2N$. The restriction $2N-N_{\rm f}-2|\kappa| \geq 0$ was obtained in~\cite{Intriligator-ml-1997pq} under the condition that the whole Coulomb branch can be described as an $\SU(N)$ gauge theory. However, if we allow the possibility of phase transitions on the Coulomb branch, we may still have theories with $2N-N_{\rm f}-2|\kappa| < 0$. What values of $N_{\rm f}$ and $\kappa$ can be consistent? The instanton operators are supposed to correspond to generators of the symmetry $G_{\rm UV}$ of the UV superconformal theory which contains $\SU(N_{\rm f})$ as a subgroup. Then, we should never get representations $\wedge^3, \cdots, \wedge^{N_{\rm f}-3}$ since there is no symmetry breaking $G_{\rm UV} \to \SU(N_{\rm f})$ under which the adjoint representation of $G_{\rm UV}$ produces these representations (when $N_{\rm f} >8)$. Therefore, the parameter region $ | \pm N - \kappa| \leq N_{\rm f}/2 -3 $ is excluded by using~\eqref{eq:rootp} and~\eqref{eq:rootm}. Assuming that the Chern-Simons level is not too large, e.g., $|\kappa| \leq N$, the excluded region is $N_{\rm f} \geq 2N-2|\kappa|+6$. However, there is a possibility that theories with \beq N_{\rm f} \leq 2N-2|\kappa|+4 \label{eq:const} \eeq might exist. (Recall that $N_{\rm f} \equiv 2\kappa$ mod 2 due to parity anomaly, so $N_{\rm f}=2N-2|\kappa|+5$ cannot occur.) The existence of such theories is really argumed recently~\cite{Kim-ml-2015jba}. As illustrative examples, let us consider the cases $\kappa=0$, $N_{\rm f}=2N+2$ or $N_{\rm f}=2N+4$. As before, $\ket{E_\pm} $ has charge $(H_+, H_-)=(1,0)$ and $(0,1)$ respectively, so $(H_\pm, E_\pm, (E_\pm)^*)$ form $\SU(2)_\pm$ sub-algebras. However, they also transform in nontrivial representations of $\SU(N_{\rm f})$. The Dynkin diagram of $\SU(N_{\rm f})$ is \begin{equation} \node{}-\node{}-\node{}-\cdots -\node{}-\node{}-\node{} \end{equation} where there are $N_{\rm f}-1$ nodes. When $N_{\rm f}=2N+2$, the $E_+$ and $E_-$ transform in the fundamental and anti-fundamental representation of $\SU(N_{\rm f})$, respectively as one can check by~\eqref{eq:rootp} and~\eqref{eq:rootm}. This means that we get a new Dynkin diagram \beq \blacknode{}-\node{}-\node{}-\node{}-\cdots -\node{}-\node{}-\node{} - \blacknode{} \label{eq:2n2} \eeq where the two black nodes come from the $\SU(2)$ subalgebras of the instanton operators $(H_+, E_+, (E_+)^*)$ and $(H_-, E_-, (E_-)^*)$, respectively. Therefore the symmetry is enhanced to $\SU(N_{\rm f}+2)$. When $N_{\rm f}=2N+4$, it was predicted~\cite{Kim-ml-2015jba} that the theory has a UV completion as $S^1$ compactification of a 6d SCFT. In this case we get an affine Dynkin diagram \beq \node{}-\node{\blackver}-\node{}-\cdots -\node{}-\node{\blackver}-\node{} \,. \eeq As discussed in~\cite{Tachikawa-ml-2015mha}, when the Dynkin diagram becomes affine, the theory is in fact $S^1$ compactification of a 6d SCFT. Thus the symmetry enhancement is consistent with the claim of~\cite{Kim-ml-2015jba} that this is indeed a 6d SCFT. The symmetry corresponding to this affine Dynkin diagram is $\SO(2N_{\rm f})=\SO(4N+8)$, so this should be the symmetry