We consider constraints on higher-dimensional operators for supersymmetric
effective field theories. In four dimensions with maximal supersymmetry and
Article funded by SCOAP3
4$}. For $n=4$, we have shifted numerator in $\langle 3\hat{4}\rangle$, and momentum conservation allows cancellation to yield a local polynomial. Next let's consider $D^{2p}F^n$ operators, again for the case of MHV, it requires 2 $\lambda_1$'s, 2 $\lambda_2$'s, and 2 $\tilde\lambda_k$'s for $k \neq 1,2$, but now with $2p$ additional pairs of $|j\rangle[j|$ where $j$ can be arbitrary legs. This implies $2+p$ spinor brackets, and thus can only cancel the poles if $p>1$, i.e.\ $D^2F^n$ operators do not have a SUSY local completion for MHV matrix elements. Let us see if $D^4F^6$ exists. Now the requirement is that \eq \frac{A_{n}(-,-,+,+,+,+)}{\langle12\rangle^4} \eqe to be local. This implies that it must be (here perm indicates permutation in $\{3,4,5,6\}$) \eq \left([12]^2[34]^2[56]^2+{\mathrm{perm} }\right) \eqe All other possibilities are equivalent via Schouten identities. Thus in conclusion, we have ruled out $F^n$ and $D^2F^n$ matrix elements as having MHV local matrix elements, but the S-matrix of $D^mF^n$ with $m \geq 4$ has a valid SUSY completion. \paragraph{NMHV matrix elements:} as discussed in~\cite{ElvangSUSYW}, for an N$^k$MHV matrix element, the amplitude is given by \eq \mathcal{A}_n=\frac{\delta^8(Q)}{\langle nn{-}1\rangle^4}\sum_{I} c_{I}\prod_{a=1}^k(X_a)_{(i_a)_1,(i_a)_2,(i_a)_3,(i_a)_4} \eqe where $I$ labels the distinct Young tableaux where the indices $\{(i_a)_1,(i_a)_2,(i_a)_3,(i_a)_4\}$ for a given $a$ populates a row and are hence symmetrize while each column is antisymmetrized. The function $X$ is defined as: \eq (X_a)_{(i_a)_1,(i_a)_2,(i_a)_3,(i_a)_4}=\prod_{A=1}^4\frac{\eta^A_{i_A}[n{-}2,n{-}3]+\eta^A_{n{-}2}[n{-}3,i_A]+\eta^A_{n{-}3}[i_A,n{-}2]}{[n{-}2,n{-}3]^4} \eqe and $c_I$s are component amplitudes with legs $n{-}3$, $n{-}2$, $n{-}1$, $n$ taking helicity $(+,+,-,-)$ respectively, while the remaining legs are determined by the $4k$ set of indices $\{(i_a)_1$,$(i_a)_2$, $(i_a)_3$, $(i_a)_4\}$. The important point is the presence of the fourth order pole $\langle nn{-}1\rangle$. To test whether the singularity of this pole is a true singularity, we perform a three-line shift on legs $n{-}3$, $n{-}2$, and $n{-}1$, such that $\langle nn{-}1\rangle\rightarrow \langle n \widehat{n{-}1}\rangle$. Next, we project out the component matrix element $A_n(-,-,\cdots,+,+,+,-)$. Note that this component amplitude was chosen such that we have plus helicity on legs $n{-}3$, $n{-}2$, and $n{-}1$. This is advantageous as this isolates the shifted angle brackets to be completely contained in $c_{I}$, which are the component amplitudes. For NMHV $D^{2p}F^n$ operators, we must have 6 $\lambda_i$s and $(n{-}3)$ $\tilde{\lambda}_i$s, with additional $p$ angle and square brackets respectively. Thus besides $p=0$, we will in principle have enough angle brackets to cancel the fourth order pole. However, for $p=0$, the matrix element is proportional to $3$ angle brackets constructed from the three negative helicity legs, and vanishes under permutation symmetry. One might wonder if other component operators can have non-zero matrix elements, and thus we would have a non-local SUSY element. In appendix~\ref{SUSYWard}, we will show that using permutation symmetry, we can express all possible components in terms of that of $F^n$ for $n=6$, and thus if the S-matrix of $F^6$ vanishes, so must each the