Computing the Haar state on ${\mathcal{O}(SL_{q}(3))}$

Not every monomial has a non-zero Haar state value. In this section, we will give a criterion to determine whether the Haar state of a monomial is zero or not.

Let $D_{n}$ be the diagonal subgroup of $SL_{q}(n)$. Recall that the coordinate Hopf algebra $\mathcal{O}(D_{n})$ is the commutative algebra $\mathbb{C}[t_{1},t_{1}^{-1},\cdots,t_{n},t_{n}^{-1}]$ of all Laurent polynomials in $n$ indeterminates $t_{1},t_{2},\dots,t_{n}$ with comultiplication $\Delta(t_{i})=t_{i}\otimes t_{i}$ and counit $\varepsilon(t_{i})=1$. The surjective homomorphism $\pi_{D_{n}}:\mathcal{O}(SL_{q}(n))\mapsto\mathcal{O}(D_{n})$ is given by $\pi_{D_{n}}(x_{ij})=\delta_{ij}t_{i}$. Since we have $D_{q}^{k}=1_{\mathcal{O}(SL_{q}(n))}$ for all $k\in\mathbb{N}^{+}$, $\pi_{D_{n}}$ tells us $1_{\mathcal{O}(D_{n})}=\pi_{D_{n}}(1_{\mathcal{O}(SL_{q}(n))})=\pi_{D_{n}}(D_{q}^{k})=(\Pi_{i=1}^{n}t_{i})^{k}$.

The right and left action of $\mathcal{O}(SL_{q}(n))$ on $\mathcal{O}(D_{n})$, denoted as $L_{D_{n}}$ and $R_{D_{n}}$, is defined as:

$$\begin{split}L_{D_{n}}&=(\pi_{D_{n}}\otimes Id)\circ\Delta,\\ R_{D_{n}}&=(Id\otimes\pi_{D_{n}})\circ\Delta.\end{split}$$

Given vector $v=(v_{1},v_{2},\dots,v_{n})\in\mathbb{R}^{n}$, we write $t^{v}=\Pi_{i=1}^{n}t_{i}^{v_{i}}$. If $x$ is a monomial, we have:

$$\begin{split}L_{D_{n}}(x)&=t^{\alpha(\theta(x))}\otimes x,\\ R_{D_{n}}(x)&=x\otimes t^{\beta(\theta(x))}.\end{split}$$

The next theorem is a generalization of Klimyk and Schmudgen’s observation [6]. It gives the necessary condition such that $h(x)\neq 0$ for $x\in\mathcal{O}(SL_{q}(n))$:

Theorem 1 a): Let $x$ be a monomial. Then $h(x)\neq 0$ implies that there exist $k\in\mathbb{N}^{+}$ such that $\theta(x)$ is a ${k}$-doubly stochastic matrix.

Proof: Consider $(\pi_{D_{n}}\otimes h)\circ\Delta(x)$. There are two ways to compute this object:

$$\begin{split}(\pi_{D_{n}}\otimes h)\circ\Delta(x)&=\pi_{D_{n}}\circ(Id\otimes h)\circ\Delta(x)=\pi_{D_{n}}(h(x)\cdot 1_{\mathcal{O}(SL_{q}(n))})\\ &=h(x)\cdot 1_{\mathcal{O}(D_{n})},\\ (\pi_{D_{n}}\otimes h)\circ\Delta(x)&=(id\otimes h)\circ(\pi_{D_{n}}\otimes id)\circ\Delta(x)=(id\otimes h)\circ L_{D_{n}}(x)\\ &=(id\otimes h)(t^{\alpha(\theta(x))}\otimes x)\\ &=h(x)\cdot t^{\alpha(\theta(x))}.\end{split}$$

Thus, $h(x)\cdot 1_{\mathcal{O}(D_{n})}=h(x)\cdot t^{\alpha(\theta(x))}$. Since $h(x)\neq 0$, we get $1_{\mathcal{O}(D_{n})}=t^{\alpha(\theta(x))}$. This means that we can find integer $k_{1}>0$ such that $t^{\alpha(\theta(x))}=(\Pi_{i=1}^{n}t_{i})^{k_{1}}$. Thus, $\alpha(\theta(x))=(k_{1})_{i=1}^{n}$.

