modules is obtained from

$\sum_{\overline{p}\in\mathbb{H}/\ell\mathbb{Z}}\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$

by projecting onto the subspace $\mathcal{F}(C_{N})$ and evaluating $v=1$.
Furthermore if $[\Delta_{\mathrm{q}}(\mu)]$ and $[\Delta_{\mathrm{q}}(\nu)]$
appear in the decomposition with $\mu$ and $\nu$ linked, then there exists a
unique $\overline{p}\in\mathbb{H}/\ell\mathbb{Z}$ such that $\underline{\mu}$
and $\underline{\nu}$ appear in
$\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$.

###### Proof.

The action is the one from Proposition 4.11. That the decomposition is given
by the sum of all operators is Proposition 4.1 together with Proposition 2.5
to obtain the translation of the decomposition into weight combinatorics and
then Lemma 4.2 together with the definition of the operators themselves.

That classes of two Weyl modules with linked weights are obtained from a
unique operator $\hat{\mathrm{B}}_{\overline{p}}$ is then Lemmas 4.4, 4.5,
4.7, and 4.6, . ∎

We want to address now the question of embedding into a single Fock space for
different ranks.

In type $A$, every $\mathcal{F}(A_{N})$ can be embedded into
$\mathcal{F}_{0}$. This was natural on a combinatorial level due to the choice
of $\mathfrak{gl}$ instead of $\mathfrak{sl}$ and hence the ability to choose
the element $\rho$. The choice was made such that the weight $0$ always gets
mapped to the sequence with all $1$’s in the non-positive half and $0$’s in
the positive half. From the point of view of the affine operators and the
action, any shift of the origin, i.e. an embedding into a different
$\mathcal{F}_{k}$ makes no difference, since the operators only involves
entries that are congruent mod $\ell$. Hence an embedding into a different
$\mathcal{F}_{k}$ is just related to a shift in the used operators. On the Lie
theory side this would just correspond to a different choice of $\rho$.

We can make a similar construction in type $C$ if we restrict to certain $N$.
Although this is more artificial since on the Lie theory side, the element
$\rho$ is fixed.

######  Definition 4.13.

Let $N=m\ell+k$ for $m\geq 0$ and $0\leq k<\ell$. Then we define a map from
$\mathcal{F}_{N}$ to $\mathcal{F}_{k}$ via
$\underline{a}^{(m)}(i)=\underline{a}(i+m\ell)$ for
$\underline{a}\in\mathcal{S}_{\mathbb{Z},N}$. We call $\underline{a}^{(m)}$
the sequence shifted by $m$ $\ell$-steps.

In this way we can view $\mathcal{F}(C_{N})$ as a subspace of
$\mathcal{F}_{k}$, which we call the shifted embedding of
$\mathcal{F}(C_{N})$.

That the map in Definition 4.13 is well-defined follows immediately with a
simple calculation for the charge. Since it intertwines the action of
$\mathrm{e}_{i}$ and $\mathrm{e}_{i-m\ell}$ (and similar of $\mathrm{f}_{i}$
and $\mathrm{f}_{i-m\ell}$), the affine operators
$\hat{\mathrm{B}}_{\overline{p}}$ and $\hat{\mathrm{L}}_{\overline{p}}$
defined on $\mathcal{F}_{N}$ and $\mathcal{F}_{k}$ as restrictions from
$\mathcal{F}$ commute with the map of shifting a sequence by $m$ steps. In
terms of weights this is the same as looking at the weight
$\lambda+\rho-m\ell(1,\ldots,1)$ which for most $m$ is not dominant.

Thus using this embedding for a fixed $0\leq k<\ell$ we can formulate this as
follows.

######  Proposition 4.14.

Let $\underline{a}\in\mathcal{S}_{\mathbb{Z},k}$ and $r\geq 0$. Then for $m\gg
0$, $\underline{a}=\underline{\lambda}^{(m)}$ for $\lambda$ a dominant weight
for $U_{\mathrm{q}}$ of type $C_{m\ell+k}$. Furthermore there are Laurent
polynomials $d_{\lambda,\mu}(v)$ with non-negative integer coefficients for
$\mu$ dominant for $U_{q}$ such that

$\sum_{(\mathbb{H}/\ell\mathbb{Z})^{r}}\hat{\mathrm{B}}_{\overline{p}_{1}}\cdots\hat{\mathrm{B}}_{\overline{p}_{r}}\underline{\lambda}^{(m)}=\sum_{\mu\text{
dominant for }U_{q}}d_{\lambda,\mu}(v)\underline{\mu}^{(m)},$

and

$[\Delta_{\mathrm{q}}(\lambda)\otimes\Delta_{\mathrm{q}}(\omega_{1})^{\otimes
r}]=\sum_{\mu\text{ dominant for
}U_{q}}d_{\lambda,\mu}(1)[\Delta_{\mathrm{q}}(\mu)].$

###### Proof.

For $\underline{a}$ we consider the left most $0$ in the non-positive part of
the sequence. If this zero is at position $-n^{\prime}$ then using operators
of the form $\mathrm{e}_{i}$ this can be reduced to a sequence $\underline{b}$
that has no $0$’s in the non-positive part and exactly $k$ $1$’s at the
positions $1,\ldots,k$. Then $\underline{b}$ is the image of $\underline{0}$
under the shifted embedding of $\mathcal{F}(C_{m^{\prime}\ell+k})$ for
$m^{\prime}\ell>n^{\prime}$. Then by applying operators $\mathrm{f}_{i}$ in
the reverse order to the sequence of operators $\mathrm{e}_{i}$ before one
obtains a dominant weight $\lambda$ with
$\underline{a}=\underline{\lambda}^{(m^{\prime})}$. If there was no $0$ in the
non-positive part then one can just use $m^{\prime}=0$.

A dominant weight $\lambda$ for $U_{\mathrm{q}}$ of rank $m^{\prime}\ell+k$
can be viewed as a dominant weight for $U_{\mathrm{q}}$ of rank $m\ell+k$ for
$m>m^{\prime}$, by just filling it up with $0$ at the last
$(m-m^{\prime})\ell$ entries. Choose $m>m^{\prime}$ such that
$m\ell>n^{\prime}+r$. By our discussion about which $\mathrm{f}_{i}$ can leave
the embedded subspace $\mathcal{F}_{N}$, after Lemma 4.2, we see that no
product
$\hat{\mathrm{B}}_{\overline{p}_{1}}\cdots\hat{\mathrm{B}}_{\overline{p}_{r}}$
applied to $\underline{a}$ can create a weight that is not in the shifted
embedding of $\mathcal{F}(C_{m\ell+k})$, since at most the $1$’s at the
position $-n^{\prime}-1,\ldots,-n^{\prime}-r$ can be moved to the right. Hence
the statement then follows from Theorem 4.12. ∎

Note that the proof of Proposition 5.13 gives a clear bound for what $m$ needs
to be, it is just necessary that $m\ell>n^{\prime}+r$ for $n^{\prime}$ the
position of the left most entry $0$.

######  Remark 4.15.

Note that the definition of the operators $\hat{\mathrm{B}}_{\overline{p}}$ is
not uniquely determined. As already mentioned in Remark 3.15, one can change
some coefficients in the definition of the generators and obtain a slightly
modified algebra.

Since the definition in type $A$ is not unique as well, compare for example
the definition in [RT10, Theorem 3.1] and in [Ari02, Section 10.1], one can
make similar modifications in type $C$.

##  5\. Type B

In type $B_{N}$ ($N>1$), we choose $\varepsilon_{i}\in\mathfrak{h}^{\ast}$ the
projection onto the $i$-th diagonal entries for the Cartan subalgebra of
diagonal matrices inside $\mathfrak{so}_{2n+1}(\mathbb{C})$. The $W$-invariant
bilinear form $(-,-)$ is given such that the $\varepsilon_{i}$ form an
orthonormal basis. For the positive roots we choose

$\Phi^{+}=\\{\beta_{i,j}^{\pm}=\varepsilon_{i}\pm\varepsilon_{j}\mid 1\leq
i<j\leq N\\}\cup\\{\beta_{i}=\varepsilon_{i}\mid 1\leq i\leq N\\}$

and similar to type $C$, the simple roots are $\alpha_{i}=\beta_{i,i+1}^{-}$
for $1\leq i<N$, but here $\alpha_{N}=\beta_{N}$. The corresponding coroots
are

$(\beta_{i,j}^{\pm})^{\vee}=\beta_{i,j}^{\pm}\text{ and
}\beta_{i}^{\vee}=2\varepsilon_{i}.$

Following Section 2.3 we have $d_{i}=2$ for $1\leq i<N$ and $d_{N}=1$, since
$\alpha_{N}$ is the only simple short root in this case. This gets extended to
all positive roots via $d_{\beta}=2$ for $\beta=\beta_{i}$ for some $i$ and
$d_{\beta}=1$ otherwise.

The noticeable change is in the integral weight lattice. The fundamental
weights are $\omega_{i}=\varepsilon_{1}+\ldots+\varepsilon_{i}$ for $i<N$ and
$\omega_{N}=\nicefrac{{1}}{{2}}(\varepsilon_{1}+\ldots+\varepsilon_{N})$.
Hence for the integral weights $X=\bigoplus_{i+1}^{N}\mathbb{Z}\omega_{i}$,
the dominant weights $X^{+}$ that can be naturally divided into two subsets

$\displaystyle X^{\nicefrac{{1}}{{2}},+}$
$\displaystyle=\\{(\lambda_{1},\ldots,\lambda_{N})\mid\lambda_{i}\in\mathbb{Z},\lambda_{i}\geq\lambda_{i+1},\text{
and }\lambda_{N}\geq 0\\}\text{ and}$ $\displaystyle X^{1,+}$
$\displaystyle=\\{(\lambda_{1},\ldots,\lambda_{N})\mid\lambda_{i}\in\mathbb{H},\lambda_{i}\geq\lambda_{i+1},\text{
and }\lambda_{N}\geq 0\\},$

written as row vectors with respect to $\\{\varepsilon_{i}\\}_{1\leq i\leq N}$
inside $\mathfrak{h}^{\ast}$. We call $X^{\nicefrac{{1}}{{2}},+}$ the integer
weights, not to be confused with the integral weights, and $X^{1,+}$ the half-
integer weights.

