*3.4. Counting Paths in the Auxiliary Lattice Path Model*

Consider the lattice path model for the set of paths on L descending from (0, 0) to (*M*, *N*) with steps S in the presence of restrictions W*<sup>L</sup>* <sup>0</sup> , F<sup>1</sup> *<sup>l</sup>*−1, F<sup>2</sup> *nl*−1, *n* ∈ N, *n* ≥ 2. Such set is denoted as

$$L\_N((0,0)\to(M,N);\mathbb{B}\mid\mathcal{W}\_0^L,\mathcal{F}\_{l-1}^1,\{\mathcal{F}\_{nl-1}^2\}\_{n=2}^\infty).$$

and such a model is called the auxiliary lattice path model. The main theorem of [13] gives an explicit formula for weighted numbers of paths in the auxiliary lattice path model. Then, in [9], periodicity conditions (*M* + 2*l*, *N*)=(*M*, *N*), *M*, *N* ≥ *l* − 1 were applied, resulting in a folded Brattelli diagram. For such a diagram, recursion on the weighted numbers of paths coincides with recursion on multiplicities of indecomposable *uq*(*sl*2)-modules in tensor product decomposition of *T*(1)⊗*N*. Note that due to properties of the category **Rep**(*uq*(*sl*2)), we mostly considered odd values of *l*; however, the results remain to be true for even values of *l* as well.

Before coming to modifications of the auxiliary lattice path model relevant to the representation theory of *Uq*(*sl*2) at the roots of unity, we need to slightly tweak it. We are interested in paths descending from (0, 0) to (*M*, *N*) with steps S in the presence of restrictions W*<sup>L</sup>* <sup>0</sup> , F<sup>1</sup> *nl*−1, *n* ∈ N, instead of filters of type 2. Such lattice path model is depicted in Figure 4.

**Figure 4.** Arrangement of steps for points of L in the considered, slightly tweaked version of the auxiliary lattice path model. Here, we depict the case, where *l* = 5.

**Definition 3.** *We denote by multiplicity function in the j-th strip M<sup>j</sup>* (*M*,*N*) *the weighted number of paths in set*

$$L\_N((0,0)\to(M,N);\mathbb{S}\mid\mathcal{W}\_0^L,\{\mathcal{F}\_{nl-1}^1\},n\in\mathbb{N})$$

*with the endpoint* (*M*, *N*) *that lies within* (*j* − 1)*l* − 1 ≤ *M* ≤ *jl* − 2

$$M\_{(M,N)}^j = Z(L\_N((0,0)\to(M,N); \mathbb{S}\mid \mathcal{W}\_0^L, \{\mathcal{F}\_{nl-1}^1\}, n \in \mathbb{N})),\tag{5}$$

*where M* ≥ 0 *and j* = @ *M*+1 *<sup>l</sup>* + 1 A *.*

Now consider the version of the main theorem in [13] corresponding to this model.

**Theorem 1** ([13])**.** *The multiplicity function in the j-th strip is given by*

$$\begin{aligned} M\_{(M,N)}^{j} &= \sum\_{k=0}^{\left[\frac{N-(j-1)l+1}{4l}\right]} P\_{\bar{j}}(k) F\_{M+4kl}^{(N)} + \sum\_{k=0}^{\left[\frac{N-jl}{4l}\right]} P\_{\bar{j}}(k) F\_{M-4kl-2jl}^{(N)} - \\ &- \sum\_{k=0}^{\left[\frac{N-(j+1)l+1}{4l}\right]} Q\_{\bar{j}}(k) F\_{M+2l+4kl}^{(N)} - \sum\_{k=0}^{\left[\frac{N-jl-2l}{4l}\right]} Q\_{\bar{j}}(k) F\_{M-4kl-2(j+1)l}^{(N)} \end{aligned}$$

*where*

$$P\_j(k) = \sum\_{i=0}^{\left[\frac{j}{2}\right]} \binom{j-2}{2i} \binom{k-i+j-2}{j-2}, \qquad Q\_j(k) = \sum\_{i=0}^{\left[\frac{j}{2}\right]} \binom{j-2}{2i+1} \binom{k-i+j-2}{j-2}, \tag{6}$$

$$F\_M^{(N)} = \binom{N}{\frac{N-M}{2}} - \binom{N}{\frac{N-M}{2}-1}.$$

**Proof.** The proof is the same as the proof of the main theorem in [13], except that instead of Lemma 4.9 in [13], for the slightly tweaked model one should use Lemma 3.

From now on, when mentioning the auxiliary lattice path model, we mean its slightly tweaked version. This model will be further modified in subsequent sections. Instead of applying periodicity conditions, as for *uq*(*sl*2), we enhance this model with long steps, source and target points which are located near filters. As a result, recursion for the weighted numbers of paths on the resultant Bratteli diagram recreates recursion for multiplicities of indecomposable *Uq*(*sl*2)-modules in the decomposition of *T*(1)⊗*N*.

