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The compact open topology is essential for getting compact sets in your function space–especially a version of the Arzelà-Ascoli theorem holds for spaces of continuous functions to uniform spaces equipped with the compact open topology ($C(X,Y)$ with the compact open topology where $X$ is an appropriate topological space and $Y$ is an appropriate uniform space).

In your case (Pontryagin duality) the uniform space $Y$ is just the circle group $\mathbb{T}$ and $X$ is your group $G$. Consider a compact neighbourhood $U$ of the neutral element $e\in G$, an open neighbourhood $I:=\exp((-2\pi i\epsilon, 2\pi i\epsilon))\subset\mathbb{T}$ of 1 where $0<\epsilon<1$. Now there is a maximal set of irreducible representations $S\subset \hat{G}\subset C(G,\mathbb{T})$ such that $\forall \pi\in S\ \pi(U)\subset I$. The definition of the compact open topology guarantees that $S$ is open (it is just the definition). But $S$ is also equicontinuous: For $0\lt n\in\mathbb{N}$ define $U_n:=\{x\in G\mid n\cdot x\in C\}$. Clearly for $\pi\in S$ we have $\pi(U_n)\subset \exp((-2\pi i \epsilon/n,2\pi i \epsilon/n))$. Now we can apply Arzelà-Ascoli: $S$ is in fact precompact and we get a compact neighbourhood $\bar{S}$ of the neutral element of $\hat{G}$ thus $\hat{G}$ is locally compact.

There is also a categorical motivation for this topology: The category of topological spaces is not cartesian closed, thus in this category (and many usual related topologies) the adjunction mentioned by David White in the comment above (currying) does not exist—it is impossible to choose the right topology. However, in the category of compactly generated spaces it works: It is cartesian closed and the topology for function spaces is the compact open topology. Again, the compact open topology guarantees that there are enough compact sets such that the topology is actually compactly generated (I guess that this is also a consequence of a version of Arzelà-Ascoli using even continuity, which does not need a uniformity, instead of equicontinuity, but I am not sure).

The compact open topology is essential for getting compact sets in your function space–especially a version of the Arzelà-Ascoli theorem holds for spaces of continuous functions to uniform spaces equipped with the compact open topology ($C(X,Y)$ with the compact open topology where $X$ is an appropriate topological space and $Y$ is an appropriate uniform space).

In your case (Pontryagin duality) the uniform space $Y$ is just the circle group $\mathbb{T}$ and $X$ is your group $G$. Consider a compact neighbourhood $U$ of the neutral element $e\in G$, an open neighbourhood $I:=\exp((-2\pi i\epsilon, 2\pi i\epsilon))\subset\mathbb{T}$ of 1 where $0<\epsilon<1$. Now there is a maximal set of irreducible representations $S\subset \hat{G}\subset C(G,\mathbb{T})$ such that $\forall \pi\in S\ \pi(U)\subset I$. The definition of the compact open topology guarantees that $S$ is open (it is just the definition). But $S$ is also equicontinuous: For $n\in\mathbb{N}$ define $U_0:=U$,$U_{n+1}:=\{x\in U_n\mid x+x\in U_n\}$. Clearly for $\pi\in S$ we have $\pi(U_n)\subset \exp((-2\pi i \epsilon/2^n,2\pi i \epsilon/2^n))$. It suffices to prove equicontinuity at the point $e$ (shifting does not change the situation). Now we can apply Arzelà-Ascoli: $S$ is in fact precompact and we get a compact neighbourhood $\bar{S}$ of the neutral element of $\hat{G}$ thus $\hat{G}$ is locally compact.

There is also a categorical motivation for this topology: The category of topological spaces is not cartesian closed, thus in this category (and many usual related topologies) the adjunction mentioned by David White in the comment above (currying) does not exist—it is impossible to choose the right topology. However, in the category of compactly generated spaces it works: It is cartesian closed and the topology for function spaces is the compact open topology. Again, the compact open topology guarantees that there are enough compact sets such that the topology is actually compactly generated (I guess that this is also a consequence of a version of Arzelà-Ascoli using even continuity, which does not need a uniformity, instead of equicontinuity, but I am not sure).