Apply the same argument to $(h\otimes\pi_{D_{n}})\circ\Delta(x)$, we get $1_{\mathcal{O}(D_{n})}=t^{\beta(\theta(x))}$. Thus, we can find $k_{2}>0$ such that $\beta(\theta(x))=(k_{2})_{i=1}^{n}$. But we must have $k_{1}=k_{2}$ since

$$nk_{1}=\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}=nk_{2}.$$

Let $\mathcal{NZ}$ be the linear subspace spanned by monomials with non-zero Haar states. In this section, we give a criterion to pick a basis for $\mathcal{NZ}$. We write $A_{n}(m),\ m\in\mathbb{N}^{+}$ as the set of $n\times n$ $m$-doubly stochastic matrices and $B_{n}(m)$ as the set of monomials on $\mathcal{O}(SL_{q}(n))$ whose counting matrices belong to $A_{n}(m)$.

First, we introduce a total order ’$<$’ on $A_{n}(m)$. For every $C=(c_{ij})_{i,jk=1}^{n}\in A_{n}(m)$, we associate a vector

$$\mathcal{V}(C)=(c_{11},c_{12},\dots,c_{1n},c_{21},c_{22},\dots,c_{nn}),$$

and we compare such vectors in lexicographic order. We say matrices $C<D$ if $\mathcal{V}(C)<\mathcal{V}(D)$. With this total order, we have the following observation:

If $x=P\cdot x_{ik}x_{jl}\cdot Q\in B_{n}(m)$ $(i<j,k<l)$ where $P,Q$ are two monomials and we switch the order of $x_{ik}x_{jl}$ so that:

$$x=y+(q-q^{-1})z,$$

where $y=P\cdot x_{jl}x_{ik}\cdot Q$ and $z=P\cdot x_{il}x_{jk}\cdot Q$, then we have: $\theta(z)\in A_{n}(m)$ and $\theta(z)<\theta(x)=\theta(y)$.

Based on the observation, we get the following lemma:
Lemma 1: For each $A\in A_{n}(m)$, we fix monomial $x_{A}\in B_{n}(m)$ such that $\theta(x_{A})=A$. If $\phi\in B_{n}(m)$ is a monomial with counting matrix $M$, then we can decompose $\phi$ as:

$$\phi=c_{M}\cdot x_{M}+\sum_{\begin{subarray}{c}P<M\\ P\in A_{n}(m)\end{subarray}}c_{P}\cdot x_{P}.$$

(1)

Proof: Since $\phi$ and $x_{M}$ have the same counting matrix, we can permute the generators in $\phi$ to the same order as in $x_{M}$. We denote this process as a chain:

$$\phi=\phi_{0}\rightarrow\phi_{1}\rightarrow\phi_{2}\rightarrow\cdots\rightarrow\phi_{k}=x_{M},$$

where each $\phi_{i}$ is a reordering of $\phi$ and we get $\phi_{i+1}$ by switching the order of two adjacent generators in $\phi_{i}$. From $\phi_{i}$ to $\phi_{i+1}$, we may get a new term $\varphi_{i+1}$. As discussed before, $\theta(\varphi_{i+1})\in A_{n}(m)$ and $\theta(\varphi_{i+1})<\theta(\phi_{i+1})=M$. We can permute these newly generated $\varphi_{i}$’s to their corresponding $x_{\theta(\varphi_{i+1})}$’s, and we may get new terms in this process as well. However, each time we repeat this permuting process to a monomial $y$, the counting matrix of the newly generated monomial is always smaller than $\theta(y)$. Since the counting matrix of the newly generated monomial is always descending, we can finish this permuting process in finite steps. In other words, we will get a chain on which every transposition does not generate new monomials. Then, every monomial appearing in the summation will be in the desired form, and we get Equation (1). $\blacksquare$

Lemma 1 provides a criterion for picking a basis for $\mathcal{NZ}$. Let $S_{n}(m)=\{x_{M},M\in A_{n}(m)\}$. Then, we can write

$$\mathcal{NZ}=\left\langle\bigcup_{i=1}^{\infty}S_{n}(i)\right\rangle.$$

Theorem 1 b): Every monomial with non-zero Haar state value can be written as a linear combination of standard monomials.

