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Math 139 041125 The Dual Group is an Orthonormal Basis
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The setting: inner product space of functions on a finite abelian group. We want an orthonormal basis of V (so that we can do Fourier analysis). The characters are what we want.
First: cancellation lemma for characters. Theorem: the dual group (the group of characters) is orthonormal. So it suffices to show that the order of the dual group equals that of the group. We do this by invoking the Spectral Theorem for unitary transformations which has, as corollary that if one has commuting unitary transformations, then one can diagonalize them simultaneously.
Now we can prove the main result. If we consider the set of (unitary) transformations T_a on V, where T_a f(x) := f(ax), then we can simultaneously diagonalize them: i.e., there exists a basis of functions that are eigenvectors for all of the T_a. If we normalize those vectors appropriately, the normalized functions are easily seen to be characters (trivial to see that they cannot vanish, and *by construction of the basis,* they are multiplicative. The earlier lemma about non-vanishing, multiplicative functions then yields that they must be characters.
First: cancellation lemma for characters. Theorem: the dual group (the group of characters) is orthonormal. So it suffices to show that the order of the dual group equals that of the group. We do this by invoking the Spectral Theorem for unitary transformations which has, as corollary that if one has commuting unitary transformations, then one can diagonalize them simultaneously.
Now we can prove the main result. If we consider the set of (unitary) transformations T_a on V, where T_a f(x) := f(ax), then we can simultaneously diagonalize them: i.e., there exists a basis of functions that are eigenvectors for all of the T_a. If we normalize those vectors appropriately, the normalized functions are easily seen to be characters (trivial to see that they cannot vanish, and *by construction of the basis,* they are multiplicative. The earlier lemma about non-vanishing, multiplicative functions then yields that they must be characters.