Summing up the fundamental weights we get

$\rho=\sum_{i=1}^{N}\omega_{i}=(N-\nicefrac{{1}}{{2}})\varepsilon_{1}+(N-\nicefrac{{3}}{{2}})\varepsilon_{2}+\ldots+\nicefrac{{1}}{{2}}\varepsilon_{N}\in
X^{\nicefrac{{1}}{{2}},+}.$

Thus adding $\rho$ we obtain the sets to define sequences, namely

$\displaystyle X^{\nicefrac{{1}}{{2}},+}_{\rho}$
$\displaystyle=\\{(\underline{\lambda}_{1},\ldots,\underline{\lambda}_{N})\mid\underline{\lambda}_{i}\in\mathbb{H},\,\underline{\lambda}_{i}>\underline{\lambda}_{i+1},\text{
and }\underline{\lambda}_{N}>0\\},$ $\displaystyle X^{1,+}_{\rho}$
$\displaystyle=\\{(\underline{\lambda}_{1},\ldots,\underline{\lambda}_{N})\mid\underline{\lambda}_{i}\in\mathbb{Z},\,\underline{\lambda}_{i}>\underline{\lambda}_{i+1},\text{
and }\underline{\lambda}_{N}>0\\},$

again recalling that $\underline{\lambda}=\lambda+\rho$ for $\lambda\in
X^{+}$.

######  Remark 5.1.

This makes clear our convention of naming the set of integer weights
$X^{\nicefrac{{1}}{{2}},+}$ and the set of half-integer weights $X^{1,+}$ (and
not the other way around). After adding $\rho$, elements in
$X^{\nicefrac{{1}}{{2}},+}_{\rho}$ have entries in $\mathbb{H}$ and elements
of $X^{1,+}_{\rho}$ have entries in $\mathbb{Z}$. Thus the naming convention
makes it easy to recognise what the domains for the sequences in the different
cases are.

Again the parity of $\ell$ is important. For $\ell$ odd, $W_{\ell}$ is the
affine Weyl group of type $B_{N}$, scaled by the factor $\ell$. In case $\ell$
even, $W_{\ell}$ is the affine Weyl group of type $C_{N}$ scaled by a factor
$\ell$ for the dual root system, i.e. where one would replace
$\beta_{ij}^{\pm}$ in the root system by
$\nicefrac{{1}}{{2}}\beta_{ij}^{\pm}$.

For the action on $U_{\mathrm{q}}\text{-}\mathrm{mod}$ we use the functor
$-\otimes_{\mathbb{C}}\Delta_{\mathrm{q}}(\omega_{1})$, again the tensor
product with the the specialization of the quantum analogue of the natural
representation.

######  Remark 5.2.

Note that in contrast to the case of $U_{v}(\mathfrak{g})$ of type $A$ or $C$,
$\Delta_{v}(\omega_{1})$ is not a tensor generator of
$U_{v}(\mathfrak{g})\text{-}\mathrm{mod}$, hence there are other possible
finite dimensional representations, like the specialization of the spin
representation $\Delta_{\mathrm{q}}(\omega_{N})$, that can give interesting
actions on $U_{\mathrm{q}}\text{-}\mathrm{mod}$.

In type $B$ the natural representation is not minuscule. Hence the tensor
product decomposition has a slight complication.

######  Proposition 5.3.

[HK02, Proposition 8.6.3] For $\lambda\in X^{+}$ it holds in
$U_{v}(\mathfrak{g})\text{-}\mathrm{mod}$

$\Delta_{v}(\lambda)\otimes\Delta_{v}(\omega_{1})\cong\bigoplus_{i:\lambda+\varepsilon_{i}\in
X^{+}}\Delta_{v}(\lambda+\varepsilon_{i})\oplus\bigoplus_{i:\lambda-\varepsilon_{i}\in
X^{+}}\Delta_{v}(\lambda-\varepsilon_{i})\oplus\Delta_{v}(\lambda)^{\oplus\delta_{\mathrm{pos}}},$

where $\delta_{\mathrm{pos}}=0$ if $\lambda_{N}=0$ and
$\delta_{\mathrm{pos}}=1$ if $\lambda_{N}>0$.

Thus in type $B$ the tensor product decomposition for weights in $X^{1,+}$ is
straight-forward with the only difference to type $C$ being the appearance of
a summand $\Delta_{v}(\lambda)$ itself. Even the sequences are looking very
similar to the type $C$ case.

For weights in $X^{\nicefrac{{1}}{{2}},+}$ on the other hand, the tensor
product decomposition rule depends on the explicit weight of the Weyl module,
namely the summand $\Delta_{v}(\lambda)$ only appears if $\lambda_{N}>0$. In
addition we are dealing with sequences on half-integers in this case.

###  5.1. Fock space of sequences and operators

We decompose the Fock space as

$\mathcal{F}(B_{N})=\mathcal{F}^{1}(B_{N})\oplus\mathcal{F}^{\nicefrac{{1}}{{2}}}(B_{N}),$

with $\mathcal{F}^{1}(B_{N})$ having basis $\underline{\lambda}\in
X^{1,+}_{\rho}$ and $\mathcal{F}^{\nicefrac{{1}}{{2}}}(B_{N})$ having basis
$\underline{\lambda}\in X^{\nicefrac{{1}}{{2}},+}_{\rho}$. For
$\mathcal{F}^{1}(B_{N})$ we can use the same definition of sequences and Fock
spaces as in type $C$, and embed $\mathcal{F}^{1}(B_{N})$ into
$\mathcal{F}_{N}$ as in Section 4.1.

For the case of $\mathcal{F}^{\nicefrac{{1}}{{2}}}(B_{N})$ we use the
sequences $\mathcal{S}_{\mathbb{H}}$ and corresponding Fock space
$\mathcal{F}^{\nicefrac{{1}}{{2}}}$ from Section 3. Hence we map
$\underline{\lambda}\in X^{\nicefrac{{1}}{{2}},+}_{\rho}$ to the sequence
$\underline{a}_{\underline{\lambda}}$ with

$\underline{a}_{\underline{\lambda}}(i)=\begin{cases}1&\text{if }i<0,\\\
1&\text{if there exists }1\leq j\leq N\text{ s.t.
}i=\underline{\lambda}_{j},\\\ 0&\text{otherwise}\end{cases}$

to embed $\mathcal{F}^{\nicefrac{{1}}{{2}}}(B_{N})$ into
$\mathcal{F}^{\nicefrac{{1}}{{2}}}$. As before it is clear that the image is
contained in the subspace $\mathcal{F}^{\nicefrac{{1}}{{2}}}_{N}$ of sequences
of charge $N$. We continue to write $\underline{\lambda}$ for the sequence as
well.

The analogue of Lemma 4.2 holds in type $B$ as well, i.e. $\mathrm{e}_{i}$
preserves the subpsaces $\mathcal{F}^{1}(B_{N})$ and
$\mathcal{F}^{\nicefrac{{1}}{{2}}}(B_{N})$, $\mathrm{f}_{i}$ does so for
$i\notin\\{0,\nicefrac{{1}}{{2}}\\}$, and $\mathrm{f}_{i}$ does not preserve
the subpsaces for $i\in\\{0,\nicefrac{{1}}{{2}}\\}$.

In contrast to type $C$, the cases of $\ell$ odd and even have to be treated
separately.

###  5.2. Linkage and operators for odd $\ell$

Throughout this section we assume that

$\ell>3\text{ is odd.}$

The proofs in this section have a very similar flavour to the ones in Section
4.2. Unfortunately there are some subtle differences. We start with the
statements that work for both the integer and the half-integer weight cases.

######  Lemma 5.4.

Let $\underline{\lambda}\in X^{+}_{\rho}$ and $\ell$ odd. Assume that
$\underline{\mu}=\mathrm{e}_{r}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{e}_{s}\underline{\lambda}$ are defined for $r\neq s$.
Then $\mu$ and $\nu$ are linked if and only if $r\in s+\ell\mathbb{Z}$.

Assume that $\underline{\mu^{\prime}}=\mathrm{f}_{r}\underline{\lambda}$ and
$\underline{\nu^{\prime}}=\mathrm{f}_{s}\underline{\lambda}$ are defined for
$r\neq s$. Then $\mu^{\prime}$ and $\nu^{\prime}$ are linked if and only if
$r\in s+\ell\mathbb{Z}$.

###### Proof.

As in Lemma 4.4, $\underline{\mu}=\underline{\lambda}-\varepsilon_{i}$ with
$\underline{\lambda}_{i}=r+\nicefrac{{1}}{{2}}$ and
$\underline{\nu}=\underline{\lambda}-\varepsilon_{j}$ with
$\underline{\lambda}_{j}=s+\nicefrac{{1}}{{2}}$ and we assume that $i<j$.

Assume now that $\underline{\mu}$ and $\underline{\nu}$ are linked. Use Lemma
3.12 to obtain alcoves $A_{\underline{\mu}}$ and $A_{\underline{\nu}}$ with a
minimal gallery only crossing walls strictly between the alcoves. The absolute
value of $(\underline{\nu}-\underline{\mu},\beta^{\vee})$ can be $2$ in case
of $\beta\in\\{\beta_{ij}^{-},\beta_{i},\beta_{j}\\}$. In case that a
hyperplane of the form $H_{\beta_{ij}^{-},m}$ is strictly between the alcoves,
it follows as in Lemma 4.4 that $r\in s+\ell\mathbb{Z}$.

Hence assume that there is no hyperplane of the form $H_{\beta_{ij}^{-},m}$
strictly between the alcoves. Since
$\underline{\nu}-\underline{\mu}=\varepsilon_{i}-\varepsilon_{j}$ there has to
be both a hyperplane of the form $H_{\beta_{i},m^{\prime}}$ and of the form
$H_{\beta_{j},m^{\prime\prime}}$ strictly between the alcoves since we assume
that the weights are linked. This implies
$2\underline{\lambda}_{i}=m^{\prime}\ell$ and
$2\underline{\lambda}_{j}=m^{\prime\prime}\ell$. Hence
$2(\underline{\lambda}_{i}-\underline{\lambda}_{j})=(m^{\prime}-m^{\prime\prime})\ell$.
Since $\ell$ is odd $(m^{\prime}-m^{\prime\prime})$ has to be even and so
$H_{\beta_{ij}^{-},(m^{\prime}-m^{\prime\prime})/2}$ is strictly between the
alcoves, a contradiction.

Conversely assume that $r\in s+\ell\mathbb{Z}$. Then
$\underline{\lambda}_{i}-\underline{\lambda}_{j}$ is a multiple of $\ell$.
Hence
$(\underline{\lambda},(\beta_{ij}^{-})^{\vee})=\underline{\lambda}_{i}-\underline{\lambda}_{j}$
is a multiple of $\ell$ and so
$(\underline{\nu},(\beta_{ij}^{-})^{\vee})=1+m\ell$ for some $m\in\mathbb{Z}$.
Thus $s_{\beta_{ij}^{-},m}(\underline{\nu})=\underline{\mu}$ and so the
weights are linked.

