Determinant twists translate GL_r highest weights

Determinant scaling of the alternant. Put $\tau =\lambda +(k^r)$ with the zero-removal convention in (i). Since $k\ge -{\lambda }_r$, the shifted coordinates are weakly decreasing and nonnegative, so $\tau$ is a partition with at most $r$ parts. After padding its coordinates back to length $r$, ${\tau }_j={\lambda }_j+k$. Thus every entry in row $i$ of the alternant matrix for $\lambda +{\rho }_r$ is multiplied by $x_i^k$ to obtain the matrix for $\tau +{\rho }_r$, giving $a_{\tau +{\rho }_r}=(x_1\cdots x_r{)}^k{a}_{\lambda +{\rho }_r}\mathrm{.}$ Dividing by $a_{{\rho }_r}$ in the Laurent rational function field and applying [F2] yields $s_{\tau }(x_1,\dots ,x_r)=(x_1\cdots x_r{)}^k{s}_{\lambda }(x_1,\dots ,x_r)\mathrm{.}$ This identity is valid also for negative $k$; the left side is a polynomial because the shifted coordinates are nonnegative.

By [F1] and step 1.1, $\mathrm{ch}(S_{\lambda }(V)\otimes {\det }^k)=s_{\lambda }(x)(x_1\cdots x_r{)}^k=s_{\tau }(x)=\mathrm{ch}{S}_{\tau }(V)$. Tensoring with the one-dimensional character ${\det }^k$ preserves invariant subspaces: each representing operator is multiplied by a nonzero scalar. Hence $S_{\lambda }(V)\otimes {\det }^k$ is irreducible, and its highest weight is $({\lambda }_1+k,\dots ,{\lambda }_r+k)$, since a highest weight vector is multiplied on the diagonal torus by $(x_1\cdots x_r{)}^k$ and ${\det }^k$ is trivial on the upper unipotent subgroup. The polynomial irreducible $S_{\tau }(V)$ has the same padded highest weight by [F3]. Highest-weight uniqueness in [F3] gives the isomorphism in (i).

By [F3] every irreducible rational module is $S_{\lambda }(V)\otimes {\det }^k$ for a partition $\lambda$ of at most $r$ parts and $k\in \mathbb{Z}$, and conversely each such twist is irreducible of highest weight $({\lambda }_1+k,\dots ,{\lambda }_r+k)$. For a dominant integral highest weight $\eta$, take $k={\eta }_r$ and form $\bar{\eta }$ by deleting the trailing zeros of $({\eta }_1-{\eta }_r,\dots ,{\eta }_r-{\eta }_r)$. This is a partition, including $\mathrm{\varnothing }$ when all coordinates vanish, and its padded coordinates satisfy ${\bar{\eta }}_j+k={\eta }_j$. Thus $S_{\bar{\eta }}(V)\otimes {\det }^{{\eta }_r}$ has highest weight $\eta$ and is the required irreducible by [F3]. The parametrisation with padded last coordinate ${\lambda }_r=0$ is unique: then $k={\eta }_r$ and the remaining positive coordinates determine $\lambda$. Arbitrary pairs $(\lambda ,k)$ need not be unique.