Proof: By the Birkhoff-Von Neumann Theorem [2] [16], every $M\in A_{n}(m)$ can be decomposed into $M=m_{1}\sigma_{1}+m_{2}\sigma_{2}+\cdots+m_{n!}\sigma_{n!}$, where $\sigma_{i}$’s are matrix in $A_{n}(1)$ and $m_{i}$’s are non-negative integers whose sum is $m$. Notice that each matrix $\sigma_{i}$ can be identified with a permutation on $n$ letters. We denote the corresponding permutation as $\sigma_{i}$ as well. Then, the counting matrix of the monomial $\prod_{\sigma_{i}\in S_{n}}(x_{\sigma_{i}})^{m_{i}}$ is $M$. This implies that for every $M\in A_{n}(m)$, we can choose $x_{M}$ in form $\prod_{\sigma_{i}\in S_{n}}(x_{\sigma_{i}})^{m_{i}}$. Combining Lemma 1, the statement in Main Theorem b) is clear. $\blacksquare$

Notice that the set of all standard monomials contains a basis of $\mathcal{NZ}$, but the set itself is not a basis of $\mathcal{NZ}$. The reason is that the different standard monomials could have the same counting matrix. In this case, standard monomials with the same counting matrix are linearly dependent(see Appendix A, Equation (22) and (23)). To find a basis of $\mathcal{NZ}$, for each $M\in A(m)$, we have to preserve only one standard monomial corresponding to $M$ and filter out ’unnecessary’ standard monomials.

Once we solve the Haar state of standard monomials of each order, we can find the Haar state for other monomials according to their linear combination. We will use the defining relation $((id\otimes h)\circ\Delta)(x)=h(x)\cdot 1=((h\otimes id)\circ\Delta)(x)$ to solve the Haar state of every standard basis. We start with the investigation of the comultiplication of a monomial.

Lemma 2: Let $x$ be a monomial and we write:

$$\Delta(x)=\sum_{i\in I}z_{i}\otimes y_{i},$$

with $I$ an index set and $y_{i},z_{i}$ monomials. Then we have the following equations:

$$\alpha(\theta(x))=\alpha(\theta(z_{i}))\ \ \ \beta(\theta(x))=\beta(\theta(y_{i}))\ \ \ \beta(\theta(z_{i}))=\alpha(\theta(y_{i})).$$

Proof: Recall that $\Delta(x_{ij})=\sum_{k=1}^{n}x_{ik}\otimes x_{kj}$ and $\Delta$ is a morphism of algebra. If $x=\Pi_{l=1}^{p}x_{i_{l}j_{l}}$, then

$$\Delta(x)=\Delta(\Pi_{l=1}^{p}x_{i_{l},j_{l}})=\Pi_{l=1}^{p}\Delta(x_{i_{l},j_{l}})=\prod_{l=1}^{p}\left(\sum_{k=1}^{n}x_{i_{l},k}\otimes x_{k,j_{l}}\right).$$

For each $z_{i}$ the $l$-th generator is in the same row as the $l$-th generator in $x$, and for each $y_{i}$ the $l$-th generator is in the same column as the $l$-th generator in $x$. The column index of the $l$-th generator in $z_{i}$ is the same as the row index of the $l$-th generator in $y_{i}$. Thus, the row sum of $x$ equals the row sum of $z_{i}$; the column sum of $x$ equals the column sum of $y_{i}$, and the column sum of $z_{i}$ equals the row sum of $y_{i}$. $\blacksquare$

With Lemma 2, we have the following result:

Lemma 3: If $\theta(x)\in A_{n}(m)$ then $h(y_{i})\neq 0$ (or $h(z_{i})\neq 0$) if and only if $\theta(y_{i})\in A_{n}(m)$ ( or $\theta(z_{i})\in A_{n}(m)$). Moreover, $\theta(y_{i})\in A_{n}(m)$ if and only if $\theta(z_{i})\in A_{n}(m)$.

Proof: Use Theorem 1 a) and Lemma 2. $\blacksquare$

Theorem 1 c): Let $\{s_{l}\}_{l\in\mathcal{I}_{m}}$, be the set of standard monomials of order $m$. Then, we can write $(Id\otimes h)\circ\Delta(s_{l})$ and $(h\otimes Id)\circ\Delta(s_{l})$ as linear combinations of $s_{j}$’s and the coefficient of each $s_{j}$ is a linear combination of $h(s_{i})$’s.

