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What is the general solution to $f(\vec{x_1}+\vec{x_2})=f(\vec{x_1})f(\vec{x_2})$ where $\vec{x}$'s are in discrete vector space $x\in \{n_1\vec{e_1}+n_2\vec{e_2}+n_3\vec{e_3},n_1,n_2,n_3 \in Z\}$?

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Is $f$ required to be linear? –  AndreasT Mar 7 '13 at 10:11
    
$f:Z\oplus Z\oplus Z\to {\rm where?}$ –  Boris Novikov Mar 7 '13 at 10:14
    
@ Boris, Thanks. To $C$ –  richard Mar 7 '13 at 10:18
    
@ Andreas, I think No. –  richard Mar 7 '13 at 10:22

2 Answers 2

$f\in {\rm Hom}(\bigoplus_{i=1}^3Z,C(\cdot))\cong \bigoplus_{i=1}^3{\rm Hom}(Z,C(\cdot))\cong \bigoplus_{i=1}^3C(\cdot)$

Here $C(\cdot)$ is the multiplicative group of $C$.

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Thanks Boris... –  richard Mar 7 '13 at 12:56

For $\vec x_1=\vec x_2=\vec 0$ you have $$ f(\vec 0+\vec 0) = f(\vec 0)f(\vec 0) \quad\text{so}\quad f(\vec 0)=f(\vec 0)^2 $$ Therefore, either $f(\vec 0)=0$ or $f(\vec 0)=1$.

  1. If $f(\vec 0)=0$, then for every $\vec x$ you have $f(\vec x)=f(\vec x+\vec 0)=f(\vec x)f(\vec 0)=0$. So in this case $f\equiv 0$ (note that such $f$ satisfies the assumptions).

  2. If $f(\vec 0)=1$, then for every $\vec x$ you have $$ 1=f(\vec x-\vec x)=f(\vec x)f(-\vec x) $$ therefore $\forall x~f(\vec x)\neq 0$. If $\vec x=n_1\vec e_1+n_2\vec e_2+n_3\vec e_3$ then $$ f(\vec x) ~=~ f(\vec e_1)^{n_1}\cdot f(\vec e_2)^{n_2}\cdot f(\vec e_3)^{n_3} $$ so $f$ is fully determined by the image of a basis. Conversely, chosing 3 real numbers (or complex, or whatever the codomain of $f$ is) not null $f_1,f_2,f_3$ you have that $$\tag{F} f(n_1\vec e_1+n_2\vec e_2+n_3\vec e_3) ~=~ f_1^{n_1}\cdot f_2^{n_2}\cdot f_3^{n_3} $$ works.

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Andreas, Perfect –  richard Mar 7 '13 at 13:05

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