Calculate powers of sums

Question

Assuming we have an unital associative complex algebra with generatorn $a_i,b_i$, $i_1,...,n$ such that they anticommute, that is $$a_ia_j=-a_ja_i,\quad b_ib_j=-b_jb_i,\quad a_ib_j=-b_ja_i$$. Consider the element $$F_n:=\sum_{i=1}^n2a_ib_i.$$

Is it possible to give, for fixed $n$, expanded expressions for $F_n^k$ for $k=2,...,n$? Considering the case $n=2$ and using that powers of the generator vanishes as they anticommute we get $$F_2^2=4a_1b_1a_2b_2+4a_2b_2a_1b_1=8a_1b_1a_2b_2.$$

But what happens in general?

Answer 1

Okay, so you have $$a_ib_ia_jb_j=-a_ia_jb_ib_j=a_ia_jb_jb_i=\cdots=a_jb_ja_ib_i.$$ Therefore, setting $c_i=2a_ib_i$ we get $c_ic_j=c_jc_i$ and $c_i^2=0$, and you want to compute $$(c_1+\cdots+c_n)^k.$$

Note that in the polynomial ring $\mathbb{C}[x_1,\ldots,x_n]$ we have $$ (x_1+\cdots+x_n)^k=\sum_{\substack{r_1,\ldots,r_n\\r_1+\cdots+r_n=k}}{k\choose {r_1,\ldots,r_n}}x_1^{r_1}\cdots x_n^{r_n} $$ where $${k\choose {r_1,\ldots,r_n}}=\frac{k!}{r_1!\cdots r_n!}.$$

Since $c_i^2=0$, we deduce that $(c_1+\cdots c_n)^k=0$ for $k>n$ (since some $r_i>1$) and, for $k<n$, $$ (c_1+\cdots+c_n)^k=\sum_{i_1<i_2<\cdots <i_k}k!c_{i_1}c_{i_2}\cdots c_{i_k}=\sum_{i_1<i_2<\cdots i_k}k!2^ka_{i_1}b_{i_1}a_{i_2}b_{i_2}\cdots a_{i_k}b_{i_k}. $$

Answer 2

We observe that the given rules for multiplying the $a_i$ and $b_j$, which state that any pair of them anticommute, imply

$a_i b_i a_j b_j = -a_i a_j b_i b_j = a_j a_i b_i b_j = -a_j a_i b_j b_i = a_j b_j a_i b_i; \tag 1$

thus, the paired products

$c_i = a_ib_i \tag 2$

all commute with one another. Furthermore, from

$a_i a_j = -a_j a_i \tag 3$

we may write in the usual fashion, since we are in a $\Bbb C$-algebra,

$a_i^2 = -a_i^2 \Longrightarrow 2a_i^2 = 0 \Longrightarrow a_i^2 = 0, \tag 4$

and likewsie,

$b_i^2 = 0; \tag 5$

therefore,

$c_i^2 = a_ib_ia_ib_i = - a_ia_ib_ib_i = -a_i^2 b_i^2 = 0; \tag 6$

thus, in calculating the powers of

$F_n = \displaystyle \sum_1^n 2a_ib_i = 2\sum_1^n a_ib_i = 2\sum_1^n c_i, \tag 7$

we may treat the $c_i$ as nilpotent elements of a commutative algebra over $\Bbb C$; this greatly simplifies operations involved. We note that

$F_n^k = 2^k \left ( \displaystyle \sum_1^n c_i \right )^k; \tag 8$

setting

$G_n = \displaystyle \sum_1^n c_i, \tag 9$

we have

$F_n^k = 2^k G_n^k; \tag{10}$

but $G_n^k$ is easy to compute, since we have

$G_n^k = \left ( \displaystyle \sum_1^n c_i \right )^k = (c_1 + c_2 + \ldots + c_n)^k; \tag{11}$

when we expand the expression on the right we obtain a collection of terms, each a $k$-fold product selected from the $c_i$, so each is of degree $k$ in the $c_i$; furthermore, in the light of (6), the only terms which do not vanish are those in which any given $c_i$ in one of them occurs as a factor precisely once; that is, products of the first powers of $k$ of the $c_i$, $1 \le i \le n$; thus the computation becomes essentially combinatorial in nature, since we need simply count the occurrances of the relevant products of the $c_i$, of which there are a total of $n$, and we will be choosing subsets of the totality having $k$ distinct elements each; and each such subset will contain the $c_i$ for precisely one non-vanishing term of $G_n^k$ if we write it with strictly increasing indices, viz.

$c_{i_1}c_{i_2} c_{i_3} \ldots c_{i_k}; \; i_1 < i_2 < i_3 < \ldots < i_k; \tag{12}$

but in point of fact, there will be $k!$ terms in the product

$(c_1 + c_2 + \ldots + c_n)^k \tag{13}$

drawn from a given $k$-fold subset of the $n$ $c_i$, one for each permutation of the indicies $i_1, i_2, \ldots, i_k$ occurring in (12); therefore we may write

$G_n^k = \displaystyle \sum_{i_1, i_2, \ldots i_k = 1; i_1 < i_2 < \ldots < i_k}^n k!c_{i_1}c_{i_2} \ldots c_{i_k} = k! \left ( \sum_{i_1, i_2, \ldots i_k = 1; i_1 < i_2 < \ldots < i_k}^n c_{i_1}c_{i_2} \ldots c_{i_k} \right ); \tag{14}$

if follows then from (10) that

$F_n^k = 2^k G_n^k = 2^k k! \left ( \sum_{i_1, i_2, \ldots i_k = 1; i_1 < i_2 < \ldots < i_k}^n c_{i_1}c_{i_2} \ldots c_{i_k} \right ). \tag{15}$

For example, with $k = n = 2$ we find

$F_2^2 = 2^2 \cdot 2 c_1c_2 = 8c_1c_2, \tag{16}$

and also,

$G_2^2 = (c_1 + c_2)^2 = c_1^2 + c_1 c_2 + c_2 c_1 + c_2^2 = 2c_1c_2, \tag{17}$

validating

$F_2^2 = 4G_2^2; \tag{18}$

with $n = 3$, $k = 2$ we find

$F_3^2 = 8(c_1c_2 + c_1c_3 + c_2c_3), \tag{19}$

whereas

$G_3^2 = (c_1 + c_2 + c_3)^2 = c_1^2 + c_2^2 + c_3^2 + c_1c_2 + c_1c_3 + c_2c_3 + c_2c_1 + c_3c_1 + c_3c_2$ $= 2(c_1c_2 + c_1c_3 + c_2c_3); \tag{20}$

again,

$F_3^2 = 4G_3^2; \tag{21}$

finally,

$F_3^3 = 8 \cdot 6 c_1c_2c_3 = 48c_1c_2c_3; \tag{22}$

also,

$G_3^3 = (c_1 + c_2 + c_3)^3 = (c_1 + c_2 + c_3)^2(c_1 + c_2 + c_3)$ $= 2(c_1c_2 + c_1c_3 + c_2c_3)(c_1 + c_2 + c_3) = 2(c_1c_2c_3 + c_1c_3c_2 + c_2c_3c_1)$ $= 2(3c_1c_2c_3) = 6c_1c_2c_3; \tag{23}$

and so at last

$F_3^3 = 48c_1c_2c_3 = 8(6c_1c_2c_3) = 8G_3^3. \tag{24}$

Of course, one can back-substitute $a_ib_i = c_i$ into these formulas if so desired.