent entries is given simply by the sum of the numbers S (µ i , n).

A list of tensors T (a)
which transforms under a direct sum of irreducible representations µ i , where each µ i is repeated r i times -(µ 1 ) r 1 + (µ 2 ) r 2 + · · · -can be reconstructed from just a few of the T (a) . Specifically, only max (r i ) linear combinations of the T (a) are needed for that.
These two observations can be used to count the number of operators and the minimum number of terms in an effective field theory.

Application to operators with repeated fields
We are now in a position to go back to the discussion of operators with a repeated field χ which has two indices: a group index g, and a flavor index f going from 1 to some number n. The product of m χ's is either completely symmetric (if χ is a boson) or completely anti-symmetric (if χ is a fermion), In the Lagrangian, the {g} indices contract with some numeric tensors c The numerical tensors c (α) {g} always obey a symmetry relation of the type where the P matrices form a representation (perhaps reducible) of the permutation group S m , and they can be made real and unitary. As a consequence we see that the parameter tensor κ (α) {f } possesses the following symmetry: The product (±1) π P is itself a representation P of S m , so we can write This representation P can be decomposed into irreducible representations µ m of S m : P = µ m r µ µ, where r µ is the multiplicity of µ in P . 10 The main point of the earlier discussion on tensors is that κ free parameters, so this is the total number of operators, and furthermore they all can be written with max (r µ ) terms. 11 These comments apply to operators χ m with a single type of field χ, however the generalization is trivial for cases where there are more fields, χ m χ m χ m · · · , namely the above considerations apply separately to each group of repeated fields. The fact that generally there is more than one group index g is not a problem either. For example, the Standard Model fields carry an index from each of the four groups SU (3) C , SU (2) L , SU (2) l and SU (2) r . These indices contract with some symmetry P C , P L , P l and P r so, taken together as a single index g, the relevant matrix P is given by the Kronecker product P C ⊗ P L ⊗ P l ⊗ P r , and P = (±1) π P C ⊗ P L ⊗ P l ⊗ P r .
Take the operators of type QQQL: there are n L n Q 2n 2 Q + 1 /3 of them for n Q,L flavors of the fields Q, L, and they can all be written with one term. These two conclusions can be obtained by tracking the permutation symmetry of the contraction of the Q's, as shown in table 2 (L is not repeated, so it is pointless to track its permutation symmetry, which is trivial). In particular, the color indices of the Q's contract anti-symmetrically ({1, 1, 1}), while the SU (2) L indices are contracted with the mixed symmetry {2, 1}. The same is true for the SU (2) l part of the Lorentz group. Since the Q's are left-handed fields, they are singlets of SU (2) r so these indices contract trivially in a symmetric fashion ({3}). Finally, we should take into account that these are fermionic fields, so they anti-commute ({1, 1, 1}). The full symmetry under permutations of the QQQ contractions is given by the product of these S 3 representations, which decomposes into the irreducible components However, we do not need to write down 4 terms in the Lagrangian: since each irreducible representation of S 3 appears only one, we infer that a single term is enough. Furthermore, the total number of operators of the type QQQL is given by the number For n Q = n L = 3, this adds up to 57 complex operators [8]. Historically, the dimension six baryon number violating operators were written as 6 terms in [1] and [2], although it is possible to do so with only 4 [3]. In the particular case of QQQL type operators, they were written as 2 terms in [1] and in [7] 12 but, as stated above, these operators require only 1 term. Notice that Fierz identities were not used to reach this conclusion; spinors and any other field with Lorentz indices are viewed as representations of an SU (2) l × SU (2) r group, and in turn this group is treated in exactly the same way as the gauge group. Consider another example: the interaction of 4 right-handed neutrinos mentioned in the introduction (N c N c N c N c ). Each neutrino transforms as a doublet of the SU (2) l group (which is part of the Lorentz group), and we have seen that the two invariant contractions of four doublets have a permutation symmetry {2, 2}. Furthermore, the components of the N c 's are anti-commuting fields, so there is a total anti-symmetry {1, 1, 1, 1} to be taken into consideration. Overall, under permutations, the N c N c N c N c interactions have a symmetry