The compact open topology is essential for getting compact sets in your function space–especially a version of the Arzelà-Ascoli theorem holds for spaces of continuous functions to uniform spaces equipped with the compact open topology ($C(X,Y)$ with the compact open topology where $X$ is a topological space and $Y$ is a uniform space).

In your case (Pontryagin duality) the uniform space $Y$ is just the circle group $\mathbb{T}$ and $X$ is your group $G$. Consider a compact neighbourhood $U$ of the neutral element $e\in G$, an open neighbourhood $I:=\exp((-2\pi i\epsilon, 2\pi i\epsilon))\subset\mathbb{T}$ of 1 where $0<\epsilon<1$. Now there is a maximal set of irreducible representations $S\subset \hat{G}\subset C(G,\mathbb{T})$ such that $\forall \pi\in S\ \pi(U)\subset I$. The definition of the compact open topology guarantees that $S$ is open (it is just the definition). But $S$ is also equicontinuous: For $0\lt n\in\mathbb{N}$ define $U_n:=\{x\in G\mid n\cdot x\in C\}$. Clearly for $\pi\in S$ we have $\pi(U_n)\subset \exp((-2\pi i \epsilon/n,2\pi i \epsilon/n))$. Now we can apply Arzelà-Ascoli: $S$ is in fact precompact and we get a compact neighbourhood $\bar{S}$ of the neutral element of $\hat{G}$ thus $\hat{G}$ is locally compact.

There is also a categorical motivation for this topology: The category of topological spaces is not cartesian closed, thus in this category (and many usual related topologies) the adjunction mentioned by David White in the comment above (currying) does not exist—it is impossible to choose the right topology. However, in the category of compactly generated spaces it works: It is cartesian closed and the topology for function spaces is the compact open topology. Again, the compact open topology guarantees that there are enough compact sets such that the topology is actually compactly generated (I guess that this is also a consequence of a version of Arzelà-Ascoli using even continuity, which does not need a uniformity, instead of equicontinuity, but I am not sure).

The compact open topology is essential for getting compact sets in your function space–especially a version of the Arzelà-Ascoli theorem holds for spaces of continuous functions to uniform spaces equipped with the compact open topology ($C(X,Y)$ with the compact open topology where $X$ is an appropriate topological space and $Y$ is an appropriate uniform space).

In your case (Pontryagin duality) the uniform space $Y$ is just the circle group $\mathbb{T}$ and $X$ is your group $G$. Consider a compact neighbourhood $U$ of the neutral element $e\in G$, an open neighbourhood $I:=\exp((-2\pi i\epsilon, 2\pi i\epsilon))\subset\mathbb{T}$ of 1 where $0<\epsilon<1$. Now there is a maximal set of irreducible representations $S\subset \hat{G}\subset C(G,\mathbb{T})$ such that $\forall \pi\in S\ \pi(U)\subset I$. The definition of the compact open topology guarantees that $S$ is open (it is just the definition). But $S$ is also equicontinuous: For $0\lt n\in\mathbb{N}$ define $U_n:=\{x\in G\mid n\cdot x\in C\}$. Clearly for $\pi\in S$ we have $\pi(U_n)\subset \exp((-2\pi i \epsilon/n,2\pi i \epsilon/n))$. Now we can apply Arzelà-Ascoli: $S$ is in fact precompact and we get a compact neighbourhood $\bar{S}$ of the neutral element of $\hat{G}$ thus $\hat{G}$ is locally compact.

There is also a categorical motivation for this topology: The category of topological spaces is not cartesian closed, thus in this category (and many usual related topologies) the adjunction mentioned by David White in the comment above (currying) does not exist—it is impossible to choose the right topology. However, in the category of compactly generated spaces it works: It is cartesian closed and the topology for function spaces is the compact open topology. Again, the compact open topology guarantees that there are enough compact sets such that the topology is actually compactly generated (I guess that this is also a consequence of a version of Arzelà-Ascoli using even continuity, which does not need a uniformity, instead of equicontinuity, but I am not sure).