The analogous statement for $\mathrm{f}_{i}$ and $\mathrm{f}_{j}$ holds with
the analogous proof. ∎

For the case of mixed operators we get accordingly.

######  Lemma 5.5.

Let $\underline{\lambda}\in X^{+}_{\rho}$ and $\ell$ odd. Assume that
$\underline{\mu}=\mathrm{e}_{r}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{f}_{s}\underline{\lambda}$ are defined for $r+1\neq
s$. Then $\mu$ and $\nu$ are linked if and only if $r\in-s+\ell\mathbb{Z}$.

###### Proof.

The proof is nearly identical to Lemma 5.4. Fix $i\neq j$ via
$\underline{\mu}=\underline{\lambda}-\varepsilon_{i}$ and
$\underline{\nu}=\underline{\lambda}+\varepsilon_{j}$. Then the roots that
give absolute values strictly bigger than $1$ for
$(\underline{\nu}-\underline{\mu},\beta^{\vee})=(\varepsilon_{i}+\varepsilon_{j},\beta^{\vee})$
are $\beta\in\\{\beta_{ij}^{+},\beta_{i},\beta_{j}\\}$.

Assume that the weights are linked. Fixing the alcoves as usual and the
corresponding minimal gallery, we see that if $H_{\beta_{ij}^{+},m}$ is
strictly between the alcoves for some $m$ then the statement follows
immediately. If one assumes that such a hyperplane does not exist, the
contradiction is obtained as in the proof of Lemma 5.4.

If it holds $r\in-s+\ell\mathbb{Z}$, then
$(\underline{\lambda},(\beta_{ij}^{+})^{\vee})=\underline{\lambda}_{i}+\underline{\lambda}_{j}$
is a multiple of $\ell$ and so
$(\underline{\nu},(\beta_{ij}^{-})^{\vee})=1+m\ell$ for some $m\in\mathbb{Z}$.
Thus $s_{\beta_{ij}^{+},m}(\underline{\nu})=\underline{\mu}$ and so the
weights are linked. ∎

For the analogue of Lemmas 4.6 and 4.7 we get the following.

######  Lemma 5.6.

Let $\ell$ odd. Assume $\underline{\lambda}\in X^{1,+}_{\rho}$ and that
$\underline{\mu}=\mathrm{e}_{r-1}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ are defined. Then $\mu$
and $\nu$ are linked if and only if $r\in\nicefrac{{1}}{{2}}+\ell\mathbb{Z}$.

Assume $\underline{\lambda}\in X^{\nicefrac{{1}}{{2}},+}_{\rho}$ and that
$\underline{\mu}=\mathrm{e}_{r-1}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ are defined. Then $\mu$
and $\nu$ are linked if and only if
$r\in\nicefrac{{(\ell+1)}}{{2}}+\ell\mathbb{Z}$.

###### Proof.

Fix $i$ via $\underline{\mu}=\underline{\lambda}-\varepsilon_{i}$ respectively
$\underline{\nu}=\underline{\lambda}+\varepsilon_{i}$, i.e. with
$\underline{\lambda}_{i}=r-\nicefrac{{1}}{{2}}$.

Assume that $\mu$ and $\nu$ are linked. Again use Lemma 3.12 to obtain alcoves
$A_{\underline{\mu}}$ and $A_{\underline{\nu}}$ with a minimal gallery only
crossing walls strictly between the alcoves. The possible positive roots such
that the absolute value $|(\underline{\nu}-\underline{\mu},\beta^{\vee})|>1$
are $\beta=\beta_{i}$ for the value $4$ and for the value $2$ it is
$\beta=\beta_{ij}^{\pm}$ for $i<j$ or $\beta=\beta_{ji}^{\pm}$ for $j<i$ .

Let $A_{0},\ldots,A_{t}$ be the alcoves in the minimal gallery connecting
$A_{\underline{\mu}}$ and $A_{\underline{\nu}}$ and $H_{1},\ldots,H_{t}$ the
crossed hyperplanes in order, all strictly between the alcoves by Lemma 3.12.
Fix positive roots and integers such that $H_{p}=H_{\gamma_{p},m_{p}}$. Recall
that for all these it holds
$|(\underline{\nu}-\underline{\mu},\gamma_{p}^{\vee})|>1$.

Case: $\gamma_{1}=\beta_{i}$ and $r=1$: Then
$(\underline{\nu},\beta_{i}^{\vee})=2+m_{1}\ell$ since we need to have
$s_{\beta_{i},m_{1}}(\underline{\nu})=\underline{\mu}$. Thus
$2\underline{\lambda}_{i}+2=2+m_{1}\ell$. In case that $\underline{\lambda}\in
X^{1,+}_{\rho}$, it holds that $m_{1}$ must be even (since $\ell$ is odd) and
so $\underline{\lambda}_{i}\in\ell\mathbb{Z}$ or equivalently
$r\in\nicefrac{{1}}{{2}}+\ell\mathbb{Z}$. If on the other hand
$\underline{\lambda}\in X^{\nicefrac{{1}}{{2}},+}_{\rho}$, then $m_{1}$ must
be odd and we obtain
$\underline{\lambda}_{i}\in\nicefrac{{\ell}}{{2}}+\ell\mathbb{Z}$ or
equivalently $r\in\nicefrac{{(\ell+1)}}{{2}}+\ell\mathbb{Z}$.

Case: $\gamma_{1}=\beta_{ij}^{-}$ for some $j$: Then it holds that
$(\underline{\nu},(\beta_{ij}^{-})^{\vee})=1+m_{1}\ell$,
$s_{\beta_{ij}^{-},m_{1}}(\underline{\nu})=\underline{\lambda}+\varepsilon_{j}\neq\underline{\mu}$,
and $\underline{\lambda}_{i}-\underline{\lambda}_{j}=m_{1}\ell$. Now for $q>1$
it holds
$(\underline{\lambda}+\varepsilon_{j},\gamma_{q}^{\vee})\in\ell\mathbb{Z}$ and
thus
$s_{\gamma_{q},m_{q}}(\underline{\lambda}+\varepsilon_{j})=\underline{\lambda}+\varepsilon_{j}$
unless $\gamma_{q}\in\\{\beta_{i},\beta_{ij}^{+}\\}$. Hence either $\beta_{i}$
or $\beta_{ij}^{+}$ has to appear as a $\gamma_{q}$ for $q>1$. Let $p>1$ be
the minimal such that $\gamma_{p}\in\\{\beta_{i},\beta_{ij}^{+}\\}$.

If $\gamma_{p}=\beta_{ij}^{+}$ then

$(\underline{\lambda}+\varepsilon_{j},(\beta_{ij}^{+})^{\vee})=(\underline{\nu},(\beta_{ij}^{+})^{\vee})=1+m_{p}\ell.$

Hence $\underline{\lambda}_{i}+\underline{\lambda}_{j}=m_{p}\ell$ and so
$2\underline{\lambda}_{i}=(m_{1}+m_{p})\ell$. As above, for
$\underline{\lambda}\in X^{1,+}_{\rho}$, it follows that
$r\in\nicefrac{{1}}{{2}}+\ell\mathbb{Z}$ and for $\underline{\lambda}\in
X^{\nicefrac{{1}}{{2}},+}_{\rho}$ it follows that
$r\in\nicefrac{{(\ell+1)}}{{2}}+\ell\mathbb{Z}$. Furthermore, we see that
$\gamma_{q}\neq\beta_{i}$ for all $q$ since we already have
$s_{\beta_{ij}^{+},m^{\prime}}(\underline{\lambda}+\varepsilon_{j})=\underline{\mu}$
in this case.

If $\gamma_{p}=\beta_{i}$ then

$(\underline{\lambda}+\varepsilon_{j},(\beta_{i})^{\vee})=(\underline{\nu},(\beta_{i})^{\vee})-2.$

But $H_{\beta_{i},m_{p}}$ is assumed to be strictly between the alcoves, thus
$(\underline{\nu},(\beta_{i})^{\vee})=k+m_{p}\ell$ for $k\in\\{1,2,3\\}$
($k=4$ is not possible since $\underline{\mu}$ would lie on
$H_{\beta_{i},m_{p}}$ in that case). Then
$(\underline{\lambda}+\varepsilon_{j},(\beta_{i})^{\vee})=k-2+m_{p}\ell$.

  1. (1)

If $k=1$, then $H_{\beta_{i},m_{p}}$ was already crossed, which cannot be the
case.

  2. (2)

If $k=2$ then $\underline{\lambda}+\varepsilon_{j}$ is on
$H_{\beta_{i},m_{p}}$ and invariant under $s_{\beta_{i},m_{p}}$, hence
$\beta_{ij}^{+}$ has to appear in the sequence of $\gamma$’s that follow and
one argues as above.

  3. (3)

If $k=3$, it holds
$s_{\beta_{i},m_{p}}(\underline{\lambda}+\varepsilon_{j})=\underline{\lambda}+\varepsilon_{j}-\varepsilon_{i}$.
But
$(\underline{\lambda}+\varepsilon_{j}-\varepsilon_{i},\gamma_{q}^{\vee})=(\underline{\mu},\gamma_{q}^{\vee})$
for $q>p$ except for $\gamma_{q}=\beta_{ij}^{+}$ and for $\beta_{ij}^{+}$ it
holds
$(\underline{\lambda}+\varepsilon_{j}-\varepsilon_{i},(\beta_{ij}^{+})^{\vee})=(\underline{\lambda},(\beta_{ij}^{+})^{\vee})$.
Hence $\underline{\lambda}+\varepsilon_{j}-\varepsilon_{i}$ is already in the
same half-space as $\underline{\mu}$ for all $H_{q}$ with $q>p$ and
$\gamma_{q}\neq\beta_{ij}^{+}$ and it is on the hyperplane $H_{q}$ for $q>p$
and $\gamma_{q}=\beta_{ij}^{+}$. Thus the only hyperplane left to cross is
$H_{p+1}$ and it must hold that $\gamma_{p+1}=\beta_{ij}^{+}$, but
$s_{\beta_{ij}^{+},m_{p+1}}(\underline{\lambda}+\varepsilon_{j}-\varepsilon_{i})=\underline{\lambda}+\varepsilon_{j}-\varepsilon_{i}\neq\underline{\mu}$,
which is a contradiction to the construction of the gallery.

Remaining cases: The remaining cases for $\gamma_{1}$ are all done in the same
way as the previous case with slight modifications in the signs that appear,
but the same general arguments.

