Weakly dense C*-algebra in a commutative von Neumann algebra and order convergence












2












$begingroup$


Let $H$ be a Hilbert space and $mathscr{A}$ a commutative norm-closed unital $*$-subalgebra of $mathcal{B}(H)$. Let $mathscr{M}$ be the weak operator closure of $mathscr{A}$.



Question: For given a projection $Pinmathscr{M}$, is the following true? $$P=infleft{Ainmathscr{A}:Pleq Aleq 1right}$$



It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $P$. Also, if the above is true, what happens if $mathscr{A}$ is non-commutative?






operator-theory c-star-algebras von-neumann-algebras







share|cite|improve this question











$endgroup$












  • $begingroup$
    By $P=inf U$ do you mean $Pleq A$ for all $Ain U$ and if $Qleq A$ for all $Ain U$ then $Qleq P$?
    $endgroup$
    – SmileyCraft
    Jan 4 at 5:57












  • $begingroup$
    @SmileyCraft Yes, and if we assume $U$ is norm-bounded, then $inf U$ is the same as $lim U$ when we regard $U$ as a decreasing net, in the weak operator topology.
    $endgroup$
    – Junekey Jeon
    Jan 4 at 6:02


















2












$begingroup$


Let $H$ be a Hilbert space and $mathscr{A}$ a commutative norm-closed unital $*$-subalgebra of $mathcal{B}(H)$. Let $mathscr{M}$ be the weak operator closure of $mathscr{A}$.



Question: For given a projection $Pinmathscr{M}$, is the following true? $$P=infleft{Ainmathscr{A}:Pleq Aleq 1right}$$



It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $P$. Also, if the above is true, what happens if $mathscr{A}$ is non-commutative?






operator-theory c-star-algebras von-neumann-algebras







share|cite|improve this question











$endgroup$












  • $begingroup$
    By $P=inf U$ do you mean $Pleq A$ for all $Ain U$ and if $Qleq A$ for all $Ain U$ then $Qleq P$?
    $endgroup$
    – SmileyCraft
    Jan 4 at 5:57












  • $begingroup$
    @SmileyCraft Yes, and if we assume $U$ is norm-bounded, then $inf U$ is the same as $lim U$ when we regard $U$ as a decreasing net, in the weak operator topology.
    $endgroup$
    – Junekey Jeon
    Jan 4 at 6:02
















2












2








2


1



$begingroup$


Let $H$ be a Hilbert space and $mathscr{A}$ a commutative norm-closed unital $*$-subalgebra of $mathcal{B}(H)$. Let $mathscr{M}$ be the weak operator closure of $mathscr{A}$.



Question: For given a projection $Pinmathscr{M}$, is the following true? $$P=infleft{Ainmathscr{A}:Pleq Aleq 1right}$$



It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $P$. Also, if the above is true, what happens if $mathscr{A}$ is non-commutative?






operator-theory c-star-algebras von-neumann-algebras







share|cite|improve this question











$endgroup$




Let $H$ be a Hilbert space and $mathscr{A}$ a commutative norm-closed unital $*$-subalgebra of $mathcal{B}(H)$. Let $mathscr{M}$ be the weak operator closure of $mathscr{A}$.



Question: For given a projection $Pinmathscr{M}$, is the following true? $$P=infleft{Ainmathscr{A}:Pleq Aleq 1right}$$



It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $P$. Also, if the above is true, what happens if $mathscr{A}$ is non-commutative?






operator-theory c-star-algebras von-neumann-algebras




operator-theory c-star-algebras von-neumann-algebras






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 8:31







Junekey Jeon

















asked Jan 4 at 5:34









Junekey JeonJunekey Jeon

25118




25118












  • $begingroup$
    By $P=inf U$ do you mean $Pleq A$ for all $Ain U$ and if $Qleq A$ for all $Ain U$ then $Qleq P$?
    $endgroup$
    – SmileyCraft
    Jan 4 at 5:57












  • $begingroup$
    @SmileyCraft Yes, and if we assume $U$ is norm-bounded, then $inf U$ is the same as $lim U$ when we regard $U$ as a decreasing net, in the weak operator topology.
    $endgroup$
    – Junekey Jeon
    Jan 4 at 6:02




















