# If $(f(x_1,…,x_n))^2$ is symmetric in $x_1,..,x_n$, is $f(x_1,…,x_n)$ necessarily symmetric?

Consider a polynomial in $n$ variables: $f(x_1,...,x_n)$. If the polynomial $(f)^2$ is symmetric in $x_1,..,x_n$, does it follow that $f$ is also symmetric in $x_1,..,x_n$? Generalize for higher exponents.

I just made up this problem. I think it is closely related to the fundamental theorem of elementary symmetric polynomials, but I have as of now no idea how to approach it.

Interesting question, but your conjecture is not right (but almost!). The correct theorem is:

Is $\require{cancel} f \cancel{\equiv} 0$ a polynomial in $n$ variables and $(f)^k$ is symmetric, then

$f \text{ is }\begin{cases} \text{symmetric} & \text{for k odd} \\ \text{symmetric or antisymmetric} & \text{for k even} \end{cases}$

It is clear that for any antisymmetric polynomial $a$, then $a^{2l}$ is symmetric for $l \in \mathbb N$.

I won't post here tedious calculations in order to show the theorem, because I don't think this is the right place for it. Instead I'll provide a sketch of the proof and examples for convenience.

Lemma: Any polynomial $f$ can be represented as the sum of a symmetric $s$ and an antisymmetric $a$ part.

It is easy to prove this lemma. The symmetric part is $s=\frac{S(f)}{n!}$, where $S$ is the symmetrization operator. One has to show that $a = f - s$ is antisymmetric, which is true if and only if $S(a) = 0$, but this is a trivial task, since the $S$ operator is linear and $S(g) = n! \ g$ for any symmetric polynomial $g$ in $n$ variables.

The theorem is shown by expanding $(s+a)^k$ and setting to $0$ the antisymmetric part. For example:

$$(s+a)^2 = \underbrace{s^2}_{\text{symmetric}} + \underbrace{2sa}_{\text{antisymmetric}} + \underbrace{a^2}_{\text{symmetric}}$$

Thus $2sa = 0$, which directly implies that $f$ is either symmetric ($a=0$) or antisymmetric ($s=0$).

This analysis becomes harder for higher powers. For example, the antisymmetric part of $(s+a)^3$ and $(s+a)^4$ is $3 s^2 a + a^3 = a(3 s^2 + a^2)$ and $4 s^3 a + 4 s a^3 = 4 s a (s^2 + a^2)$ respectively. In general, from setting this part to $0$ follows that for odd/even exponents ($a=0$)/($a=0$ or $s=0$) is a sufficient condition. The challenge is to show that the remaining factor cannot be $0$, taking the examples above, that $3 s^2 + a^2 = 0$ and $s^2 + a^2 = 0$ have no solutions.

The antisymmetric part of $(s+a)^k$, as we saw, can be factored in two or three factors. One will be always $a$, for $k$ even there will be a factor $s$, and another factor that we will call $t$. Note that $t$ is a homogeneous polynomial of degree $m$, where $m$ is $k-1$ or $k-2$ for $k$ odd or even respectively. Note that $t$ is a sum of monomials, where one and only one of these is $c a^m$ and another one and only one is $d s^m$, with $c,d$ being constants. All other monomials will be "mixed" (i.e. $\propto s^j a ^{m-j}$). Considering $a(x_1;x_2,x_3,...)$ and $s(x_1;x_2,x_3,...)$ as polinomials in only the first variable, $x_1$, we can proof that for any root $r$ of $a$, we get $(s(r))^m=0$, and viceversa, so, in principle $s$ and $a$ have the same roots. Moreover, by substituting $a = a' \frac{a}{(x_1-r)^u}$, we can show that the roots have the same multiplicity in both $a$ and $s$ by using the same principle. Then $s=a$, thus $s=a=0$, which contradicts our initial assumption that $f$ is not $0$.

For example:

$$f^4 = (s+a)^5 = \underbrace{s^5}_{\text{sym}} + \underbrace{5 s^4 a}_{\text{asym}} + \underbrace{10 s^3 a^2}_{\text{sym}} + \underbrace{10 s^2 a^3}_{\text{asym}} + \underbrace{10 s a^4}_{\text{sym}} + \underbrace{a^5}_{\text{asym}}$$

$$\rightarrow 5 s^4 a + 10 s^2 a^3 + a^5 = a(5 s^4 + 10 s^2 a^2 + a^4) = 0$$

From $5 s^4 + 10 s^2 a^2 + a^4 = 0$ follows that $s=a$, a contradiction, thus we have $a=0$.

• Oops, sorry, this was totally wrong. – darij grinberg Mar 15 '15 at 16:01
• On the other hand, here is a non-tedious proof. Let $f$ be a polynomial in $n$ variables $x_1, x_2, \ldots, x_n$ over a field of characteristic $0$, and $k$ be a positive integer such that $f^k$ (the $k$-th power of $f$) is symmetric. We need to show that $f$ is symmetric or antisymmetric, and actually is symmetric if $k$ is odd. For every $i \in \left\{1,2,\ldots,n-1\right\}$, let $f_i$ denote the polynomial obtained from $f$ by switching the variables $x_i$ and $x_{i+1}$. Then, it is ... – darij grinberg Mar 15 '15 at 16:04
• ... enough to show that $f = \lambda f_i$ for all $i$, where $\lambda$ is an element of $\left\{1,-1\right\}$ independent of $i$. We know that $f^k$ is symmetric; thus, $f^k = f_i^k$ for all $i$. Thus, in the field of rational functions, we have $\left(f/f_i\right)^k = 1$ for all $i$. Hence, for all $i$, the rational function $f/f_i$ is a constant; call this constant $\lambda_i$. We then have $f = \lambda_i f_i$. Switching $x_i$ and $x_{i+1}$ in this equality, we obtain $f_i = \lambda_i f$. Combined with $f = \lambda_i f_i$, this yields $\lambda_i \in \left\{1,-1\right\}$ (here we ... – darij grinberg Mar 15 '15 at 16:07
• ... WLOG assume that $f \neq 0$, since in this case everything is obvious). So we know that $f = \lambda_i f_i$ for all $i$. It remains to prove that the $\lambda_i$ are all equal, and that they all equal $1$ if $k$ is odd. Here is a way to do this: For each $i \in \left\{1,2,\ldots,n-2\right\}$, it is easy to see that $\left(\left(f_i\right)_{i+1}\right)_i = \left(\left(f_{i+1}\right)_i\right)_{i+1}$, where we generally use the notation $g_k$ for the result of switching $x_k$ with $x_{k+1}$ in the polynomial $g$. Now, rewriting this using the ... – darij grinberg Mar 15 '15 at 16:10
• ... formulas $f_i = \lambda_i f$ and $f_{i+1} = \lambda_{i+1} f$, we obtain $\lambda_i \lambda_{i+1} \lambda_i f = \lambda_{i+1} \lambda_i \lambda_{i+1} f$. After cancelling, this becomes $\lambda_i = \lambda_{i+1}$. Since this holds for all $i \leq n-2$, we thus obtain $\lambda_1 = \lambda_2 = \cdots = \lambda_{n-1}$. I'll let you figure out the rest. – darij grinberg Mar 15 '15 at 16:11

No, take $f(x_1,x_2)=x_1-x_2$ or more generally $\prod_{1\leq i<j\leq n}(x_i-x_j)$. You always have a sign change which disappears upon squaring.

• Oops I should have thought for more than 1 second about this... – Joshua Benabou Mar 13 '15 at 4:15