of the 6d~SCFT. Other cases of $N_{\rm f}$ and $\kappa$ are listed in the table~\ref{tab:nflarge}. The argument of this paper alone does not guarantee the existence of these theories, but at least their Dynkin diagrams are consistent. \begin{table} \centering \begin{tabular}{|Sc|Sc|Sc|Sc|Sc|} \hline &$N_{\rm f}= 2N+1$ &$N_{\rm f}= 2N+2$ &$N_{\rm f}= 2N+3$ &$N_{\rm f}= 2N+4$ \\ \hline $|\kappa|=0$ & &$\SU(N_{\rm f}+2)$ && $\widehat{\SO(2N_{\rm f})}$ \\ \hline $|\kappa|=1/2$ &$\SU(N_{\rm f}+1) \times \SU(2)$&&$\SO(2N_{\rm f}+2)$& \\ \hline $|\kappa|=1$ &&$\SO(2N_{\rm f}) \times \SU(2)$&& \\ \hline $|\kappa|=3/2$ & $\SO(2N_{\rm f})$ &&& \\ \hline \end{tabular} ]]>

0$, and denote the spinor representations of $\SO(2w)$ as $S_\pm$. If we neglect the conditions imposed by the invariance under $-1 \in \SU(2)$, we get gauge invariant states \beq \ket{-M/2, S_+},\qquad \ket{M/2, S_+},\qquad \ket{-M/2, S_-},\qquad \ket{M/2, S_-} \eeq where $\mp M/2$ is the baryon charge associated to the $\SU(M)$ gauge group. We need to impose the condition under $-1 \in \SU(2)$. Under the assumption $w>0$, the parity of ${\rm O}(2w)$ can exchange the two values of the discrete theta angle, so we just set $\theta=0$ without loss of generality. Then the surviving states are the ones on which the center of $\SU(2)$ gauge group acts trivially. Depending on whether $M$ is odd or even, the surviving state is \begin{alignat}{2} &\ket{ \mp M/2, S_\pm}&\qquad (M&=\text{odd}),\\ &\ket{\mp M/2, S_+}&\qquad (M&=\text{even}). \end{alignat} Now let us study it case-by-case. When $M$ is odd and $w=1$, we get two states $\ket{ \mp M/2, \pm 1/2}$, where $\pm 1/2$ is the $\U(1) \cong \SO(2)$ charge. In this case, the $\SU(2)$ instanton gives two subalgebras $\SU(2)_+$ and $\SU(2)_-$ which are part of ${\rm Dyn}_+$ and ${\rm Dyn}_-$, respectively. When $M=3$, the $\SU(2)$ node has $N_{\rm f}=M+w=4$ and the result here is exactly the same as in section~\ref{sec:less}. When $M$ is odd and $w=2$, the flavor group is $\SO(4) \cong \SU(2) \times \SU(2)$ and we get two states $\ket{ \mp M/2, S_\pm}$ where $S_+$ is the doublet representation of the first $\SU(2)$ and $S_-$ is that of the second $\SU(2)$. Then, one can see that the instanton operators mix with $\SO(4) \cong \SU(2) \times \SU(2)$ to give subalgebras $\SU(3)_+ \times \SU(3)_-$. These subalgebras $\SU(3)_+ $ and $ \SU(3)_-$ are further connected to the rest of ${\rm Dyn}_+$ and ${\rm Dyn}_-$, respectively. The result here is simply summarized as follows. We formally consider a quiver \beq \SU(1)-\SU(2)-\SU(M=3)-\cdots. \eeq Then we formally consider ``$\SU(1)$ instanton'' and naively apply the result of section~\ref{sec:less}. The ``$\SU(1)$ instanton'' plays the role of supplying one doublet of $\SU(2)$ in addition to the bifundamental of $\SU(1)-\SU(2)$. This was already observed in e.g.,~\cite{Bergman-ml-2014kza,Hayashi-ml-2014hfa,Tachikawa-ml-2015mha}. When $M$ is odd and $w=3$, the flavor group is $\SO(6) \cong \SU(4)$ and we get two states $\ket{ \mp M/2, S_\pm}$ where $S_+$ and $S_-$ are the fundamental and anti-funamental representations