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Edit: I just realized that this question is related to Andreas Thom's very interesting question here. I think the question below is more crude...

Michael's question here reminded me of the first lemma of this paper of Kadison, establishing that those $C^{*}$-algebras $\mathcal{A}\subset B(\mathcal{H})$ for which the set of self-adjoint elements $\mathcal{A_{s.a.}}$ is strong-operator closed under taking monotone increasing limits are, in fact, von Neumann algebras. I've often wondered about the following

Question: What are necessary and sufficient conditions on a $*$-subalgebra $\mathcal{A}\subset B(\mathcal{H})$ such that strong operator closure of $\mathcal{A_{s.a.}}$ under monotone increasing nets guarantees that $\mathcal{A}$ is closed in the strong operator topology (i.e. is actually a von Neumann algebra)?

I'd like to know if there are weaker conditions than $\mathcal{A}$ being a $C^{\ast}$-algebra for which Kadison's conclusion holds. For example, conditions including things like "$\mathcal{A}$ is closed under continuous functional calculus" and such, which are essentially equivalent to $\mathcal{A}$ being a $C^{*}$-algebra in the (singly-generated) abelian case, but not in general.

(My impression is that this question is really tough, but I hope I'm wrong. It would be nice even to have an expert's digression on precisely why this question should be tough...as I'd learn some new things from that insight!)

It is also possible to ask about analogues of Pedersen's "up-down" theorem, which says that any self-adjoint element in the strong closure of a $C^{\ast}$-algebra $\mathcal{A}$ on a separable Hilbert space $\mathcal{H}$ is the strong limit of a monotone decreasing sequence of self-adjoint elements each of which is a strong limit of a monotone increasing sequence of self-adjoint elements. Can one weaken the $C^{\ast}$-condition on $\mathcal{A}$ and still get this result? (I've tried to modify the argument in Pedersen's $C^{*}$-algebras and their automorphism groups but if I remember correctly this seems to use the definition of a $C^{*}$-algebra in an essential way. Is there a way around this?!)

Edit: I just realized that this question is related to Andreas Thom's very interesting question here. I think the question below is more crude...

Michael's question here reminded me of the first lemma of this paper of Kadison, establishing that those $C^{*}$-algebras $\mathcal{A}\subset B(\mathcal{H})$ for which the set of self-adjoint elements $\mathcal{A_{s.a.}}$ is strong-operator closed under taking monotone increasing limits are, in fact, von Neumann algebras. I've often wondered about the following

Question: What are necessary and sufficient conditions on a $*$-subalgebra $\mathcal{A}\subset B(\mathcal{H})$ such that strong operator closure of $\mathcal{A_{s.a.}}$ under monotone increasing nets guarantees that $\mathcal{A}$ is closed in the strong operator topology (i.e. is actually a von Neumann algebra)?

I'd like to know if there are weaker conditions than $\mathcal{A}$ being a $C^{\ast}$-algebra for which Kadison's conclusion holds. For example, conditions including things like "$\mathcal{A}$ is closed under continuous functional calculus" and such, which are essentially equivalent to $\mathcal{A}$ being a $C^{*}$-algebra in the (singly-generated) abelian case, but not in general.

(My impression is that this question is really tough, but I hope I'm wrong. It would be nice even to have an expert's digression on precisely why this question should be tough...as I'd learn some new things from that insight!)

It is also possible to ask about analogues of Pedersen's "up-down" theorem, which says that any self-adjoint element in the strong closure of a $C^{\ast}$-algebra $\mathcal{A}$ on a separable Hilbert space $\mathcal{H}$ is the strong limit of a monotone decreasing sequence of self-adjoint elements each of which is a strong limit of a monotone increasing sequence of self-adjoint elements. Can one weaken the $C^{\ast}$-condition on $\mathcal{A}$ and still get this result? (I've tried to modify the argument in Pedersen's $C^{*}$-algebras and their automorphism groups but if I remember correctly this seems to use the definition of a $C^{*}$-algebra in an essential way. Is there a way around this?!)

Edit: I just realized that this question is related to Andreas Thom's very interesting question here but is much more crude, since it is likely that if $\mathcal{A}$ in the question below is unital and acts separably, then it must be a $C^{\ast}$-algebra. (A friend suggested the following strategy: Show that $\mathcal{A}_{s.a.}$ is also closed under decreasing sequences to conclude that whenever $A \in \mathcal{A} _{s.a.}$

then the positive and negative parts $A_{+}$ and $A_{-}$ both lie in $\mathcal{A}$, and hence $\mathcal{A}$ is norm closed. I need a reference for this last characterization of a $C^{\ast}$-algebra. Provided this is valid, the idea seems likely to work...)

