Reconstruction from the inertia component

The space $W$ is nonzero and $I$-stable. Its inclusion into $V{\mathrm{\mid }}_I$ has a corresponding $G$-map under adjunction. In the stated function model this map is $\mathrm{\Phi }$: replacing $t$ by $ti$ leaves $tif(ti)=tf(t)$ unchanged. For $g\in G$, write $g^{-1}t=t^{\mathrm{\prime }}i$; then $t(gf)(t)=t{i}^{-1}f(t^{\mathrm{\prime }})=g{t}^{\mathrm{\prime }}f(t^{\mathrm{\prime }})$, so reindexing gives $\mathrm{\Phi }(gf)=g\mathrm{\Phi }(f)$.

By transversal evaluation, the functions supported on $tI$ form a copy of $W$, and $\mathrm{\Phi }$ restricts there to the invertible linear map $w\mapsto tw$ onto $tW=V_{{}^t\theta }$. Distinct left cosets give distinct types, and the orbit result says these are all components of $V{\mathrm{\mid }}_N$. Their sum is direct, so $\mathrm{\Phi }$ is bijective before any irreducibility of $W$ is asserted.

Let $U\subseteq W$ be an $I$-submodule. The direct sum ${\sum }_{t\in T}tU$ is $G$-stable: if $gt=t^{\mathrm{\prime }}i$ then $g(tU)=t^{\mathrm{\prime }}U$. For $U\mathrm{\ne }0$ it is nonzero and hence equals $V$. Its dimension is $[G:I]\dim U$, whereas step 2.1 gives $\dim V=[G:I]\dim W$. Therefore $U=W$, proving irreducibility. This argument allows $T=\{1\}$ and $I=N$ without change.