  • $begingroup$
    By $P=inf U$ do you mean $Pleq A$ for all $Ain U$ and if $Qleq A$ for all $Ain U$ then $Qleq P$?
    $endgroup$
    – SmileyCraft
    Jan 4 at 5:57












  • $begingroup$
    @SmileyCraft Yes, and if we assume $U$ is norm-bounded, then $inf U$ is the same as $lim U$ when we regard $U$ as a decreasing net, in the weak operator topology.
    $endgroup$
    – Junekey Jeon
    Jan 4 at 6:02


















$begingroup$
By $P=inf U$ do you mean $Pleq A$ for all $Ain U$ and if $Qleq A$ for all $Ain U$ then $Qleq P$?
$endgroup$
– SmileyCraft
Jan 4 at 5:57






$begingroup$
By $P=inf U$ do you mean $Pleq A$ for all $Ain U$ and if $Qleq A$ for all $Ain U$ then $Qleq P$?
$endgroup$
– SmileyCraft
Jan 4 at 5:57














$begingroup$
@SmileyCraft Yes, and if we assume $U$ is norm-bounded, then $inf U$ is the same as $lim U$ when we regard $U$ as a decreasing net, in the weak operator topology.
$endgroup$
– Junekey Jeon
Jan 4 at 6:02






$begingroup$
@SmileyCraft Yes, and if we assume $U$ is norm-bounded, then $inf U$ is the same as $lim U$ when we regard $U$ as a decreasing net, in the weak operator topology.
$endgroup$
– Junekey Jeon
Jan 4 at 6:02












1 Answer
1






active

oldest

votes


















2












$begingroup$

I think I found a counterexample.



Take $mathscr{M}=L^{infty}[0,1]$ and $mathscr{A}=C[0,1]$.



Let $mathbb{Q}cap[0,1]=left{r_{n}right}_{n=1}^{infty}$ be an enumeration of rationals in $[0,1]$, and define $$E:=bigcup_{n=1}^{infty}left(r_{n}-frac{epsilon}{2^{n}},r_{n}+frac{epsilon}{2^{n}}right)cap[0,1]$$ for some small $epsilon>0$ so that $m(E)$ is strictly less than $1$. Clearly, the only continuous function $f:[0,1]rightarrow[0,1]$ satisfying $mathbb{1}_{E}leq fleq 1$ is $fequiv1$, but the projection $mathbb{1}_{E}$ is not equal to $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, that gives you a counterexample. You might struggle a bit if you want to make it concrete, it's not obvious to me how to choose $epsilon$, and the order of the enumeration probably makes a difference. It might be easier working with a the complement of a fat cantor set.
    $endgroup$
    – Martin Argerami
    Jan 4 at 17:21










  • $begingroup$
    Maybe I overlook something, but the only projections in $mathscr A$ are $0$ and $1$. If you take $P = chi_{[0,1/2]}$, then $P$ is not the infimum of as described in the question, since the infimum is $1$.
    $endgroup$
    – user42761
    Jan 4 at 18:33












  • $begingroup$
    @AndréS. If $P$ is the indicator of $[0,1/2]$, then the set described in the question is the set of all continuous functions $f:[0,1]rightarrow[0,1]$ such that $f(x)=1$ for all $xin[0,1/2]$. So in this case, $P$ is indeed the infimum of that set. The fact that there is no nontrivial projections in $mathscr{A}$ does not matter here.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 10:24












  • $begingroup$
    Ah okay, I had projections in $mathscr A$ in mind. So is the pointwise infimum of those continuous functions that majorize the indicator, call it $P$, then the infimum in the sense of operators on $L^2([0,1])$ ?
    $endgroup$
    – user42761
    Jan 5 at 12:45










  • $begingroup$
    @AndréS. Yes, and the pointwise infimum as functions on $[0,1]$ is equal to the infimum in the sense of operators in this case, because the pointwise infimum can be obtained by a countable sequence of approximations.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 14:47












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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

votes









2












$begingroup$

I think I found a counterexample.



Take $mathscr{M}=L^{infty}[0,1]$ and $mathscr{A}=C[0,1]$.