of $\SU(4)$, respectively. When $M=3$, the $\SU(2)$ node has $N_{\rm f}=6$ flavors. Here we encounter a slight modification of the result of section~\ref{sec:more}. In that section, the Dynkin diagrams ${\rm Dyn}_+$ and ${\rm Dyn}_-$ are connected by the Dynkin diagram of $\SU(w=3)$. However, here they are connected by $\SU(4) \cong \SO(6)$. We get one additional node. The case $M$ is odd and $w \geq 4$ can be analyzed in a similar way. However, generically they give an inconsistent Dynkin diagram when ${\rm Dyn}_\pm$ are large enough. Let us turn to the case of even $M (=4)$. When $M$ is even and $w=1$, we get two states $\ket{ \mp M/2, 1/2}$, where $1/2$ is the $\U(1) \cong \SO(2)$ charge. In this case we seem to still get ${\rm Dyn}_\pm$, but the relation between Cartan subalgebras and the $\U(1)$ charges should be changed. We do not work out the detail. When $M$ is even and $w=2$, the flavor group is $\SO(4) \cong \SU(2) \times \SU(2)$ and we get two states $\ket{ \mp M/2, S_+}$ where $S_+$ is the doublet representation of the first $\SU(2)$. The two Dynkin diagrams ${\rm Dyn}_\pm$ are connected by the Dynkin diagram of the first $\SU(2)$ in $\SO(4) \cong \SU(2) \times \SU(2)$. When $M=4$, the $\SU(2)$ node has the flavor number $N_{\rm f}=6$. Thus, the result here is the same as in section~\ref{sec:more}, except the fact that the $\U(1)$ part of $\U(w=2)=\SU(2) \times \U(1)$ is actually the Cartan of the second $\SU(2)$ in $\SO(4)$. The case $M$ is even and $w \geq 3$ can be analyzed in a similar way. However, generically they give an inconsistent Dynkin diagram when ${\rm Dyn}_\pm$ are large enough. When $w=0$, the result will depend on the value of the discrete theta angle. When $M$ is even and $\theta=0$, the nodes in ${\rm Dyn}_\pm$ corresponding to the $\SU(2)$ gauge node survive, but when $\theta=\pi$, both of them are eliminated. This result is consistent with the results of section~\ref{sec:less} if we replace $\kappa \to \theta/\pi$ mod $2$ (see also the above footnote in this subsection). When $M$ is odd, the corresponding node of one of ${\rm Dyn}_+$ or ${\rm Dyn}_-$ is eliminated depending on whether $\theta=0$ or $\theta=\pi$. This is again consistent with the results of section~\ref{sec:less} by $\kappa \to \theta/ \pi - 1/2$ mod $2$. We also note that when all the nodes are $\SU(2)$, more drastic change can happen. For example, the theory~\eqref{eq:nazoquiver1} actually has affine $\SO(2(2n+6))$ symmetry when $N=2$ instead of $\SU(2n+6) \subset \SO(2(2n+6))$. Thus the corresponding 6d theory has the $\SO(4n+12)$ symmetry. This can be analyzed in the same way~\cite{Tachikawa-ml-2015mha}. We leave the detailed analysis to the reader. ]]>

