Contragredient (dual) rational representation

Let $(V,r)$ be a finite-dimensional rational representation of an affine group scheme $G$ over $k$ (Rational representations and comodules of an affine group scheme). The contragredient representation $r^{\vee }$ is the representation on the algebraic dual $V^{\ast }={\mathrm{Hom}}_k(V,k)$ (Linear functionals and the algebraic dual $V^{\ast }=\mathcal{L}(V,F)$) defined by $(r^{\vee }(g)f)(v)=f(r(g{)}^{-1}v)$ for $g\in G(R)$, $f\in V_R^{\ast }$ and $v\in V_R$. When $V$ is finite-dimensional, $r^{\vee }$ is a rational representation and its comodule is the dual comodule of $(V,r)$; moreover the weights of $(V^{\ast },r^{\vee })$ are the negatives of the weights of $(V,r)$.

• Left action. For $R$-points $g,h$ and $f\in V_R^{\ast }$ one has $(r^{\vee }(g)(r^{\vee }(h)f))(v)=(r^{\vee }(h)f)(r(g{)}^{-1}v)=f(r(h{)}^{-1}r(g{)}^{-1}v)=f(r(gh{)}^{-1}v)=(r^{\vee }(gh)f)(v)$ for all $v\in V_R$, since $r$ is a representation and $(gh{)}^{-1}=h^{-1}{g}^{-1}$; hence $r^{\vee }$ is a left action by $R$-linear automorphisms of $V_R^{\ast }$. The inverse in the formula is what makes the action a left action, and it is also the reason that the contragredient of a contragredient recovers the original representation.
• Rationality in the finite-dimensional case. Choose a basis $e_1,\dots ,e_n$ of $V$, its dual basis $e_1^{\ast },\dots ,e_n^{\ast }$, and write $\rho (e_j)={\sum }_i{e}_i\otimes a_{ij}$. The dual coaction is ${\rho }^{\vee }(e_i^{\ast })={\sum }_j{e}_j^{\ast }\otimes S(a_{ij}),$ where $S$ is the antipode of $O(G)$. Evaluating at $g\in G(R)$ gives the transpose of $r_R(g{)}^{-1}$, so its action is the displayed formula. The inverse and transpose matrix identities give the group law naturally in $R$, hence the comodule identities by the representation/comodule dictionary. All matrix entries are regular functions on $G$.
• Weights. If $V$ is finite-dimensional and $T$ is a diagonalizable group acting on $V$ with weight spaces $V_{\chi }$, then the dual basis of a basis of $V_{\chi }$ spans the weight space $(V^{\ast }{)}^{-\chi }$: for $f\in (V^{\ast }{)}^{-\chi }$ and $v\in V_{\chi }$ the pairing is compatible with the dual action, so the weights of $V^{\ast }$ are exactly the $-\chi$ with $V_{\chi }\mathrm{\ne }0$. This is the fact used for the contragredient of a simple module.
• Infinite-dimensional case. For arbitrary $V$, the formula $f\mapsto f\circ r_k(g{)}^{-1}$ defines an action of the abstract group $G(k)$ on the full algebraic dual. It need not be rational and need not extend to the module $V^{\ast }{\otimes }_{k}R$ for every $k$-algebra $R$. For example, let $G={\mathbf{G}}_m$, $V={⨁}_{n\ge 0}k{e}_n$ with $e_n$ of weight $n$, and $f(e_n)=1$. Over $R=k[t,t^{-1}]$, precomposition by the universal point gives values $t^{-n}$, which cannot lie in $V^{\ast }{\otimes }_{k}R$: values of any element of that tensor product span a finite-dimensional $k$-subspace of $R$. Thus the rational contragredient above is stated for finite-dimensional representations, exactly the range used by its consumers.