Alternating function condition is transitive

Statement

Suppose $A$ and $B$ are abelian groups. Suppose $f:A^{n}\to B$ is a multihomomorphism, i.e., it is additive in each variable. Suppose, further, that $f$ is alternating in the $i^{th}$ and $j^{th}$ variable, and also separately is alternating in the $j^{th}$ and $k^{th}$ variable. Then, $f$ is alternating in the $i^{th}$ and $k^{th}$ variable.

Related facts

Similar facts

• Polarization trick
• Alternating and skew-symmetric in pairs with a common variable implies alternating in all three variables: A somewhat stronger version
• Symmetric or skew-symmetric function condition needs to be checked only on a generating set for the symmetric group: This could actually provide an alternative proof if we were in characteristic zero.

Applications

• Equivalence of definitions of alternative ring
• Equivalence of definitions of 2-Engel Lie ring

Facts used

1. Polarization trick: We use the alternating implies skew-symmetric case.
2. Alternating and skew-symmetric in pairs with a common variable implies alternating in all three variables

Proof

Direct proof without using other facts

For simplicity of notation and without loss of generality, we take $i=1,j=2,k=3,n=3$.

Given: $A,B$ abelian groups, $f:A\times A\times A\to B$ multi-additive with $f(x,x,y)=f(x,y,y)=0\ \forall \ x,y\in A$

To prove: $f(x,y,x)=0\ \forall \ x,y\in A$

Proof: By multi-additivity, we have:

$f(x,x+y,x+y)=f(x,x,x)+f(x,x,y)+f(x,y,x)+f(x,y,y)$

The left side $f(x,x+y,x+y)$ is zero by the alternation in the second and third variable. $f(x,x,x)=(f,x,x,y)=0$ by alternation in the first two variables. $f(x,y,y)=0$ by alternation in the second and third variable. This forces $f(x,y,x)=0$ as desired.

Proof based on other facts

We use Fact (1) to show that $f$ is skew-symmetric in the $j^{th}$ and $k^{th}$ variable, then Fact (2) to complete the proof.