Let $G$ be a finite group and let $\phi$ be an irreducible character of $G$ over $\mathbb C$. How does the Wedderburn component of $\mathbb C[G]$ corresponding to the contragredient character of $\phi$ look like? Is there an explicit description?
Thank you.
I'm not sure what kind of answer you are looking for. Over $\mathbb{C}$, all the components are just matrix algebras, so they are determined by their degree. But a representation and its dual have the same degree. So the two components are isomorphic as $\mathbb{C}$-algebras.
The question is : what does it mean to describe a component of $\mathbb{C}[G]$ ? Does it mean to describe it as an algebra ? The strongest thing I can see is to describe it as an algebra equipped with its canonical projection from $\mathbb{C}[G]$.
Now if you write $(V_i,\rho_i)_{i\in I}$ for the distinct irreducible representations of $G$, you get $\psi_i : \mathbb{C}[G]\to Hom_\mathbb{C}(V_i)$ induced by $\rho_i$ for all $i\in I$, and the $\psi_i$ give by definition of a product a morphism $\Psi: \mathbb{C}[G]\to \prod_{i\in I} Hom_\mathbb{C}(V_i)$. You can say that most of the basic theory of linear representations of finite groups over $\mathbb{C}$ is contained in the following statement : $\Psi$ is an isomorphism of $\mathbb{C}$-algebras.
Now if $(V,\phi)$ is one of the $(V_i,\rho_i)$, then you have (as for any vector space) an anti-isomorphism of $\mathbb{C}$-algebras $\Phi: Hom_\mathbb{C}(V)\to Hom_\mathbb{C}(V^*)$ given by $u\mapsto ( ^t u : \varphi\mapsto \varphi\circ u)$. Also denote $\psi: \mathbb{C}[G]\to Hom_\mathbb{C}(V)$ and $\psi': \mathbb{C}[G]\to Hom_\mathbb{C}(V^*)$ the projections induced by $\rho$ and $\phi$ and $\phi^*$, and $\tau$ the canonical involution of $\mathbb{C}[G]$ given by $g\mapsto g^{-1}$.
My claim is that $\Phi\circ \psi = \psi'\circ \tau$. So not only is $\Phi$ an anti-isomorphism of $\mathbb{C}$-algebras, it also conjugates the canonical projections of $\mathbb{C}[G]$ onto $Hom_\mathbb{C}(V)$ and $Hom_\mathbb{C}(V^*)$, so it's as close to an "isomorphism of Wedderburn components" as you can expect.