Conversely for $\underline{\lambda}\in X^{1,+}_{\rho}$ and
$r\in\nicefrac{{1}}{{2}}+\ell\mathbb{Z}$, $\underline{\lambda}_{i}=m\ell$ for
some $m$. Hence
$(\underline{\lambda}+\varepsilon_{i},\beta_{i}^{\vee})=2\underline{\lambda}_{i}+2=2+2m\ell$.
Thus $s_{\beta_{i},2m}(\underline{\nu})=\underline{\mu}$ and so $\nu$ and
$\mu$ are linked.

For $\underline{\lambda}\in X^{\nicefrac{{1}}{{2}},+}_{\rho}$ and
$r\in\nicefrac{{(\ell+1)}}{{2}}+\ell\mathbb{Z}$,
$\underline{\lambda}_{i}=\nicefrac{{\ell}}{{2}}+m\ell$. Hence
$(\underline{\lambda}+\varepsilon_{i},\beta_{i}^{\vee})=2\underline{\lambda}_{i}+2=2+(2m+1)\ell$.
Thus $s_{\beta_{i},2m+1}(\underline{\nu})=\underline{\mu}$ and so $\nu$ and
$\mu$ are linked. ∎

In contrast to type $C$, $\lambda$ can be linked to a weight $\mu$ obtained by
applying an operator.

######  Lemma 5.7.

Let $\underline{\lambda}\in X^{1,+}_{\rho}$ and $\ell$ is odd. Assume that
$\underline{\mu}=\mathrm{e}_{r}\underline{\lambda}$ is defined. Then $\mu$ and
$\lambda$ are linked if and only if
$r\in\nicefrac{{\ell}}{{2}}+\ell\mathbb{Z}$.

Assume that $\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ is defined.
Then $\nu$ and $\lambda$ are linked if and only if
$r\in\nicefrac{{\ell}}{{2}}+\ell\mathbb{Z}$.

###### Proof.

Fix $i$ via $\underline{\mu}=\underline{\lambda}-\varepsilon_{i}$ with
$\underline{\lambda}_{i}=r+\nicefrac{{1}}{{2}}$.

Assume now that $\mu$ and $\lambda$ are linked. The absolute value of
$(\underline{\lambda}-\underline{\mu},\beta^{\vee})$ can only be $2$ for
$\beta=\beta_{i}$. Hence there exists a hyperplane $H_{\beta_{i},m}$ strictly
between and so
$(\underline{\lambda},\beta_{i}^{\vee})=2\underline{\lambda}_{i}=1+m\ell$.
Since $2\underline{\lambda}_{i}$ is an even integer and $\ell$ is odd, $m$ is
odd as well. So we get
$r=\nicefrac{{\ell}}{{2}}+\frac{m-1}{2}\ell\in\nicefrac{{\ell}}{{2}}+\ell\mathbb{Z}$.

If $r\in\nicefrac{{\ell}}{{2}}+\ell\mathbb{Z}$ then
$\underline{\lambda}_{i}=\nicefrac{{(\ell+1)}}{{2}}+m\ell$ for some
$m\in\mathbb{Z}$. Hence $(\underline{\lambda},\beta_{i}^{\vee})=1+(2m+1)\ell$
and so $\lambda$ and $\mu$ are linked.

The case of $\underline{\nu}$ follows, considering that in this case
$\mathrm{e}_{r}\underline{\nu}=\underline{\lambda}$. ∎

The proof for the next Lemma is completely analogous to Lemma 5.7.

######  Lemma 5.8.

Let $\underline{\lambda}\in X^{\nicefrac{{1}}{{2}},+}_{\rho}$ and $\ell$ is
odd. Assume that $\underline{\mu}=\mathrm{e}_{r}\underline{\lambda}$ is
defined. Then $\mu$ and $\lambda$ are linked if and only if
$r\in\ell\mathbb{Z}$.

Assume that $\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ is defined.
Then $\nu$ and $\lambda$ are linked if and only if $r\in\ell\mathbb{Z}$.

###  5.3. Quantum symmetric pair action for odd $\ell$

As in the previous section we are assuming throughout the section that

$\ell>3\text{ is odd.}$

Recall the various operators and counting statistics from Section 3.1. In
contrast to type $C$, we need to use operators for both types of index sets.

The definition of the linear operators is very close to the ones of type $C$
in Section 4.3. with one difference for the elements of
$\mathbb{H}/\ell\mathbb{Z}$ respectively $\mathbb{Z}/\ell\mathbb{Z}$ that are
fixed by $\Theta$ as defined in Section 3.4. Note that for $\ell$ odd,
$\Theta(\overline{i})=\overline{i}$ implies that
$\overline{i}=\overline{\nicefrac{{\ell}}{{2}}}$ for
$\overline{i}\in\mathbb{H}/\ell\mathbb{Z}$ and $\overline{i}=\overline{0}$ for
$\overline{i}\in\mathbb{Z}/\ell\mathbb{Z}$. This corresponds to the two fixed
nodes in the odd Dynkin diagrams in Section 3.4.

######  Definition 5.9.

Assume $\ell$ odd. For $\underline{a}\in\mathcal{S}_{\mathbb{Z}}$ let
$\overline{p}\in\mathbb{H}/\ell\mathbb{Z}$ and for
$\underline{a}\in\mathcal{S}_{\mathbb{H}}$ let
$\overline{p}\in\mathbb{Z}/\ell\mathbb{Z}$. If
$\Theta(\overline{p})\neq\overline{p}$ define

$\displaystyle\hat{\mathrm{B}}_{\overline{p}}\,\underline{a}$
$\displaystyle=v^{T_{\ell}^{e-f}(-\overline{p},\underline{a})}\sum_{j\in\overline{p}}v^{R_{\ell}^{e-f}(j,\underline{a})}\mathrm{e}_{j}\,\underline{a}+\sum_{j\in-\overline{p}}v^{L_{\ell}^{f-e}(j,\underline{a})}\mathrm{f}_{j}\,\underline{a}.$
$\displaystyle\hat{\mathrm{L}}_{\overline{p}}\underline{a}$
$\displaystyle=v^{T_{\ell}^{f-e}(\overline{p},\underline{a})}v^{T_{\ell}^{e-f}(-\overline{p},\underline{a})}\underline{a}.$

If $\Theta(\overline{p})=\overline{p}$ define

$\hat{\mathrm{B}}_{\overline{p}}\underline{a}=v^{-1}v^{T_{\ell}^{e-f}(\overline{p},\underline{a})}\sum_{j\in\overline{p}}v^{R_{\ell}^{e-f}(j,\underline{a})}\mathrm{e}_{j}\underline{a}+\sum_{j\in\overline{p}}v^{L_{\ell}^{f-e}(j,\underline{a})}\mathrm{f}_{j}\underline{a}+v^{T_{\ell}^{e-f}(\overline{p},\underline{a})}\underline{a}.$

Finally, for $\underline{a}\in\mathcal{S}_{\mathbb{H}}$ and
$z\in\nicefrac{{1}}{{2}}+\ell\mathbb{Z}$ define

$\hat{\mathrm{B}}_{\overline{0}}^{[z]}\underline{a}=v^{-1}v^{T_{\ell}^{e-f}(\overline{0},\underline{a})}\sum_{j\in\overline{0}}v^{R_{\ell}^{e-f}(j,\underline{a})}\mathrm{e}_{j}\underline{a}+\sum_{j\in\overline{0}}v^{L_{\ell}^{f-e}(j,\underline{a})}\mathrm{f}_{j}\underline{a}+\delta_{\underline{a}(z),0}v^{T_{\ell}^{e-f}(\overline{0},\underline{a})}\underline{a},$

where $\delta_{\underline{a}(z),0}=1-\underline{a}(z)$. This defines linear
operators on $\mathcal{F}$ and $\mathcal{F}^{\nicefrac{{1}}{{2}}}$ that
restrict to $\mathcal{F}_{N}$ and $\mathcal{F}_{N}^{\nicefrac{{1}}{{2}}}$ for
any $N$.

The linear operator $\hat{\mathrm{B}}_{\overline{0}}^{[z]}$ is not contained
in the image of the affine quantum symmetric pair. It is needed to describe
the relationship with the tensor product decomposition. Comparing this with
the statements in Lemma 4.9, the same identifications hold except for the case
$\Theta(\underline{p})=\underline{p}$.

######  Lemma 5.10.

Let $\underline{a}\in\mathcal{S}_{\mathbb{Z}}$ (respectively
$\underline{a}\in\mathcal{S}_{\mathbb{H}}$) and
$\overline{p}\in\mathbb{H}/\ell\mathbb{Z}$ (respectively
$\overline{p}\in\mathbb{Z}/\ell\mathbb{Z}$) with
$\Theta(\underline{i})=\underline{i}$ then

$\hat{\mathrm{B}}_{\overline{p}}\,\underline{a}=v^{-1}\hat{\mathrm{E}}_{\overline{p}}\hat{\mathrm{K}}_{\overline{p}}^{-1}\,\underline{a}+\hat{\mathrm{F}}_{\overline{p}}\,\underline{a}+\hat{\mathrm{K}}_{\overline{p}}^{-1}\underline{a}$

and for $\underline{a}\in\mathcal{S}_{\mathbb{H}}$ and
$z\in\nicefrac{{1}}{{2}}+\ell\mathbb{Z}$

$\hat{\mathrm{B}}_{\overline{p}}^{[z]}\,\underline{a}=v^{-1}\hat{\mathrm{E}}_{\overline{p}}\hat{\mathrm{K}}_{\overline{p}}^{-1}\,\underline{a}+\hat{\mathrm{F}}_{\overline{p}}\,\underline{a}+\delta_{\underline{a}(z),0}\hat{\mathrm{K}}_{\overline{p}}^{-1}\underline{a}$

######  Remark 5.11.

In [Kol14] a quantum symmetric pair using a generator of the form
$v^{-1}\hat{\mathrm{E}}_{\overline{p}}\hat{\mathrm{K}}_{\overline{p}}^{-1}+\hat{\mathrm{F}}_{\overline{p}}+\hat{\mathrm{K}}_{\overline{p}}^{-1}$
is called a non-standard quantum symmetric pair. While the one that uses only
generators of the form that appeared in type $C$, i.e. without the extra
summand $\hat{\mathrm{K}}_{\overline{p}}^{-1}$ are called standard quantum
symmetric pairs.

Acting with the generators on $\mathcal{F}^{1}(B_{N})$ respectively
$\mathcal{F}^{1}(B_{N})$ has the same possibility of having a single summand
that is not in the subspace.