Operators with derivatives Handling operator redundancies
The values of a function and the value of its derivatives do not need to be correlated in any way, so a field ψ and its space-time derivatives ∂ n ψ can be treated as different fields altogether. There is however one problem: some operators obtained in this way might be redundant (let us call them O 0 i ), and two Lagrangians differing by such operators will be equivalent.
Faced with this situation, one should work with classes of equivalent Lagrangians i } rather than individual Lagrangians. But these C L are rather abstract constructions, so it might be better to look at redundancies in effective fields theories as a linear algebra problem. For a given model, the Lagrangian can be viewed as a vector This way of selecting non-redundant operators will always work when there is a single redundancy; however, repeating this simple procedure for multiple redundancies might be problematic, so these more complicated scenarios require caution. 13 Reference [25] is particularly relevant for this linear algebra view on operator redundancies.
Two types of potentially redundant operators are often considered: (1) those which are zero when the classical equations of motion (EOM) of the fields are applied, and (2) those which are a divergence of a vector field (∂ µ O µ ). Both of these kinds of operators can be ignored only under some assumptions, whose merits will not be assess in this work. Rather, I will just mention what needs to be done if one wants to factor out these operators. The solution to these problems was already given in the papers [16] and [18]; the following two sub-sections elaborate on the proposal in [16] to remove EOM redundancies, and they contain also a discussion of a simple adaptation to the procedure mentioned in [18] for factoring out operators of the type ∂ µ O µ .
It should be stressed at this point that derivatives in a gauge theory always appear through the combination ∂ µ +igT a A a µ ≡ D µ , where D µ is the well-known covariant derivative. For this reason, all derivatives should be seen as being covariant. However, because [D µ , D ν ] can be written down with field strength tensors F µν , in order not to over-count operators one should only consider the completely symmetric part of D µ 1 D µ 2 · · · applied to some field ψ, in which case the D µ 's can be seen as commuting with each other -just like normal partial derivatives. To highlight that the term igT a A a µ is not important, in this work the symbol ∂ µ will be used instead of D µ .