#### **4. Boundary Points and Congruent Regions**

In this section, we consider notions, which are convenient for counting paths in the auxiliary lattice path model in the presence of long steps. We will see, that multiplicities on the boundary of a region uniquely define multiplicities in the rest of the region. For proving identities between multiplicities in two congruent regions, it is sufficient to prove such identities for their boundary points.

**Definition 4.** *Consider the lattice path model, defined by a set of steps* S *and a set of restrictions* C *on lattice* L*. Subset* L<sup>0</sup> ⊂ L *with steps* S *and restrictions* C *is called a region of the lattice path model under consideration.*

Intuitively, region L<sup>0</sup> ⊂ L is a restriction of the lattice path model defined by S, C on lattice L to the subset L0. The word 'restriction' is overused, so we consider regions of lattice path models instead.

**Definition 5.** *Consider* L<sup>0</sup> ⊂ L *a region of the lattice path model defined by steps* S *and restrictions* C*. Point B* ∈ L<sup>0</sup> *is called a boundary point of* L<sup>0</sup> *if there exists B* ∈ L*, B* ∈ L / <sup>0</sup> *such that step B* → *B is allowed in* L *by a set of steps* S *and restrictions* C*. The union of all such points is a boundary of* L<sup>0</sup> *and is denoted by ∂*L0*.*

The Definition 5 introduces a notion, reminiscent of the outer boundary in graph theory. Note that boundary points are defined with respect to some lattice path models under consideration. For brevity, we assume that this lattice path model is known from the context, and mentioning it will be mostly omitted.

**Example 1.** *For a strip in the auxiliary lattice path model, its boundary is in the left filter. It is depicted in Figure 5.*

**Figure 5.** Region L0, highlighted with blue dashed lines, is a 2nd strip for *l* = 5. Its boundary *∂*L<sup>0</sup> is a set of points in the left filter restriction F<sup>1</sup> *<sup>l</sup>*−1, which is highlighted with purple dashed lines.

**Example 2.** *Consider region* L<sup>0</sup> *of the unrestricted lattice path model, as depicted in Figure 6 and highlighted with blue dashed lines. Its boundary is a set of points highlighted with purple dashed lines.*

**Figure 6.** Region L<sup>0</sup> is highlighted with blue dashed lines. Its boundary *∂*L<sup>0</sup> is a set of points highlighted with purple dashed lines.

**Lemma 4.** *Consider region* L<sup>0</sup> *of a lattice path model defined by* S*,* C *on lattice* L*. Weighted numbers of paths Z*(*LN*((0, 0) → (*M*, *N*); ...) *for* (*M*, *N*) ∈ L<sup>0</sup> *are uniquely defined by weighted numbers of paths for its boundary points ∂*L0*.*

**Proof.** Suppose weighted numbers of paths for *∂*L<sup>0</sup> are known. Suppose that there exists some point *A* ∈ L0, such that its weighted number of paths cannot be expressed in terms of weighted numbers of paths for points in *∂*L0.

The first case is that recursion for a weighted number of paths for *A* involves some point *A* ∈ L0, a weighted number of paths for which cannot be expressed in terms of such for points in *∂*L0. In this case, we need to consider *A* and recursion on the weighted number of paths for such a point instead of *A*.

The second case is that recursion for a weighted number of paths for *A* involves a weighted number of paths for some point *A* ∈ L / 0. Then, *A* ∈ *∂*L<sup>0</sup> by definition of a boundary point and weighted number of paths for such point is known by the initial supposition of the lemma.

Note that due to the fact that we consider descending paths, *M* and *N*, to be finite, the first case can be iterated finitely many times at most.

**Definition 6.** *Consider two lattice path models with steps* S1*,* S<sup>2</sup> *and restrictions* C1*,* C<sup>2</sup> *defined on lattice* L*. Subset* L<sup>1</sup> ⊂ L *is a region in the lattice path model defined by* S1, C1*. Subset* L<sup>2</sup> ⊂ L *is a region in the lattice path model defined by* S2, C2*. Regions* L<sup>1</sup> *and* L<sup>2</sup> *are congruent if there exists a translation T in* L *such that*


The second condition can be written down explicitly. Firstly, for each (*M*, *N*) ∈ L<sup>1</sup> and each step (*M*, *N*) *<sup>w</sup>* −→ (*P*, *Q*) in S<sup>1</sup> obeying C<sup>1</sup> such that (*P*, *Q*) ∈ L1, there is a step (*M* , *N* ) *<sup>w</sup>* −→ (*P* , *Q* ) in S<sup>2</sup> obeying C2, where *T*(*M*, *N*)=(*M* , *N* ), *T*(*P*, *Q*)=(*P* , *Q* ). Secondly, for each (*M* , *N* ) ∈ L<sup>2</sup> and each step (*M* , *N* ) *<sup>w</sup>* −→ (*P* , *Q* ) in S<sup>2</sup> obeying C<sup>2</sup> such that (*P* , *Q* ) ∈ L2, there is a step (*M*, *<sup>N</sup>*) *<sup>w</sup>* −→ (*P*, *Q*) in S<sup>1</sup> obeying C1, where *T*−1(*M* , *N* ) = (*M*, *N*), *T*−1(*P* , *Q* )=(*M*, *N*). To put it simply, if we forget about lattice path models outside L<sup>1</sup> and L2, these two regions will be indistinguishable. Due to translations in L being invertible, it is easy to see that congruence defines an equivalence relation.

