Let $H$ be a subgroup of $\textrm{Isom}(\mathbb{R^n})$ And let $O(n)$ be the orthogonal group. Let $T_v$ be the translation by $v$.
If we have the following application : $\Phi : H \rightarrow O(n)$ s.t. $h \mapsto T_{-v} ∘ h ∘ T_v$
I want to see that it is well defined. My professor simply wrote the following:
$(T_{-v}∘h∘T_v)(0) = T_{-v}(h(v)) = v-v = 0$
My question: How is it enough to show that $\Phi(0) = 0$ in order to see that $\Phi$ is well defined?
the composition of isometric are an isometric, so it is of the form $v\rightarrow Av+w$ for som fixed $w$ and $A\in O(n)$, so this isometric is an element of $O(n)$ if and if $w=0$ , this equivalente that this isometric fixe $0$.
what happen if $h=T_v$ if you suppose that $ v\not=0$? this answer that $\Phi $ not always defined.
if $H$ is a subgroup of euclidien groupe that fixed $v$ then for $h\in H$ we have $(T_{-v}ohoT_v)(0)=0$ and so $(T_{-v}ohoT_v)\in O(n)$ and $\Phi $ well defined.
$\newcommand{\Reals}{\mathbf{R}}$Loosely, a relation $F$ is "well-defined" (or "defines a function") if $x_{1} = x_{2}$ implies $F(x_{1}) = F(x_{2})$. (Compare carefully with the definition of an "injective mapping".)
Usually, the term "well-defined" arises when you have a mapping $f:X \to Y$ and an equivalence relation $R$ on $X$, and wish to define an "induced mapping" $F:X/R \to Y$ via $$ F([x]) = f(x). \tag{1} $$ The issue is to show that if $[x_{1}] = [x_{2}]$, i.e., if $x_{1}$ and $x_{2}$ are in the same $R$-equivalence class, then $f(x_{1}) = f(x_{2})$, so that (1) is well-defined.
Here, the situation is a bit different: You have a mapping $\Phi$ from $H$ to the isometry group of $\Reals^{n}$, and are writing$\Phi:H \to O(n)$. This mapping (conjugation by a translation) is obviously well-defined in the preceding sense, but it's not a priori obvious that the image of $\Phi$ is contained in the orthogonal group.
As you surely know, a Euclidean isometry that fixes the origin is an orthogonal transformation, so showing $\Phi(h)(0) = 0$ is sufficient to show $\Phi(h) \in O(n)$. It looks to me that's what your instructor meant by "well-defined". (I don't, however, understand why $h(v) = v$ based on what's stated in the question.)
