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Suppose we have a noncompact Hausdorff space $S$ and a Banach algebra $A \subset C^*(S,\mathbb R)$. For niceness let's assume $A$ separates the points of $S$. If it makes things any easier we can also assume $S = \mathbb R$.

By Gelfand duality, the inclusion $A \to C^*(S,\mathbb R)$ induces a continuous map $f: \mathbb \beta S \to \gamma S$ between compact spaces. Here $\mathbb \beta S$ is the Stone Cech compactification and $\gamma S$ is the Gelfand Spectrum of $A$.

Is there a way to check in terms of $A$ whether the remainder $\gamma S /S$ is closed in $\gamma S$? Equivalently when is it compact?

It would be enough for the restriction $f: S \to S$ to be a homeomorphism. This is because in that case $f(\beta S/ S) = \gamma S/ S$ which is compact by continuity. To see this let $u \in \beta S/ S$ be arbitrary and let $x_\alpha$ be a net in $S$ tending to $u$. Since $x_\alpha$ does not converge in $S$ we know $f(x_\alpha)$ does not converge in $f(S)$. Since $\gamma S$ is compact the only option is for $f(x_\alpha)$ to converge to a point in the remainder. hence by continuity $f(u)$ is that point of the remainder.

The reason I am asking is because I learnt through this question that the Bohr compactification $G \to b G$ of an Abelian locally compact group is not a compactification in the topological sense. i.e the injection is not an embedding. From this it can be shown that the image of the group is not open and so the remainder is not closed.

Despite this, the Bohr compactification comes about as the Gelfand Spectrum of the Algebra of almost periodic functions.

I wonder is there a way to see the noncompactness directly from the function algebra.

Suppose we have a noncompact Hausdorff space $S$ and a Banach algebra $A \subset C^*(S,\mathbb R)$ of the space of real-valued bounded functions on $S$. For niceness let's assume $A$ separates the points of $S$. If it makes things any easier we can also assume $S = \mathbb R$.

By Gelfand duality, the inclusion $A \to C^*(S,\mathbb R)$ induces a continuous map $f: \mathbb \beta S \to \gamma S$ between compact spaces. Here $\mathbb \beta S$ is the Stone Cech compactification and $\gamma S$ is the Gelfand Spectrum of $A$.

Is there a way to check in terms of $A$ whether the remainder $\gamma S /S$ is closed in $\gamma S$? Equivalently when is it compact?

It would be enough for the restriction $f: S \to S$ to be a homeomorphism. This is because in that case $f(\beta S/ S) = \gamma S/ S$ which is compact by continuity. To see this let $u \in \beta S/ S$ be arbitrary and let $x_\alpha$ be a net in $S$ tending to $u$. Since $x_\alpha$ does not converge in $S$ we know $f(x_\alpha)$ does not converge in $f(S)$. Since $\gamma S$ is compact the only option is for $f(x_\alpha)$ to converge to a point in the remainder. hence by continuity $f(u)$ is that point of the remainder.

The reason I am asking is because I learnt through this question that the Bohr compactification $G \to b G$ of an Abelian locally compact group is not a compactification in the topological sense. i.e the injection is not an embedding. From this it can be shown that the image of the group is not open and so the remainder is not closed.

Despite this, the Bohr compactification comes about as the Gelfand Spectrum of the Algebra of almost periodic functions.

I wonder is there a way to see the noncompactness directly from the function algebra.

Suppose we have a noncompact Hausdorff space $S$ and a Banach algebra $A \subset C^*(S,\mathbb R)$. For niceness let's assume $A$ separates the points of $A$. If it makes things any easier we can also assume $S = \mathbb R$.

By Gelfand duality, the inclusion $A \to C^*(S,\mathbb R)$ induces a continuous map $f: \mathbb \beta S \to \gamma S$ between compact spaces. Here $\mathbb \beta S$ is the Stone Cech compactification and $\gamma S$ is the Gelfand Spectrum of $S$.

Is there a way to check in terms of $A$ whether the remainder $\gamma S /S$ is closed in $\gamma S$? Equivalently when is it compact?

It would be enough for the restriction $f: S \to S$ to be a homeomorphism. This is because in that case $f(\beta S/ S) = \gamma S/ S$ which is compact by continuity. To see this let $u \in \beta S/ S$ be arbitrary and let $x_\alpha$ be a net in $S$ tending to $u$. Since $x_\alpha$ does not converge in $S$ we know $f(x_\alpha)$ does not converge in $f(S)$. Since $\gamma S$ is compact the only option is for $f(x_\alpha)$ to converge to a point in the remainder. hence by continuity $f(u)$ is that point of the remainder.

The reason I am asking is because I learnt through this question that the Bohr compactification $G \to b G$ of an Abelian locally compact group is not a compactification in the topological sense. i.e the injection is not an embedding. From this it can be shown that the image of the group is not open and so the remainder is not closed.

Despite this, the Bohr compactification comes about as the Gelfand Spectrum of the Algebra of almost periodic functions.