Now we must prove the main theorem of this subsection.

**Theorem 2.** *Consider two lattice path models with steps* S1*,* S<sup>2</sup> *and restrictions* C1*,* C<sup>2</sup> *defined on lattice* L*. Region* L<sup>1</sup> *of the lattice path model defined by* S1*,* C<sup>1</sup> *is congruent to region* L<sup>2</sup> *of the lattice path model defined by* S2*,* C2*, where T*L<sup>1</sup> = L2*. If equality*

$$Z(L\_N((0,0)\to(M,N);\mathbb{S}\_1\mid\mathcal{C}\_1))=Z(L\_N((0,0)\to T(M,N);\mathbb{S}\_2\mid\mathcal{C}\_2))\tag{7}$$

*holds for all* (*M*, *N*) ∈ *∂*L<sup>1</sup> ∪ *T*−1(*∂*L2)*, then it holds for all* (*M*, *N*) ∈ L1*.*

Note, that if (*M*, *N*) ∈ *∂*L<sup>1</sup> it does not necessarily follow that *T*(*M*, *N*) ∈ *∂*L2, due to C<sup>1</sup> and C<sup>2</sup> being different. So, it is natural to ask Formula (7) to hold for *∂*L<sup>1</sup> ∪ *T*−1(*∂*L2).

**Proof.** We need to prove that Formula (7) is true for (*M*, *N*) ∈ L1. The l.h.s. can be uniquely expressed in terms of its values at *∂*L<sup>1</sup> ∪ *T*−1(*∂*L2), following procedure in Lemma 4. Due to the congruence between L<sup>1</sup> and L2, recursion for the r.h.s. of (7) coincides with the one for the l.h.s., so we can obtain the same expression on the r.h.s., but with values of weighted numbers of paths for *T*(*∂*L<sup>1</sup> ∪ *T*−1(*∂*L2)) = *T*(*∂*L1) ∪ *∂*L<sup>2</sup> instead of *∂*L<sup>1</sup> ∪ *T*−1(*∂*L2). We can compare the l.h.s. and the r.h.s. term by term, for points related by translation *T*. All of such terms have the same values due to the initial supposition of the theorem.

**Corollary 1.** *Consider lattice path models with steps* S1*,* S2*,* S<sup>3</sup> *and restrictions* C1*,* C2*,* C<sup>3</sup> *defined on lattice* L*. Region* L<sup>1</sup> *is congruent to* L<sup>2</sup> *and* L3*, where T*1(*M*1, *N*1)=(*M*2, *N*2)*, T*2(*M*1, *N*1) = (*M*3, *N*3) *for* (*M*1, *N*1) ∈ L1*. If equality*

$$Z(L\_N((0,0)\to(M,N);\mathbb{S}\_1\mid\mathcal{E}\_1))=$$

$$Z = Z(L\_N((0,0)\to T\_1(M,N);\mathbb{S}\_2\mid\mathcal{E}\_2)) + Z(L\_N((0,0)\to T\_2(M,N);\mathbb{S}\_3\mid\mathcal{E}\_3))\tag{8}$$

*holds for all* (*M*, *N*) ∈ *∂*L<sup>1</sup> ∪ *T*−<sup>1</sup> <sup>1</sup> (*∂*L2) <sup>∪</sup> *<sup>T</sup>*−<sup>1</sup> <sup>2</sup> (*∂*L3)*, then it holds for all* (*M*, *N*) ∈ L1*.*

**Proof.** Due to linearity of the r.h.s. of Formula (8), the proof repeats the one of Theorem 2.

The moral of this section is that for two congruent regions, weighted numbers of paths are defined by values of such at the boundary of the considered regions. For proving identities, it is sufficient to establish equality for weighted numbers of paths at boundary points, while equality for the rest of the region will follow due to the congruence.

## **5. The Auxiliary Lattice Path Model in the Presence of Long Steps**

*Long Steps in Lattice Path Models with Filter Restrictions*

Long step is a step (*x*, *y*) *<sup>w</sup>* −→ (*x* , *y* + 1) in L such that |*x* − *x* | > 1. We denote the sequence of long steps as

$$\mathbb{E}[M\_1, M\_2] = \{ (M\_1, M\_1 + 2m) \to (M\_2, M\_1 + 1 + 2m) \}\_{m = 0 \prime}^{\infty}$$

where *x* = *M*<sup>1</sup> is the source point for the sequence and *x* = *M*<sup>2</sup> is the target point, |*M*<sup>1</sup> − *M*2| > 1. For the purposes of this paper, we are mainly interested in sequences

$$\mathbb{S}(k) \equiv \mathbb{S}[l(k+2) - 2, lk - 1] = \{ (l(k+2) - 2, lk - 2 + 2m) \to (lk - 1, lk - 2 + 1 + 2m) \}\_{m = 0}^{\infty}$$

where *k* ∈ N and C consists of F<sup>1</sup> *lk*−<sup>1</sup> and F<sup>1</sup> *<sup>l</sup>*(*k*+2)−1. We need such sequences of long steps for modification of the auxiliary lattice path model, relevant to the representation theory of *Uq*(*sl*2) at roots of unity.