######  Remark 5.12.

In case of $\mathcal{F}_{N}$ the operators $\hat{\mathrm{B}}_{\overline{i}}$
leave the subspace $\mathcal{F}^{1}(B_{N})$ invariant, except for
$-\nicefrac{{1}}{{2}}\in\overline{i}$. For the operator
$\hat{\mathrm{B}}_{\overline{\nicefrac{{1}}{{2}}}}$ the reason is again that
it contains a summand $\mathrm{f}_{\nicefrac{{1}}{{2}}}$ that does not leave
$\mathcal{F}^{1}(B_{N})$ invariant. While for
$\mathcal{F}_{N}^{\nicefrac{{1}}{{2}}}$ the operators
$\hat{\mathrm{B}}_{\overline{0}}$ and
$\hat{\mathrm{B}}_{\overline{0}}^{[\nicefrac{{1}}{{2}}]}$ do not leave the
subspace invariant because of the summand $\mathrm{f}_{0}$.

For the relations of the operators we then get the following.

######  Proposition 5.13.

Assume $\ell$ odd and
$\mathrm{I}\in\\{\mathbb{Z}/\ell\mathbb{Z},\mathbb{H}/\ell\mathbb{Z}\\}$. The
linear operators
$\\{\hat{\mathrm{B}}_{\overline{p}}\mid\overline{p}\in\mathrm{I}\\}$ and
$\\{\hat{\mathrm{L}}_{\overline{p}}\mid\overline{p}\in\mathrm{I},\Theta(\overline{p})\neq\overline{p}\\}$
satisfy the relations of $B_{v}(\mathrm{I},\Theta)$ given in Definition 3.14
by substituting the operators for the generators with the same name.

###### Proof.

All relations in Definition 3.14 not involving the operator
$\hat{\mathrm{B}}_{\overline{p}}$ for $\overline{p}$ fixed are precisely the
same as in type $C$ and thus hold by Proposition 4.11. Note that for
$\overline{p}$ fixed, the new operator $\hat{\mathrm{B}}_{\overline{p}}$ only
differs from the description in Lemma 4.9 by adding a term of the form
$\hat{\mathrm{K}}_{\overline{p}}^{-1}$. The commutator relation with all
$\hat{\mathrm{L}}_{\overline{q}}$ is then obvious. The commutator relation
with “distant” $\hat{\mathrm{B}}_{\overline{q}}$ is also obvious. Hence the
only one left are the non-trivial quantum Serre relations. There are only two
that involve a fixed index, which are both a simple few line calculation to
verify. ∎

As in type $C$ this all combines then to the following. For the case of
weights in $X^{1,+}$ the statement is the complete analogue to Theorem 4.12
with precisely the same proof.

######  Theorem 5.14.

Assume $\ell$ odd. There exists an action of
$B_{v}(\mathbb{H}/\ell\mathbb{Z},\Theta)$ on $\mathcal{F}_{N}$ such that for
$\lambda\in X^{1,+}$ the decomposition of
$[\Delta_{\mathrm{q}}(\lambda)\otimes\Delta_{\mathrm{q}}(\omega_{1})]$ in
$[U_{\mathrm{q}}\text{-}\mathrm{mod}]$ with respect to the classes of Weyl
modules is obtained from

$\sum_{\overline{p}\in\mathbb{H}/\ell\mathbb{Z}}\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$

by projecting onto the subspace $\mathcal{F}(C_{N})$ and evaluating $v=1$.

Furthermore if $[\Delta_{\mathrm{q}}(\mu)]$ and $[\Delta_{\mathrm{q}}(\nu)]$
appear in the decomposition for $\mu$ and $\nu$ linked, then there exists a
unique $\overline{p}\in\mathbb{H}/\ell\mathbb{Z}$ such that $\underline{\mu}$
and $\underline{\nu}$ appear in
$\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$.

In case of $\lambda\in X^{\nicefrac{{1}}{{2}},+}$, the dependence of the
tensor product decomposition rule on $\lambda_{N}$ makes this less clean.

######  Theorem 5.15.

Assume $\ell$ odd. There exists an action of
$B_{v}(\mathbb{Z}/\ell\mathbb{Z},\Theta)$ on
$\mathcal{F}_{N}^{\nicefrac{{1}}{{2}}}$ such that for $\lambda\in
X^{\nicefrac{{1}}{{2}},+}$ the decomposition of
$[\Delta_{\mathrm{q}}(\lambda)\otimes\Delta_{\mathrm{q}}(\omega_{1})]$ in
$[U_{\mathrm{q}}\text{-}\mathrm{mod}]$ with respect to the classes of Weyl
modules is obtained from

$\sum_{\overline{p}\in\mathbb{Z}/\ell\mathbb{Z}}\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$

if $\lambda_{N}>0$ and from

$\sum_{\overline{p}\in\mathbb{Z}/\ell\mathbb{Z},\overline{p}\neq\overline{0}}\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}+\hat{\mathrm{B}}_{\overline{0}}^{[\nicefrac{{1}}{{2}}]}\underline{\lambda}$

if $\lambda_{N}=0$, by projecting onto the subspace $\mathcal{F}(C_{N})$ and
evaluating $v=1$.

If $[\Delta_{\mathrm{q}}(\mu)]$ and $[\Delta_{\mathrm{q}}(\nu)]$ appear in the
decomposition for $\mu$ and $\nu$ linked, then there exists a unique
$\overline{p}\in\mathbb{Z}/\ell\mathbb{Z}$ such that $\underline{\mu}$ and
$\underline{\nu}$ appear in
$\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$.

###### Proof.

The only difference to Theorem 5.14 is that in case of $\lambda\in
X^{\nicefrac{{1}}{{2}},+}$ the class of the tensor product
$[\Delta_{\mathrm{q}}(\lambda)\otimes\Delta_{\mathrm{q}}(\omega_{1})]$
contains a summand $[\Delta_{\mathrm{q}}(\lambda)]$ if and only if
$\lambda_{N}>0$, But $\hat{\mathrm{B}}_{\overline{0}}\underline{\lambda}$
always contain a non-zero multiple of $\underline{\lambda}$ as a summand,
hence one has to apply
$\hat{\mathrm{B}}_{\overline{0}}^{[\nicefrac{{1}}{{2}}]}$ instead in case
$\lambda_{N}=0$. ∎

Let $0\leq k<\ell$ and $N=m\ell+k$ for $m\geq 0$. One can use the map from
Definition 4.13 to embed $\mathcal{F}^{1}(B_{m\ell+k})$ into
$\mathcal{F}^{1}_{k}$. In this case the analogue of Proposition 4.14 follows
immediately.

In case of $\underline{a}\in\mathcal{S}_{\mathbb{H},N}$ the map from
Definition 4.13 embeds $\mathcal{F}^{\nicefrac{{1}}{{2}}}(B_{m\ell+k})$ into
$\mathcal{F}^{\nicefrac{{1}}{{2}}}_{k}$. To formulate the corresponding
statement, define
$\hat{\mathrm{B}}_{\overline{p}}^{[z]}=\hat{\mathrm{B}}_{\overline{p}}$ if
$\overline{p}\neq\overline{0}$ for $z\in\nicefrac{{1}}{{2}}+\ell\mathbb{Z}$.

######  Proposition 5.16.

Let $\underline{a}\in\mathcal{S}_{\mathbb{H},k}$ and $r\geq 0$. Then for $m\gg
0$, $\underline{a}=\underline{\lambda}^{(m)}$ for $\lambda\in
X^{\nicefrac{{1}}{{2}},+}$ for $U_{\mathrm{q}}$ of type $B_{m\ell+k}$.
Furthermore there are Laurent polynomials $d_{\lambda,\mu}(v)$ with non-
negative integer coefficients for $\mu\in X^{\nicefrac{{1}}{{2}},+}$ such that

$\sum_{(\mathbb{Z}/\ell\mathbb{Z})^{r}}\hat{\mathrm{B}}_{\overline{p}_{1}}^{[z]}\cdots\hat{\mathrm{B}}_{\overline{p}_{r}}^{[z]}\underline{\lambda}^{(m)}=\sum_{\mu\text{
dominant for }U_{\mathrm{q}}}d_{\lambda,\mu}(v)\underline{\mu}^{(m)},$

with $z=\nicefrac{{1}}{{2}}-m\ell$ and

$[\Delta_{\mathrm{q}}(\lambda)\otimes\Delta_{\mathrm{q}}(\omega_{1})^{\otimes
r}]=\sum_{\mu\text{ dominant for
}U_{\mathrm{q}}}d_{\lambda,\mu}(1)[\Delta_{\mathrm{q}}(\mu)].$

###### Proof.

Note that the only difference to Proposition 4.14 is that instead of
$\hat{\mathrm{B}}_{\overline{0}}$ one has to use
$\hat{\mathrm{B}}_{\overline{0}}^{[z]}$. We already know that
$\sum_{(\mathbb{Z}/\ell\mathbb{Z})^{r}}\hat{\mathrm{B}}_{\overline{p}_{1}}\cdots\hat{\mathrm{B}}_{\overline{p}_{r}}\underline{\lambda}^{(m)}$
is contained in the shifted embedding of
$\mathcal{F}^{\nicefrac{{1}}{{2}}}(B_{m\ell+k})$ for $m\gg 0$, hence also the
summand
$\sum_{(\mathbb{Z}/\ell\mathbb{Z})^{r}}\hat{\mathrm{B}}_{\overline{p}_{1}}^{[z]}\cdots\hat{\mathrm{B}}_{\overline{p}_{r}}^{[z]}\underline{\lambda}^{(m)}$.

In case $m\gg 0$ all weights $\nu$ such that $[\Delta_{\mathrm{q}}(\nu)]$
appears in
$[\Delta_{\mathrm{q}}(\lambda)\otimes\Delta_{\mathrm{q}}(\omega_{1})^{\otimes
s}]$ (for $0\leq s<r$) have the property that $\nu_{k+m\ell}=0$, hence
$\underline{\nu}(\nicefrac{{1}}{{2}})=1$. Hence the
$[\Delta_{\mathrm{q}}(\nu)\otimes\Delta_{\mathrm{q}}(\omega_{1})]$ never
contain the summand $[\Delta_{\mathrm{q}}(\nu)]$. Thus
$\hat{\mathrm{B}}_{\overline{0}}^{[\nicefrac{{1}}{{2}}]}\underline{\nu}$ has
to be used to get the tensor product decomposition by Theorem 5.15. After the
shift this becomes the operator
$\hat{\mathrm{B}}_{\overline{0}}^{[z]}\underline{\nu}$. ∎

###  5.4. Linkage and operators for even $\ell$

The first difference in case $\ell$ even is that $\ell_{\beta}=2$ for the
roots of the form $\beta=\beta_{ij}^{\pm}$. These were responsible in the odd
case that all indices had to be taken modulo $\ell$ to obtain the operators
and that one had to group $\mathrm{e}_{p}$ and $\mathrm{f}_{q}$ for
$p+q\in\ell\mathbb{Z}$. Now this is replaced by the analogue statements with
$\nicefrac{{\ell}}{{2}}$ everywhere. Thus we have to make the following
assumption throughout the section

$\ell\text{ is even and }\nicefrac{{\ell}}{{2}}>3.$

As before this is needed to avoid quantum symmetric pairs for small
$\nicefrac{{\ell}}{{2}}$, the general arguments for linkage work the same.

