How to make an expression manifestly symmetric

Question

Believe it or not, the following expression is symmetric under the exchange of the indices $j$ and $k$, i.e. $R_{kj}=R_{jk}$: $$ R_{jk}=j s_js_k-\sum_{n=1}^{\min(N-k,j)}(k-j+2n)s_{j-n}s_{k+n} $$ Where $1\leq j\leq N$ and $1\leq k\leq N$. The symmetry is far from obvious to me. Is there a way of rewriting an expression like this into a manifestly symmetric form?

EDIT: I know $R_{jk}$ is symmetric, I don't need to prove it. Moreover, the symmetrization "by force" (i.e. writing $R_{jk}=\frac12(R_{jk}+R_{kj})$) does not seem to give me a simpler formula, but I might just not be doing it right...

Answer 1

Let $s=k+j,d=k-j$, then we want to show that $R_{s,d}=R_{s,-d}$
Let $P_{s,d}=s_js_k=s_ks_j=P_{s,-d}$
$k=\frac{s+d}{2},j=\frac{s-d}{2}$, let $m=(d/2)+n$ $$R_{s,d}=\frac{s-d}{2}P_{s,d}-\sum_{m=\frac{d}{2}+1}^{\min(N-s/2,s/2)}2mP_{s,2m}$$ Replace $d$ by $-d$, $$R_{s,-d}=\frac{s+d}{2}P_{s,-d}-\sum_{m=-\frac{d}{2}+1}^{\min(N-s/2,s/2)}2mP_{s,2m}$$ $$R_{s,-d}-R_{s,d}=dP_{s,d}-\sum_{m=-\frac{d}{2}+1}^{\frac{d}{2}}2mP_{s,2m}$$ By symmetry about zero, most terms in the sum cancel, leaving just the $m=d/2$ term, $dP_{s,d}$

Answer 2

There's a simple way to do this: define

$$ \tilde{R}_{jk} = \frac{1}{2}\left( R_{jk} + R_{kj} \right) $$

Then $\tilde{R}_{jk}$ is manifestly symmetric. And if $R_{jk}$ truly is symmetric too, then you will have $R_{jk} = \tilde{R}_{jk}$.

A common way to prove that $R_{jk}$ is symmetric is to compute its antisymmetric part:

$$ \hat{R}_{jk} = \frac{1}{2}\left( R_{jk} - R_{kj} \right) $$

and show that $\hat{R}_{jk} = 0$.

($\tilde{R}$ and $\hat{R}$ are just decorations I've added for the purposes of this answer; they are not standard notations for the symmetrization and antisymmetrization of $R$)