**Lemma 5.** *Fix k* ∈ N*. Let*

$$Z\_{(M,N)} \equiv Z(L\_N((0,0)\to(M,N)); \mathbb{S} \mid \mathcal{F}\_{lk-1\prime}^1 \mathcal{F}\_{l(k+2)-1}^1)$$

*be the weighted number of lattice paths from* (0, 0) *to* (*M*, *N*) *with filter restrictions* F<sup>1</sup> *lk*−1*,* F<sup>1</sup> *<sup>l</sup>*(*k*+2)−<sup>1</sup> *and set of unrestricted elementary steps* S*. Let*

$$Z\_{(M,N)}^{\prime} \equiv Z(L\_N((0,0)\to(M,N)); \mathbb{S}\cup\mathbb{S}(k)\mid \mathcal{F}\_{lk-1\prime}^{1}\mathcal{F}\_{l(k+2)-1}^{1})$$

*be the weighted number of lattice paths from* (0, 0) *to* (*M*, *N*) *with the same restrictions, with steps* S ∪ S(*k*)*. Then for lk* − 1 ≤ *M* ≤ *l*(*k* + 2) − 2 *we have*

$$Z\_{(M,N)}' = Z\_{(M,N)'} \quad \text{if } N \le M + 2l - 2,\tag{9}$$

$$Z\_{(M,N)}' = Z\_{(M,N)} + Z\_{(M+2l,N)'} \quad \text{if } M+2l \le N \le l(k+4)-2. \tag{10}$$

**Proof.** In Figure 7, we depict the setting of the Lemma 5. Long steps do not impact region I, so Formula (9) is true.

**Figure 7.** By square and circle we denote points, where long steps first appear. Regions I and II highlighted with blue dashed lines correspond to cases *N* ≤ *M* + 2*l* − 2, as in (9), and *M* + 2*l* ≤ *N* ≤ *l*(*k* + 4) − 2, as in (10).

Consider Formula (10). The weighted number of paths in the l.h.s. involves points from region II. Its boundary contains points of the left cathetus of region II, of the form (*lk* − 1, *N*) for *l*(*k* + 2) − 1 ≤ *N* ≤ *l*(*k* + 4) − 2, and points of the hypotenuse of the region II, of the form (*lk* − 1 + *j*, *l*(*k* + 2) − 1 + *j*) for *j* = 1, ... , 2*l* − 1. Denote this set by *∂*L*I I*. The r.h.s. of (10) has two terms. The first involves region II, the boundary of which we have already considered. The second term involves points of the region congruent to region II, as they are related by translation *T*(*M*, *N*)=(*M* + 2*l*, *N*), satisfying Definition 6. Its boundary consists of the image of the left cathetus of region II under translation *T*. Denote this set by *∂*L *I I*. By Corollary 1, it is sufficient to prove Formula (10) for *<sup>∂</sup>*L*I I* <sup>∪</sup> *<sup>T</sup>*−1(*∂*L *I I*) = *∂*L*I I*.

We proceed by induction over *n*, where *N* = *l*(*k* + 2) − 1 + 2*n*. For *n* = 0 from recursion we have

$$Z\_{(lk-1,l(k+2)-1)}^{l} = Z\_{(lk-2,l(k+2)-2)} + 2Z\_{(lk,l(k+2)-2)} + Z\_{(l(k+2)-2,l(k+2)-2)} \tag{11}$$

which, taking into account that

$$Z\_{(lk-2,l(k+2)-2)} + 2Z\_{(lk,l(k+2)-2)} = Z\_{(lk-1,l(k+2)-1)},$$

$$Z\_{(l(k+2)-2,l(k+2)-2)} = Z\_{(l(k+2)-1,l(k+2)-1)}.$$

gives us

$$Z\_{(lk-1,l(k+2)-1)}^{l} = Z\_{(lk-1,l(k+2)-1)} + Z\_{(l(k+2)-1,l(k+2)-1)}.\tag{12}$$

We obtained the base of induction.

In a similar manner, it also follows, that Formula (10) is true for boundary points of the hypotenuse of region II. In order to show this, one must consider recursion explicitly and use the fact that

$$Z\_{(j,j)} = Z\_{(k,k)}, \quad \text{for all } j,k \gg 0. \tag{13}$$

Now it is sufficient to prove Formula (10) for boundary points, situated in the left cathetus of region II.

