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Seva
Seva
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The map does not have to be surjective.

Suppose thatLet $N$ is divisible by $2$, but not by $4$be even. In this case, the groupChoose $\mathbb Z/N\mathbb Z$ has exactly one involution; hence$\delta\in(0,1)$ arbitrarily, the dual groupand suppose that $\widehat{\mathbb Z/N\mathbb Z}$ has two real characters. Choose$\Gamma\subseteq(N/2)^\perp$; that is, $\Gamma$ to consist of these twocontains only those characters, and let with $\delta\in(0,1)$$N/2$ in the kernel. ThenIn this case $\mathrm{Bohr}(\Gamma;\delta)$ contains $0$ and $N/2$, while all other elements of $\mathrm{Bohr}(\Gamma;\delta)$ come in pairs: $g\in\mathrm{Bohr}(\Gamma;\delta)$ if and only if $-g\in\mathrm{Bohr}(\Gamma;\delta)$. Therefore, $|\mathrm{Bohr}(\Gamma;\delta)|$ is even.

On the other hand, the intersection $\Lambda\cap[-\delta N/4,\delta N/4]^k$ has an odd size (it contains $0$ and is symmetric). It follows that $\mathrm{Bohr}(\Gamma;\delta)$ and $\Lambda\cap[-\delta N/4,\delta N/4]^k$ are not isomprphic.

The map does not have to be surjective.

Suppose that $N$ is divisible by $2$, but not by $4$. In this case, the group $\mathbb Z/N\mathbb Z$ has exactly one involution; hence, the dual group $\widehat{\mathbb Z/N\mathbb Z}$ has two real characters. Choose $\Gamma$ to consist of these two characters, and let $\delta\in(0,1)$. Then $\mathrm{Bohr}(\Gamma;\delta)$ contains $0$ and $N/2$, while all other elements of $\mathrm{Bohr}(\Gamma;\delta)$ come in pairs: $g\in\mathrm{Bohr}(\Gamma;\delta)$ if and only if $-g\in\mathrm{Bohr}(\Gamma;\delta)$. Therefore, $|\mathrm{Bohr}(\Gamma;\delta)|$ is even.

On the other hand, the intersection $\Lambda\cap[-\delta N/4,\delta N/4]^k$ has an odd size (it contains $0$ and is symmetric). It follows that $\mathrm{Bohr}(\Gamma;\delta)$ and $\Lambda\cap[-\delta N/4,\delta N/4]^k$ are not isomprphic.

The map does not have to be surjective.

Let $N$ be even. Choose $\delta\in(0,1)$ arbitrarily, and suppose that $\Gamma\subseteq(N/2)^\perp$; that is, $\Gamma$ contains only those characters with $N/2$ in the kernel. In this case $\mathrm{Bohr}(\Gamma;\delta)$ contains $0$ and $N/2$, while all other elements come in pairs: $g\in\mathrm{Bohr}(\Gamma;\delta)$ if and only if $-g\in\mathrm{Bohr}(\Gamma;\delta)$. Therefore, $|\mathrm{Bohr}(\Gamma;\delta)|$ is even.

On the other hand, the intersection $\Lambda\cap[-\delta N/4,\delta N/4]^k$ has an odd size (it contains $0$ and is symmetric). It follows that $\mathrm{Bohr}(\Gamma;\delta)$ and $\Lambda\cap[-\delta N/4,\delta N/4]^k$ are not isomprphic.

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Seva
Seva
  • 23k
  • 2
  • 59
  • 141

The map does not have to be surjective.

Suppose that $N$ is divisible by $2$, but not by $4$. In this case, the group $\mathbb Z/N\mathbb Z$ has exactly one involution; hence, the dual group $\widehat{\mathbb Z/N\mathbb Z}$ has two real characters. Choose $\Gamma$ to consist of these two characters, and let $\delta\in(0,1)$. Then $\mathrm{Bohr}(\Gamma;\delta)$ contains $0$ and $N/2$, while all other elements of $\mathrm{Bohr}(\Gamma;\delta)$ come in pairs: $g\in\mathrm{Bohr}(\Gamma;\delta)$ if and only if $-g\in\mathrm{Bohr}(\Gamma;\delta)$. Therefore, $|\mathrm{Bohr}(\Gamma;\delta)|$ is even.

On the other hand, the intersection $\Lambda\cap[-\delta N/4,\delta N/4]^k$ has an odd size (it contains $0$ and is symmetric). It follows that $\mathrm{Bohr}(\Gamma;\delta)$ and $\Lambda\cap[-\delta N/4,\delta N/4]^k$ are not isomprphic.