Michael's question here reminded me of the first lemma of this paper of Kadison, establishing that those $C^{*}$-algebras $\mathcal{A}\subset B(\mathcal{H})$ for which the set of self-adjoint elements $\mathcal{A_{s.a.}}$ is strong-operator closed under taking monotone increasing limits are, in fact, von Neumann algebras. I've often wondered about the following

Question: What are necessary and sufficient conditions on a $*$-subalgebra $\mathcal{A}\subset B(\mathcal{H})$ such that strong operator closure of $\mathcal{A_{s.a.}}$ under monotone increasing nets guarantees that $\mathcal{A}$ is closed in the strong operator topology (i.e. is actually a von Neumann algebra)?

I'd like to know if there are weaker conditions than $\mathcal{A}$ being a $C^{\ast}$-algebra for which Kadison's conclusion holds. For example, conditions including things like "$\mathcal{A}$ is closed under continuous functional calculus" and such, which are essentially equivalent to $\mathcal{A}$ being a $C^{*}$-algebra in the (singly-generated) abelian case, but not in general.

(My impression is that this question is really tough, but I hope I'm wrong. It would be nice even to have an expert's digression on precisely why this question should be tough...as I'd learn some new things from that insight!)

It is also possible to ask about analogues of Pedersen's "up-down" theorem, which says that any self-adjoint element in the strong closure of a $C^{\ast}$-algebra $\mathcal{A}$ on a separable Hilbert space $\mathcal{H}$ is the strong limit of a monotone decreasing sequence of self-adjoint elements each of which is a strong limit of a monotone increasing sequence of self-adjoint elements. Can one weaken the $C^{\ast}$-condition on $\mathcal{A}$ and still get this result? (I've tried to modify the argument in Pedersen's $C^{*}$-algebras and their automorphism groups but if I remember correctly this seems to use the definition of a $C^{*}$-algebra in an essential way. Is there a way around this?!)

Edit: I just realized that this question is related to Andreas Thom's very interesting question here. I think the question below is more crude...

Michael's question here reminded me of the first lemma of this paper of Kadison, establishing that those $C^{*}$-algebras $\mathcal{A}\subset B(\mathcal{H})$ for which the set of self-adjoint elements $\mathcal{A_{s.a.}}$ is strong-operator closed under taking monotone increasing limits are, in fact, von Neumann algebras. I've often wondered about the following

Question: What are necessary and sufficient conditions on a $*$-subalgebra $\mathcal{A}\subset B(\mathcal{H})$ such that strong operator closure of $\mathcal{A_{s.a.}}$ under monotone increasing nets guarantees that $\mathcal{A}$ is closed in the strong operator topology (i.e. is actually a von Neumann algebra)?

I'd like to know if there are weaker conditions than $\mathcal{A}$ being a $C^{\ast}$-algebra for which Kadison's conclusion holds. For example, conditions including things like "$\mathcal{A}$ is closed under continuous functional calculus" and such, which are essentially equivalent to $\mathcal{A}$ being a $C^{*}$-algebra in the (singly-generated) abelian case, but not in general.

(My impression is that this question is really tough, but I hope I'm wrong. It would be nice even to have an expert's digression on precisely why this question should be tough...as I'd learn some new things from that insight!)

It is also possible to ask about analogues of Pedersen's "up-down" theorem, which says that any self-adjoint element in the strong closure of a $C^{\ast}$-algebra $\mathcal{A}$ on a separable Hilbert space $\mathcal{H}$ is the strong limit of a monotone decreasing sequence of self-adjoint elements each of which is a strong limit of a monotone increasing sequence of self-adjoint elements. Can one weaken the $C^{\ast}$-condition on $\mathcal{A}$ and still get this result? (I've tried to modify the argument in Pedersen's $C^{*}$-algebras and their automorphism groups but if I remember correctly this seems to use the definition of a $C^{*}$-algebra in an essential way. Is there a way around this?!)

Edit: I just realized that this question is extremely close to Andreas Thom's very interesting question here. If the moderators see fit, the present question should be closed as a duplicate.