Suppose

$$Z\_{(lk-1,l(k+2)-1+2n)}' = Z\_{(lk-1,l(k+2)-1+2n)} + Z\_{(l(k+2)-1,l(k+2)-1+2n)} \tag{14}$$

is true. For the sake of brevity, we rewrite this expression as

$$Z\_{(p,q+2n)}^{\prime} = Z\_{(p,q+2n)} + Z\_{(q,q+2n)\prime} \tag{15}$$

where *p* = *lk* − 1, *q* = *l*(*k* + 2) − 1, *q* = *p* + 2*l*. By Theorem 2, it follows that Formula (10) is true for the region, corresponding to boundary points, covered by the inductive supposition. In particular, this region includes points (*p* + *j*, *q* + 2*n* + *j*) for *j* = 0, ... , 2*l* − 1. Need to prove that

$$Z\_{(p,q+2(n+1))}^{\\'} = Z\_{(p,q+2(n+1))} + Z\_{(q,q+2(n+1))} \tag{16}$$

Taking into account, that

$$\begin{aligned} Z'\_{(p,q+2n+2)} &= Z\_{(p-1,q+2n+1)} + 2Z'\_{(p+1,q+2n+1)} + Z\_{(q-1,q+2n+1)}, \\ Z\_{(p,q+2n+2)} &= Z\_{(p-1,q+2n+1)} + 2Z\_{(p+1,q+2n+1)}, \\ Z\_{(q,q+2n+2)} &= Z\_{(q-1,q+2n+1)} + 2Z\_{(q+1,q+2n+1)}. \end{aligned}$$

after getting rid of the factors, we obtain

$$Z\_{(p+1,q+2n+1)}^{!} = Z\_{(p+1,q+2n+1)} + Z\_{(q+1,q+2n+1)}.\tag{17}$$

However, this is true from the inductive supposition.

Note that Formula (10) is not true for greater values of *N*. Region II indeed can be made into a parallelogram, similar to region I, since the set of boundary points will remain the same. However, the region corresponding to this parallelogram being translated by *T* contains new boundary points, where (10) does not hold and Corollary 1 cannot be used further, even though these regions are congruent to each other. The formula for greater values of *N* needs to include some new terms. In this parallelogram-like region, we need to take into account the reflection of paths, induced by the term *Z*(*M*+2*l*,*N*) in *Z* (*M*,*N*) , from the filter restriction F<sup>1</sup> *<sup>l</sup>*(*k*+2)−1. This is achieved by means of the first part of Lemma 3. Now consider the triangular region, which, similarly to region II being below region I, is below the parallelogram-like region considered previously. There, we need to take into account long steps, acting on paths induced by the term *Z*(*M*+2*l*,*N*), which have descended to (*l*(*k* + 2) − 2, *N*) and were acted upon by long steps for the second time. This is being conducted in a similar fashion to Corollary 1, where *Z*(*M*+2*l*,*N*) is assumed to be known from the second part of Lemma 3. This situation for the case of the auxiliary lattice path model in the presence of long steps will be elaborated upon later.

**Corollary 2.** *Fix j*, *k* ∈ N*, j* ≤ *k. Let*

$$Z\_{(M,N)} \equiv Z(L\_N((0,0)\to(M,N)); \mathbb{S} \mid \mathcal{W}\_0^L \; \{\mathcal{F}\_{nl-1}^1\}\_{n=j}^\infty),$$

*be the weighted number of lattice paths from* (0, 0) *to* (*M*, *N*) *with filter restrictions* {F<sup>1</sup> *nl*−1}<sup>∞</sup> *n*=*j and set of unrestricted elementary steps* S*. Let*

$$Z\_{(M,N)}^{\\'} \equiv Z(L\_N((0,0)\to(M,N)); \mathbb{S}\cup\mathbb{S}(k)\mid\mathcal{W}\_0^L,\{\mathcal{F}\_{nl-1}^1\}\_{n=j}^\infty)$$

*be the weighted number of lattice paths from* (0, 0) *to* (*M*, *N*) *with the same restrictions, with steps* S ∪ S(*k*)*. Then, for lk* − 1 ≤ *M* ≤ *l*(*k* + 2) − 2 *we have*

$$Z\_{(M,N)}' = Z\_{(M,N)}, \quad \text{if } N \le M + 2l - 2,\tag{18}$$

$$Z\_{(M,N)}' = Z\_{(M,N)} + Z\_{(M+2l,N)}, \quad \text{if } M+2l \le N \le l(k+4)-2. \tag{19}$$

**Proof.** The proof is the same, as for Lemma 5. When proving the inductive step, we still can apply Corollary 1 as region II is still congruent to the one, translated by *T*.

Note, that Formula (19), unlike (10), is true for greater values of *N*, as making region II into a parallelogram-like region will not add new boundary points. The manifestation of this fact is that we do not need to take into account the reflection of paths, as they have already been dealt with in term *Z*(*M*+2*l*,*N*) due to the periodicity of filter restrictions. So, for such a region Formula (19) holds. However, for the triangular region below the same problem remains.

Consider the auxiliary lattice path model in the presence of the sequence of steps S(*k*).