whole super-matrix element. Thus there exists no local super-matrix elements for NMHV with mass-dimensions $n$, at least for $n=6$. For any non-zero $p$'s, there can be NMHV local amplitudes. For example, an explicit form of local SUSY completion of $D^2F^6$ was given in ~\cite{ElvangCT}. Beyond NMHV, in general there will be sufficient number of angle brackets to cancel the higher order poles. It is intriguing to see what conclusions one can draw with reduced supersymmetry. Such scenario may arise if one considers corrections due to BPS objects. For $\mathcal{N}=2$ and $\mathcal{N}=1$, the on-shell degrees of freedom are encapsulated in two distinct superfields, $\Phi,\,\Psi$, which contains the positive and negative helicity vector respectively~\cite{ElvangH}. Thus there are distinct $\frac{n!}{(k{+}2)!(n{-}k{-}2)!}$ super-matrix elements for N$^2$MHV configuration. For MHV, the super-matrix element that contains $m(-,-,+,\cdots,+)$ for $F^n$ can be written as: \eq \mathcal{A}_n=\frac{\delta^{2\mathcal{N}}(Q)}{\langle12\rangle^{\mathcal{N}}}A_n(-,-,+,\cdots,+)\,, \eqe where $\mathcal{N}=1,2$. Following the previous analysis we can see that local super-matrix elements \emph{can} exists for $\mathcal{N}=1,2$. For example, one has: \eqa \nonumber\mathcal{N}=2&:& \; \delta^{4}(Q)\left([34]^2[56]^2\cdots[nn{-}1]^2{+}{\rm perm}\right)\,\\ \quad \mathcal{N}=1&:& \; \delta^{2}(Q)\langle12\rangle\left([34]^2[56]^2\cdots[nn{-}1]^2{+}{\rm perm}\right)\,. \eqae Note that for $\mathcal{N}=1$ the super-matrix element is anti-symmetric in $1,2$. This is valid since $\Psi$ is fermionic for $\mathcal{N}=1$. ]]>
5$ with N$^{\frac{1}{2}}$MHV amplitudes. In such case, the only possible angle brackets stem from $q_{i}$, which is simply the component matrix element $A_n(\cdots, \phi(i),+,+,-,-)$ for $(F_-)^2(F^+)^{2q}\phi$. This will have exactly 2 angle brackets, and thus will not be sufficient to cancel all the poles. For $n=5$ it is straight forward to see from eq.~(\ref{Sp45pt}) that the poles cannot be canceled. Thus we conclude that there are no local $\mathcal{N}=4$ SUSY completion for $(F_-)^2(F^+)^{2q}\phi$. Note that the above result can be straightforwardly generalized to operators of \linebreak $(F_-)^2(F^+)^{2q}\phi^m$. These simply correspond to matrix elements of the form \eq \frac{\delta^8(Q)}{\langle n{-}1n\rangle^4}q_{i_{1},,\cdots,i_{m}}\prod_{a=1}^mY_{i_{1}}\cdots Y_{i_{a}}\,, \eqe where again the coefficients $q_{i_{1},\cdots,i_{m}}$ are given by linear combinations of component amplitudes $A_n(\cdots,\phi(i_1) ,\cdots,\phi(i_m),+,+,-,-)$ of $(F_-)^2(F^+)^{2q}\phi^m$. Since the latter still has two angle brackets and cannot cancel against the fourth order pole, this implies that its coefficient will again be iteratively determined by $(F_-)^2(F^+)^{2}\phi^m$, which in fact again is one-loop exact~\cite{DineSeiberg} and can be obtained from eq.~(\ref{eq:DineSeiberg}). This proportionality can be easily fixed simply by considering the analogous diagrams as in \SU(4) case, with summing over all possible insertions of extra scalar lines at each vertex. We will use $c_0^{1,q}(\phi)$ as an example. Beginning at seven points, we have two ways to assign the scalar line to one of the two vertices. As a result, relative to the six-point vertex in eq.