Invariant Tensors in Gauge Theories

Invariant tensors play an important role in gauge theories, for example, in dualities of N=1 gauge theories. However, for theories with fields in representations larger than the fundamental, the full set of invariant tensors is often difficult to construct. We present a new approach to the construction of these tensors, and use it to find the complete set of invariant tensors of a theory of SO(3) with fields in the symmetric tensor representation.


I. INTRODUCTION
The invariant objects in a gauge theory play an important role. In a gauge theory, fields transform under various representations of the symmetry. However, the observables are gauge invariant, and, hence, are invariant combinations of the fields. Similarly, in a confining theory, the physical spectrum is composed of gauge-invariant bound states of the fundamental fields.
Similarly, duality of supersymmetric gauge theories was initiated by [1]; here two or more supersymmetric gauge theories were conjectured to be equivalent at low energies. Testing these duality conjectures requires detailed knowledge of the theory being studied. This is the case for matching the flat directions, matching the spectra or the chiral rings, or for checking 't Hooft's anomaly matching conditions [2]. For all these tests, a knowledge of the gauge invariant operators is necessary. Indeed, one might expect an exact equivalence of the operators in these theories [3][4][5].
The invariant tensors of the fundamental representations of the classical groups are known 1 . However, the tensors for the other representations are not completely classified. For the exceptional groups such as E 6 , it is not even known if the full set of tensors for the fundamental representation have been found [18].
Our goal in this paper is to present an approach to finding the fundamental polynomial invariants of gauge theories (this is sometimes known as a Hilbert basis). The elements of the Hilbert basis are said to be (polynomially) independent. We, therefore, wish to find the invariant tensors for physically relevant groups and representations, with the goal of eventually being able to classify the tensors for the various representations of the exceptional groups.
Our approach is to successively reduce the symmetry group to a tractable subgroup for which the invariant structure is known. We then find the necessary invariants in the full theory that are required to reproduce the invariants of the subgroup. This then yields all invariants of the larger group.
The method is more easily described by example. We begin by applying it to two simple cases for which the answer is known. These are (1) the group SO(2) with fields in the fundamental representation, and (2) the group SO(3) with fields in the fundamental. The invariants of these theories are well known, and we rederive them by our method below.
Finally, we apply our method to a nontrivial case: SO(3) with fields in the symmetric tensor representation. The invariants here are not well known. However, we are able to prove that there are invariants of degree 2, 3, 4, and 5, and no others. We are also able to explicitly find the form of the invariants.

II. SO(2) WITH FUNDAMENTALS
SO (2) is the group of 2 by 2 orthogonal matrices (we are here using the physicist approach to groups and algebras). A general element can be represented as where 0 ≤ θ < 2π. The corresponding algebra is generated by the single matrix The general group element is related to the algebra element as g = e iθm . The fundamental representation can be written as a column vector of two elements (One could also impose a condition on the norm e.g. a 2 + b 2 = 1, but this is not necessary). The group acts on the representation as while the infinitesimal transformations act on the representation as The so(2) algebra is isomorphic to a U(1) algebra. If we define Φ = a + ib, then the action of the group is just Φ → Φe iθ . Invariants are, therefore, of the form Re(Φ 1 Φ 2 ) or Im(Φ 1 Φ 2 ) (or products thereof). These can be written in SO(2) language as r i r i or ij r i r j .
There are thus two invariants of degree 2 in this theory. As we have seen, this is easy to show directly. Nevertheless, we rederive this using a more generalizable method.
Accordingly, we take the approach where we try to reduce the symmetry. We do this by choosing one particular fundamental and giving it a fixed value (i.e. what is called spontaneous symmetry breaking in quantum field theory). We choose the fundamental r (0) and set where C is a constant. We will refer to this fixed value as the vev (short for vacuum expectation value). This vev is fixed; therefore, the remaining symmetries are those that leave the vev fixed. In this case, there are no remaining symmetries. The vev has broken the symmetry from SO(2) to nothing non-trivial. Now consider a second fundamental Now the symmetry is completely broken; and hence the SO(2) structure has no meaning. This is really two independent fields c and d with no symmetry linking them. The invariants are therefore immediate; they are any combinations of c, d. If we add a further fundamental the invariants would be combinations of c, d, e, and f . Hence when the symmetry is broken, the invariants are of degree 1 and are arbitrary combinations of the fields. What does this tell us about the original theory, assuming that we knew nothing of the invariants of SO(2)? We must have that the invariants of so(2) are such that after r (0) gets its fixed vev, they reduce to the degree 1 invariants of the broken theory.
The simplest way this could happen is if there were degree 1 invariants in the SO(2) theory; it is easy to check that there are not.
The next possibility is a degree 2 invariant such as r which would become a degree 1 invariant once r (0) is replaced by a fixed value. Indeed, this is an invariant, and becomes Cd once evaluated. Here C is just a constant, and is not relevant for the purposes of matching invariants.
We still need something that will give us the field c. This is the other degree 2 invariant ij r These two invariants r are therefore capable of generating all invariants of the broken theory.
We can therefore assert that r j generate the invariants of the full SO(2) theory, where r

