**6. On Decomposition of** *T***(1)***⊗<sup>N</sup>* **for** *Uq***(***sl***2) at Roots of Unity**

Consider the auxiliary lattice path model with filter restrictions of type 1, in the presence of steps

$$\mathbb{B}^{\scriptscriptstyle{U}} \equiv \mathbb{B} \cup \left( \bigcup\_{k=1}^{\infty} \mathbb{B}(k) \right),$$

and denote it as L*U*. The arrangement of steps for points of L*<sup>U</sup>* is depicted in Figure 11.

**Figure 11.** Arrangement of steps for points of the lattice path model L*U*. Here, we depict the case, where *l* = 5.

Now, fix *q* = *e πi <sup>l</sup>* and *l* is odd. Category **Rep**(*Uq*(*sl*2)) is the category of representations of *Uq*(*sl*2), a quantized universal enveloping algebra of *sl*<sup>2</sup> with divided powers. Consider tensor product decomposition of a tensor power of fundamental *Uq*(*sl*2)-module

$$T(1)^{\otimes N} = \bigoplus\_{k=0}^{N} M\_{T(k)}^{(l)}(N)T(k), \quad T(1), \ T(k) \in \mathbf{Rep}(\mathcal{U}\_q(sl\_2)), \tag{28}$$

where *M*(*l*) *T*(*k*) (*N*) is the multiplicity of *T*(*k*) in tensor product decomposition. We consider the tensor powers of a tilting module, and as a category of tilting modules is closed under finite tensor products, it can be decomposed into a direct sum of tilting modules. The highest weight of *T*(*k*) can be written as *k* = *lk*<sup>1</sup> + *k*0. The Grothendieck ring of the category of tilting modules over *Uq*(*sl*2) at odd roots of unity gives the following tensor product rules ([9,17])

$$\begin{aligned} T(k\_0) \otimes T(1) &= T(k\_0 + 1) \oplus T(k\_0 - 1), \quad 0 \le k\_0 \le l - 2; \\\\ T(lk\_1 + k\_0) \otimes T(1) &= T(lk\_1 + k\_0 + 1) \oplus T(lk\_1 + k\_0 - 1), \quad 1 \le k\_0 \le l - 3, k\_1 \ge 1; \end{aligned}$$

$$T(lk\_1 + l - 2) \otimes T(1) = T(l(k\_1 + 1) - 3) \oplus T((k\_1 + 1)l - 1) \oplus T((k\_1 - 1)l - 1), \quad k\_1 \ge 1;$$

$$T(lk\_1 - 1) \otimes T(1) = T(lk\_1), \quad k\_1 \ge 1;$$

$$T(lk\_1) \otimes T(1) = T(lk\_1 + 1) \oplus 2T(lk\_1 - 1), \quad k\_1 \ge 1.$$

**Theorem 3** ([9])**.** *The multiplicity of the tilting Uq*(*sl*2)*-module T*(*k*) *in the decomposition of T*(1)⊗*<sup>N</sup> is equal to the weighted number of lattice paths on* L*<sup>U</sup> connecting* (0, 0) *and* (*k*, *N*) *with weights given by multiplicities of elementary steps* S*U.*

**Proof.** Tensor product rules allow the following recursive description of multiplicities

$$\mathcal{M}\_{T(0)}^{(l)}(N+1) = \mathcal{M}\_{T(1)}^{(l)}(N);$$

$$M\_{T(lk\_1+k\_0)}^{(l)}(N+1) = M\_{T(lk\_1+k\_0-1)}^{(l)}(N) + M\_{T(lk\_1+k\_0+1)}^{(l)}(N), \quad 1 \le k\_0 \le l-3,\ k\_1 \ge 0;$$

$$M\_{T(lk\_1-2)}^{(l)}(N+1) = M\_{T(lk\_1-3)}^{(l)}(N), \quad k\_1 \ge 1;$$

$$M\_{T(lk\_1-1)}^{(l)}(N+1) = M\_{T(lk\_1-2)}^{(l)}(N) + 2M\_{T(lk\_1)}^{(l)}(N) + M\_{T((lk\_1+2)l-2)}^{(l)}(N), \quad k\_1 \ge 1;$$

$$M\_{T(lk\_1)}^{(l)}(N+1) = M\_{T(lk\_1-1)}^{(l)}(N) + M\_{T(lk\_1+1)}^{(l)}(N), \quad k\_1 \ge 1.$$

This recursion coincides with the recursion for weighted numbers of paths descending from (0, 0) to (*k*, *N*) in lattice path model L*U*. The latter is depicted in Figure 11.