Equations of motion
It can be shown that the classical equations of motion of a field χ can be used to reduce the number of non-renormalizable operators in an effective field theory [26,27]. That is because non-renormalizable operators proportional to the quantity or its derivatives, where L 4 is the renormalizable part of a Lagrangian, do not affect the S-matrix. The standard approach to these redundancies is to remove the operators with the highest number of derivatives (the second term in expression (55)), since these can be traded by other operators with fewer derivatives (the first term in expression (55)). The authors of [16] pointed out that these redundancies due to the equations of motion are best seen if we decompose the field derivatives ∂ i χ into irreducible representations of the Lorentz group. 14 For example, the 16 second derivatives of a scalar φ transform under SU (2) l × SU (2) r as ∂ 2 φ = (0, 0) + (0, 1) + (1, 0) + (1, 1) ,  . This procedure works well not just for Hilbert series calculations [16] but also for the more straightforward approach to operator counting being described in this work. In practice, the EOM degeneracies are taken into account by introducing a tower of new fields representing the non-redundant parts of ∂χ,∂ 2 χ, · · · , ∂ i χ, for every standard model field χ. They have the same gauge quantum numbers as χ but different Lorentz representations. Since we are expanding significantly the number of fields in the effective field theory, this has the adverse effect of increasing the computational complexity of the calculations. Nevertheless, the procedure to handle the equations of motion is conceptually very simple.
In any case, one must know what are the components of the field derivatives to be kept. For a generic field χ we just saw that there are two considerations to have in mind: • We want only the ∂ i χ components which cannot be written with field strength tensors, hence they correspond to those components which are completely symmetric under permutations of the derivatives. We may represent them by the expression {∂ i } χ which, in general, transforms as a reducible representation of the Lorentz group -for example {∂ 4 } φ = (2, 2) + (1, 1) + (0, 0).
• Some components of {∂ i } χ are also redundant for another reason. If we represent the equation of motion of χ with the notation ∂ x χ = · · · (x = 1 for fermions and x = 2 for bosons), the operator ∂ x χ and its derivatives should also be discarded.
However, there is a difficulty with this two-stage reduction of operators: we cannot just take the irreducible Lorentz representations in {∂ i } χ and remove all those in ∂ i−x ∂ x χ , because some of the latter components are anti-symmetric under permutations of the derivatives, and therefore can be written with the field strength tensor F µν . So in reality the terms to be retained are Figuring out what are the components of ∂ i χ to keep is not a trivial exercise; it turns out that for a scalar φ = (0, 0), a left-hand fermion ψ = 1 2 , 0 and a field strength tensor F = {1, 0} (recall that F = F + F * ) only the highest spin component should be retained, i.e.
The rest of this subsection discusses this result (see also [28]). Let us consider first a scalar φ. It is not hard to show that On the other hand, in order to calculate the Lorentz transformation properties of the components ∂ i−2 ∂ 2 φ which need to be removed, we may want to apply the i − 2 derivatives in a completely symmetric way: It is then tempting to subtract these Lorentz irreducible representations from the ones in equation (64) and conclude that the only non-redundant piece of {∂ i } φ is the one transforming as i 2 , i 2 . The last statement is true, but one should keep in mind that this argument only works because none of the components of {∂ i−2 } ∂ 2 φ is proportional to F µν ; in other words, all components of {∂ i−2 } ∂ 2 φ have a part which is completely symmetric under all permutations of all i derivatives. 15 For derivatives of a left-handed fermion, the transformation properties under Lorentz transformations are as follows: Assuming again that all components of {∂ i−1 } ∂ψ are, at least in part, fully symmetric, the only non-redundant component of ∂ i ψ is i+1 2 , i 2 , with maximum spin. This is the only part of the right-hand side of equation (66) which remains after subtracting the right-hand side of equation (67).
Finally, we have to consider what happens to derivatives of field strength tensors F. As before, one might try to compute the irreducible Lorentz representations associated to the components {∂ i } F (let us call these components ALL) and remove those which appear in the equations of motion {∂ i−1 } ∂F (we may call this set of components EOM ). The trouble is that in some cases the i + 1 derivatives in {∂ i } F and {∂ i−1 } ∂F 16 are not contracted in a fully symmetric way (they form a set AS ). Given that all derivatives are in reality covariant, these {∂ i } F components can be written with two or more field strength tensors F, so they are redundant. We wish to calculate the elements of ALL which are 15 We know that the two derivatives in ∂ 2 φ = η µν ∂ µ ∂ ν are contracted symmetrically, and so are the remaining ones in ∂ i−2 (obviously). Therefore the (i + 1)!/ [3! (i − 2)!] components of ∂ i−2 ∂ 2 φ transform as a representation {i − 2} × {2} of the group S i−2 × S 2 under permutations of the derivatives. From this information alone, one cannot conclude that under permutations of the bigger group S i ⊃ S i−2 ×S 2 all these components transform as a fully symmetric representation {i} (it is conceivable that they could also transform as {i − 1, 1} or {i − 2, 2}). 16 Note that F already contains one derivative. neither in EOM nor AS, but this is complicated by the fact that these last two sets intersect (see figure 1). The number of elements of each set, as a function of i, is the following: The sets EOM and AS cannot be disjoint simply from the fact that #AS + #EOM can be larger than #ALL. Up to large values of i, it is possible to check explicitly that the intersection of EOM and AS contains elements, and from here we can conclude that there are components of {∂ i } F which are not redundant. The maximum spin components, transforming under the Lorentz representation i+2 2 , i 2 + i 2 , i+2 2 , are certainly not contained in either EOM nor AS, and they are precisely 2 (i + 3) (i + 1) in total. Hence, one can conclude that these are the only components of {∂ i } F to keep.