3$, since the anti-symmetric representation for $N=3$ is just an anti-fundamental representation. For the moment, we also assume $N>4$, since the anti-symmetric representation in the $N=4$ case is a real representation and hence an additional enhancement occurs. First, we consider the case of a single $\SU(N)$ gauge group with one anti-symmetric representation and $N_{\rm f}$ fundamental representations. Under the decomposition $\SU(N) \supset \SU(2) \times \SU(N-2) \times \U(1)$, the anti-symmetric representation is decomposed into (i) a bifundametal of $\SU(2) \times \SU(N-2)$ with $\U(1)$ charge $(N-4)/(N-2)$ and (ii) singlets of $\SU(2)$. Then, as in section~\ref{sec:2}, it is straightforward to see that the anti-symmetric representation contributes $\SU(N-2)$ invariant states given as \beq \ket{ \pm (N-4)/2, 0, \pm (N-2)/2 }_a \eeq where $\pm (N-4)/2$ is the $\U(1)$ gauge charge, $0$ is the $\U(1)_B$ baryon charge which assings charge $1$ only to the hypermultiplets in the fundamental representation, and $\pm (N-2)/2$ is the charge under the global symmetry $\U(1)_A$ which assigns charge $1$ only to the hypermultiplet in the anti-symmetric representation. The possible gauge invariant states are, using the notation of section~\ref{sec:2}, given as \beq \ket{E_+} &:= \ket{(ij)} \otimes \ket{(- N+\kappa) }_g \otimes \ket{\wedge^{+ N/2+2-\kappa+N_{\rm f}/2}, ( + N/2+2-\kappa), ( + N/2+2-\kappa)}_h ~\nonumber \\[\jot] &\enspace\hphantom{:=} \otimes \ket{ + (N-4)/2, 0, + (N-2)/2 }_a\qquad (\text{only if ~} |+ N/2+2-\kappa| \leq N_{\rm f}/2) . \label{eq:antip} \\[\jot] \ket{E_-} &:= \ket{(ij)} \otimes \ket{(+ N+\kappa) }_g \otimes \ket{\wedge^{- N/2-2-\kappa+N_{\rm f}/2}, ( - N/2-2-\kappa), ( - N/2-2-\kappa)}_h ~\nonumber \\[\jot] &\enspace\hphantom{:=} \otimes \ket{ - (N-4)/2, 0, - (N-2)/2 }_a\qquad (\text{only if ~} |- N/2-2-\kappa| \leq N_{\rm f}/2) .\label{eq:antim} \eeq and also for small $N$, \beq \ket{F_+} &:= \ket{(ij)} \otimes \ket{(- N+\kappa) }_g \otimes \ket{\wedge^{+ 3N/2-2-\kappa+N_{\rm f}/2}, ( + 3N/2-2-\kappa), ( + 3N/2-2-\kappa)}_h ~\nonumber \\[\jot] &\enspace\hphantom{:=} \otimes \ket{ - (N-4)/2, 0, - (N-2)/2 }_a\qquad (\text{only if ~} |+ 3N/2-2-\kappa| \leq N_{\rm f}/2) . \\[\jot] \ket{F_-} &:= \ket{(ij)} \otimes \ket{(+ N+\kappa) }_g \otimes \ket{\wedge^{- 3N/2+2-\kappa+N_{\rm f}/2}, ( - 3N/2+2-\kappa), ( - 3N/2 + 2-\kappa)}_h ~\nonumber \\[\jot] &\enspace\hphantom{:=} \otimes \ket{ + (N-4)/2, 0, - (N-2)/2 }_a\qquad (\text{only if ~} |- 3N/2+2-\kappa| \leq N_{\rm f}/2) .\eeq Denoting the $\U(1)_A$ charge as $A$, the charges of $\ket{E_\pm} $ under $(I, B+\kappa I, A)$ are given as $(1, \pm (N +4)/2, \pm (N-2)/2)$, and those of $\ket{F_\pm} $ are given as $(1, \pm 3(N +4)/2, \mp (N-2)/2)$. Note that if $N$ is large enough, the states $\ket{F_\pm} $ cannot exist because of the constraint $|\pm (3N/2-2)-\kappa| \leq N_{\rm f}/2$. In the present case, there are several $\U(1)$ symmetries, and hence it is more nontrivial to determine the Cartan generators of enhanced $\SU(2)_\pm$. For this purpose, we use the argument of section~\ref{sec:pre} which uses RG flows between different theories. We concentrate our attention to $\ket{E_\pm}$, since $\ket{F_\pm}$ do not generically exist. First, we take $N+4-2\kappa=N_{\rm f}$, in which case $\ket{E_+}$ is a singlet of $\SU(N_{\rm f})$ and hence we get enhanced $\SU(2)_+$ symmetry. Second, we give mass to the hypermultiplet in the anti-symmetric representation and integrate it out. We assume that $\SU(2)_+$ symmetry