III. SO(3) WITH FUNDAMENTALS
SO (3) is the group of 3 by 3 orthogonal matrices. The corresponding algebra is generated by the three matrices The fundamental representation can be written as a column vector of three real elements The infinitesimal transformations act on the representation as Again we try to reduce the symmetry. We do this by choosing one particular fundamental and giving it a fixed value. We choose the fundamental r (0) and give it a fixed vev The remaining symmetries are those that leave the vev fixed. In this case, this would be a choice of α i such that It is fairly straightforward to go through the exercise and show that the only possibility is α 2 = α 3 = 0, while α 1 is anything. This is a residual SO(2) symmetry. The vev has broken the symmetry from SO(3) to SO(2). Now consider a second fundamental Now the symmetry is broken to SO(2). The d and e fields form a fundamental of SO (2), while f is a singlet. There are degree 2 invariants involving the fundamental, as described in the previous section, as well as degree 1 invariants from the singlets. More precisely, if we introduce a further fundamental the invariants would be generated by the singlets f, k the bilinear (dg+eh) and the bilinear (dh-gf). We must have that the invariants of SO(3) are such that after r (0) gets its fixed vev, they reduce to these invariants. A quick check shows that we need a bilinear invariant r k . (One needs to check that these are actually invariants, but that is also well known.) Specifically, the bilinears r generate the degree 1 invariants after r (0) gets a vev, the bilinear r generates the bilinear (dg+eh), and the trilinear ijk r k generates the bilinear (dh-gf). We can therefore assert that r j generate the invariants of the full SO(3) theory with fundamentals.