Namely one obtains the two statements.

######  Lemma 5.17.

Let $\underline{\lambda}\in X^{+}_{\rho}$ and $\ell$ even. Assume that
$\underline{\mu}=\mathrm{e}_{r}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{e}_{s}\underline{\lambda}$ are defined for $r\neq s$.
Then $\mu$ and $\nu$ are linked if and only if $r\in
s+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

Assume that $\underline{\mu^{\prime}}=\mathrm{f}_{r}\underline{\lambda}$ and
$\underline{\nu^{\prime}}=\mathrm{f}_{s}\underline{\lambda}$ are defined for
$r\neq s$. Then $\mu^{\prime}$ and $\nu^{\prime}$ are linked if and only if
$r\in s+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

######  Lemma 5.18.

Let $\underline{\lambda}\in X^{+}_{\rho}$ and $\ell$ even. Assume that
$\underline{\mu}=\mathrm{e}_{r}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{f}_{s}\underline{\lambda}$ are defined for $r+1\neq
s$. Then $\mu$ and $\nu$ are linked if and only if
$r\in-s+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

The analogue of Lemma 5.6 now depend on whether $\nicefrac{{\ell}}{{2}}$ is
itself odd or even. This makes sense if one looks at the corresponding Dynkin
diagrams in Section 2.3 for $r=\nicefrac{{\ell}}{{2}}$. In case of
$X_{\rho}^{1,+}$ the Dynkin diagram with $\nicefrac{{\ell}}{{2}}$ nodes always
has the property that $\overline{\nicefrac{{1}}{{2}}}$ is $\Theta$-linked, but
it depends on the parity of $\nicefrac{{\ell}}{{2}}$ whether
$\overline{\nicefrac{{\ell}}{{4}}}$ is fixed in case $\nicefrac{{\ell}}{{2}}$
is even or $\overline{\nicefrac{{\ell-2}}{{4}}}$ is also $\Theta$-linked in
case $\nicefrac{{\ell}}{{2}}$ is odd. Similarly for
$X_{\rho}^{\nicefrac{{1}}{{2}},+}$ the Dynkin diagram with
$\nicefrac{{\ell}}{{2}}$ nodes always has $\overline{0}$ is fixed, but it
depends on the parity of $\nicefrac{{\ell}}{{2}}$ whether the “opposite” side
of the Dynkin diagram has another fixed label or a pair of $\Theta$-linked
labels. Thus the analogue of Lemma 5.5 has four possible combinations.

######  Lemma 5.19.

Let $\ell$ even.

  1. (1)

Assume $\nicefrac{{\ell}}{{2}}$ odd, $\underline{\lambda}\in X^{1,+}_{\rho}$,
and that $\underline{\mu}=\mathrm{e}_{r-1}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ are defined. Then $\mu$
and $\nu$ are linked if and only if
$r\in\nicefrac{{1}}{{2}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

  2. (2)

Assume $\nicefrac{{\ell}}{{2}}$ even, $\underline{\lambda}\in X^{1,+}_{\rho}$,
and that $\underline{\mu}=\mathrm{e}_{r-1}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ are defined. If $\mu$ and
$\nu$ are linked then
$r\in\nicefrac{{1}}{{2}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$ or
$r\in\nicefrac{{\ell+2}}{{4}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

  3. (3)

Assume $\nicefrac{{\ell}}{{2}}$ odd, $\underline{\lambda}\in
X^{\nicefrac{{1}}{{2}},+}_{\rho}$, and that
$\underline{\mu}=\mathrm{e}_{r-1}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ are defined. If $\mu$ and
$\nu$ are linked then
$r\in\nicefrac{{\ell+2}}{{4}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

  4. (4)

Assume $\nicefrac{{\ell}}{{2}}$ even, $\underline{\lambda}\in
X^{\nicefrac{{1}}{{2}},+}_{\rho}$, and that
$\underline{\mu}=\mathrm{e}_{r-1}\underline{\lambda}$ and
$\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ are defined. Then $\mu$
and $\nu$ are not linked.

###### Proof.

We only sketch the proof here, since for each situation one has to go through
the same procedure as in the proof of Lemma 5.5.

For $\nicefrac{{\ell}}{{2}}$ odd and $\underline{\lambda}\in X^{1,+}_{\rho}$
the proof is nearly word for word the same, hence the result is also the same.

For $\nicefrac{{\ell}}{{2}}$ even and $\underline{\lambda}\in X^{1,+}_{\rho}$
one obtains that $r\in\nicefrac{{1}}{{2}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$
in case of the minimal gallery having length one and $\gamma_{1}=\beta_{i}$.
But in the remaining cases that $\gamma_{1}$ is different one obtains either
$r\in\nicefrac{{1}}{{2}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$ if $m_{1}+m_{p}$
appearing in the proof of Lemma 5.5 is even or
$r\in\nicefrac{{(\ell+2)}}{{4}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$ if
$m_{1}+m_{p}$ is odd. In case that
$r\in\nicefrac{{(\ell+2)}}{{4}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$ one does
not obtain the converse. The argument using a reflection at a hyperplane for
$\beta_{i}$ does not work here, since $\ell_{\beta_{i}}=\ell$.

In the case $\nicefrac{{\ell}}{{2}}$ odd and $\underline{\lambda}\in
X^{\nicefrac{{1}}{{2}},+}_{\rho}$ the case of a minimal gallery of length $1$
gives a contradiction, while the other cases give
$r\in\nicefrac{{\ell+2}}{{4}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$. Note that
if $\nicefrac{{\ell}}{{2}}$ is odd, $\nicefrac{{\ell+2}}{{4}}$ is an integer,
so the $r$ is well-defined.

Finally, the case $\nicefrac{{\ell}}{{2}}$ even and $\underline{\lambda}\in
X^{\nicefrac{{1}}{{2}},+}_{\rho}$ with the assumption of linked weights always
gives a contradiction. ∎

Thus we are left with comparing $\underline{\lambda}$ and the image under one
of the operators $\mathrm{e}_{i}$ or $\mathrm{f}_{i}$. In the case of
$\underline{\lambda}\in X^{1,+}_{\rho}$ the situation becomes quite easy.

######  Lemma 5.20.

Let $\underline{\lambda}\in X^{1,+}_{\rho}$ and $\ell$ is even. Assume that
$\underline{\mu}=\mathrm{e}_{r}\underline{\lambda}$ is defined. Then $\mu$ and
$\lambda$ are not linked. Assume that
$\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ is defined. Then $\nu$ and
$\lambda$ are not linked.

###### Proof.

Fix $i$ via $\underline{\mu}=\underline{\lambda}-\varepsilon_{i}$ and
$r+\nicefrac{{1}}{{2}}=\underline{\lambda}_{i}$. Then as in the proof of Lemma
5.7 one only has to consider the root $\beta_{i}$. And one assumes that the
weights are linked with hyperplane $H_{\beta_{i},m}$ strictly between the
alcoves having to exist. Hence one obtains

$(\underline{\lambda},\beta_{i}^{\vee})=2\underline{\lambda}_{i}=1+m\ell$

for some $m\in\mathbb{Z}$. But $2\underline{\lambda}_{i}$ is an even integer,
while $1+m\ell$ is always odd, hence this is a contradiction.

Again the second case follows since if
$\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ is defined, then
$\mathrm{e}_{r}\underline{\nu}=\underline{\lambda}$ holds. ∎

In case of $\underline{\lambda}\in X^{\nicefrac{{1}}{{2}},+}_{\rho}$, the
situation resembles the case of $\ell$ odd.

######  Lemma 5.21.

Let $\underline{\lambda}\in X^{\nicefrac{{1}}{{2}},+}_{\rho}$ and $\ell$ is
even. Assume that $\underline{\mu}=\mathrm{e}_{r}\underline{\lambda}$ is
defined. Then $\mu$ and $\lambda$ are linked if and only if
$r\in(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

Assume that $\underline{\nu}=\mathrm{f}_{r}\underline{\lambda}$ is defined.
Then $\nu$ and $\lambda$ are linked if and only if
$r\in(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

###### Proof.

We proceed like in the proof of Lemma 5.20 and obtain the same equation

$(\underline{\lambda},\beta_{i}^{\vee})=2\underline{\lambda}_{i}=1+m\ell$

for some $m\in\mathbb{Z}$. But now $2\underline{\lambda}_{i}$ is an odd
integer, hence
$\underline{\lambda}=\nicefrac{{1}}{{2}}+m\nicefrac{{\ell}}{{2}}$ which then
implies $r\in(\nicefrac{{\ell}}{{2}})\mathbb{Z}$.

Conversely if $r\in(\nicefrac{{\ell}}{{2}})\mathbb{Z}$, then
$(\underline{\lambda},\beta_{i}^{\vee})=2\underline{\lambda}_{i}=1+m\ell$ for
some $m\in\mathbb{Z}$, hence
$s_{\beta_{i},m}(\underline{\lambda}=\underline{\mu}$ and $\lambda$ and $\mu$
are linked.

As before the case for $\mathrm{f}_{r}$ is done by using that
$\mathrm{e}_{r}\underline{\nu}=\underline{\lambda}$ under the assumptions. ∎

###  5.5. Quantum symmetric pair action for even $\ell$

Throughout the section we make the assumption that

$\ell\text{ is even and }\nicefrac{{\ell}}{{2}}>3.$

In this situation the action of the quantum symmetric pair depends on two
different choices. First whether $\nicefrac{{\ell}}{{2}}$ is even or odd and
second whether one is looking at weights in $X_{\rho}^{1,+}$ or
$X_{\rho}^{\nicefrac{{1}}{{2}},+}$. We just list the definitions of the linear
operators in the different situations. Note that we obtain all possible
quantum symmetric pairs for the affine type $A$ Dynkin diagram that we
described before, i.e. we can have either two fixed labels, two
$\Theta$-linked label pairs, or one of each and in that case we will see that
they differ in the sense that one of them is a standard quantum symmetric
pair, the other one is a non-standard quantum symmetric pair.