~(\ref{greyL}), there is an additional factor 2 in the seven-point vertex. Furthermore seven-point vertex can be viewed as a contraction of two four-point vertices with a overall factor 2 which is the number of possible ways for the scalar insertion. The procedure iteratively continues to higher-point vertices as what has been done before. In general, ($2n_v+3$)-point diagrams (except for the diagram from contact term) can be categorized into $n_v$ 4-point vertices with different numbers of contractions by black and grey lines and one scalar line assigned to the one of the $n_v$ vertices. In other words, for each category of ($2n_v+3$)-point diagrams, it produces $n_v$ number of scalar insertion diagrams. It turns out there is an overall $n_v$ factor for every category from the scalar insertion and the remaining contractions between 4-point vertices give the same result as in analysis of MHV matrix element. With $n_v=(n-3)/2$, this implies that \eq\label{Predict2} c_0^{1,q}(\phi)=q \, 4^{q-1}\left(c_0^{1,1}\right)^{q-1}c_0^{1,1}(\phi)\,. \eqe Since $c_0^{1,1}$ is protected and unrenormalized, this implies that $c_0^{1,q}(\phi)$ is completely determined by $c_0^{1,1}(\phi)$, which turns out is also one-loop exact as we will see shortly. It is known that four derivative terms in $\mathcal{N}=4$ SYM given in the following expression is all-loop exact and does not receive any non-perturbative corrections~\cite{DineSeiberg}, \eq \label{eq:DineSeiberg} S_4 \sim \int d^4 \theta d^4 \bar{\theta} \ln \left( { \Psi \over \Lambda }\right) \ln \left( { \bar{\Psi} \over \Lambda }\right) \, , \eqe with $\Psi$ and $\bar{\Psi}$ are written in a $\mathcal{N}=2$ superspace, schematically, \eq \Psi = \phi + \sqrt{2} \lambda \theta + F_+ \theta^2 + \ldots \, , \quad \bar{\Psi} = \bar{\phi} + \sqrt{2} \bar{\lambda} \bar{\theta} + F_- \bar{\theta}^2 + \ldots \, . \eqe Now if we give a vev to the scalar field $\phi \rightarrow v + \phi$, and expand the action $S_4$ to fifth order we find the action contains, \eq S_4 \ni {(F_+)^2 (F_-)^2 \over v^4} -2 \left[ {(F_+)^2 (F_-)^2 \phi \over v^5}+ {(F_+)^2 (F_-)^2 \bar{\phi} \over v^5} \right] \, . \eqe From this simple expansion we find the relative coefficient between operator $F^4$ and $F^4\phi$ is simply $-2$, that is $c_0^{1,1}(\phi)=-2c_0^{1,1}$, thus is also one-loop exact. This is consistent with expanding \SO(6) invariant $F^4/|X \bar{X}|^2$ around a vev $v$. In the next section, we will explicitly compute the one-loop contribution to $c_0^{1,1}(\phi)$ and $c_0^{1,1}$ from the amplitude point of view. ]]>
6$. Since now all bosonic states are equivalent to scalars, there is no corresponding helicity categorization. The on-shell representation of $\mathcal{N}={\rm even}$ supersymmetry is furnished by breaking \SO($\mathcal{N}$) to \U($\mathcal{N}/2$). Thus we introduce $\eta^I$ as fundamentals of \U($\mathcal{N}/2$), and represent the \SO($\mathcal{N}$) generators as \eq\label{3DGen} R^{IJ}=\eta^I\eta^J \, ,\quad R^I\,_J=\eta^I\partial_{\eta^J}-\frac{1}{2} \delta^I_J \, , \quad R_{IJ}=\partial_{\eta^I} \partial_{\eta^J}\,. \eqe Due to the fact that the R-symmetry generators that are part of \SO($\mathcal{N}$)/\U($\mathcal{N}/2$) are non-linear, solutions to their constraints are rather involved. For $\mathcal{N}$=8 theories, the on-shell degrees of freedom are encoded in a single scalar superfield. In the case where only \SO(7) is present, the R-symmetry generators are~\cite{LipsteinMason}: \eq R^{IJ}=\eta^I\eta^J+\frac{1}{2}\epsilon^{IJKL}\partial_{\eta^K}\partial_{\eta^L},\quad R^I\,_J=\eta^I\partial_{\eta^J}-\frac{1}{4}\delta^I_J \eta^K\partial_{\eta^K}\,, \eqe where now $I=1,\cdots,4$. For $\mathcal{N}=6$, it is contained in a scalar and a fermionic superfield, $\Phi$ and $\overline{\Psi}$ respectively, with their degrees of freedom contained as, \eqa \nonumber \Phi&=&\phi^4+\eta^I\psi_I+\eta^{I}\eta^J\phi_{IJ}+\eta^3 \psi_4 \, ,\\ \overline{\Psi}&=&\bar{\psi}^4+\eta^I\bar{\phi}_I+\eta^{I}\eta^J\bar{\psi}_{IJ}+\eta^3 \bar{\phi}_4\,. \eqae Note that due to the constant $\frac{3}{2}$ for the \U(1) generator in eq.