The resubmission aims to clarify the question of $U(n)$ versus $SU(n)$ symmetry in the spin chain, the loop model and the CFT, raised by the reviewers. In short, our model has $PSU(n) = U(n)/U(1)$ symmetry, because $U(1)$ acts trivially on our spectrum. In our particular case, it is possible to make sense of PSU(n) symmetry at generic $n$, because only a particular set of representations of SU(n) appear in the spectrum of the CFT. It is however easier to write the spectrum in terms of representations of $U(n)$, on which $U(1)$ acts trivially.

Below is a copy of the more detailed answer which we gave in a note attached as an answer to the reviews.

$U(n)$ versus $SU(n)$ global symmetry

In our spectrum for the $U(n)$ CFT, there are $n^2-1$ currents, which is the dimension of $PSU(n) = U(n) / U(1)$.

In the alternated spin chain $([1] \otimes \overline{[1]})^{\otimes L}$, while $U(n)$ commutes with the hamiltonian, the $U(1) \subset U(n)$ acts trivially. To call a group a symmetry of a theory all elements of the group need to act non-trivially, except for the identity element. Otherwise, one could let an arbitrarily large group act trivially on the theory and declare that it is a symmetry. Hence it is incorrect to say that the spin chain has $U(n)$ global symmetry, it exactly has $PSU(n)$ global symmetry.

This appears to be a problem since it is noted in [@br19] that no Deligne category exists for the group $SU(n)$, and the known constructions cannot be simply extended to construct one. However, in our spectrum [@rjrs24 eq. (4.1-4.3)], all of the irreducible representations of $SU(n)$ that appear can be represented by Young diagrams of length a multiple of $n$. This is not evident from [@rjrs24 eq (4.6)] since this adopts a $U(n)$-notation for these representations, but is shown in the new equation (2.21), in the new paragraph summarizing the relation between $U(n)$ and $SU(n)$ representations. As we explain in the new paragraph "The category $\underline{\operatorname{Rep}}_0 SU(n)$" at the end of section 2, this particular subset of all representations of $SU(n)$ does admit a Deligne category. This is because these representations admit a canonical bijective mapping to a family of representations of $U(n)$ on which $U(1)$ acts trivially, which we call $\underline{\operatorname{Rep}}_0 U(n)$. Since these representations are closed under tensor product, they have an associated subcategory in the Deligne category $\underline{\operatorname{Rep}}_0 U(n)$, which allows us to define $\underline{\operatorname{Rep}}_0 SU(n)$.

There remains the question of whether we should change the name of the model to the $SU(n)$ model.

If we were to write the representations in eq. (4.6) in terms of representations of $SU(n)$ instead of $U(n)$, the reasonably simple equations (4.6) would involve many complicated diagrams all depending on $n$. We prefer to keep the notation of $U(n)$, keeping in the back of our minds that $U(1)$ acts trivially and so everything could be equivalently written in terms of representations of $SU(n)$ if it was needed. For that reason we chose to keep the name $U(n)$ model.

# Answer to the first anonymous referee's questions and remarks

#### Section 2.

- See the new paragraph below \"Phase diagrams\"

- The ranges of summations are infinite, however the coefficients are
zero for large diagrams.

- Defects is the standard term in the theory of diagram algebras.
There is a mapping between loop model and RSOS (height) models. In
this case inserting defects causes the height field to have a
non-trivial monodromy, which is what one expects with QFT defects.
However in our case we don't think of the defects as QFT defects.

- we added a reference for the walled-Brauer algebra, and a definition
for Specht modules. Note that the paragraph actually defines the
walled-Brauer algebra, no prior knowledge is assumed. we gave
details for (2.35) (former (2.26)).

#### Section 4.

- we added a definition of the conformal algebra in the intro of the
paper and in the beginning of the section, also for $r \wedge 0$.

- $\omega_1 = [1]$ is true, but in your computation you used
$[1] \otimes \overline{[1]} = [1]\overline{[1]}$ while
$[1]\otimes \overline{[1]} = [1]\overline{[1]} + [\,]$, see (2.10b).

- we detailed the comment on the logarithmic structure.

- factor of $\frac12$ corrected.

- our expressions are more convenient and more naturally written in
terms of the polynomials $\tilde T = 2T(X/2)$ where $T$ are the
usual Chebyshev polynomials of the first kind. We changed the
notation from $U$ to $\tilde T$ to avoid confusion with polynomials
of the second kind.

- we added a diagram for $u^2$.

- The interchiral algebra was introduced in [@js18], and we don't want
to say much about it here. We simply need to know that it is the
algebra generated by the modes of the stress tensor and the
degenerate field $V_{\langle 1, 2 \rangle}$ and consequently its
modules include all shifts of the $s$ index, as we explain in
(4.37 - 4.38).

- $e_1$ acting on the 7th diagram indeed gives the first diagram. What
we meant to say is that a vector is in $W^r_{(r, s)}$ iff it is in
$\bigcap_{i=1}^L \operatorname{ker} e_i$, which is a complementary
space to $\bigoplus \operatorname{Im}e_i$, because the $e_i$ are
(unnormalised) projectors. The space
$\bigoplus \operatorname{Im}e_i$ is generated by the first 6
diagrams, hence the complementary has the same dimension as the
number of the other diagrams. The resulting representation can be
obtained from the diagrams in $\ell^{r, r}_{|\lambda|,|\mu|}$ by
imposing that the action of $e_i$ is 0 (remembering that these
diagrams are actually just labels for a certain combination of
diagrams in $d^{r, r}_{|\lambda|,|\mu|}$) which is in the kernel of
all $e_i$s.

- As for the modular transform of $Z(g)$, it should be interpreted as
inserting a vertical $g$-twist, i.e. twisting the Hilbert space of
the theory by a group element $g$. This is what we explain in
section 5. This description is not useful for the computation of the
partition function, hence we didn't include it in section 4.

# Answer to the second anonymous referee

For more detail, see the discussion in the first section above.

- This is correct, there is a $U(n)/U(1) = PSU(n)$ symmetry, hence
$n^2-1$ currents.

- In particular for $n=1$ there is no global symmetry since
$PSU(1) = {1}$.

- Indeed for $n=1$ there are no operators with dimension 1 other than
$V_{(1, 1)}$ which has multiplicity 0.

# Answer to Bernard Nienhuis

1. we modified this

2. modified

3. In a spin chain, if two neighbouring sites are identical and cannot
interact between each other, they can equivalently be replaced by a
single site, since the hamiltonian on the spin chain with two
identical sites is the one on the spin chain with a single site
instead, just shifted by $id_{i, i+1}$.

4. What is meant is that the CFT that describes the dilute and
completely packed models is the same. Said otherwise, the critical
behaviour of the models is the same.

5. Indeed the previous paragraph was badly formulated, to the point
where it was wrong. We addressed this.

6. see the new formatting of references.