**Definition 7.** *We denote by multiplicity function in the j-th strip M*˜ *<sup>j</sup>* (*M*,*N*) *the weighted number of paths in set*

$$L\_N((0,0)\to(M,N);\mathbb{S}\cup\mathbb{S}\mid\mathcal{W}\_0^L,\{\mathcal{F}\_{nl-1}^1\},n\in\mathbb{N})$$

*with the endpoint* (*M*, *N*) *that lies within* (*j* − 1)*l* − 1 ≤ *M* ≤ *jl* − 2

$$\bar{M}\_{(M,N)}^{\dot{l}} = Z(L\_N((0,0)\to(M,N); \mathbb{S}\cup\mathbb{S}\mid\mathcal{W}\_0^L,\{\mathcal{F}\_{nl-1}^1\},n\in\mathbb{N})),\tag{20}$$

*where* S˜ *is a set of some additional steps and M* ≥ 0 *and j* = @ *M*+1 *<sup>l</sup>* + 1 A *.*

In this subsection, S˜ = S(*k*) if not stated otherwise.

**Lemma 6.** *For fixed k* ∈ N

$$\tilde{M}\_{(M,N)}^{k+1} = \sum\_{j=0}^{\lfloor \frac{N-\lfloor k+1 \rfloor}{2l} \rfloor} M\_{(M+2jl,N)'}^{k+1+2j} \tag{21}$$

$$
\tilde{M}\_{(M,N)}^{k+3} = \sum\_{j=0}^{\lfloor \frac{N-l(k+2)+1}{2l} \rfloor} M\_{(M+2jl,N)'}^{k+3+2j} \tag{22}
$$

*where M*˜ *<sup>j</sup>* (*M*,*N*) *is the multiplicity function for j-th strip in the auxiliary model with steps* <sup>S</sup> <sup>∪</sup> <sup>S</sup>˜*, Mj* (*M*,*N*) *is the multiplicity function for j-th strip in the auxiliary model with steps* S*.*

**Proof.** We proceed by induction over *n*, where *n* = [ *<sup>N</sup>*−*l*(*k*+2)+<sup>2</sup> <sup>2</sup>*<sup>l</sup>* ], first proving (22), then (21). For *n* = 0, Formula (22) follows immediately from the Theorem 2, as long steps do not impact this region. Formula (21) follows from Corollary 2. As was discussed, Formula (19) is true for greater values of *N*, mainly, it is true for a parallelogram-like region, satisfying *n* = 0. So, we obtained the base of induction.

Suppose, that

$$
\bar{\mathcal{M}}\_{(M,N)}^{k+1} = \sum\_{j=0}^{n+1} \mathcal{M}\_{(M+2jl,N)}^{k+1+2j} \tag{23}
$$

$$
\tilde{M}\_{(M,N)}^{k+3} = \sum\_{j=0}^{n} M\_{(M+2jl,N)}^{k+3+2j} \tag{24}
$$

is true.

Need to prove the inductive step for (24) first, thus we need to prove (22) for *M* + 2*ln* ≤ *N* ≤ *M* + 2*l*(*n* + 1), where *l*(*k* + 2) − 1 ≤ *M* ≤ *l*(*k* + 3) − 2. Denote this region as L1. The l.h.s. of (22) is a weighted number of paths, *∂*L<sup>1</sup> consists of points (*l*(*k* + 2) − 1, *N*) for *l*(*k* +2) −1+2*ln* ≤ *N* ≤ *l*(*k* +2) −1+2*l*(*n* +2) and (*l*(*k* +2) −1+ *j*, *l*(*k* +2) −1+2*ln* + *j*) for *j* = 0, ... , *l* − 1. We divide the r.h.s. of (22) into two terms. The first corresponds to the sum given by inductive supposition in (24). It is also a weighted number of paths defined for L<sup>2</sup> = L<sup>1</sup> with the same boundary points *∂*L<sup>2</sup> = *∂*L1, as the l.h.s. of (22). These two regions are congruent, *T*<sup>1</sup> = *id*. The second is an additional term, which we expect to appear during an inductive step. It is given by

$$M\_{(M+2(n+1)l,N)}^{k+3+2(n+1)} = Z(L\_N((0,0)\to (M+2(n+1)l,N)); \mathbb{S} \mid \mathcal{W}\_0^l, \{\mathcal{F}\_{ml-1}^1\}\_{m=1}^\infty)$$

for region *l*(*k* + 2 + 2(*n* + 1)) − 1 ≤ *M* ≤ *l*(*k* + 3 + 2(*n* + 1)) − 2 and *M* ≤ *N* ≤ *M* + 2*l*. Denote it by L3. Its boundary *∂*L<sup>3</sup> consists of points (*l*(*k* + 2 + 2(*n* + 1)) − 1, *N*) for *l*(*k* + 2 + 2(*n* + 1)) − 1 ≤ *N* ≤ *l*(*k* + 2 + 2(*n* + 1)) − 1 + 2*l*. This region is an image of L<sup>1</sup> under translation *T*2(*M*, *N*)=(*M* + 2*l*(*n* + 1), *N*), they are congruent. By Corollary 1, it is sufficient to prove inductive step at points (*l*(*k* + 2) − 1, *N*) for *l*(*k* + 2) − 1 + 2*ln* ≤ *N* ≤ *l*(*k* +2) −1+2*l*(*n* +2) and points (*l*(*k* +2) −1+ *j*, *l*(*k* +2) −1+2*ln* + *j*) for *j* = 0, ... , *l* −1. These are drawn in Figure 8.