We are considering now $\mathbb{Z}/(\nicefrac{{\ell}}{{2}})\mathbb{Z}$ and
$\mathbb{H}/(\nicefrac{{\ell}}{{2}})\mathbb{Z}$ with the usual automorphism
$\Theta$.

######  Definition 5.22.

Assume $\ell$ even. For $\underline{a}\in\mathcal{S}_{\mathbb{Z}}$ let
$\overline{p}\in\mathbb{H}/(\nicefrac{{\ell}}{{2}})\mathbb{Z}$ and for
$\underline{a}\in\mathcal{S}_{\mathbb{H}}$ let
$\overline{p}\in\mathbb{Z}/(\nicefrac{{\ell}}{{2}})\mathbb{Z}$. Then define
for $\overline{p}\neq\Theta(\overline{p})$

$\displaystyle\hat{\mathrm{B}}_{\overline{p}}\,\underline{a}$
$\displaystyle=v^{T_{\nicefrac{{\ell}}{{2}}}^{e-f}(-\overline{p},\underline{a})}\sum_{j\in\overline{p}}v^{R_{\nicefrac{{\ell}}{{2}}}^{e-f}(j,\underline{a})}\mathrm{e}_{j}\,\underline{a}+\sum_{j\in-\overline{p}}v^{L_{\nicefrac{{\ell}}{{2}}}^{f-e}(j,\underline{a})}\mathrm{f}_{j}\,\underline{a}.$
$\displaystyle\hat{\mathrm{L}}_{\overline{p}}\,\underline{a}$
$\displaystyle=v^{T_{\nicefrac{{\ell}}{{2}}}^{f-e}(\overline{p},\underline{a})}v^{T_{\nicefrac{{\ell}}{{2}}}^{e-f}(-\overline{p},\underline{a})}\underline{a}.$

For $\overline{p}=\Theta(\overline{p})$, but $0\notin\overline{p}$ define

$\hat{\mathrm{B}}_{\overline{p}}\underline{a}=v^{-1}v^{T_{\nicefrac{{\ell}}{{2}}}^{e-f}(\overline{p},\underline{a})}\sum_{j\in\overline{p}}v^{R_{\nicefrac{{\ell}}{{2}}}^{e-f}(j,\underline{a})}\mathrm{e}_{j}\underline{a}+\sum_{j\in\overline{p}}v^{L_{\nicefrac{{\ell}}{{2}}}^{f-e}(j,\underline{a})}\mathrm{f}_{j}\underline{a},$

while for $0\in\overline{i}$
$z\in\nicefrac{{1}}{{2}}+(\nicefrac{{\ell}}{{2}})\mathbb{Z}$ define

$\displaystyle\hat{\mathrm{B}}_{\overline{p}}\underline{a}$
$\displaystyle=v^{-1}v^{T_{\nicefrac{{\ell}}{{2}}}^{e-f}(\overline{p},\underline{a})}\sum_{j\in\overline{p}}v^{R_{\nicefrac{{\ell}}{{2}}}^{e-f}(j,\underline{a})}\mathrm{e}_{j}\underline{a}+\sum_{j\in\overline{p}}v^{L_{\nicefrac{{\ell}}{{2}}}^{f-e}(j,\underline{a})}\mathrm{f}_{j}\underline{a}+v^{T_{\nicefrac{{\ell}}{{2}}}^{e-f}(\overline{p},\underline{a})}\underline{a},\text{
and}$ $\displaystyle\hat{\mathrm{B}}_{\overline{0}}^{[z]}\underline{a}$
$\displaystyle=v^{-1}v^{T_{\nicefrac{{\ell}}{{2}}}^{e-f}(\overline{0},\underline{a})}\sum_{j\in\overline{0}}v^{R_{\nicefrac{{\ell}}{{2}}}^{e-f}(j,\underline{a})}\mathrm{e}_{j}\underline{a}+\sum_{j\in\overline{0}}v^{L_{\nicefrac{{\ell}}{{2}}}^{f-e}(j,\underline{a})}\mathrm{f}_{j}\underline{a}+\delta_{\underline{a}(z),0}v^{T_{\nicefrac{{\ell}}{{2}}}^{e-f}(\overline{0},\underline{a})}\underline{a}.$

This defines linear operators on $\mathcal{F}$ and
$\mathcal{F}^{\nicefrac{{1}}{{2}}}$ that restrict to $\mathcal{F}_{N}$ and
$\mathcal{F}_{N}^{\nicefrac{{1}}{{2}}}$ for any $N$.

As one can see, in case that $\underline{a}\in\mathcal{S}_{\mathbb{Z}}$ and
$\nicefrac{{\ell}}{{2}}$ is even there is no fixed index
$\overline{p}\in\mathbb{H}/(\nicefrac{{\ell}}{{2}})\mathbb{Z}$, instead there
are two pairs of indices that are $\Theta$-linked, namely
$\pm\overline{\nicefrac{{1}}{{2}}}$ and $\overline{\nicefrac{{\ell\pm
2}}{{4}}}$.

In case that $\underline{a}\in\mathcal{S}_{\mathbb{Z}}$ and
$\nicefrac{{\ell}}{{2}}$ is odd, the pair $\pm\overline{\nicefrac{{1}}{{2}}}$
is still $\Theta$-linked, but $\overline{\nicefrac{{\ell}}{{4}}}$ is a fixed
index. But the quantum symmetric pair is a standard quantum symmetric pair,
like in type $C$.

In case that $\underline{a}\in\mathcal{S}_{\mathbb{H}}$ and
$\nicefrac{{\ell}}{{2}}$ is even, both $\overline{0}$ and
$\overline{\nicefrac{{\ell}}{{4}}}$ are fixed, but the generator for
$\overline{0}$ makes this a non-standard quantum symmetric pair like in type
$B$ for odd $\ell$.

In case that $\underline{a}\in\mathcal{S}_{\mathbb{H}}$ and
$\nicefrac{{\ell}}{{2}}$ is odd, we again have a non-standard operator for the
fixed index $\overline{0}$ and a pair of $\Theta$-linked indices with
$\overline{\nicefrac{{\ell}}{{4}}}$.

We skip the analogue of Lemma 5.10 in this case, it is just a combination of
Lemmas 4.9 and 5.10 depending on whether a generator for a fixed index is
defined as in type $C$ or as in type $B$. Thus we get the corresponding
actions

######  Proposition 5.23.

Assume $\ell$ even and
$\mathrm{I}\in\\{\mathbb{Z}/(\nicefrac{{\ell}}{{2}})\mathbb{Z},\mathbb{H}/(\nicefrac{{\ell}}{{2}})\mathbb{Z}\\}$.
The linear operators
$\\{\hat{\mathrm{B}}_{\overline{p}}\mid\overline{p}\in\mathrm{I}\\}$ and
$\\{\hat{\mathrm{L}}_{\overline{p}}\mid\overline{p}\in\mathrm{I},\Theta(\overline{p})\neq\overline{p}\\}$
satisfy the relations of $B_{v}(\mathrm{I},\Theta)$ given in Definition 3.14
by substituting the operators for the generators with the same name.

Depending on the definition of the operators this follows either from the
situation in type $C$ or from the case $\ell$ odd in type $B$. Then the
corresponding theorems for the action and the decomposition of the tensor
product can then be immediately formulated as follows.

######  Theorem 5.24.

Assume $\ell$ even. There exists an action of
$B_{v}(\mathbb{H}/(\nicefrac{{\ell}}{{2}})\mathbb{Z},\Theta)$ on
$\mathcal{F}_{N}$ such that for $\lambda\in X^{1,+}$ the decomposition of
$[\Delta_{\mathrm{q}}(\lambda)\otimes\Delta_{\mathrm{q}}(\omega_{1})]$ in
$[U_{\mathrm{q}}\text{-}\mathrm{mod}]$ with respect to the classes of Weyl
modules is obtained from

$\sum_{\overline{p}\in\mathbb{H}/(\nicefrac{{\ell}}{{2}})\mathbb{Z}}\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}+\underline{\lambda}$

by projecting onto the subspace $\mathcal{F}(C_{N})$ and evaluating $v=1$.

Furthermore if $[\Delta_{\mathrm{q}}(\mu)]$ and $[\Delta_{\mathrm{q}}(\nu)]$
appear in the decomposition for $\mu$ and $\nu$ linked, then there exists a
unique $\overline{p}\in\mathbb{H}/\ell\mathbb{Z}$ such that $\underline{\mu}$
and $\underline{\nu}$ appear in
$\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$.

###### Proof.

This is the exact analogue of Theorem 5.14. But in case of
$\nicefrac{{\ell}}{{2}}$ even there is no operator that produces a multiple of
$\underline{\lambda}$ and in case of $\nicefrac{{\ell}}{{2}}$ odd the linear
operator for the fixed index is standard, i.e. also does not produce a
multiple of $\underline{\lambda}$. But since $\lambda\in X^{1,+}$, the tensor
product always includes the class of $[\Delta_{\mathrm{q}}(\lambda)]$ in the
Grothendieck group, hence one has to naively add it. ∎

And the analogue of Theorem 5.15 is the following.

######  Theorem 5.25.

Assume $\ell$ even. There exists an action of
$B_{v}(\mathbb{Z}/(\nicefrac{{\ell}}{{2}})\mathbb{Z},\Theta)$ on
$\mathcal{F}_{N}^{\nicefrac{{1}}{{2}}}$ such that for $\lambda\in
X^{\nicefrac{{1}}{{2}},+}$ the decomposition of
$[\Delta_{\mathrm{q}}(\lambda)\otimes\Delta_{\mathrm{q}}(\omega_{1})]$ in
$[U_{\mathrm{q}}\text{-}\mathrm{mod}]$ with respect to the classes of Weyl
modules is obtained from

$\sum_{\overline{p}\in\mathbb{Z}/(\nicefrac{{\ell}}{{2}})\mathbb{Z}}\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$

if $\lambda_{N}>0$ and from

$\sum_{\overline{p}\in\mathbb{Z}/(\nicefrac{{\ell}}{{2}})\mathbb{Z},\overline{p}\neq\overline{0}}\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}+\hat{\mathrm{B}}_{\overline{0}}^{[\nicefrac{{1}}{{2}}]}\underline{\lambda}$

if $\lambda_{N}=0$, by projecting onto the subspace $\mathcal{F}(C_{N})$ and
evaluating $v=1$.