Proof: If $s_{l}\in B_{n}(m)$ is a standard basis, Lemma 3 implies that

$$(id\otimes h)\circ\Delta(s_{l})=\sum_{\begin{subarray}{c}y\in B_{n}(m)\\ z\in B_{n}(m)\end{subarray}}h(y)\cdot z=\sum_{i}h(y_{i})\cdot z_{i}$$

(2)

Then, by Lemma 1, we can decompose each $y_{i}$ and $z_{i}$ as:

$$y_{i}=\sum_{j=1}^{k}d_{j}^{y_{i}}\cdot s_{j},$$

(3)

$$z_{i}=\sum_{j=1}^{k}d_{j}^{z_{i}}\cdot s_{j}.$$

(4)

Substitute Equation (3) and Equation (4) into Equation (2), we get:

$$(id\otimes h)\circ\Delta(s_{l})=\sum_{j=1}^{k}\left(\sum_{i=1}^{k}c_{ij}h(s_{i})\right)\cdot s_{j}.$$

(5)

$\blacksquare$

In Equation (2), we call $y_{i}$ the relation component and call $z_{i}$ the comparing component. We will say $\boldsymbol{z_{i}}$ (or $\boldsymbol{y_{i}}$) contains $\boldsymbol{s_{j}}$ if $d_{j}^{z_{i}}\neq 0$ (or $d_{j}^{y_{i}}\neq 0$).

Since we can identify $1$ with $D_{q}^{m}$, we get $(id\otimes h)\circ\Delta(s_{l})=h(s_{l})\cdot D_{q}^{m}$. Notice that we can decompose $D_{q}^{m}$ as a linear combination of standard monomials of order $m$. Thus, by comparing the coefficient of the same standard monomial on both sides of $(id\otimes h)\circ\Delta(s_{l})=h(s_{l})\cdot D_{q}^{m}$, we can find a linear relation consisting of the Haar states of standard monomials of order $m$ (See section 2.5 for more detail). We call such a linear relation linear relation of order $\boldsymbol{m}$. We call a linear system consisting of linear relations of order $m$ a system of order $\boldsymbol{m}$.

In this section, we will prove Theorem 1 d). The standard basis for $B_{n}(1)$ is in the form of $x_{\tau_{i}}=\Pi_{k=1}^{n}x_{k,\tau_{i}(k)}$ where $\tau_{i}$ is a permutation on $n$ letters. We have:

$$\Delta(x_{\tau_{i}})=\Delta(\Pi_{k=1}^{n}x_{k,\tau_{i}(k)})=\displaystyle{\prod_{k=1}^{n}\left(\sum_{p=1}^{n}x_{k,p}\otimes x_{p,\tau_{i}(k)}\right)}.$$

By Lemma 3, after apply $(id\otimes h)$ to $\Delta(x_{\tau_{i}})$, we get:

$$\begin{split}(id\otimes h)\circ\Delta(x_{\tau_{i}})=\sum_{\sigma_{j}\in S_{n}}h(\Pi_{k=1}^{n}x_{\sigma_{j}(k),\tau_{i}(k)})\cdot\Pi_{k=1}^{n}x_{k,\sigma_{j}(k)}.\end{split}$$

(6)

On the other hand, recall that

$$1=D_{q}=\sum_{\sigma_{j}\in S_{n}}(-q)^{l(\sigma_{j})}\prod_{k=1}^{n}x_{k,\sigma_{j}(k)},$$

(7)

where $l(\sigma_{j})$ is the inverse number of $\sigma_{j}$.

Thus, using $(id\otimes h)\circ\Delta(x_{\tau_{i}})=h(x_{\tau_{i}})\cdot 1$ and comparing the coefficients of each standard basis, we get for every $\sigma_{j}\in S_{n}$:

$$h(\Pi_{k=1}^{n}x_{\sigma_{j}(k),\tau_{i}(k)})=(-q)^{l(\sigma_{j})}h(x_{\tau_{i}}).$$

(8)