Consider points of a form (*l*(*k* + <sup>2</sup>) − 1, *N*). At *n*-th iteration we added *Mk*+1+2(*n*+1) (*M*+2(*n*+1)*l*,*N*) to *M*˜ *<sup>k</sup>*+<sup>1</sup> (*M*,*N*) . This term induces paths, which further descend from (*k* + 1)-th strip to boundary points of (*k* + 3)-th strip. The region in which induced paths descend is congruent to the region, where paths corresponding to *Mk*+1+2(*n*+1) (*M*+2(*n*+1)*l*,*N*) continue to descend to the boundary of (*k* + 3 + 2(*n* + 1))-th strip in the auxiliary lattice path model. This is due to the periodicity of filter restrictions. Here, we can apply Theorem 2 to conclude that the weighted number of induced paths arriving at the boundary of (*k* + 3)-th strip is equal to *Mk*+3+2(*n*+1) (2(*n*+1)*l*,*N*) .

Consider points of a form (*l*(*k* + 2) − 1 + *j*, *l*(*k* + 2) − 1 + 2*ln* + *j*). For such points, the proof is the same as for the Formula (10) for the hypotenuse of region II.

Now that we proved the inductive step for the boundary of the considered region, by Corollary 1, it follows that

$$\tilde{M}\_{(M,N)}^{k+3} = \sum\_{j=0}^{n} M\_{(M+2jl,N)}^{k+3+2j} + M\_{(M+2(n+1)l,N)}^{k+3+2(n+1)} = \sum\_{j=0}^{n+1} M\_{(M+2jl,N)}^{k+3+2j} \tag{25}$$

is true for the whole region, which proves the inductive step for Formula (24).

**Figure 8.** Region L1, for which it is sufficient to prove (22) consists of points *M* + 2*ln* ≤ *N* ≤ *M* + 2*l*(*n* + 1) where *l*(*k* + 2) − 1 ≤ *M* ≤ *l*(*k* + 3) − 2. It is highlighted with blue dashed lines. Union of boundaries for all terms of the considered expression *<sup>∂</sup>*L<sup>1</sup> ∪ *<sup>T</sup>*−<sup>1</sup> <sup>1</sup> (*∂*L2) <sup>∪</sup> *<sup>T</sup>*−<sup>1</sup> <sup>2</sup> (*∂*L3) consists of points (*l*(*k* + 2) − 1, *N*) for *l*(*k* + 2) − 1 + 2*ln* ≤ *N* ≤ *l*(*k* + 2) − 1 + 2*l*(*n* + 2) and points (*l*(*k* + 2) − 1 + *j*, *l*(*k* + 2) − 1 + 2*ln* + *j*) for *j* = 0, ... , *l* − 1. It is highlighted with purple dashed lines. Here, we depict the case, where *l* = 5.

This process for the first iterations is shown in Figure 9.

**Figure 9.** Color emphasizes the number of iterations in the induction. Paths induced at the boundary of (*k* + 1)-th strip during (*j* − 1)-th iteration descend in the region, highlighted with color, corresponding to *j*-th iteration. Colored lines outline regions congruent to each other. Dashed colored arrows denote weighted numbers of induced paths, inflicted to (*k* + 3)-th strip once they have descended, and their equivalents in strips of the auxiliary lattice path model.

This figure also shows how long steps act on descended paths corresponding to dashed colored arrows, inducing paths at boundary points of (*k* + 1)-th strip, highlighted with the dashed arrow of the same color. These induced paths, in turn, descend in the region, highlighted with a color corresponding to the next, (*j* + 1)-th iteration. Proving that long steps induce paths at boundary points of (*k* + 1)-th strip following this scenario amounts to proving the inductive step for Formula (23).

Now we need to prove the inductive step for Formula (23), which amounts to proving (21) for *lk* − 1 ≤ *M* ≤ *l*(*k* + 1) − 2 and *M* + 2*ln* ≤ *N* ≤ *M* + 2*l*(*n* + 1). It is being conducted in a fashion similar to the proof of (22). Again, we divide the r.h.s. of (21) in two terms. The first one corresponds to the sum given by inductive supposition in (23). The second one is an additional term, which we expect to appear during an inductive step. It is given by

$$M\_{(M+2(n+1)l,N)}^{k+1+2(n+1)} = Z(L\_N((0,0)\to (M+2(n+1)l,N)); \mathbb{S} \mid \mathcal{W}\_0^L, \{\mathcal{F}\_{ml-1}^1\}\_{m=1}^\infty)$$

for region *l*(*k* + 2(*n* + 1)) − 1 ≤ *M* ≤ *l*(*k* + 1 + 2(*n* + 1)) − 2 and *M* ≤ *N* ≤ *M* + 2*l*. By Corollary 1, it is sufficient to prove inductive step at points (*lk* − 1, *N*) for *lk* − 1 + 2*ln* ≤ *N* ≤ *l*(*k* + 1) − 1 + 2*l*(*n* + 2) and points (*lk* − 1 + *j*, *lk* − 1 + 2*ln* + *j*) for *j* = 0, ... , *l* − 1. It is shown in Figure 10. This region is the same, as depicted in Figure 8, but translated by *T*(*M*, *N*)=(*M* − 2*l*, *N*).

