Possible values of symmetric functions evaluated on quaternions

$$\DeclareMathOperator\sym{sym}$$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: $$e^{(i+j+k)/\sqrt3}=\sum_n (\sqrt{3}^n/n!)\sum_{a,b,c} \sym_{a,b,c}(i,j,k) =(i+j+k)\sin(90)=i+j+k$$.

• These are not what are normally called "symmetric polynomials" (but I don't know a better term for them). – Sam Hopkins Apr 22 '19 at 0:27
• @SamHopkins Ok I edited it. – Name displayed Apr 22 '19 at 15:31
• 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. – Kimball Apr 22 '19 at 16:52
• @Name displayed I think Sam Hopkins' point was that they're not what is usually called symmetric (they are polynomials) – Jules Lamers Apr 22 '19 at 22:21
• In $\operatorname{sym}_{a, 0, 0}(i, j, k) = i^n$, $n$ and $a$ should be the same. – 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.