Let $mathbb{Q}cap[0,1]=left{r_{n}right}_{n=1}^{infty}$ be an enumeration of rationals in $[0,1]$, and define $$E:=bigcup_{n=1}^{infty}left(r_{n}-frac{epsilon}{2^{n}},r_{n}+frac{epsilon}{2^{n}}right)cap[0,1]$$ for some small $epsilon>0$ so that $m(E)$ is strictly less than $1$. Clearly, the only continuous function $f:[0,1]rightarrow[0,1]$ satisfying $mathbb{1}_{E}leq fleq 1$ is $fequiv1$, but the projection $mathbb{1}_{E}$ is not equal to $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, that gives you a counterexample. You might struggle a bit if you want to make it concrete, it's not obvious to me how to choose $epsilon$, and the order of the enumeration probably makes a difference. It might be easier working with a the complement of a fat cantor set.
    $endgroup$
    – Martin Argerami
    Jan 4 at 17:21










  • $begingroup$
    Maybe I overlook something, but the only projections in $mathscr A$ are $0$ and $1$. If you take $P = chi_{[0,1/2]}$, then $P$ is not the infimum of as described in the question, since the infimum is $1$.
    $endgroup$
    – user42761
    Jan 4 at 18:33












  • $begingroup$
    @AndréS. If $P$ is the indicator of $[0,1/2]$, then the set described in the question is the set of all continuous functions $f:[0,1]rightarrow[0,1]$ such that $f(x)=1$ for all $xin[0,1/2]$. So in this case, $P$ is indeed the infimum of that set. The fact that there is no nontrivial projections in $mathscr{A}$ does not matter here.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 10:24












  • $begingroup$
    Ah okay, I had projections in $mathscr A$ in mind. So is the pointwise infimum of those continuous functions that majorize the indicator, call it $P$, then the infimum in the sense of operators on $L^2([0,1])$ ?
    $endgroup$
    – user42761
    Jan 5 at 12:45










  • $begingroup$
    @AndréS. Yes, and the pointwise infimum as functions on $[0,1]$ is equal to the infimum in the sense of operators in this case, because the pointwise infimum can be obtained by a countable sequence of approximations.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 14:47
















2












$begingroup$

I think I found a counterexample.



Take $mathscr{M}=L^{infty}[0,1]$ and $mathscr{A}=C[0,1]$.



Let $mathbb{Q}cap[0,1]=left{r_{n}right}_{n=1}^{infty}$ be an enumeration of rationals in $[0,1]$, and define $$E:=bigcup_{n=1}^{infty}left(r_{n}-frac{epsilon}{2^{n}},r_{n}+frac{epsilon}{2^{n}}right)cap[0,1]$$ for some small $epsilon>0$ so that $m(E)$ is strictly less than $1$. Clearly, the only continuous function $f:[0,1]rightarrow[0,1]$ satisfying $mathbb{1}_{E}leq fleq 1$ is $fequiv1$, but the projection $mathbb{1}_{E}$ is not equal to $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, that gives you a counterexample. You might struggle a bit if you want to make it concrete, it's not obvious to me how to choose $epsilon$, and the order of the enumeration probably makes a difference. It might be easier working with a the complement of a fat cantor set.
    $endgroup$
    – Martin Argerami
    Jan 4 at 17:21










  • $begingroup$
    Maybe I overlook something, but the only projections in $mathscr A$ are $0$ and $1$. If you take $P = chi_{[0,1/2]}$, then $P$ is not the infimum of as described in the question, since the infimum is $1$.
    $endgroup$
    – user42761
    Jan 4 at 18:33












  • $begingroup$
    @AndréS. If $P$ is the indicator of $[0,1/2]$, then the set described in the question is the set of all continuous functions $f:[0,1]rightarrow[0,1]$ such that $f(x)=1$ for all $xin[0,1/2]$. So in this case, $P$ is indeed the infimum of that set. The fact that there is no nontrivial projections in $mathscr{A}$ does not matter here.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 10:24












  • $begingroup$
    Ah okay, I had projections in $mathscr A$ in mind. So is the pointwise infimum of those continuous functions that majorize the indicator, call it $P$, then the infimum in the sense of operators on $L^2([0,1])$ ?
    $endgroup$
    – user42761
    Jan 5 at 12:45










  • $begingroup$
    @AndréS. Yes, and the pointwise infimum as functions on $[0,1]$ is equal to the infimum in the sense of operators in this case, because the pointwise infimum can be obtained by a countable sequence of approximations.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 14:47














2












2








2





$begingroup$

I think I found a counterexample.