Integration by parts
There is another kind of operator redundancy. The action is calculated by integrating the Lagrangian density over all spacetime, so what matters are not the operators O themselves but rather their spacetime integral. If O is a divergence, O = ∂ µ O µ , and the value of O µ drops to zero fast enough at large distances/times, then by the divergence theorem where M is some infinitely large volume of spacetime. Putting aside the possibility that the above integral might not always vanish, operators of the form O = ∂ µ O µ should be discarded. This is often seen as an integration by parts redundancy: if the sum of two operators O (1) µ and O (2) µ is a total divergence, an integration by parts reveals that O (1) µ is equivalence to −O (2) µ . We may track the operators which are total derivatives by introducing a dummy field D in the effective field theory representing total derivatives, so that operators of the form D (· · · ) should be factored out. However, some of these latter operators are redundant [18], because there are linear combinations of them which add up to zero. Unfortunately, it does not seem easy to detect these relations among operators which are total divergences.
Nevertheless, the authors of [18] managed to count them with the Hilbert series method and conformal field theory; they also offered an interesting interpretation of their calculation which does not rely on these particular theoretical tools. It goes as follows, using the language of differential forms. The redundant operators O = ∂ µ O µ are associated to those differential 4-forms ω (4),red which are exact, meaning that they are the differential of some 3-form ω (3) : ω (4),red = dω (3) . Nevertheless, one cannot simply consider all ω (3) because the differential of some 3-forms is identically zero, i.e. dω (3),red = 0. In turn, according to Poincaré's lemma, these 'redundancies of the redundancies' are associated to all those 3-forms which are the differential of a 2-form ω (2) : ω (3),red = dω (2) . This recursive argument goes on, but it eventually stops because of the dimensionality of spacetime (four) when we reach 0-forms, which are pure functions.
Let us rephrase this argument in the language of operators. In general, a differential i-form ω (i) can be written as where O µ 1 µ 2 ···µ i is a completely anti-symmetric tensor under an exchange of indices; to highlight this fact, from now on square brackets will be applied to its indices. Applying the differential operator 4 − i times to ω (i) yields In summary, all linear combinations of the operators {O} are non-zero, but we wish to remove the degeneracies {∂ µ O µ }. Unfortunately, this last set includes some operators which are identically zero, hence the need to include the remaining sets in expression (76), containing only null operators. If all we want is to count the number of non-redundant operators up to some dimension d, then the result is 17 . (77) In order to generate not just the scalar operators O but also the tensors O [µ 1 µ 2 ···µ i ] , one can introduce the derivative D as a stand-alone field as mentioned earlier, with all the expected properties, except that it anti-commutes (making it a Grassmann field). As an example, table 3 #O [µνρ] = 1 6 n 2 (n + 1)(n + 2) (operators with 3 D's) , #O [µνρσ] = 1 24 n(n + 1)(n + 2)(n + 3) (operators with 4 D's) .
It follows that there is a total of 1 8 n (n 3 + 2n 2 + 3n + 2) operators of the generic form ∂ 4 S 4 .

Implementation in a computer code
The approach described in this work makes it possible to count operators of an effective field theory, and also to extract some extra information about them. However, in order to implement it, for a representation R of some Lie group G one must be able to decompose   the tensor product R m = R × R × · · · × R into irreducible representations of G × S m . It is not enough to know the decomposition of such products into irreducible representations of the Lie group G alone. There is a widely known technique involving Young tableaux to extract this information when G is a special unitary group SU (p) and R is its fundamental representation. Any SU (p) representation can be labeled by a partition or a Young diagram, much like the representations of the permutation group, with two caveats: 1. Columns with p rows can be ignored, so two Young diagrams differing only by such columns stand for the same SU (p) representation; 2. Diagrams with more than p rows are not associated to any SU (p) representation.
If R is the fundamental representation of SU (p), it turns out that the tensor product R m = R × R × · · · × R decomposes into the sum of all representations λ λ of the group SU (p) Sm (λ is a partition of m with at most p rows). For example, in the case of four SU (2) doublets, λ can be {4}, {3, 1} or {2, 2}, so we obtain the decomposition  (83) there is a total of 1 × 5 + 3 × 3 + 2 × 1 = 16 = 2 4 components, as expected. One can alternatively express the product of four doublets as but in doing so we are erasing critical information from equation (83). The tight connection between SU (p) and the permutation group which was just mentioned is all that is needed to count operators of some models (including SMEFT). However, it is worth pointing out that this type of decomposition can be calculated for any representation of any group, including discrete ones. The LiE program [29] does this efficiently for any simple Lie group, referring to it as a computation of a plethysm. The algorithm described in LiE's manual was implemented in the Susyno package for Mathematica [19]. This latter program also contains several other functions related to Lie algebras and the permutation group S n . The first version of the Sym2Int package [20] uses them to list the operators without derivatives (nor field strength tensors) of any effective field theory. This can be done up to some arbitrary mass dimension. Adapting the solutions in the literature [16,18] to tackle the problems inherent to derivatives, the latest version of Sym2Int can list automatically all operators in an effective field theory, including those with derivatives and field strength tensors.