is preserved under this RG flow if the sign of the mass is taken appropriately. By integrating out the anti-symmetric matter, the Chern-Simons level is shifted as $\kappa \to \kappa'= \kappa+ (N-4)/2$ if the sign of the mass is taken appropriately. In the low energy, we get a theory with $N_{\rm f}$ fundamental flavors and Chern-Simons level $\kappa'=\kappa+ (N-4)/2=N-N_{\rm f}/2$. This is the correct Chern-Simons level for the enhancement of $\SU(2)_+$ in the low energy theory. Therefore, in the low energy theory the Cartan of $\SU(2)_+$ is given by \beq H_+ = \frac{1}{2} I+ \frac{B + (\kappa+(N-4)/2) I}{2N} +c\left(A -\frac{1}{2}(N-2) I\right), \eeq where we have taken into account the fact that all the states in the low energy theory have $A-\frac{1}{2}(N-2)I=0$ and hence we have ambiguity represented by the constant $c$. This is because all the low energy fields have $A=0$, and only instantons have charge $A=\frac{1}{2}(N-2)I$ which can be shown in the same way as~\eqref{eq:BCS}. By doing the same consideration for $\SU(2)_-$, we get \beq H_\pm = \frac{1}{2} I \pm \frac{B + (\kappa \pm (N-4)/2 )I}{2N} \pm c \left(A \mp \frac{1}{2}(N-2) I\right), \eeq where we have used the fact that the coefficient of $A$ must be $\pm c$ by parity and charge conjugation (spurious) symmetries. We also use a different type of RG flows. We can go to the subspace of the Higgs branch by giving vevs to two of the fundamental quarks so that the gauge group is broken as $\SU(N) \to \SU(N-1)$. We lose two of the fundamental flavors by the Higgs mechanism, but get one additional fundamental from the anti-symmetric matter. Thus we get a theory in which $N$ and $N_{\rm f}$ are reduced by one, and the condition of the enhanced $\SU(2)_+$ is still preserved if this symmetry exists before Higgsing. In terms of gauge invariant operators, this RG flow is triggered just by meson vevs, and hence we can assume that the $\SU(2)_+$ is unchanged under this RG flow. The low energy baryon symmetry is a mixture of the high energy baryon symmetry and $\U(1) \subset \SU(N)$ charge such that the nonzero vevs of the quarks are neutral under the new baryon symmetry. The $\U(1) \subset \SU(N)$ charge is given as \beq \diag\left(-1, \frac{1}{N-1},\frac{1}{N-1}, \cdots, \frac{1}{N-1}\right). \eeq Then the charges under $(B,A)$ of the low energy fields are the following. The $N_{\rm f}-2$ quarks have charge $(B,A)=(\frac{N}{N-1}, 0)$. One quark which comes from the anti-symmetric matter has charge $(B,A)=(-\frac{N-2}{N-1},1)$. The anti-symmetric matter has charge $(B,A)=(\frac{2}{N-1},1)$. Now, we require that the charges of the $N_{\rm f}-2$ quarks and one quark are the same under $H_+$. This is because $H_+$ should be insensitive to quark flavors.\footnote{If we consider a flow from $N=4$ to $N=3$, the low energy $\SU(2)_+$ would be mixed with the flavor symmetry and the argument here need modification.