If $[\Delta_{\mathrm{q}}(\mu)]$ and $[\Delta_{\mathrm{q}}(\nu)]$ appear in the
decomposition for $\mu$ and $\nu$ linked, then there exists a unique
$\overline{p}\in\mathbb{Z}/\ell\mathbb{Z}$ such that $\underline{\mu}$ and
$\underline{\nu}$ appear in
$\hat{\mathrm{B}}_{\overline{p}}\underline{\lambda}$.

Note in case of $\nicefrac{{\ell}}{{2}}$ even and for $\lambda\in
X^{\nicefrac{{1}}{{2}},+}$, there exist the operator
$\hat{\mathrm{B}}_{\overline{\nicefrac{{\ell}}{{4}}}}$ that has a fixed index,
but it is of the same form as in type $C$, hence does not produce a multiple
$\underline{\lambda}$. Hence only
$\hat{\mathrm{B}}_{\overline{0}}\underline{\lambda}$ needs to be replaced by
$\hat{\mathrm{B}}_{\overline{0}}^{[\nicefrac{{1}}{{2}}]}\underline{\lambda}$.

We skip the discussion of an analogues of Proposition 4.14 and Proposition
5.16. The construction is analogous to type $C$ for the space
$\mathcal{F}_{k}$ and analogous, including the same manipulations, as in type
$B$ (with $\ell$ odd) for $\mathcal{F}_{k}^{\nicefrac{{1}}{{2}}}$.

##  6\. Type D and beyond

The type $D$ case is part of the second authors Master thesis and the details
will be published separately. We give a rough summary of what happens in this
situation.

In type $D_{N}$ ($N>3$), the choices of $\varepsilon_{i}$ and the invariant
bilinear form are precisely the same as in type $B$. The positive roots can be
chosen as

$\Phi^{+}=\\{\beta_{i,j}^{\pm}=\varepsilon_{i}\pm\varepsilon_{j}\mid 1\leq
i<j\leq N\\}.$

Note that all roots are short and hence equal to their own coroots and
$\ell_{\beta}=\ell$ for all positive roots. Integral weights have a similar
structure as in type $B$ with

$\displaystyle X^{1,+}$
$\displaystyle=\\{(\lambda_{1},\ldots,\lambda_{N})\mid\lambda_{i}\in\mathbb{Z},\lambda_{i}\geq\lambda_{i+1},\text{
and }\lambda_{N-1}\geq|\lambda_{N}|\\}\text{ and}$ $\displaystyle
X^{\nicefrac{{1}}{{2}},+}$
$\displaystyle=\\{(\lambda_{1},\ldots,\lambda_{N})\mid\lambda_{i}\in\mathbb{H},\lambda_{i}\geq\lambda_{i+1},\text{
and }\lambda_{N-1}\geq|\lambda_{N}|\\}.$

With
$\rho=(N-1)\varepsilon_{1}+(N-2)\varepsilon_{2}+\ldots+\varepsilon_{N-1}\in
X^{\nicefrac{{1}}{{2}},+}$ we thus get

$\displaystyle X^{1,+}_{\rho}$
$\displaystyle=\\{(\underline{\lambda}_{1},\ldots,\underline{\lambda}_{N})\mid\underline{\lambda}_{i}\in\mathbb{H},\,\underline{\lambda}_{i}>\underline{\lambda}_{i+1},\text{
and }\lambda_{N-1}\geq|\lambda_{N}|\\},$ $\displaystyle
X^{\nicefrac{{1}}{{2}},+}_{\rho}$
$\displaystyle=\\{(\underline{\lambda}_{1},\ldots,\underline{\lambda}_{N})\mid\underline{\lambda}_{i}\in\mathbb{Z},\,\underline{\lambda}_{i}>\underline{\lambda}_{i+1},\text{
and }\lambda_{N-1}\geq|\lambda_{N}|\\}.$

In analogy to type $B$, the combinatorial Fock space of type $D$ decomposes
into two summands $\mathcal{F}^{\nicefrac{{1}}{{2}}}(D_{N})$ and
$\mathcal{F}^{1}(D_{N})$. Furthermore the tensor product decomposition in type
$D$ is the same as in type $C$ in Proposition 4.1, especially there is no
complication with a $\Delta_{v}(\lambda)$ summand as in type $B$.

######  Remark 6.1.

The condition for $\lambda$ to be dominant does not depend on $\lambda_{N}$
itself, but rather on its absolute value. Thus
$(\lambda_{1},\ldots,\lambda_{N-1},\lambda_{N})+\varepsilon_{N}$ is dominant
if and only if
$(\lambda_{1},\ldots,\lambda_{N-1},-\lambda_{N})-\varepsilon_{N}$ is dominant.
In terms of sequences this can be interpreted as saying that a $1$ should be
placed at position $|\underline{\lambda}_{N}|$ and the sign of $\lambda_{N}$
needs to be recorded as well.

In case of $X^{\nicefrac{{1}}{{2}},+}_{\rho}$, following Remark 6.1 one sees
that one can embed $\mathcal{F}^{\nicefrac{{1}}{{2}}}(D_{N})$ into
$\mathcal{F}_{N}^{\nicefrac{{1}}{{2}},+}\oplus\mathcal{F}_{N}^{\nicefrac{{1}}{{2}},-}$,
where both spaces are isomorphic to $\mathcal{F}_{N}^{\nicefrac{{1}}{{2}}}$.
For this let $\underline{\lambda}^{\prime}$ be equal to $\underline{\lambda}$
except $\underline{\lambda}^{\prime}_{N}=|\underline{\lambda}_{N}|$. Then
$\underline{\lambda}$ is mapped to $(\underline{a}_{\underline{\lambda}},0)$
if $\underline{\lambda}_{N}>0$ and to
$(0,\underline{a}_{\underline{\lambda}^{\prime}})$ otherwise. The moving
operators from Definition 3.2 are defined on
$\mathcal{F}_{N}^{\nicefrac{{1}}{{2}},+}\oplus\mathcal{F}_{N}^{\nicefrac{{1}}{{2}},-}$,
since they are defined on $\mathcal{F}_{N}^{\nicefrac{{1}}{{2}}}$.

The interplay between moving operators and linkage, is analogous to types $C$.
Hence the operators $\hat{\mathrm{B}}_{\overline{p}}$ and
$\hat{\mathrm{L}}_{\overline{p}}$ are given as in Definition 4.8, up to
changing the index set from $\mathbb{H}/\ell\mathbb{Z}$ to
$\mathbb{Z}/\ell\mathbb{Z}$, and give an action of
$B_{v}(\mathbb{Z}/\ell\mathbb{Z},\Theta)$.

In addition, one introduces an operator $\hat{\mathrm{B}}$ that goes between
$\mathcal{F}_{N}^{\nicefrac{{1}}{{2}},+}$ and
$\mathcal{F}_{N}^{\nicefrac{{1}}{{2}},-}$, since
$(\lambda_{1},\ldots,\lambda_{N-1},\nicefrac{{1}}{{2}})-\varepsilon_{N}=(\lambda_{1},\ldots,\lambda_{N-1},-\nicefrac{{1}}{{2}})$,
but both are represented by the same sequence, but in different components.
One defines $\hat{\mathrm{B}}(\underline{a},0)=(0,\underline{a})$ if
$\underline{a}(\nicefrac{{1}}{{2}})=1$ and the operator is zero otherwise, and
similarly $\hat{\mathrm{B}}(0,\underline{a})=(\underline{a},0)$ if
$\underline{a}(\nicefrac{{1}}{{2}})=1$ and zero otherwise. It follows that
only $\hat{\mathrm{B}}$ and $\hat{\mathrm{B}}_{\overline{0}}$ can produce
linked weights, hence one needs to consider
$\hat{\mathrm{B}}_{\overline{0}}+\hat{\mathrm{B}}$.

The analogue of Theorem 5.15 holds, with the difference that
$\hat{\mathrm{B}}_{\underline{0}}$ is replaced by
$\hat{\mathrm{B}}_{\overline{0}}+\hat{\mathrm{B}}$. The limit construction
from Proposition 4.14 does not work in type $D$ as one cannot regard a
dominant weight $\lambda$ with $\lambda_{N}<0$ as a dominant weight for a
bigger rank. If one restricts to this case then one essentially recovers the
combinatorics of type $C$.

The case of $X^{1,+}_{\rho}$ brings additional complications with it, since
$\lambda_{N}=0$ does not allow for a “good” embedding into a sum of Fock
spaces $\mathcal{F}^{1,+}\oplus\mathcal{F}^{1,-}$, where these spaces are
defined as in the half-integer case. One instead embeds $\underline{\lambda}$
with $\lambda_{N}=0$ diagonally. This allows for an analogue of Theorem 4.12,
but depending on the starting weight, $(\underline{\lambda},0)$,
$(0,\underline{\lambda})$ or $(\underline{\lambda},\underline{\lambda})$
represents the class of the Weyl module $[\Delta_{\mathrm{q}}(\lambda)]$ and
similar for the occurring Weyl modules in the decompositions. The limit
construction does not work for the same reason as above.

######  Remark 6.2.

As a concluding remark in type $D$ one can note that the combinatorial
calculations are easier than in type $B$ and $C$, since it is of simply-laced
type. But the action of the quantum symmetric pair is further away from the
combinatorics of the tensor product multiplicities and the limit construction
is not possible.

This kind of complication in type $D$ has some analogy in [ES18]. Here the
type $D$ situation is investigated in terms of category $\mathcal{O}$ and
certain generators of the quantum symmetric pair do not act as indecomposable
translation functors, while in the analogous situation in type $B$ they would.

We focused in this paper on non-exceptional Lie algebras. The general ideas
can of course be transferred to exceptional cases as well. There is in general
no clear analogue of the representation $\Delta_{\mathrm{q}}(\omega_{1})$,
i.e. a “natural” representation. For example, in type $G_{2}$ one can use the
specialization and quantization of the natural representation of
$\mathfrak{so}_{8}(\mathbb{C})$, since $\mathfrak{so}_{8}(\mathbb{C})$
contains the Lie algebra of type $G_{2}$ as a fixed point Lie algebra. But
there is nothing analogous for the other exceptional types.

As for the analogue of the quantum symmetric pair, it is reasonable to expect
that in type $G_{2}$ the acting algebra has some relationship to the quantum
symmetric pair and an automorphism of order $3$. This is in analogy to the
fact that the quantum symmetric pair is constructed from the quantum affine
algebra using an automorphism of order $2$. It is a priori not clear what the
corresponding object in other exceptional cases would look like or come from.

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