I wonder is there a way to see the noncompactness directly from the function algebra.

Suppose we have a noncompact Hausdorff space $S$ and a Banach algebra $A \subset C^*(S,\mathbb R)$. For niceness let's assume $A$ separates the points of $S$. If it makes things any easier we can also assume $S = \mathbb R$.

By Gelfand duality, the inclusion $A \to C^*(S,\mathbb R)$ induces a continuous map $f: \mathbb \beta S \to \gamma S$ between compact spaces. Here $\mathbb \beta S$ is the Stone Cech compactification and $\gamma S$ is the Gelfand Spectrum of $A$.

Is there a way to check in terms of $A$ whether the remainder $\gamma S /S$ is closed in $\gamma S$? Equivalently when is it compact?

It would be enough for the restriction $f: S \to S$ to be a homeomorphism. This is because in that case $f(\beta S/ S) = \gamma S/ S$ which is compact by continuity. To see this let $u \in \beta S/ S$ be arbitrary and let $x_\alpha$ be a net in $S$ tending to $u$. Since $x_\alpha$ does not converge in $S$ we know $f(x_\alpha)$ does not converge in $f(S)$. Since $\gamma S$ is compact the only option is for $f(x_\alpha)$ to converge to a point in the remainder. hence by continuity $f(u)$ is that point of the remainder.

The reason I am asking is because I learnt through this question that the Bohr compactification $G \to b G$ of an Abelian locally compact group is not a compactification in the topological sense. i.e the injection is not an embedding. From this it can be shown that the image of the group is not open and so the remainder is not closed.

Despite this, the Bohr compactification comes about as the Gelfand Spectrum of the Algebra of almost periodic functions.

I wonder is there a way to see the noncompactness directly from the function algebra.

Suppose we have a noncompact Hausdorff space $S$ and a Banach algebra $A \subset C^*(S,\mathbb R)$. For niceness let's assume $A$ separates the points of $A$. If it makes things any easier we can also assume $S = \mathbb R$.

By Gelfand duality, the inclusion $A \to C^*(S,\mathbb R)$ induces a continuous map $f: \mathbb \beta S \to \gamma S$ between compact spaces. Here $\mathbb \beta S$ is the Stone Cech compactification and $\gamma S$ is the Gelfand Spectrum of $S$.

Is there a way to check in terms of $A$ whether the remainder $\gamma S /S$ is closed in $\gamma S$?

It would be enough for the restriction $f: S \to S$ to be a homeomorphism. This is because in that case $f(\beta S/ S) = \gamma S/ S$ which is compact by continuity. To see this let $u \in \beta S/ S$ be arbitrary and let $x_\alpha$ be a net in $S$ tending to $u$. Since $x_\alpha$ does not converge in $S$ we know $f(x_\alpha)$ does not converge in $f(S)$. Since $\gamma S$ is compact the only option is for $f(x_\alpha)$ to converge to a point in the remainder. hence by continuity $f(u)$ is that point of the remainder.

The reason I am asking is because I learnt through this question that the Bohr compactification $G \to b G$ of a group is not a compactification in the topological sense. i.e the injection is not an embedding. From this it can be shown that the image of the group is not open and so the remainder is not closed.

Despite this, the Bohr compactification comes about as the Gelfand Spectrum of the Algebra of almost periodic functions.

Suppose we have a noncompact Hausdorff space $S$ and a Banach algebra $A \subset C^*(S,\mathbb R)$. For niceness let's assume $A$ separates the points of $A$. If it makes things any easier we can also assume $S = \mathbb R$.

By Gelfand duality, the inclusion $A \to C^*(S,\mathbb R)$ induces a continuous map $f: \mathbb \beta S \to \gamma S$ between compact spaces. Here $\mathbb \beta S$ is the Stone Cech compactification and $\gamma S$ is the Gelfand Spectrum of $S$.

Is there a way to check in terms of $A$ whether the remainder $\gamma S /S$ is closed in $\gamma S$? Equivalently when is it compact?

It would be enough for the restriction $f: S \to S$ to be a homeomorphism. This is because in that case $f(\beta S/ S) = \gamma S/ S$ which is compact by continuity. To see this let $u \in \beta S/ S$ be arbitrary and let $x_\alpha$ be a net in $S$ tending to $u$. Since $x_\alpha$ does not converge in $S$ we know $f(x_\alpha)$ does not converge in $f(S)$. Since $\gamma S$ is compact the only option is for $f(x_\alpha)$ to converge to a point in the remainder. hence by continuity $f(u)$ is that point of the remainder.

The reason I am asking is because I learnt through this question that the Bohr compactification $G \to b G$ of an Abelian locally compact group is not a compactification in the topological sense. i.e the injection is not an embedding. From this it can be shown that the image of the group is not open and so the remainder is not closed.

Despite this, the Bohr compactification comes about as the Gelfand Spectrum of the Algebra of almost periodic functions.

I wonder is there a way to see the noncompactness directly from the function algebra.