# Construction of polynomials with non-commutative elements.

I have a simple set of polynomials which I know how to construct for each integer $n$, but I havn't been able to write them down in terms of concrete sums and products.

For $n\in\mathbb N_+$, we have the elements $\{\alpha_m\}_{m=1}^n$ and $\{\beta_m\}_{m=2}^n$, where everything in non-commutative. The polynomial $P_n(\alpha, \beta)$ is constructed by the following rules:

1. We must have all combination of products of $\alpha$ and $\beta$'s, where we start with $\alpha_n$ or $\beta_n$ and multiply successively by $\alpha_m$ or $\beta_m$ on the right, where $m<n$.
2. To the right of any $\alpha_m$ we can only have $\alpha_{m-1}$ or $\beta_{m-1}$ (index goes down by 1).
3. To the right of any $\beta_m$ we can only have $\alpha_{m-2}$ or $\beta_{m-2}$ (index goes down by 2).
4. The product stops with $\alpha_1$ or $\beta_2$. All these terms are summed together.

For example

$$P_1(\alpha,\beta) = \alpha_1$$ $$P_2(\alpha,\beta) = \alpha_2\alpha_1 + \beta_2$$ $$P_3(\alpha,\beta) = \alpha_3\alpha_2\alpha_1 + \beta_3 \alpha_1 + \alpha_3\beta_2$$ $$P_4(\alpha,\beta) = \alpha_4\alpha_3\alpha_2\alpha_1 + \alpha_4\alpha_3\beta_2 + \alpha_4\beta_3\alpha_1 + \beta_4\alpha_2\alpha_1 + \beta_4\beta_2$$ $$P_5(\alpha,\beta) = 8-terms$$

I haven't been able to write this polynomial in terms of concrete sums and product. I am only able to do it if point 3 is changed such that the index goes down with one. Do you have any ideas?

EDIT: Let me clarify what I want to achieve. Lets add $\beta_1$ and change point (3) such that the index only goes down by 1 at each product. In this case I think one can express the Polynomial as

$$Q_n(\alpha,\beta) = \sum_{\alpha=1}^n\prod_{i=1}^{\alpha}\left(\prod_{j=1}^i\sum_{M_{j-1}}\sum_{N_{j}}\right)\left(\prod_{m_i=N_i+1}^{M_i}\beta_{m_i}\prod_{n_i=M_{i-1}+1}^{N_i}\alpha_{n_i}\right)$$

where $M_0 = 0$, $M_n = n$ and $$0\leq N_i \leq n-2(\alpha-1),$$ $$M_{i-1}+1\leq N_i \leq n-2(\alpha-i),\;\;\;\;\text{for}\;\; i=2, \dots, \alpha,$$ $$N_i+1\leq M_i \leq n-2(\alpha-i)+1,\;\;\;\;\text{for}\;\; i=1, \dots, \alpha.$$

I would like to have a formula like this for $P(\alpha,\beta)$, but I don't seem to be able to do that. It seems that the parity of $n$ plays an annoying role.

• What's the motivation here? – Qiaochu Yuan Mar 26 '12 at 19:35
• It's part of a calculations I am doing regarding correlation functions in AdS/CFT correspondence around a rotating black hole background, more concretely I am working with many coupled PDE's. Even more concretely, I'm trying to find a solution for the recurrence relation in my last question (math.stackexchange.com/questions/123477/…). – Heidar Mar 26 '12 at 19:43
• What do you mean by "concrete" sums? Does it help to observe that $P_n$ satisfies the recursion $$P_n(\alpha, \beta) = \alpha_n\cdot P_{n-1} + \beta_n\cdot P_{n-2}$$ with base cases $P_{-1}(\alpha, \beta) = 0$ and $P_0(\alpha, \beta) = 1$ ? – MJD Mar 26 '12 at 19:51
• @Mark, If you follow the link in my comment above you will see that this is the exact recursion relation I am trying to solve. I will edit my question and make it more clear what I am trying to achieve. – Heidar Mar 26 '12 at 20:03