The main goal of the following section is to obtain the explicit formula by combinatorial means, mainly counting lattice paths in modification L*<sup>U</sup>* of the auxiliary lattice path model.

#### **7. Counting Paths**

Consider the lattice path model L*U*. From now on, following 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

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

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

$$\tilde{M}\_{(M,N)}^{j} = Z(L\_N((0,0)\to(M,N); \mathbb{S}\_{lI} \mid \mathcal{W}\_0^{L}, \{\mathcal{F}\_{nl-1}^{1}\}, n \in \mathbb{N})),\tag{29}$$

where

$$\mathbb{B}\_{\mathrm{II}} = \mathbb{B} \cup \left( \bigcup\_{k=1}^{\infty} \mathbb{B}(k) \right).$$

and *M* ≥ 0 and *j* = @ *M*+1 *<sup>l</sup>* + 1 A . The main goal of this section is to derive an explicit formula for *M*˜ *<sup>j</sup>* (*M*,*N*) .

**Lemma 7.** *For the lattice path model* L*<sup>U</sup>*

$$
\bar{M}^1\_{(M,N)} = M^1\_{(M,N)'} \tag{30}
$$

*and for k* ∈ N*,*

$$\mathcal{M}\_{(M,N)}^{k+1} = \sum\_{j=0}^{\lceil \frac{N-\lfloor k+1 \rfloor}{2l} \rceil} F\_{k-1}^{(k-1+2j)} \mathcal{M}\_{(M+2jl,N)'}^{k+1+2j} \tag{31}$$

**Proof.** The formula for the 1st strip follows immediately as long steps have no impact and multiplicity is the same as in the auxiliary lattice path model.

This lemma follows from gradually adding each S(*k*) for *k* = 1, 2, ... to the initial set of steps S and applying results of the Corollary 3 repeatedly. Let us start with *k* = 1. From Corollary 3, it follows that in case of having one series of long steps, for (*m* + 1)-th strip we would simply have

$$\left| \tilde{M}\_{(M,N)}^{m+1} \right|\_{k=1} = \sum\_{j=0}^{\left[ \frac{N-\lfloor lm+1 \rfloor}{2l} \right]} M\_{(M+2jl,N)}^{m+1+2j} \tag{32}$$

where *m* ∈ N. This is a summation of multiplicities in the auxiliary lattice path model with trivial coefficients. This situation is depicted in Figure 12.

**Figure 12.** In each strip, triangles with coefficients correspond to terms in Formula (32). Each term is given by the multiplicity of a strip in the auxiliary lattice path model, situated to the far right of the considered triangle. As an example, we show how this mnemonic rule works for *M*˜ <sup>5</sup> (*M*,*N*) .

In each strip, triangles with coefficients correspond to terms in Formula (32). Each term is given by the multiplicity function of a strip in the auxiliary lattice path model, situated to the far right of the considered triangle. The coefficient in a triangle tells us how many terms corresponding to this multiplicity function are in Formula (32). This mnemonic rule comes from considerations in Figure 9. The proofs of Theorem 6 and Corollary 3 define a recursion on the coefficients near multiplicity functions from the auxiliary lattice path model in Formula (32). This recursion is depicted in Figure 13.

**Figure 13.** Numbers near vertices of the lattice are the coefficients in Figure 12. Blue arrows denote steps in the recursion, which were added by the long steps S(1) in the lattice path model. Length of paths *N*, descending to a considered vertex of a lattice gives the number of the strip in the auxiliary lattice path model, the multiplicity function of which is being added to (32) as a term. As an example, we show formula for *M*˜ <sup>5</sup> (*M*,*N*) |*<sup>k</sup>* <sup>=</sup> 1.

Blue arrows correspond to steps in the recursion on the coefficients, which were added as a consequence of the presence of long steps S(1). In the black frame, it is noted that although long steps have length 2*l*, as shown, for example, in Figure 12, their source and target points belong to two adjacent strips, so when dealing with the coefficients it is convenient to denote blue arrows as in the Figure 13. Long steps, following the idea of the proof of Formula (22) depicted in Figure 9, induce paths that descend further, giving the result as in Corollary 3. Similarly, blue arrows induce paths in the lattice, which descend further, adding new terms in (32).

