# scalar map

Given a ring $R$, a left $R$-module $U$, a right $R$-module $V$ and a two-sided $R$-module $W$ then a map $b:U\times V\to W$ is an $R$-scalar map if

1. 1.

$b$ is biadditive, that is $b(u+u^{\prime},v)=b(u,v)+b(u^{\prime},v)$ and $b(u,v+v^{\prime})=b(u,v)+b(u,v^{\prime})$ for all $u,u^{\prime}\in U$ and $v,v^{\prime}\in V$;

2. 2.

$b(ru,v)=rb(u,v)$ and $b(u,vr)=b(u,v)r$ for all $u\in U$, $v\in V$ and $r\in R$.

Such maps can also be called outer linear.

Unlike bilinear maps, scalar maps do not force a commutative multiplication on $R$ even when the map is non-degenerate and the modules are faithful. For example, if $A$ is an associative ring then the multiplication of $A$, $b:A\times A\to A$ is a $A$-outer linear:

 $b(xy,z)=(xy)z=x(yz)=xb(y,z)$

and likewise $b(x,yz)=b(x,y)z$. Using a non-commutative ring $A$ confirms the claim.

It is immediate however that $\langle b(U,V)\rangle$ is in fact an $R$-bimodule. This is because:

 $s(b(u,v)r)=sb(u,vr)=b(su,vr)=sb(u,vr)=(sb(u,v))r$

for all $u\in U$, $v\in V$ and $s,r\in R$. Therefore it is not uncommon to require that indeed all of $W$ be an $R$-bimodule.

Title scalar map ScalarMap 2013-03-22 17:24:22 2013-03-22 17:24:22 Algeboy (12884) Algeboy (12884) 7 Algeboy (12884) Definition msc 13C99 outer linear BilinearMap scalar map