**Figure 10.** Points *M* + 2*ln* ≤ *N* ≤ *M* + 2*l*(*n* + 1) where *l*(*k* + 2) − 1 ≤ *M* ≤ *l*(*k* + 3) − 2 are highlighted with blue dashed lines. Union of boundaries for all terms of the considered expression consists of points (*l*(*k* + 2) − 1, *N*) for *l*(*k* + 2) − 1 + 2*ln* ≤ *N* ≤ *l*(*k* + 2) − 1 + 2*l*(*n* + 2) and points (*l*(*k* + 2) − 1 + *j*, *l*(*k* + 2) − 1 + 2*ln* + *j*) for *j* = 0, ... , *l* − 1. They are highlighted with purple dashed lines. Here, we depict the case where *l* = 5.

Consider points of a form (*lk* − 1, *N*). Above, we have seen that Formula (22) receives term *Mk*+3+2(*n*+1) (*M*+2(*n*+1)*l*,*N*) during the inductive step. By inductive supposition (23), it is left to account for the action of long steps, acting on paths, induced by this term. Denote the weighted number of paths, corresponding to this term as

$$\begin{aligned} Z\_{(M,N)} \equiv Z(\mathcal{L}\_N((-2l(n+1),0) \to (M,N)); \mathbb{S} \mid \mathcal{W}^{\mathcal{L}}\_{-2l(n+1)}, \{\mathcal{F}^{\mathcal{L}}\_{ml-1-2l(n+1)}\}\_{m=1}^{\infty}) &= \\ &= M\_{(M+2(n+1)lN)}^{k+3+2(n+1)} \end{aligned}$$

where *lk* − 1 ≤ *M* ≤ *l*(*k* + 2) − 1, *M* + 2*ln* ≤ *N* ≤ *M* + 2*l*(*n* + 1). Now, we want to calculate

$$Z'\_{(M,N)} \equiv Z(L\_N((-2l(n+1),0)\to(M,N)); \mathbb{S}\cup\mathbb{\tilde{S}}\mid\mathcal{W}^L\_{-2l(n+1)},\{\mathcal{F}^1\_{ml-1-2l(n+1)}\}\_{m=1}^\infty).$$

From Corollary 2, it is given by

$$Z'\_{(M,N)} = Z\_{(M,N)} + Z\_{(M+2l,N)'}$$

where

$$Z\_{(M+2l,N)} = M^{k+3+2(n+2)}\_{(M+2(n+2)l,N)}$$

Consider points of a form (*lk* − 1 + *j*, *lk* − 1 + 2*ln* + *j*). For such points, the proof is the same as for Formula (10) for the hypotenuse of region II.

Now that we have proven the inductive step for the boundary of the considered region, by Corollary 1, it follows that

$$\tilde{M}\_{(M,N)}^{k+1} = \sum\_{j=0}^{n+1} M\_{(M+2\vec{j},N)}^{k+1+2\vec{j}} + M\_{(M+2(n+2)l,N)}^{k+1+2(n+2)} = \sum\_{j=0}^{n+1} M\_{(M+2\vec{j},N)}^{k+1+2\vec{j}} \tag{26}$$

.

is true for the whole region, which proves the inductive step for Formula (23).

**Corollary 3.** *For fixed k* ∈ N *and m* ≥ *k*

$$\bar{M}\_{(M,N)}^{m+1} = \sum\_{j=0}^{\lfloor \frac{M-\lfloor m+1 \rfloor}{2l} \rfloor} M\_{(M+2jl,N)'}^{m+1+2j} \tag{27}$$

**Proof.** The result of Lemma 6 can be extended to other strips in a similar fashion to the proof of (22). Each new term *<sup>M</sup>k*+1+2*<sup>j</sup>* (*M*+2*jl*,*N*) in *<sup>M</sup>*˜ *<sup>k</sup>*+<sup>1</sup> (*M*,*N*) induces paths, which further descend from (*k* + 1)-th strip to boundary points of each consequent (*k* + 1 + *m*)-th strip. The region in which these induced paths descend is congruent to the region, where they would continue to descend in the auxiliary path model due to the periodicity of filter restrictions. Hence, each *M*˜ *<sup>k</sup>*+1+*<sup>m</sup>* (*M*,*N*) acquires term *<sup>M</sup>k*+1+*m*+2*<sup>j</sup>* (*M*+2*jl*,*N*) , which proves the statement.

During this subsection, we introduced long steps and proved lemmas, necessary for counting weighted numbers of paths in modification of the auxiliary lattice path model, relevant for the representation theory of *Uq*(*sl*2) at roots of unity.
