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

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active

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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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Are they similar matrix

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0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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