} This requirement gives $\frac{1}{N-1}=-\frac{N-2}{(N-1)N} +2c$, so we get $c=1/N$. Therefore we finally get the Cartan generators \beq H_\pm = \frac{1}{4} I \pm \frac{B + \kappa I +2A }{2N} . \eeq This is our formula for the Cartan generators. Note that gauge invariant chiral operators constructed from one anti-symmetric matters and two anti-fundamental quarks have charge $H_\pm=0$ by using the above result, so these operators are singlets of $\SU(2)_\pm$. Now we can see the difference from the case of theories without an anti-symmetric matter. The charges of $\ket{E_+}$ and $\ket{E_-}$ under $(H_+, H_-)$ are given as $(1, - 1/2)$ and $(-1/2, 1)$, respectively. This means the following. If both of $\ket{E_+}$ and $\ket{E_-}$ exist, they form a single $\SU(3)$ algebra instead of $\SU(2)_+ \times \SU(2)_-$. In other words, the two nodes of $\SU(2)_+$ and $\SU(2)_-$ are connected. Now let us see what happens in some cases. When $ N_{\rm f} < N+4 $, nothing so special happens compared to the case without anti-symmetric matters, because only one of $\SU(2)_+$ or $\SU(2)_-$ can be enhanced. It is just like that the anti-symmetric matter contributes effectively $N-4$ flavors of quarks. When $\kappa=0$ and $N_{\rm f}=N+4$, we get the connection of $\SU(2)_+$ and $\SU(2)_-$ into $\SU(3)$ discussed above. The total non-abelian flavor symmetry is $\SU(3) \times \SU(N_{\rm f})$. When $\kappa=1/2$ and $N_{\rm f}=N+5$, the $\ket{E_+}$ is in the fundamental representation of the flavor group $\SU(N_{\rm f})$. This means that the Dynkin diagrams of $\SU(3)$ and $\SU(N_{\rm f})$ are connected to form a single Dynkin diagram of $\SU(N_{\rm f}+2)$. When $\kappa=0$ and $N_{\rm f}=N+6$, we actually get a 6d theory. The $\ket{E_+}$ and $\ket{E_-}$ are in the fundamental and anti-fundamental representations of the flavor group $\SU(N_{\rm f})$, so we get a Dynkin diagram like~\eqref{eq:2n2}. However, now the two black nodes of~\eqref{eq:2n2} are also connected, and hence we actually get an affine $\SU(N_{\rm f}+1)$ diagram. Therefore, this is a 6d theory with $\SU(N_{\rm f}+1)$ flavor symmetry. Finally let us briefly comment on the case $N=4$. In this case, the $\U(1)_A$ symmetry of the anti-symmetric matter is enhanced to $\SU(2)_A$, since the anti-symmetric representation is a real representation in this case. One can see that $\ket{E_\pm}$ and $\ket{F_\pm}$ form doublets of $\SU(2)_A$. Thus the $\SU(2)_+ - \SU(2)_A- \SU(2)_-$ are actually combined into a single $\SU(4)$. The rest of the story is similar to the above generic case. \paragraph{Quivers.} Now we consider the case that anti-symmetric matters are introduced into quiver theories. Suppose that a node $i$ has an anti-symmetric matter. When this node satisfies $w_i+(N_i-4) < \sum_j A_{ij} N_k$, nothing special happens. The anti-symmetric matter is just regarded as contributing effectively $N_i-4$ flavors as mentioned above. Then the results are similar to those of section~\ref{sec:less}. An interesting thing happens when $\kappa_i=0$ and $w_i+(N_i-4)= \sum_j A_{ij} N_k$. In this case, the nodes in ${\rm Dyn}_+$ and ${\rm Dyn}_-$ corresponding to the node $i$ is connected. This is due to the above phenomenon of connecting $\SU(2)_+$ and $\SU(2)_-$ into $\SU(3)$. The results are thus similar to those of section~\ref{sec:more}. ]]>