IV. SO(3) WITH SYMMETRIC TENSORS
The symmetric tensor representation can be written as a tensor T ij where T ij = T ji and δ ij T ij = 0.
The infinitesimal transformations act on the representation as We are looking for invariants; in this case, in fact, there is no answer in the literature for the full set of invariants. Note that we can contract the indices either using the invariant tensor δ ij or the invariant tensor ijk . However, since an element of the symmetric tensor representation has an even number of indices, there must be an even number of epsilon tensors,tr which can then be converted to delta tensors. It follows, then, that we may write a general invariant of this representation as the trace of an arbitrary (matrix) product of symmetric tensors. We will denote such an invariant as tr(AB . . . N ) = A ij B kl . . . N mn δ jk δ lm δ ni .
Again we try to reduce the symmetry. We choose one fundamental T (0) and give it a fixed vev. Now however, there are many possible choices which are physically inequivalent. We will choose That is, This breaks the SO(3) symmetry; the remaining continuous symmetry is the SO(2) symmetry where α 1 is nonzero, and the rest zero. Any other tensor T ij then decomposes into 5 fields, of charges 2, 1,0,-1,-2. The fields (with the charge labelled as a subscript) are explicitly: and so may be written as In addition there is a discrete Z 2 symmetry which interchanges all indices 1 with an index of 2, so that for example T 13 ↔ T 23 . This symmetry removes the epsilon tensor of the SO(2) symmetry. Under the discrete symmetry which interchanges the 1 and 2 indices, we have, We now want to construct all possible combinations of these fields which are invariant under both the U(1) and Z 2 symmetries.
For the U(1), the invariants combinations are those where the charges sum to zero. These are T 2 T −2 , T 1 T −1 , T 2 T −1T−1 , T −2 T 1T1 , and T 0 . However, these are not all invariant under the Z 2 symmetry. Under the symmetry, for example, T 2 T −2 → T 2 T −2 , so there are two linear combinations: T 2 T −2 + T 2 T −2 , which is invariant under Z 2 , and T 2 T −2 − T 2 T −2 , which picks up a minus sign under a Z 2 transformation. The invariants are, then, either even or odd under the Z 2 .
We then find that the following expressions are invariant under both the U(1) and the Z 2 symmetries On the other hand, there are combinations which are invariant under the U(1), but odd under the Z 2 , which are We can find new invariants by taking pairwise products of these. There are 6 possible pairwise products . In principle, all these are further invariants. However, many of these can be reexpressed in terms of the previously found O 1−4 . In fact, However, the operator cannot be expressed in terms of the O 1−4 . The ring of operators is therefore generated by the operators O 1−5 .
We would now like to use these operators to pick from the sea of possibilities the unique SO(3) invariant operators that allow us to generate the full space of SO(3) invariants. So now, the question is whether we can generate all these invariants of the broken theory from SO(3) invariant combinations of tensors when one SO(3) tensor, which will be denoted T (0) , gets a vev as in equation (17).
If one of these tensors obtains a vev i.e. B = T (0) , this is We then conclude tr(AB) must be part of the chiral ring of SO (3), and that the operator O 1 is generated from this element of the ring when a tensor gets a vev and breaks the group to SO (2). That is, The element tr(AB) also yields a SO(2) bilinear when neither obtains a vev. This can be written as Thus, the specific combination of bilinears Re(A 1 B −1 ) + 1 4 Re(A 2 B −2 ) can also be generated by combinations of the bilinear element of the SO(3) chiral ring. However, the other linear combinations of the bilinears cannot be produced by uisng the SO(3) bilinear. We need to use another element of the SO(3) chiral ring.
The next possibility is the invariant of degree 3: tr(ABC). Using the decomposition (18), we find If C gets a vev, then we can set C 0 = 1 and the other C i =0. This then generates the bilinear operator This implies that the trilinear tr(ABC) is part of the SO(3) chiral ring. Then the bilinear operator Re(A 2 B −2 ) − 2Re(A 1 B −1 ) can be generated in terms of elements of the SO(3) chiral ring. In combination with the bilinear tr(AB), all operators O 1−3 can be generated.
The trilinear tr(ABC) also yields a trilinear in the SO(2) fields if none of the tensors has a vev. This trilinear can be written as where the ellipses are combinations of O 1−3 . Hence the completely symmetric trilinear is generated. To generate the other trilinears, we must consider a new SO(3) invariant.
The next possibility is a invariant of degree 4; tr(ABCD). However, we can narrow our search a little. The completely symmetrized trilinear has already been generated, and, so, we need to look for invariants where at least two tensors are antisymmetrized. There are exactly two such possibilities; where the ellipses are combinations of O 1−3 . We therefore generate the trilinear operator (Re( We also find that the tr( We now turn to the generation of all terms of the form O 5 . It is useful, however, to first work out what structures we are looking for. The O 5 terms have the general form Im(A 1 B −1 )Im(C 2 D −2 ). There are other such forms related by a permutation acting on ABCD. There are 6 such independent permutations which are Im( It is useful to separate these into combinations which are (anti)symmetric when we interchange the two pairs of fields. In the symmetric terms we further isolate the expression which is antisymmetric under any pairwise exchange. These are If no field gets a vev, the operator tr([A, B][C, D]) generates a operator of the form O 5 : Similarly, we also have We must consider invariants of degree 5. These are of the form tr(ABCDE) where one field gets a vev, since we must generate a quadrilinear. We can take E to be the field getting a vev; there are then 24 possible permutations of the remaining field in this invariant.
We first try to generate the completely antisymmetric (in ABCD) structure. It is relatively easy to see that this requires the SO(3) chiral ring to include the structure tr(ABCDE) where all five tensors are completely antisymmetrized.
We then generate the structures in which (AB) (CD) are antisymmetric, and where the structure is odd under pairwise interchange of (AB), (CD). The only possible nonzero candidate structure is We therefore include the last candidate as part of the chiral ring, and find its decomposition once E gets a vev. This In this note, we have presented a new approach to constructing the invariant tensors for theories with arbitrary representation content. We have applied this method to simple examples and shown that the known results are reproduced. We have also applied this method to a theory where the chiral ring was not known -a SO(3) theory with symmetric tensors -and shown that we can extract the chiral ring by our methods.
While we have worked with relatively simple groups here we expect that our methods can be generalized straightforwardly to other groups and representations. In particular, it would be interesting to find invariant tensors for SU(N ) tensors with adjoint representations, which are interesting for the physics of strong interactions, and for exceptional groups, which are important for applications to dualities in supersymmetric theories.