Note, that without blue arrows we would have obtained a single diagonal path with weighted numbers of paths equal to 1. This situation would give us coefficients as in the formula for multiplicities in the auxiliary lattice path model, meaning that we would have *M*˜ *<sup>k</sup>* (...) <sup>=</sup> *<sup>M</sup><sup>k</sup>* (...) . This is exactly what we would have in case we removed the long steps S(1) in the lattice path model.

Again, following the idea of the proof of Corollary 3, induced paths descend further to each consequent strip as if they were to continue to descend in the auxiliary lattice path model, so additional terms are dependent on how many strips these induced paths will cross while they descend. In the recursion on the coefficients, it is manifested in the fact that the length of a descending path in Figure 13 gives the number of strips in the auxiliary lattice path model, to which the additional term corresponds.

Now, our main goal is to apply S(*k*) for other *k*. As the considerations above suggest, applying S(*k*) for *k* = 1, 2, ... induces other sequences of blue arrows. From Figure 14, we see that the recursion for the coefficients near multiplicity functions is satisfied by Catalan numbers. This proves Formula (31).

**Figure 14.** In each strip, numbers in vertices of a lattice correspond to terms in Formula (32). Each term is given by the multiplicity of a strip in the auxiliary lattice path model, the number of which is given by the length of a path, descending to the considered coefficient. As an example, we show formula for *M*˜ <sup>5</sup> (*M*,*N*) .

Note, that action of black arrows on terms in (31) follows from Lemma 3 and the periodicity of filter restrictions. The action of blue arrows on terms in (31) follows from Corollary 1. Now let us prove the main theorem.

**Theorem 4.** *For k* ∈ N ∪ {0} *we have*

$$\tilde{M}\_{(M,N)}^{k+1} = F\_M^{(N)} + \sum\_{j=1}^{\lfloor \frac{N-lk+1}{2l} + \frac{1}{2} \rfloor} F\_{-2lk+M-2jl}^{(N)} + \sum\_{j=1}^{\lfloor \frac{N-lk+1}{2l} \rfloor} F\_{M+2jl}^{(N)} \tag{33}$$

*where lk* − 1 ≤ *M* ≤ *l*(*k* + 1) − 2*.*

**Proof.** We proceed by induction over [ *<sup>N</sup>*−*lk*+<sup>1</sup> <sup>2</sup>*<sup>l</sup>* <sup>+</sup> <sup>1</sup> <sup>2</sup> ]. For [ *<sup>N</sup>*−*lk*+<sup>1</sup> <sup>2</sup>*<sup>l</sup>* <sup>+</sup> <sup>1</sup> <sup>2</sup> ] = 1 the Formula (33) obviously gives the same result as (31), which is the base of induction.

Suppose, that

$$
\bar{\mathcal{M}}\_{(M,N)}^{k+1} = F\_M^{(N)} + \sum\_{j=1}^n F\_{-2lk+M-2jl}^{(N)} + \sum\_{j=1}^{n-1} F\_{M+2jl}^{(N)} \tag{34}
$$

is true. We need to prove this statement for *n* + 1. It is sufficient to compare coefficients in (33) and (31) near *F*(*N*) *<sup>M</sup>*+2*nl* and *<sup>F</sup>*(*N*) −2*lk*+*M*−2(*n*+1)*l* . We focus on the term *F*(*N*) *<sup>M</sup>*+2*nl*, the rest can be performed in a similar fashion. From the structure of *M<sup>k</sup>* (*M*,*N*) , given by Theorem 1 and depicted in Figure 15, we have two cases: *n* + 1 is odd and *n* + 1 is even.

**Figure 15.** Graphical presentation of terms in the formula for *M<sup>k</sup>* (*M*,*N*) , given by Theorem 1. Each term is depicted in accordance to the domain of the lattice where it appears for the first time.

For the case of odd *n* + 1 the last term in (31) is associated with *Pj*( *<sup>n</sup>* <sup>2</sup> ), for the case of even *n* it is *Qj*( *<sup>n</sup>*+<sup>1</sup> <sup>2</sup> ). We focus on the case of odd *n* + 1, the other case can be proven in a similar manner. So, the proof boils down to a combinatorial identity

$$\sum\_{\substack{j=0\\\text{even}}}^n F\_{k-1}^{(k-1+2j)} P\_{k+1+2j} \left( \frac{n-j}{2} \right) - \sum\_{\substack{j=1\\\text{odd}}}^{n-1} F\_{k-1}^{(k-1+2j)} Q\_{k+1+2j} \left( \frac{n-1-j}{2} \right) = 1. \tag{35}$$