Comparison with other methods and computer codes
All SMEFT operators up to dimension 15 were counter and characterized in the way described here in under two hours on a standard laptop computer (details of these results can be found below, at the end of this section). All the numbers and expressions given in [18] (and in the accompanying data files) were successfully reproduced with the alternative method described here. 18 Two other noteworthy codes should be mentioned. The first one is DEFT [25]: rather than just calculate the number of operators and their symmetries, this code works with the actual operators, performing the gauge and Lorentz index contractions explicitly. Dealing with operators redundancies associated to derivatives then becomes a linear algebra problem. Having the operators written down explicitly is clearly something very useful for model builders and it opens up several possibilities, such as of implementing in different ways the operator redundancies discussed previously. One downside of the more ambitious scope of DEFT is that it takes more time to perform these calculations -as a point of reference, the authors of [25] were able to calculate SMEFT up to dimension 8 for one fermion generation, which corresponds to roughly a third of the number of SMEFT operators up to dimension 6 ,with three generations.
The second noteworthy code is BasisGen [30]. It counts operators, and does so with the basic approach of multiplying all fields together, and retaining only the gauge and Lorentz invariant solutions. This is essentially the same method of counting operators as the one 18 The only exception is the number of lepton number conserving operators of dimension 12 given in section 4.2 of [18] (it seems to be 4481 + described in this paper and implemented in Sym2Int (this was called the intuitive method in the section 2). However, repeated fields are handled differently, and this has significant impact on the computational time for models with multiple flavors. Instead of n flavors of a field ψ, BasisGen considers n distinct fields fields ψ, ψ , ψ , ... with the same quantum numbers, meaning that flavor indices are expanded. From this perspective all fields have a single flavor so, referring back to the discussion in section 3, one only needs to retain the completely symmetric contractions of repeated fields. Because each entry of the flavor tensor in front of a term is evaluated separately, the computational time of this approach increases sharply with the number of flavors. Concerning derivatives, BasisGen deals with the integration-by-parts redundancies in a way which is different than the one described in this work, however it is likely that the two methods are equivalent.

Information beyond the number of operators
It is possible to track each field flavor separately, both with the intuitive method described here (see also [30]), and with the Hilbert series method (this leads to a so-called multigraded Hiberts series). However, doing so is very time consuming therefore in the following I will analyze the information which can be extracted from both methods when the field flavor indices are not expanded. The Hilbert series approach counts operators of a certain type for an arbitrary number of generations n, telling us, for example, that there are n (n + 1) /2 operators of the type LLHH. At first sight, the intuitive method might seem to yield more information: it computes that the n × n flavor matrix κ ij in κ ij L i L j HH is symmetric, i.e. it transforms under the irreducible representation = {2} of S 2 . This implies that there are S ({2} , n) = n (n + 1) /2 operators of this type, so the second result (the permutation symmetry) implies the first one (the operator counting). But does it contain more information? For this particular example, the answer is no: there is no extra information in the permutation symmetry because from the number of operators as a function of n it follows that κ is necessarily a symmetric matrix.
More generally, we can frame the discussion as follows. The intuitive method provides the symmetry of the parameter tensors as a sum λ m r λ λ of irreducible representations λ of some permutation group S m , each with multiplicity r λ . From here one can always calculate the total number of operators as a function of the number of flavors n, while the Hilbert series method provides only the function #operators (n). 19 Knowing the left-hand-side of this expression, is it possible to retrieve the integer coefficients r λ ? In simple cases it is. The quantity S (λ, n) is a polynomial function of n of degree m (because

Counting terms with derivatives
Another important point worth mentioning is that the permutation symmetry of operators with derivatives is unclear, due to the integration-by-part redundancies. Consider the example in table 3 Finally, it is worth discussing the number of terms (as defined in section (2)) associated to operators with derivatives. In the absence of derivatives, the symmetry of the contraction of the fields can be expressed as a sum λ r λ λ of irreducible representations λ of the relevant permutation group, with r λ representing the multiplicity of λ. We have seen already that these operators can be expressed with t = max (r λ ) terms, and no less.
Integration-by-parts redundancies complicate this calculation for operators with derivatives. Nevertheless, one can easily establish bounds on the minimal number of terms t: • t cannot be smaller than the number of operators when considering only one generation of fields (n = 1).
• t does not need to be larger than the number of irreducible representations of the permutation group obtained after the procedure described a few paragraphs earlier.
• t also cannot exceed the number of terms needed to write all interactions if integrationby-parts redundancies are ignored.
For example, there is only one operator of the kind ∂ 4 S 4 if there is only one flavor of S, so t ≥ 1. On the other hand, ignoring the redundancies in table 3, one could write all operators with 3 terms, so t ≤ 3. Finally, the ∂ 4 S 4 interactions are associated to {4} + {2, 2} permutation symmetry (two irreducible components), therefore t ≤ 2. In summary, these operators require either one or two Lagrangian terms (1 ≤ t ≤ 2).