~(\ref{3DGen}), for $\mathcal{N}=6$, the matrix element must be of even multiplicity. We now consider the possible local dimension-six matrix elements. For maximal $\mathcal{N}=8$, we have:\footnote{Other invariants such as $(q^{A\alpha}_1q^B_{2\alpha}q^{C\alpha}_1q^D_{3\alpha} + {\rm perm})$ are related to the ansatz after permutation due to Schouten identities and super-momentum conservation.} \eq\label{3DAnsatz1} \delta^8(Q^{\alpha I})\epsilon_{ABCD}\left( q^{A\alpha}_1q^B_{2\alpha}q^{C\beta}_1q^D_{2\beta}+ {\rm perm} \right) \equiv \delta^8(Q)h(\lambda,\eta)\,, \eqe This ansatz manifestly vanishes under the multiplicative SUSY generator as well as having the correct permutation invariance symmetry. It is required to be degree 12 in the $\eta$s, due to the \U(1) generator $R^I_I$. It is rather non-trivial to check that it vanishes under the multiplicative R-symmetry generators. To simplify our task, we will follow~\cite{Till} and project the fermionic variables $\eta_i$ on a convenient basis. We begin by introducing the following $n/2$-set of bosonic $n$-dimensional objects \eq x^{\pm}_{\mathfrak{a}}, \quad y^{\alpha} \eqe where $\mathfrak{a}=1,2,\cdots,(n-4)/2$ and the new variables satisfies \eq\label{XDef} x_{\mathfrak{a}}^{\pm}\cdot\lambda=x_{\mathfrak{a}}^\pm \cdot x_{\mathfrak{b}}^\pm= x^\pm_{\mathfrak{a}} \cdot y^\alpha=0,\quad x_{\mathfrak{a}}^+\cdot x_{\mathfrak{b}}^-=\delta_{\mathfrak{a}\mathfrak{b}},\quad y^{\alpha}\cdot\lambda^{\beta}=\epsilon^{\alpha\beta}\,, \eqe The inner products in the above represent summing over all external leg labels, i.e.\ $\lambda\cdot\lambda=\sum_i\lambda_i\lambda_i$\,. At six points, an explicit solution to $x^\pm$ is given as: \eqa \nonumber x^{\pm}_{i}&=&\frac{\epsilon_{ijk}\langle jk\rangle}{2\sqrt{2}\sqrt{p^2_{123}}}, \quad i,j,k\in {\rm odd}\\ x^{\pm}_{i}&=&\frac{\pm i\epsilon_{ijk}\langle jk\rangle}{2\sqrt{2}\sqrt{p^2_{123}}}, \quad i,j,k\in {\rm even}\,. \eqae The explicit form of $y^\alpha$ is irrelevant as we will show that SUSY invariance dictate that the amplitude must be independent of it. The aim of introducing these variables is to separate the fermionic variables into pieces which vanish under the presence of $\delta^{\mathcal{N}}(Q^{\alpha I})=\delta^{\mathcal{N}}\left(\sum_i\lambda_i^{\alpha}\eta^I_{i}\right)$ . Defining: \eq \alpha_{\mathfrak{a}}^{\pm,I}=c^{\pm}_{\mathfrak{a}}\cdot \eta^I,\quad Y^{\alpha,I}=y^{\alpha }\cdot \eta^I\,, \eqe we see that \eq\label{Conversion} \eta_{i}^I=\sum_{\mathfrak{a}=1}^{(n-4)/2}\Big((x_i)_{\mathfrak{a}}^{+}\alpha_{\mathfrak{a}}^{-,I}+(x_i)_{\mathfrak{a}}^{-}\alpha_{\mathfrak{a}}^{+,I}\Big)+\epsilon_{\alpha\beta}\left(\lambda_i^\alpha Y^{\beta,I}-y_i^{\alpha}Q^{\beta,I}\right) \eqe and thus the last term would be the degree of freedom that is projected out by $\delta^{\mathcal{N}}(Q^{\alpha I})$. Now let's consider the constraint imposed by SUSY\@. The vanishing under the generator $Q^{\alpha I}$, is achieved in the usual way of requiring the amplitude to be of the form: \eq \mathcal{A}_n=\delta^{\mathcal{N}}(Q^{\alpha I})F(\lambda_i,\,\alpha_{a}^{\pm,I},\,Y^{\alpha,I})\,. \eqe The vanishing under $Q^{\alpha }_I=\lambda^\alpha\cdot \partial/\partial \eta^I$ the implies \eq \lambda^\alpha\cdot \frac{\partial}{\partial \eta^I}F=\lambda^\alpha\cdot\left(x^+\frac{\partial}{\partial \alpha^{+,I}}+x^-\frac{\partial}{\partial \alpha^{-,I}}+y^{\alpha}\frac{\partial}{\partial Y^{\alpha,I}}\right)F=-\frac{\partial}{\partial Y^{\alpha,I}}F=0\,, \eqe thus one concludes that $F$ is independent of $Y^{\alpha,I}$\,. Finally, the non-linear R-symmetry generators now take the form: \eq R^{IJ}=\sum_{\mathfrak{a}}\alpha^{+[I}_{\mathfrak{a}}\alpha^{-J]}_{\mathfrak{a}},\quad R_{IJ}=\sum_{\mathfrak{a}}\frac{\partial}{\partial\alpha^{+[I}_{\mathfrak{a}}}\frac{\partial}{\partial\alpha^{-J]}_{a}}\,. \eqe Thus constraints from SUSY and R-symmetry tells us that the six-point matrix element must take the form: \eq\label{Target} \mathcal{A}_6=\delta^{8}\left(Q^{\alpha I}\right)\left(\delta^{4}(\alpha^{+})f_+(\lambda)+\delta^{4}(\alpha^{-})f_-(\lambda)\right)\,. \eqe Thus our task is to show that when re-expressing eq.~(\ref{3DAnsatz1}) in terms of the new fermionic basis using eq.~(\ref{Conversion}), it must take the form of eq.~(\ref{Target}). However straight forward substitution would result in $h(\lambda, \eta)$ to contain terms proportional to $Y^m$ with $m<4$, which is forbidden under the full \SO(8) symmetry. Thus one concludes that there are no dimension-six local matrix elements, and this implies that the $c_6$ must be proportional to $c^2_4$, where $c_n$ is the coefficient of the dimension $n$ operator. We finally comment that the procedure of introducing nice variables $\alpha^{\pm}$ should be very useful for higher-point analysis, for the case of six-point amplitude, one can actually simply test the validity of the SUSY generator $\lambda_{\alpha} \partial_{ \eta^A }$, and this is the approach we will take the amplitudes in 6D. A similar analysis applies to $\mathcal{N}=8$ SYM with \SO(7). Note that the \SU(4) part of \SO(7) allows the six-point matrix element to be of degree 8, 12, and 16 in $\eta$'s. For degree $8$, mass dimension 6 would require the local matrix element to have the form $\delta^{(8)}(Q)(s_{ij}+ {\rm perm} )$ which vanishes through momentum conservation. The lack of degree 8 matrix element also implies the same fate for degree 16, since the two are related via Fourier transform in $\eta$. Finally the generator $R^{IJ}$ would then act solely on the degree 12 matrix element which can only vanish if $\sum_i\eta_i^I\eta_i^J$ and $\sum_i\partial_{\eta_i^I}\partial_{\eta_i^J}$ vanishes separately, for which the former possibility is already ruled out from the previous analysis. Thus we can conclude that for $\mathcal{N}=8$ SYM, one the effective theory on the Coulomb branch again requires $c_6$ to be proportional to $c^2_4$. It was argued in~\cite{DineSeiberg}, and computed explicitly in~\cite{bib3DInst1,bib3DInst2,PSSF43D}, that the four derivative term of the effective action for $\mathcal{N}=8$ SYM receives one-loop and non-perturbative instanton corrections. Our result then implies that the dimension-six operator receives perturbative contribution only at two loops. While the one-instanton corrections come from the cross terms of one-loop and one-instanton contributions (and the perturbation around the one-instanton background) of dimension-four operator. More generally, the $n$-instanton contributions to the operator come from the cross terms with $k$ and $l$-instanton contributions of the dimension-four operator, with $k+l=n$ ($0$-instanton contribution means the perturbative one-loop contribution.). ]]>