Sum of signed permutations of digits equals zero

Question

After playing around with signed permutations lately, (as part of studying properties of antisymmetric tensors and wedge products which are not really related to what I want to ask about), I noticed that apparently for every integer with $3$ digits or more, the signed alternating sum of its digits permutations is equal to zero.

In order to avoid abusing notation, I'll first define $P_{\pi}(n)$ to be the signed and permuted version of an $N$ digits integer via a given permutation $\pi\in{S_N}$:

$$ P_{\pi}(n) = \text{sign}(\pi)\sum_{i=1}^N{n_i10^{\pi_{i}-1}}$$

Where $n_i$ is assumed to be the $i$'th digit, and I define $n_0$ to be the least significant digit, just to ensure that the original number $n$ maintains its original sign (This way, it is "hit" with the identity permutation).

With the above, I can more clearly restate the observation, that for any $n\in\mathbb{N}$ with $N\ge3$ digits:

$$\sum_{\pi\in{S_{N}}}{P_{\pi}(n)=0}$$

Where it is assumed $\pi$ ranges over all possible elements in $S_N$.

In case this is confusing, a concrete example follows:

$$123-132+312-321+231-213=0$$

My question is whether this is a known result, and if so where can I find more information about this property? BTW, I haven't proved this, so I should also ask whether this even holds true or not (although at this point I am rather convinced it is). It should be straightforward to prove via mathematical induction I think, although it may be more enlightening to see a more direct proof (I tried to write one, but those anti symmetrized sums get really cumbersome really quickly and I gave up for now :)).

Answer 1

Nice observation! We have

\begin{align*} \sum_{\pi\in{S_{N}}} \text{sign}(\pi)\sum_{i=1}^N{n_i10^{\pi_{i}-1}} &= \sum_{i=1}^N n_i \sum_{\pi\in{S_{N}}} \text{sign}(\pi){10^{\pi_{i}-1}}\\ &= \sum_{i=1}^N n_i \sum_j 10^{j-1} \sum_{\pi\in{S_{N}| \pi(i) = j}} \text{sign}(\pi) \end{align*}

Now note that every $\pi$ with $\pi(i) = j$ can be written as $\pi = (1j) \tau (1i)$ where $(1j)$ and $(1i)$ are cycles $1\mapsto j \mapsto 1$ and $1 \mapsto i \mapsto 1$, and $\tau(1) = 1$. Then we have $\text{sign}(\pi) = \text{sign}((1j))\text{sign}(\tau)\pi((1i)) (= \text{sign}(\tau)$ unless $j=1$ or $i = 1$.) Note that $\sum_{\tau \in S_n} \text{sign}(\tau) = 0$ since there are an equal number of even and odd permutations (unless $n = 1$). We can think of $\tau$ as a permutation of $2,3, \ldots N$ or $\tau \in S_{N-1}$. So for $N >2$ and any fixed $i,j$, we have

\begin{align*} \sum_{\pi\in{S_{N}| \pi(i) = j}} \text{sign}(\pi) = \sum_{\tau\in{S_{N-1}| \pi(i) = j}} \text{sign}(1j)\text{sign}(1i) \text{sign}(\tau)\\ = \text{sign}(1j)\text{sign}(1i) \sum_{\tau\in{S_{N-1}| \pi(i) = j}}\text{sign}(\tau)=0. \end{align*}

Note that there's nothing too special about taking digits of numbers here; you can extend this for sums of the form $n_1a_1 + \ldots + n_N a_N$ for $a_1, a_2, \ldots, a_N$ arbitrary constants and $n_1, n_2, \ldots n_N$ permuted in all possible ways. In this particular case we take $a_1, a_2,\ldots, a_N = 1, 10, \ldots, 10^{N-1}$.

Answer 2

You have an expression of the form

$$F(x_1, \ldots, x_n; y_1, \ldots y_n ) = \sum_{\sigma \in S_n} \epsilon(\sigma) \sum x_i y_{\sigma(i)}$$

It is alternating separately in $(x_1, \ldots, x_n)$, and $(y_1, \ldots, y_n)$. So it has to be divisible by some products.... But the degree is $2$...

$\bf{Added}$ See what happens with sums $$\sum_{\sigma \in S_n} \epsilon(\sigma)\, (\sum_{i=1}^n x_i y_{\sigma(i)})^d$$

$\bf{Added:}$ We have

$$\sum_{\sigma \in S_n} \epsilon(\sigma)\, (\sum_{i=1}^n x_i y_{\sigma(i)})^d = 0$$ for $0\le d < \binom{n}{2}$ and for $d = \binom{n}{2}$

$$\sum_{\sigma \in S_n} \epsilon(\sigma)\, (\sum_{i=1}^n x_i y_{\sigma(i)})^d = C_n \cdot V(x_1, \ldots, x_n) \cdot V(y_1, \ldots, y_n)$$ where $V(x_i)$ is the Vandermonde polynomial, and $C_n$ is the multinomial coefficient $C_n= \frac{(1+2 + \cdot +n) !}{1! 2! \cdots n!}$.

Answer 3

With juxtaposition not meaning multiplication, here is the idea of a proof by induction. $$\color{red}{abcd}-\color{red}{acbd}+cabd-cbad+bcad-bacd+\color{red}{abdc}-\color{red}{acdb}+cadb-cbda+bcda-badc+\color{red}{adbc}-\color{red}{adcb}+cdab-cdba+bdca-bdac+dabc-dacb+dcab-dcba+dbca-dbac$$

$= \color{red}{3000\times a-3000\times a+(bcd-cbd+bdc-cdb+dbc-dcb)}$ $+ \ 3000\times b-3000\times b+(cad-acd+cda-adc+dca-dac)$ $+ \ 3000\times c-3000\times c+(abd-bad+bda-bdc+dca-dac)$ $+ \ 3000\times d-3000\times d+(abc-acb+cab-cba+bca-bac)$