From comparing coefficients near *F*(*N*) *<sup>M</sup>*+2(*n*−2)*<sup>l</sup>* in the inductive supposition (34), we know that

$$\sum\_{\substack{j=0\\n\text{ even}}}^{n-2} F\_{k-1}^{(k-1+2j)} P\_{k+1+2j} \left( \frac{n-2-j}{2} \right) - \sum\_{\substack{j=1\\odd\text{ j}}}^{n-3} F\_{k-1}^{(k-1+2j)} Q\_{k+1+2j} \left( \frac{n-3-j}{2} \right) = 1 \tag{36}$$

is true. Take into account, that

$$P\_{\vec{j}}(k) - P\_{\vec{j}}(k-1) = \binom{\vec{j} + 2k - 3}{\vec{j} - 3} \,' \,. \tag{37}$$

$$Q\_j(k) - Q\_j(k-1) = \binom{j+2k-2}{j-3}.\tag{38}$$

Subtracting (36) from (35), we obtain

$$\left(\sum\_{\substack{j=0\\r\text{even}}}^{n-2} - \sum\_{\substack{j=1\\r\text{odd}}}^{n-3} \right) F\_{k-1}^{(k-1+2j)} \binom{j+n+k-2}{2j+k-2} \tag{39}$$
 
$$+ F\_{k-1}^{(k-1+2n)} P\_{k+1+2n}(0) - F\_{k-1}^{(k-1+2n-2)} Q\_{k+1+2(n-1)}(0) = 0$$

Taking into account that *Pj*(0) = 1 and *Qj*(0) = *j* − 2 and simplifying further, we arrive at

$$\sum\_{j=0}^{n}(-1)^{j}F\_{k-1}^{(k-1+2j)}\binom{j+n+k-2}{2j+k-2} = 0.\tag{40}$$

$$\sum\_{j=0}^{n}(-1)^{j}\binom{j+n+k-2}{2j+k-2}\left(\binom{2j+k-1}{j}-\binom{2j+k-1}{j-1}\right)=0\tag{41}$$

The last identity follows from the following lemma.

**Lemma 8.** *For n*, *k* ∈ N

$$\sum\_{j=0}^{n}(-1)^{j}\binom{j+n+k-2}{2j+k-2}\binom{2j+k-1}{j} = 2(-1)^{n},\tag{42}$$

$$\sum\_{j=0}^{n}(-1)^{j}\binom{j+n+k-2}{2j+k-2}\binom{2j+k-1}{j-1}=2(-1)^{n}.\tag{43}$$

**Proof.** Let us first prove Formula (42). Denote

$$F(n,j) = \frac{(-1)^{j+n}}{2} \binom{j+n+k-2}{2j+k-2} \binom{2j+k-1}{j} \dots$$

We need to show, that *<sup>n</sup>*

$$\sum\_{j=0}^{n} F(n, j) = 1, \quad \forall n \in \mathbb{N}.$$

For *n* = 1 it is true, which gives us the base of induction. Using Zeilberger's algorithm ([18–20]), we obtain its Wilf–Zeilberger pair

$$\begin{aligned} G(n,j) &= \frac{(-1)^{j+n}j(j+k-1)(k+2n)}{2n(n+1)(j-n-1)(k+2j-1)(k^2+n(n-1)+k(2n-1))},\\ \times (1+k^2n-3n^2+k(n^2-3n-1)+j(2n^2+2kn+k-1))\binom{j+n+k-2}{2j+k-2}\binom{2j+k-1}{j},\end{aligned}$$

for which

$$-F(n+1,j) + F(n,j) = G(n,j+1) - G(n,j) \tag{44}$$

is true. Applying sum over *j* to both sides and simplifying telescopic sum to the right, we obtain that *<sup>n</sup>*

$$\sum\_{j=0}^{n} F(n+1, j) = \sum\_{j=0}^{n} F(n, j) + G(n, 0) - G(n, n+1). \tag{45}$$

Taking into account, that *G*(*n*, 0) = 0 and *G*(*n*, *n* + 1) = *F*(*n* + 1, *n* + 1), we have

$$\sum\_{j=0}^{n+1} F(n+1, j) = \sum\_{j=0}^{n} F(n, j), \tag{46}$$

which, considering inductive supposition, proves Formula (42).

