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Cartesian symmetry classes associated with certain subgroups of S_m
Let \(V\) be an \(n\)-dimensional inner product space. Assume \(G\) is a subgroup of the symmetric group of degree \(m\), and \(\lambda\) is an irreducible character of \(G\). Consider the \emph{Cartesian symmetrizer} \(C_{\lambda}\) on the Cartesian space \(\times^{m}V\) defined by \[ C_{\lambda} =...
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Published in: | arXiv.org 2023-12 |
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Main Authors: | , |
Format: | Article |
Language: | English |
Subjects: | |
Online Access: | Get full text |
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Summary: | Let \(V\) be an \(n\)-dimensional inner product space. Assume \(G\) is a subgroup of the symmetric group of degree \(m\), and \(\lambda\) is an irreducible character of \(G\). Consider the \emph{Cartesian symmetrizer} \(C_{\lambda}\) on the Cartesian space \(\times^{m}V\) defined by \[ C_{\lambda} = \frac{\lambda(1)}{|G|}\sum_{\tau\in G} \lambda(\tau) Q(\tau). \] The vector space \( V^{\lambda}(G) = C_{\lambda}(\times^{m}V) \) is called the Cartesian symmetry class associated with \(G\) and \(\lambda\). In this paper, we give a formula for the dimension of the cyclic subspace \(V^{\lambda}_{ij}\). Then we discuss the problem existing an \(O\)-basis for the Cartesian symmetry class \(V^{\lambda}(G)\). Also, we compute the dimension of the symmetry class \(V^{\lambda}(G)\) when \(G = \langle \sigma_{1} \sigma_{2} \cdots \sigma_{p} \rangle\) or \(G = \cdots \), where \(\sigma_i\) are disjoint cycles in \(S_{m}\). The dimensions are expressed in terms of the Ramanujan sum. Additionally, we provide a necessary and sufficient condition for the existence of an \(O\)-basis for Cartesian symmetry classes associated with the irreducible characters of the dihedral group \(D_{2m}\). The dimensions of these classes are also computed. |
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ISSN: | 2331-8422 |