Michael's question here reminded me of the first lemma of this paper of Kadison, establishing that those $C^{*}$-algebras $\mathcal{A}\subset B(\mathcal{H})$ for which the set of self-adjoint elements $\mathcal{A_{s.a.}}$ is strong-operator closed under taking monotone increasing limits are, in fact, von Neumann algebras. I've often wondered about the following

Question: What are necessary and sufficient conditions on a $*$-subalgebra $\mathcal{A}\subset B(\mathcal{H})$ such that strong operator closure of $\mathcal{A_{s.a.}}$ under monotone increasing nets guarantees that $\mathcal{A}$ is closed in the strong operator topology (i.e. is actually a von Neumann algebra)?

I'd like to know if there are weaker conditions than $\mathcal{A}$ being a $C^{\ast}$-algebra for which Kadison's conclusion holds. For example, conditions including things like "$\mathcal{A}$ is closed under continuous functional calculus" and such, which are essentially equivalent to $\mathcal{A}$ being a $C^{*}$-algebra in the (singly-generated) abelian case, but not in general.

(My impression is that this question is really tough, but I hope I'm wrong. It would be nice even to have an expert's digression on precisely why this question should be tough...as I'd learn some new things from that insight!)

It is also possible to ask about analogues of Pedersen's "up-down" theorem, which says that any self-adjoint element in the strong closure of a $C^{\ast}$-algebra $\mathcal{A}$ on a separable Hilbert space $\mathcal{H}$ is the strong limit of a monotone decreasing sequence of self-adjoint elements each of which is a strong limit of a monotone increasing sequence of self-adjoint elements. Can one weaken the $C^{\ast}$-condition on $\mathcal{A}$ and still get this result? (I've tried to modify the argument in Pedersen's $C^{*}$-algebras and their automorphism groups but if I remember correctly this seems to use the definition of a $C^{*}$-algebra in an essential way. Is there a way around this?!)

Edit: I just realized that this question is related to Andreas Thom's very interesting question here but is much more crude, since it is likely that if $\mathcal{A}$ in the question below is unital and acts separably, then it must be a $C^{\ast}$-algebra. (A friend suggested the following strategy: Show that $\mathcal{A}_{s.a.}$ is also closed under decreasing sequences to conclude that whenever $A \in \mathcal{A} _{s.a.}$

then the positive and negative parts $A_{+}$ and $A_{-}$ both lie in $\mathcal{A}$, and hence $\mathcal{A}$ is norm closed. I need a reference for this last characterization of a $C^{\ast}$-algebra. Provided this is valid, the idea seems likely to work...)

Michael's question here reminded me of the first lemma of this paper of Kadison, establishing that those $C^{*}$-algebras $\mathcal{A}\subset B(\mathcal{H})$ for which the set of self-adjoint elements $\mathcal{A_{s.a.}}$ is strong-operator closed under taking monotone increasing limits are, in fact, von Neumann algebras. I've often wondered about the following

Question: What are necessary and sufficient conditions on a $*$-subalgebra $\mathcal{A}\subset B(\mathcal{H})$ such that strong operator closure of $\mathcal{A_{s.a.}}$ under monotone increasing nets guarantees that $\mathcal{A}$ is closed in the strong operator topology (i.e. is actually a von Neumann algebra)?

I'd like to know if there are weaker conditions than $\mathcal{A}$ being a $C^{\ast}$-algebra for which Kadison's conclusion holds. For example, conditions including things like "$\mathcal{A}$ is closed under continuous functional calculus" and such, which are essentially equivalent to $\mathcal{A}$ being a $C^{*}$-algebra in the (singly-generated) abelian case, but not in general.

(My impression is that this question is really tough, but I hope I'm wrong. It would be nice even to have an expert's digression on precisely why this question should be tough...as I'd learn some new things from that insight!)

It is also possible to ask about analogues of Pedersen's "up-down" theorem, which says that any self-adjoint element in the strong closure of a $C^{\ast}$-algebra $\mathcal{A}$ on a separable Hilbert space $\mathcal{H}$ is the strong limit of a monotone decreasing sequence of self-adjoint elements each of which is a strong limit of a monotone increasing sequence of self-adjoint elements. Can one weaken the $C^{\ast}$-condition on $\mathcal{A}$ and still get this result? (I've tried to modify the argument in Pedersen's $C^{*}$-algebras and their automorphism groups but if I remember correctly this seems to use the definition of a $C^{*}$-algebra in an essential way. Is there a way around this?!)