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See: Symmetric group representations, General linear group representations 表示论 https://www.math4wisdom.com/wiki/Research/RepresentationTheory?action=edit
Open problem: Calculating plethysm. 影片
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Observations We have {$$\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}= \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$} However, the constant map from a group {$G$} to this matrix is not a representation. The reason is that this is not an invertible matrix. A representation takes us from {$G$} to {$GL(V)$} and the latter consists of invertible matrices. Note that a subspace {$W$} of {$V$} typically does not have a unique complement. Consider, for example, onedimensional {$W$} and twodimensional {$V$}. There are many different lines that can be the complement of {$W$}. Together {$W$} and its complement span {$V$}. Given a representation {$\theta$} and a scalar {$\alpha$}, the function {$\alpha\theta$} is not, in general, a representation because typically {$\alpha^2\neq\alpha$}. Elements of a noncyclic group can't be mapped to corresponding roots of unity if there are elements whose orders are relatively prime. For then you could combine the two roots of unity to get a generator that is a product of the two orders. Geometry and logic
Compact group Representation of a compact group is based on an integral measure that is translatable. With my orthogonal polynomials, the group is not compact, so the measure is not translatable. Topics to Learn From Amritanshu Prasad. Representation Theory: A Combinatorial Viewpoint Basic Concepts of Representation Theory
