Polarization trick

Statement

Statement for a biadditive function and a single repeated variable

Suppose $f$ is a biadditive function of two variables both from a set $T$ and:

$f(x,x)=0\ \forall \ x\in T$

Then, we have:

$f(x,y)+f(y,x)=0\ \forall \ x,y\in T$

Note that the converse implication need not be true in general, but it is true if the range of $f$ is 2-torsion-free.

This particular case is so important that it has a name. A (biadditive) function satisfying the first condition is termed alternating and a (biadditive) function satisfying the second condition is termed skew-symmetric.

Statement for a multiadditive function and a single repeated variable

Suppose $f$ is a function of $n$ variables all from a set $T$ that is additive in each variable. Suppose we have that:

$f(x,x,\dots ,x)=0\ \forall \ x\in T$

Denote by $S_{n}$ the symmetric group on the set $\{1,2,\dots ,n\}$. We have:

$\sum _{\sigma \in S_{n}}f(x_{\sigma (1)},x_{\sigma (2)},\dots ,x_{\sigma (n)})=0$

Note that the converse implication need not be true in general, but it is true if the range of $f$ has torsion-free threshold at least $n$.

General statement

Suppose $f$ is a function of $n$ variables all from a set $T$ that is additive in each variable. Consider a function:

$\{1,2,\dots ,n\}=A_{1}\sqcup A_{2}\sqcup A_{m}$

where the $A_{i}$ are pairwise disjoint nonempty subsets. Note that we can define a function:

$g:\{1,2,\dots ,n\}\to \{1,2,\dots ,m\}$

that sends a given element $i$ to the $j$ such that $x_{i}\in A_{j}$.

Suppose $f$ satisfies the identity:

$f(x_{g(1)},x_{g(2)},\dots ,x_{g(n)})=0\ \forall x_{1},x_{2},\dots ,x_{m}\in T$

Then, the following is true:

$\sum _{\sigma \in \prod _{j=1}^{m}\operatorname {Sym} (A_{j})}f(x_{\sigma (1)},x_{\sigma (2)},\dots ,x_{\sigma (n)})=0\ \forall x_{1},x_{2},\dots ,x_{n}\in T$

Related facts

• Alternating function condition is transitive
• Symmetric or skew-symmetric function condition needs to be checked only on a generating set for the symmetric group