I know that the irreducible modules of $C\_m \wr S\_n$ over $\mathbb{C}$ are parametrised by m-multipartitions. The parts of the multipartition are indexed by the elements of $C\_m$.


My question now is: If $V$ is an irreducible module over $\mathbb{C}$ with multipartition $\underline{\lambda} = (\lambda\_0, \lambda\_1, ..., \lambda\_{m-1})$, what is the multipartition of $V^\*$?


I feel like it should be a permutation of $\underline{\lambda}$ corresponding to the map $i \mapsto m-i$, but I cannot find that statement anywhere. It should be fairly easy to prove but I have not managed to do so yet.


If the proof turns out to be too complicated, any source would suffice.


Thanks in advance!



The multipartition of the dual of a $C_m \wr S_n$ module
I assume that by a multipartition you mean here that $\lambda\_i$ is a partition of $k\_i$ and $\sum\_i k\_i = n$. In this case the correspondence you described is indeed the duality correspondence. I believe the easiest way to see this is via Clifford Theory: Consider the short exact sequence $$1\to C\_m^n\to G\to S\_n\to 1$$ where $G=C\_m\wr S\_n$. 
Let $\zeta$ be an $m$-th root of unity and let $g$ be a generator of $C\_m$. Every irreducible representation of $C\_m^n$ is of the form $$\rho\_{t\_1,\ldots,t\_n}\big((g^{a\_1},\ldots,g^{a\_n})\big)=\zeta^{\sum\_i t\_ia\_i}$$ 
where $t\_i\in \{0,\ldots, m-1\}$. 


Every irreducible representation is $S\_n$-conjugate to a unique representation in which $t\_1\leq t\_2\ldots \leq t\_{n-1}$. Fix such a tuple. For $i\in \{0,\ldots, m-1\}$ write $k\_i:= |\{j| t\_j=i\}|$. The stabilizer of $\rho\_{t\_1,\ldots, t\_n}$ in $S\_n$ will then be isomorphic to $S\_{k\_0}\times\cdots\times S\_{k\_{m-1}}$. If $(\lambda\_i,\ldots, \lambda\_m)$ is a multipartition in the sense that $\lambda\_i$ is a partition of $k\_i$, then the corresponding tensor product of Specht modules $\mathbb{S}\_{\lambda\_0}\otimes\cdots\otimes \mathbb{S}\_{\lambda\_{m-1}}$ is an irreducible representation of $S\_{k\_0}\times\cdots\times S\_{k\_{m-1}}$. By letting $C\_m^n$ act via the character $\rho\_{t\_1\ldots,t\_n}$ on this vector space you get a representation $V$ of $$H:=C\_m^n\ltimes (S\_{k\_1}\times\cdots\times S\_{k\_m}).$$ By taking the induced representation $\text{Ind}^G\_H V$ you get a representation of $G$. Clifford Theory asserts that this is an irreducible representation of $G$, and that you get all the irreducible representations of $G$ this way.


Now for the duality: it holds that $$(\text{Ind}^G\_H V)^\* \cong \text{Ind}^G\_H (V^\*).$$


Since all the irreducible representations of $S\_{k\_0}\times\cdots\times S\_{k\_{m-1}}$ are self dual, you are left with the same representation of this group. However, $C\_m^n$ acts now via the dual character of $\rho\_{t\_1,\ldots, t\_n}$, which is $\rho\_{m-t\_1,\ldots, m-t\_n}$. When you calculate the numbers $k'\_i$ which correspond to this representation you will get now that $k'\_i=k\_{m-i}$. 
So the multipartition which corresponds to the dual representation is indeed $(\lambda\_0,\lambda\_{m-1},\ldots, \lambda\_1)$