In general, $\Pi_{k=1}^{n}x_{\sigma_{j}(k),\tau_{i}(k)}$ is not a standard monomial. However, if we choose $\sigma_{j}$ such that $\sigma_{j}(k)=n+1-\tau_{i}(k)$, then every generator in $\Pi_{k=1}^{n}x_{\sigma_{j}(k),\tau_{i}(k)}$ commutes with each other and $\Pi_{k=1}^{n}x_{\sigma_{j}(k),\tau_{i}(k)}=\Pi_{k=1}^{n}x_{k,n+1-k}$. Moreover, $l(\sigma_{j})=\frac{n(n-1)}{2}-l(\tau_{i})$. Thus, from Equation (8) we get:

$$h(\Pi_{k=1}^{n}x_{k,n+1-k})=(-q)^{\frac{n(n-1)}{2}-l(\tau_{i})}h(\Pi_{k=1}^{n}x_{k,\tau_{i}(k)}).$$

(9)

Therefore, using Equation (9) and Equation (7) we get:

$$\begin{split}1&=h(1)=\sum_{\sigma_{j}\in S_{n}}(-q)^{l(\sigma_{j})}h(\Pi_{k=1}^{n}x_{k,\sigma_{j}(k)})\\ &=\left(\sum_{\sigma_{j}\in S_{n}}(-q)^{2l(\sigma_{j})-\frac{n(n-1)}{2}}\right)h(\Pi_{k=1}^{n}x_{k,n+1-k}),\end{split}$$

(10)

which gives

$$h(\Pi_{k=1}^{n}x_{k,n+1-k})=\frac{(-q)^{\frac{n(n-1)}{2}}}{\sum_{\sigma_{j}\in S_{n}}(-q)^{2l(\sigma_{j})}}.$$

(11)

Then by Equation (8), notice that the inverse number for the $\tau_{i}$ corresponding to $\Pi_{k=1}^{n}x_{k,n+1-k}$ is just $\frac{n(n-1)}{2}$, we get for every $\tau_{i}\in S_{n}$:

$$h(\Pi_{k=1}^{n}x_{k,\tau_{i}(k)})=\frac{(-q)^{l(\tau_{i})}}{\sum_{\sigma_{j}\in S_{n}}(-q)^{2l(\sigma_{j})}}.$$

(12)

Now, let $I_{n}(k)$ be the number of permutations on $n$ letters with $k$ inversions. Then, the denominator of Equation (12) can be rewritten as:

$$\sum_{\sigma_{j}\in S_{n}}(-q)^{2l(\sigma_{j})}=\sum_{k=0}^{\frac{n(n+1)}{2}}I_{n}(k)q^{2k}.$$

By Andrews [1], the generating function of $I_{n}(k)$ is

$$\sum_{k=0}^{\frac{n(n+1)}{2}}I_{n}(k)x^{k}=\prod_{j=1}^{n}\frac{1-x^{j}}{1-x}.$$

So the denominator of Equation (12) can be rewritten as

$$\sum_{\sigma_{j}\in S_{n}}(-q)^{2l(\sigma_{j})}=\prod_{j=1}^{n}\frac{1-q^{2j}}{1-q^{2}}=[n]_{q^{2}}!,$$

and we get:

$$h(\Pi_{k=1}^{n}x_{k,\tau_{i}(k)})=\frac{(-q)^{l(\tau_{i})}}{[n]_{q^{2}}!}.$$

In this section, let $\{s_{l}\}_{l=1}^{K_{m}}$ be a set of linearly independent standard monomials of order $m$. Recall Equation (5):

$$\begin{split}(id\otimes h)\circ\Delta(s_{l})=\sum_{j=1}^{K_{m}}\left(\sum_{i=1}^{K_{m}}c_{ij}h(s_{i})\right)\cdot s_{j}.\end{split}$$

We can do the same thing to $h(s_{l})\cdot 1=h(s_{l})\cdot D_{q}^{m}$ and get:

$$h(s_{l})\cdot 1=h(s_{l})\cdot D_{q}^{m}=\sum_{j=1}^{K_{m}}b_{j}h(s_{l})\cdot s_{j}.$$

(13)

By comparing the coefficients of standard bases in $(id\otimes h)\circ\Delta(s_{l})$ and in $h(s_{l})\cdot 1$, we get:

$$\sum_{i=1}^{K_{m}}c_{ij}h(s_{i})=b_{j}h(s_{l})$$

(14)

for every $1\leq j\leq K_{m}$. We will call Equation (14) the linear relation derived from equation basis $\boldsymbol{s_{l}}$ and comparing basis $\boldsymbol{s_{j}}$. Each index $1\leq l\leq K_{m}$ corresponds to $K_{m}$ linear relations, so there are $K_{m}^{2}$ linear relations. Since there are $K_{m}$ unknowns, it is possible to construct more than one system of order $m$. Notice that these linear relations all have the zero right-hand side. One way to get a linear relation with the non-zero right-hand side is by decomposing $1=h(1)=h(D_{q}^{m})$ into a sum of standard monomials. Although we can construct more than one system of order $m$, not every system is invertible. We will give a more robust approach to compute the Haar state of $\mathcal{O}(SL_{q}(3))$ later.

