g=\chi^{-1}(x)g$ for all $x \in H$. Hence $hg \in I$ and is $H$–invariant and not zero. For the converse, if we have an extendible character $\chi$, we have an element $0 \neq f \in \Bbbk[G]$ that is $\chi$-semi invariant and the associated principal ideal $I=\Bbbk[G]f$ is not zero and $H$–stable. By hypothesis, we can find $hf=g \in I$ with the property that $fg=h=x\cdot h=(x\cdot f) (x
  \cdot g)=(\chi(x)f )(x\cdot g)$. If $G$ is connected we can cancel $f
  \neq 0$ and we have that $x \cdot g=\chi^{-1}(x)g$. So that $\chi^{-1}$ is extendible and in accordance with Theorem \[theo:obse2\], the group $H$ is observable. The case that $G$ is not connected can be proved following the same methods. The second assertion is basically a reformulation of the first (ideal–theoretical) characterization of observability in geometric terms.

\[obse:obse3\] If $H$ is observable in $G$ –with $G$ connected– then ${}^H[\Bbbk[G]]=[{}^H\Bbbk[G]]$ (compare with [@kn:nosotros2 Lemma 11.5.4] where this result is proved and also the converse). It is clear that in general $[{}^H \Bbbk[G]] \subseteq
  {}^H[\Bbbk[G]]$. Conversely, take $0\neq g \in {}^H[\Bbbk[G]]$ and consider the $H$–stable ideal $I_g=\Bbbk[G]g \cap \Bbbk[G]$. In accordance with Theorem \[theo:charcinv\] we can find a non zero polynomial $f_1$ in $I_g$ that 