Take $mathscr{M}=L^{infty}[0,1]$ and $mathscr{A}=C[0,1]$.



Let $mathbb{Q}cap[0,1]=left{r_{n}right}_{n=1}^{infty}$ be an enumeration of rationals in $[0,1]$, and define $$E:=bigcup_{n=1}^{infty}left(r_{n}-frac{epsilon}{2^{n}},r_{n}+frac{epsilon}{2^{n}}right)cap[0,1]$$ for some small $epsilon>0$ so that $m(E)$ is strictly less than $1$. Clearly, the only continuous function $f:[0,1]rightarrow[0,1]$ satisfying $mathbb{1}_{E}leq fleq 1$ is $fequiv1$, but the projection $mathbb{1}_{E}$ is not equal to $1$.






share|cite|improve this answer









$endgroup$



I think I found a counterexample.



Take $mathscr{M}=L^{infty}[0,1]$ and $mathscr{A}=C[0,1]$.



Let $mathbb{Q}cap[0,1]=left{r_{n}right}_{n=1}^{infty}$ be an enumeration of rationals in $[0,1]$, and define $$E:=bigcup_{n=1}^{infty}left(r_{n}-frac{epsilon}{2^{n}},r_{n}+frac{epsilon}{2^{n}}right)cap[0,1]$$ for some small $epsilon>0$ so that $m(E)$ is strictly less than $1$. Clearly, the only continuous function $f:[0,1]rightarrow[0,1]$ satisfying $mathbb{1}_{E}leq fleq 1$ is $fequiv1$, but the projection $mathbb{1}_{E}$ is not equal to $1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 14:48









Junekey JeonJunekey Jeon

25118




25118












  • $begingroup$
    Yes, that gives you a counterexample. You might struggle a bit if you want to make it concrete, it's not obvious to me how to choose $epsilon$, and the order of the enumeration probably makes a difference. It might be easier working with a the complement of a fat cantor set.
    $endgroup$
    – Martin Argerami
    Jan 4 at 17:21










  • $begingroup$
    Maybe I overlook something, but the only projections in $mathscr A$ are $0$ and $1$. If you take $P = chi_{[0,1/2]}$, then $P$ is not the infimum of as described in the question, since the infimum is $1$.
    $endgroup$
    – user42761
    Jan 4 at 18:33












  • $begingroup$
    @AndréS. If $P$ is the indicator of $[0,1/2]$, then the set described in the question is the set of all continuous functions $f:[0,1]rightarrow[0,1]$ such that $f(x)=1$ for all $xin[0,1/2]$. So in this case, $P$ is indeed the infimum of that set. The fact that there is no nontrivial projections in $mathscr{A}$ does not matter here.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 10:24












  • $begingroup$
    Ah okay, I had projections in $mathscr A$ in mind. So is the pointwise infimum of those continuous functions that majorize the indicator, call it $P$, then the infimum in the sense of operators on $L^2([0,1])$ ?
    $endgroup$
    – user42761
    Jan 5 at 12:45










  • $begingroup$
    @AndréS. Yes, and the pointwise infimum as functions on $[0,1]$ is equal to the infimum in the sense of operators in this case, because the pointwise infimum can be obtained by a countable sequence of approximations.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 14:47


















  • $begingroup$
    Yes, that gives you a counterexample. You might struggle a bit if you want to make it concrete, it's not obvious to me how to choose $epsilon$, and the order of the enumeration probably makes a difference. It might be easier working with a the complement of a fat cantor set.
    $endgroup$
    – Martin Argerami
    Jan 4 at 17:21










  • $begingroup$
    Maybe I overlook something, but the only projections in $mathscr A$ are $0$ and $1$. If you take $P = chi_{[0,1/2]}$, then $P$ is not the infimum of as described in the question, since the infimum is $1$.
    $endgroup$
    – user42761
    Jan 4 at 18:33












