Consider a Lie group $G$ and a Hilbert space $mathbb{V}$ equipped with a dot product $~~langle~,~rangle$.

Let ${^G{cal{L}}}$ be a space of square-integrable functions mapping $G$ to $mathbb{V}$:

$$

{^G{cal{L}}}=left{ f:~~ Glongrightarrowmathbb{V}qquadBig{|}quad int dg langle f(g), f(g)rangle < infty right}.

$$

We can map each $ f$ to a function of the inverse argument:

$$

f(x)longmapstovarphi(x)equiv f(x^{-1})~,quad xin G

$$

or, in short:

$$

flongmapstovarphiequiv fcirchat{zeta}~,

$$

where

$$

hat{zeta} x,=,x^{-1}

$$

is the inversion operation on $G$.

Doing this for all $f$, we obtain a new space of functions:

$$

{cal{L}}^{G}=left{~varphi:~Glongrightarrow{mathbb{V}}~~Big{|}~~varphi= fcirchat{zeta}~,~~fin{^G{cal{L}}}

right}~.

$$

To examine if $^{G}{cal{L}}$ and ${cal{L}}^{G}$ are copies of the same functional space, we check the square-integrability of the new functions:

$$

int dg langle varphi(g),varphi(g)rangle=int d(g^{-1}) langle varphi(g^{-1}) , varphi(g^{-1})rangle=

int d(g^{-1}) langle f(g),f(g)rangle

$$

For unimodular groups, the measure is invariant and $d(g^{-1}) = dg $, wherefrom

$$

int dg langle varphi(g),varphi(g)rangle=

int dg langle f(g),f(g)rangle < infty~,

$$

so the spaces $^{G}{cal{L}}$ and ${cal{L}}^{G}$ coincide.

For what nonunimodular groups would this outcome stay valid? (Say, for compact ones? For locally compact ones?)

PS.

I understand that, by Haar’s theorem, if a group is locally compact (i.e. if its identity element has a compact neighborhood), then it admits a unique left-invariant and a unique right-invariant measure (unique — up to multiplication by a positive constant). Under the transformation $gto g^{-1}$, a left-invariant measure transforms into a right-invariant one, and vice versa. Can this fact be somehow employed here?