Formula (43) can be proven in a similar fashion, the corresponding Wilf–Zilberger pair is given by

$$G(n,j) = \frac{(-1)^{j+n}(j-1)(j+k)(k+2n)}{2n(n-1)(j-n-1)(k+2j-1)(k^2+n(n-1)+k(2n-1))},$$

$$0 \times (1+k^2(n-1)+k(n^2-3n)-3n^2+j(2n^2+2kn-k-1))\binom{j+n+k-2}{2j+k-2}\binom{2j+k-1}{j-1}.$$

The proof of this lemma concludes the proof of the identity and finishes the proof of the initial theorem.

#### **8. Conclusions**

In this paper, we considered the lattice path model L*U*, which is the auxiliary lattice path model in the presence of long steps. Weighted numbers of paths in this model recreate multiplicities of *Uq*(*sl*2)-modules in tensor product decomposition of *T*(1)⊗*N*, where *Uq*(*sl*2) is a quantum deformation of the universal enveloping algebra of *sl*<sup>2</sup> with divided powers and *q* is a root of unity. Explicit formulas for multiplicities of all tilting modules in tensor product decomposition were derived by purely combinatorial means in the main theorem of this paper Theorem 4.

We found that the auxiliary lattice model defined in [13] is of great use for counting multiplicities of modules of differently defined quantum deformations of *U*(*sl*2) at *q* root of unity. For instance, in [9] we applied periodicity conditions to the auxiliary lattice path model to obtain a folded Bratteli diagram, weighted numbers of paths for which recreate multiplicities of modules in tensor product decomposition of *T*(1)⊗*N*, where *T*(1) is a fundamental module of the small quantum group *uq*(*sl*2). In this paper, we modified the auxiliary lattice path model by applying long steps to obtain multiplicities for the case of *Uq*(*sl*2) with divided powers in a similar fashion.

The model defined in [13] required analysis of combinatorial properties of filters, which we heavily relied on. In this paper, we introduced long steps and explored their combinatorial properties. In order to derive formulas for weighted numbers of paths in this setting, we also defined boundary points and congruence of regions in lattice path models. The philosophy of congruence is fairly easy to understand. Two different lattice path models can be locally indistinguishable due to coinciding recursions for weighted numbers of paths in these regions. Weighted numbers of paths at boundary points of the considered region uniquely define weighted numbers of paths for the rest of the region by recursion. So, instead of proving identities for the whole region, it is sufficient to prove such only for boundary points of the region. At boundary points, an identity can be represented as a linear combination of weighted numbers of paths from different lattice path models and one needs to take into account boundary points of congruent regions with respect to all these models.

We found that besides applying periodicity conditions to the auxiliary lattice path model, one can take *Uq*(*sl*2), consider its restriction to *u*<sup>−</sup> *<sup>q</sup> U*<sup>0</sup> *qu*<sup>+</sup> *<sup>q</sup>* , where *u*<sup>±</sup> *<sup>q</sup>* are subalgebras of the small quantum group *uq*(*sl*2), generated by *F* and *E*, respectively, and *U*<sup>0</sup> *<sup>q</sup>* is a subalgebra of *Uq*(*sl*2), generated by *<sup>K</sup>*±<sup>1</sup> and *<sup>K</sup>*; *<sup>c</sup> t* , for *t* ≥ 0, *c* ∈ Z. Then, we can restrict *u*<sup>−</sup> *<sup>q</sup> U*<sup>0</sup> *qu*<sup>+</sup> *q* to *uq*(*sl*2). This procedure defines another modification of the auxiliary lattice path model and, remarkably, gives the same result as with periodicity conditions. The lattice path model corresponding to *u*<sup>−</sup> *<sup>q</sup> U*<sup>0</sup> *qu*<sup>+</sup>

*<sup>q</sup>* will be considered in the upcoming paper. Considering other possible directions for further research, the following questions remain open:


**Funding:** This research was supported by the grant "Agreement on the provision of a grant in the form of subsidies from the federal budget for implementation of state support for the creation and development of world-class scientific centers, including international world-class centers and scientific centers, carrying out research and development on the priorities of scientific technological development No. 075-15-2019-1619" dated 8th November 2019.

#### **Data Availability Statement:** Not applicable.

**Acknowledgments:** We are grateful to Olga Postnova, Nicolai Reshetikhin, Pavel Nikitin, Fedor Petrov for fruitful discussions. We thank Lacey Lindsay for her help with corrections and support.

**Conflicts of Interest:** The authors declare no conflict of interest.

## **References**