  • $begingroup$
    @AndréS. If $P$ is the indicator of $[0,1/2]$, then the set described in the question is the set of all continuous functions $f:[0,1]rightarrow[0,1]$ such that $f(x)=1$ for all $xin[0,1/2]$. So in this case, $P$ is indeed the infimum of that set. The fact that there is no nontrivial projections in $mathscr{A}$ does not matter here.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 10:24












  • $begingroup$
    Ah okay, I had projections in $mathscr A$ in mind. So is the pointwise infimum of those continuous functions that majorize the indicator, call it $P$, then the infimum in the sense of operators on $L^2([0,1])$ ?
    $endgroup$
    – user42761
    Jan 5 at 12:45










  • $begingroup$
    @AndréS. Yes, and the pointwise infimum as functions on $[0,1]$ is equal to the infimum in the sense of operators in this case, because the pointwise infimum can be obtained by a countable sequence of approximations.
    $endgroup$
    – Junekey Jeon
    Jan 5 at 14:47
















$begingroup$
Yes, that gives you a counterexample. You might struggle a bit if you want to make it concrete, it's not obvious to me how to choose $epsilon$, and the order of the enumeration probably makes a difference. It might be easier working with a the complement of a fat cantor set.
$endgroup$
– Martin Argerami
Jan 4 at 17:21




$begingroup$
Yes, that gives you a counterexample. You might struggle a bit if you want to make it concrete, it's not obvious to me how to choose $epsilon$, and the order of the enumeration probably makes a difference. It might be easier working with a the complement of a fat cantor set.
$endgroup$
– Martin Argerami
Jan 4 at 17:21












$begingroup$
Maybe I overlook something, but the only projections in $mathscr A$ are $0$ and $1$. If you take $P = chi_{[0,1/2]}$, then $P$ is not the infimum of as described in the question, since the infimum is $1$.
$endgroup$
– user42761
Jan 4 at 18:33






$begingroup$
Maybe I overlook something, but the only projections in $mathscr A$ are $0$ and $1$. If you take $P = chi_{[0,1/2]}$, then $P$ is not the infimum of as described in the question, since the infimum is $1$.
$endgroup$
– user42761
Jan 4 at 18:33














$begingroup$
@AndréS. If $P$ is the indicator of $[0,1/2]$, then the set described in the question is the set of all continuous functions $f:[0,1]rightarrow[0,1]$ such that $f(x)=1$ for all $xin[0,1/2]$. So in this case, $P$ is indeed the infimum of that set. The fact that there is no nontrivial projections in $mathscr{A}$ does not matter here.
$endgroup$
– Junekey Jeon
Jan 5 at 10:24






$begingroup$
@AndréS. If $P$ is the indicator of $[0,1/2]$, then the set described in the question is the set of all continuous functions $f:[0,1]rightarrow[0,1]$ such that $f(x)=1$ for all $xin[0,1/2]$. So in this case, $P$ is indeed the infimum of that set. The fact that there is no nontrivial projections in $mathscr{A}$ does not matter here.
$endgroup$
– Junekey Jeon
Jan 5 at 10:24














$begingroup$
Ah okay, I had projections in $mathscr A$ in mind. So is the pointwise infimum of those continuous functions that majorize the indicator, call it $P$, then the infimum in the sense of operators on $L^2([0,1])$ ?
$endgroup$
– user42761
Jan 5 at 12:45




$begingroup$
Ah okay, I had projections in $mathscr A$ in mind. So is the pointwise infimum of those continuous functions that majorize the indicator, call it $P$, then the infimum in the sense of operators on $L^2([0,1])$ ?
$endgroup$
– user42761
Jan 5 at 12:45












$begingroup$
@AndréS. Yes, and the pointwise infimum as functions on $[0,1]$ is equal to the infimum in the sense of operators in this case, because the pointwise infimum can be obtained by a countable sequence of approximations.
$endgroup$
– Junekey Jeon
Jan 5 at 14:47




$begingroup$
@AndréS. Yes, and the pointwise infimum as functions on $[0,1]$ is equal to the infimum in the sense of operators in this case, because the pointwise infimum can be obtained by a countable sequence of approximations.
$endgroup$
– Junekey Jeon
Jan 5 at 14:47


















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Grease: Live!

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A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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