In the order 1 case, finding Equation (5) and (13) is an easy task. However, the situation is much more complicate in higher order case. To understand the difficulty to find the two equations in higher order case, we introduce the order restriction for each summand appearing in the comultiplication of a monomial:

Let $x=\prod_{k\in I}x_{i_{k},j_{k}}$ be a monomial and the comultiplication of $x$ be

$$\Delta(x)=\prod_{k\in I}\Delta(x_{i_{k},j_{k}})=\prod_{k\in I}\left(\sum_{l_{k}=1}^{n}x_{i_{k},l_{k}}\otimes x_{l_{k},j_{k}}\right)=\sum_{i\in I^{\prime}}z_{i}\otimes y_{i}.$$

Then:

• i)

  the $k$-th generator of the left component $z_{i}$ is in the $i_{k}$-th row

• ii)

  the $k$-th generator of the right component $y_{i}$ is in the $j_{k}$-th column

• iii)

  The column index of the $l$-th generator in $z_{i}$ equals to the row index of the $l$-th generator in $y_{i}$.

The order restriction is a direct consequence of the fact that the comultiplication is an algebra homomorphism. Since each index $l_{k}$ ranges from $1$ to $n$, every possible combination of $z_{i}\otimes y_{i}$ that satisfies the order restriction will appear in the summation of $\Delta(x)$. In higher order case, this means that Equation (2) includes not only summand whose left and right components are standard monomials but also summand whose left and right components are reordering of standard monomials satisfying the order restriction. As an example in $\mathcal{O}(SL_{q}(3))$, if $(x_{1,1}x_{2,3}x_{3,2})(x_{1,2}x_{2,1}x_{3,3})$ is the left component of one of the tensor products in $\Delta\left((x_{1,1}x_{2,2,}x_{3,3})^{2}\right)$ then all reordering of the left component satisfying property i) of the order restriction are:

• 1)

  $(x_{1,1}x_{2,3}x_{3,2})(x_{1,2}x_{2,1}x_{3,3})$

• 2)

  $(x_{1,2}x_{2,3}x_{3,2})(x_{1,1}x_{2,1}x_{3,3})$

• 3)

  $(x_{1,1}x_{2,1}x_{3,2})(x_{1,2}x_{2,3}x_{3,3})$

• 4)

  $(x_{1,1}x_{2,3}x_{3,3})(x_{1,2}x_{2,1}x_{3,2})$

• 5)

  $(x_{1,2}x_{2,1}x_{3,2})(x_{1,1}x_{2,3}x_{3,3})$

• 6)

  $(x_{1,2}x_{2,3}x_{3,3})(x_{1,1}x_{2,1}x_{3,2})$

• 7)

  $(x_{1,1}x_{2,1}x_{3,3})(x_{1,2}x_{2,3}x_{3,2})$

• 8)

  $(x_{1,2}x_{2,1}x_{3,3})(x_{1,1}x_{2,3}x_{3,2})$.

Thus, the comultipilcation of a standard monomial of higher order contains not only standard monomials but also variations of standard monomials satisfying the order restriction. This is the major difference between the case of order 1 and higher order cases. To find a linear relation derived from equation basis $s_{l}$ and comparing basis $s_{j}$ in higher order case, we have to:

• i)

  find all left (or right) component appearing in $\Delta(s_{l})$ that contains $s_{j}$ and compute the corresponding coefficient $d_{j}^{z}$ in Equation (4);

• ii)

  find the decomposition of the right (or left) component in $\Delta(s_{l})$ corresponding to the left (or right) component in i) and sum all such decomposition together to get a linear relation;

• iii)

  decompose every summand containing $s_{j}$ in Equation (13) to find $b_{j}$.

All 3 steps involve decomposing non-standard monomials into a linear combination of standard monomials and such decomposition is not easy in general. However, there is a simple criterion to determine whether a standard monomial appears in the decomposition of a non-standard monomial or not.