The compact open topology is essential for getting compact sets in your function space–especially a version of the Arzelà-Ascoli theorem holds for spaces of continuous functions to uniform spaces equipped with the compact open topology ($C(X,Y)$ with the compact open topology where $X$ is a topological space and $Y$ is a uniform space).

In your case (Pontryagin duality) the uniform space $Y$ is just the circle group $\mathbb{T}$ and $X$ is your group $G$. Consider a compact neighbourhood $U$ of the neutral element $e\in G$, an open neighbourhood $I:=\exp((-2\pi i\epsilon, 2\pi i\epsilon))\subset\mathbb{T}$ of 1 where $0<\epsilon<1$. Now there is a maximal set of irreducible representations $S\subset \hat{G}\subset C(G,\mathbb{T})$ such that $\forall \pi\in S\ \pi(U)\subset I$. The definition of the compact open topology guarantees that $S$ is open (it is just the definition). But $S$ is also equicontinuous: For $0\lt n\in\mathbb{N}$ define $U_n:=\{x\in G\mid n\cdot x\in C\}$. Clearly for $\pi\in S$ we have $\pi(U_n)\subset \exp((-2\pi i \epsilon/n,2\pi i \epsilon/n))$. Now we can apply Arzelà-Ascoli: $S$ is in fact precompact and we get a compact neighbourhood $\bar{S}$ of the neutral element of $\hat{G}$ thus $\hat{G}$ is locally compact.

There is also a categorical motivation for this topology: The category of topological spaces is not cartesian closed, thus in this category (and many usual related topologies) the adjunction mentioned by David White in the comment above (currying) does not exist—it is impossible to choose the right topology. However, in the category of compactly generated spaces it works: It is cartesian closed and the topology for function spaces is the compact open topology. Again, the compact open topology guarantees that there are enough compact sets such that the topology is actually compactly generated.

The compact open topology is essential for getting compact sets in your function space–especially a version of the Arzelà-Ascoli theorem holds for spaces of continuous functions to uniform spaces equipped with the compact open topology ($C(X,Y)$ with the compact open topology where $X$ is a topological space and $Y$ is a uniform space).

In your case (Pontryagin duality) the uniform space $Y$ is just the circle group $\mathbb{T}$ and $X$ is your group $G$. Consider a compact neighbourhood $U$ of the neutral element $e\in G$, an open neighbourhood $I:=\exp((-2\pi i\epsilon, 2\pi i\epsilon))\subset\mathbb{T}$ of 1 where $0<\epsilon<1$. Now there is a maximal set of irreducible representations $S\subset \hat{G}\subset C(G,\mathbb{T})$ such that $\forall \pi\in S\ \pi(U)\subset I$. The definition of the compact open topology guarantees that $S$ is open (it is just the definition). But $S$ is also equicontinuous: For $0\lt n\in\mathbb{N}$ define $U_n:=\{x\in G\mid n\cdot x\in C\}$. Clearly for $\pi\in S$ we have $\pi(U_n)\subset \exp((-2\pi i \epsilon/n,2\pi i \epsilon/n))$. Now we can apply Arzelà-Ascoli: $S$ is in fact precompact and we get a compact neighbourhood $\bar{S}$ of the neutral element of $\hat{G}$ thus $\hat{G}$ is locally compact.

There is also a categorical motivation for this topology: The category of topological spaces is not cartesian closed, thus in this category (and many usual related topologies) the adjunction mentioned by David White in the comment above (currying) does not exist—it is impossible to choose the right topology. However, in the category of compactly generated spaces it works: It is cartesian closed and the topology for function spaces is the compact open topology. Again, the compact open topology guarantees that there are enough compact sets such that the topology is actually compactly generated (I guess that this is also a consequence of a version of Arzelà-Ascoli using even continuity, which does not need a uniformity, instead of equicontinuity, but I am not sure).