$\DeclareMathOperator\sym{sym}$Let $i$, $j$, $k$ be the units of quaternions, in particular $i^2=j^2=k^2=-1$, $ijk=-1$.

We will use non commutative variables $x$, $y$, $z$. Define $\sym_{a,b,c}$ to be the polynomial made of the sum of monomials which are all possible products of $a$ variables $x$, $b$ variables $y$ and $c$ variables $z$.

For example $\sym_{2,1,0}(i,j,k)=i^2 j+i j i+j i^2$.

Considering the symmetric definition of $\sym_{a,b,c}$ and the non commutativity of quaternions I would expect a lot of simplifications on $\sym_{a,b,c}(i,j,k)$, in fact $\sym_{a,0,0}(i,j,k)=i^a$, and $\sym_{1,1,1}(i,j,k)=0$ but $\sym_{2,2,0}(i,j,k)=i^2 j^2+j^2 i^2=2$.

Is there a simple formula to determine the value of $\sym_{a,b,c}(i,j,k)$ in terms of $a$, $b$, $c$?

May/29/2019 By the way I introduced those functions we have: $e^{(i+j+k)/\sqrt3}=\sum_n (\sqrt{3}^n/n!)\sum_{a,b,c} \sym_{a,b,c}(i,j,k) =(i+j+k)\sin(90)=i+j+k$.

  • $\begingroup$ These are not what are normally called "symmetric polynomials" (but I don't know a better term for them). $\endgroup$ – Sam Hopkins Apr 22 '19 at 0:27
  • $\begingroup$ @SamHopkins Ok I edited it. $\endgroup$ – Name displayed Apr 22 '19 at 15:31
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    $\begingroup$ Note that the product of $a$ copies of $i$, $b$ copies of $j$ and $c$ copies of $k$, in any order, is determined up to sign by $a$, $b$ and $c$. Since the sign alternates with transpositions, you should be able to give a combinatorial argument to obtain the formula. $\endgroup$ – Kimball Apr 22 '19 at 16:52
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    $\begingroup$ @Name displayed I think Sam Hopkins' point was that they're not what is usually called symmetric (they are polynomials) $\endgroup$ – Jules Lamers Apr 22 '19 at 22:21
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    $\begingroup$ In $\operatorname{sym}_{a, 0, 0}(i, j, k) = i^n$, $n$ and $a$ should be the same. $\endgroup$ – LSpice Feb 24 '20 at 22:22

Careful counting gives the following formula: $$ \operatorname{sym}_{a,b,c}(i,j,k) = {\left\lfloor \frac{a}{2}\right\rfloor +\left\lfloor \frac{b}{2}\right\rfloor +\left\lfloor \frac{c}{2}\right\rfloor \choose \left\lfloor \frac{a}{2}\right\rfloor ,\left\lfloor \frac{b}{2}\right\rfloor ,\left\lfloor \frac{c}{2}\right\rfloor }\cdot \delta(a,b,c) \cdot i^a j^b k^c $$ where $$ \delta(a,b,c) = \left(1-(a\%2)\cdot(b\%2)-(b\%2)\cdot(c\%2)-(c\%2)\cdot(a\%2)+2(a\%2)(b\%2)(c\%2)\right) $$ is $0$ when at least two of $a$, $b$, $c$ are odd, and $1$ otherwise. One might further evaluate using $$ i^aj^bk^c = (-1)^{bc+\frac{c(c-1)}{2}+\left\lfloor \frac{a+c}{2}\right\rfloor +\left\lfloor \frac{b+c}{2}\right\rfloor }\cdot i^{(a+c)\%2}j^{(b+c)\%2} $$ to actually get a multiple of one of the basis elements.


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