Function Spaces on Group Decompositions

Well-Definedness.

Theorem (Well-Definedness). Let $G=AB$ with $A,B\le G$ and let $\delta :G\to {\mathbb{C}}^{\times }$ be a homomorphism. For a function $\varphi :B\to \mathbb{C}$, define a function f on $G$ by

$$f(ab):=\delta (a)\varphi (b)(a\in A,b\in B).$$

Then $f$ is well-defined on $G$ if and only if

$$\varphi (xb)=\delta (x)\varphi (b)\text{for all }x\in A\cap B,b\in B.$$

Proof. ($\Rightarrow$) If $f$ is well-defined and $x\in A\cap B$, then

$$\delta (1)\varphi (xb)=f(1\cdot xb)=f(x\cdot b)=\delta (x)\varphi (b),$$

so the relation holds. ($\Leftarrow$) Suppose $\varphi (xb)=\delta (x)\varphi (b)$ for all $x\in A\cap B$. If $ab=a^{\prime }{b}^{\prime }$ with $a^{\prime }=ax$ and $b^{\prime }=x^{-1}b$ for some $x\in A\cap B$, then

$$\delta (a^{\prime })\varphi (b^{\prime })=\delta (ax)\varphi (x^{-1}b)=\delta (a)\delta (x)\delta (x{)}^{-1}\varphi (b)=\delta (a)\varphi (b),$$

so $f$ is independent of the chosen decomposition. Uniqueness is immediate. $\square$

Structure and Basis.

Theorem (Structure and Basis). Let

$${\Phi }_{\delta }(B):=\{\varphi :B\to \mathbb{C}\mid \varphi (xb)=\delta (x)\varphi (b)\forall x\in A\cap B,b\in B\},$$

and let $V_{\delta }:=\{f:G\to \mathbb{C}\mid \exists \varphi \in {\Phi }_{\delta }(B)\text{ with }f(ab)=\delta (a)\varphi (b)\}$. The map

$$T:{\Phi }_{\delta }(B)\to V_{\delta },T(\varphi )(ab)=\delta (a)\varphi (b),$$

is a linear isomorphism. If $S\subset B$ is a set of representatives for the left cosets $(A\cap B)\backslash B$, then the family $\{{\varphi }_s{\}}_{s\in S}$ defined by

$${\varphi }_s(x{s}^{\prime })=\{\begin{array}{ll}\delta (x),& s^{\prime }=s,\\ 0,& s^{\prime }\ne s,\end{array}(x\in A\cap B,s^{\prime }\in S),$$

is a basis of ${\Phi }_{\delta }(B)$. Consequently, $\{f_s:=T({\varphi }_s){\}}_{s\in S}$ is a basis of $V_{\delta }$. In particular,

$$\dim V_{\delta }=|(A\cap B)\backslash B|(\text{finite or infinite}).$$

Proof. $T$ is well-defined and injective by the previous theorem. Surjectivity is by the definition of $V_{\delta }$. For the basis, first ${\varphi }_s\in {\Phi }_{\delta }(B)$ since for $x,x_0\in A\cap B$ and $s^{\prime }\in S$, letting $b=x_0{s}^{\prime }$, we check the condition:

$${\varphi }_s(x(x_0{s}^{\prime }))=\{\begin{array}{ll}\delta (x{x}_0),& s^{\prime }=s\\ 0,& s^{\prime }\ne s\end{array}=\delta (x){\varphi }_s(x_0{s}^{\prime }).$$

Spanning: for $\varphi \in {\Phi }_{\delta }(B)$ set $c_s:=\varphi (s)$. For $b=x{s}^{\prime }$ with $x\in A\cap B$ and $s^{\prime }\in S$, we have ${\sum }_{s\in S}{c}_s{\varphi }_s=\varphi$ because

$$\left(\sum _{s\in S}{c}_s{\varphi }_s\right)(b)=c_{s^{\prime }}\delta (x)=\varphi (s^{\prime })\delta (x)=\varphi (x{s}^{\prime })=\varphi (b).$$

Independence: if ${\sum }_{s\in S}{c}_s{\varphi }_s=0$, evaluate at $s_0\in S$ to get $c_{s_0}=0$. Hence all $c_s=0$. Mapping by $T$ preserves linear relations, giving the basis of $V_{\delta }$. $\square$