Application to specific models
The approach described in this work can be used to characterize individual interactions of a model. However, it would not be instructive to present here an exhaustive analysis of this kind. Instead, I will show some summary data of the interactions up to dimension 15 of three models: (a) the SMEFT; (b) an SU (5) model with the left-handed fermion representations 3 × 5 + 3 × 10 and a scalar transforming as a 5; (c) an SO(10) model with the left-handed fermion representations 3 × 16 and a real scalar transforming as a 10. These two latter models were picked to illustrate the effect of an enlarged symmetry group on the number of interactions, so the only fermion and scalar representations of SU (5) and SO (10) which were selected are those which contain the Standard Model fields. The total number of real operators, terms, and types of operators up to dimension d = 2, · · · , 15 in these three models is represented graphically in figures 2 and 3. Exact numbers can found in appendix. In the case of SMEFT, the number of operators up to Figure 2: Commutative number of real operators, terms, and types of operators in SMEFT (as defined in section 2) up to a given dimension. Kinetics terms are ignored. Due to the presence of derivatives in some operators, the number of required Lagrangian terms cannot be determined exactly, but fairly stringent bounds (narrow green band) can be set on this quantity. dimension 15, as well as the number of types of operators up to dimension 12 agree with the results obtained with the Hilbert series method in [18].
As mentioned earlier, the minimum number of terms which are needed to write down all operators can be calculated exactly for terms without derivatives. If there are derivatives, with the considerations made in this work it is only possible to derive bounds on this number. One can see from the figures 2 and 3 that these bounds are fairly strict. Furthermore, the lower limit is close to (but never below) the number of operators in each of the models if they had a single fermion family. Figure 3: Commutative number of real operators, terms, and types of operators up to a given dimension in three models with different gauge symmetries -SU (3) × SU (2) × U (1), SU (5) and SO (10). The fermions and scalars content of each model are analogous (details can be found on the main text). As the symmetry group is enlarged, the number of operators, terms, and types of operators is reduced, as expected.

Summary
Operators in an effective field theory can be counted with the Hilbert series method. This technique is very different from the intuitive one of simply multiplying together a model's fields in all possible ways, and retaining the combinations which are invariant under all relevant symmetries (such as the ones associated to the Lorentz and gauge groups). However, such seemingly straightforward approach -which we may call the intuitive method -is complicated by the fact that many operators have repeated fields, as well as derivatives.
This work describes a systematic and efficient way of addressing the problem of repeated fields. Derivatives can be handled with simple adaptations of the techniques known to work for the Hilbert series method. With these two difficulties solved, it becomes possible to count all operators of an effective field theory up to a high mass dimension, without replying on the Hilbert series technique.
The intuitive method is potentially useful for various reasons. An obvious one is that it makes it possible to cross check results obtained in other ways. For example, the operators of the Standard Model effective field theory (SMEFT) were previously counted up to dimension 15 with the Hilbert series technique, but in the literature there was only confirmation of these numbers up to dimension 8. With the procedure described in this work, it is possible to verify the number of operators up to dimension 15, for an arbitrary number for fermion families.
Another interesting feature of the intuitive method is that it provides more information beyond a simple operator counting. Among other things, the extra information can be used to count systematically the minimum number of Lagrangian terms required to encode a model's interactions. For instance, it indicates that all QQQL operators in SMEFT can be written down as a single term in the Lagrangian (in the past, this was not always clear), while the significantly more complicated interactions with twice as many fields, of the type Q 6 L 2 , need 11 terms.
The method described in this work was implemented in the program Sym2Int, which can readily be used to calculate the above operator properties in any field theory.