an element in $prod_n M_n(Bbb C)$
$begingroup$
I want to find an element $x=(x_n)in prod_nM_n(Bbb C)$ such that $lim operatorname{tr}_n(x_n)=0$ but $lim operatorname{tr}_n(x_n^*x_n)not to 0$,where $tr$ is the unique tracial state on $M_n(Bbb C)$.But I cannot think of an example for a while.I'll appreciate it anyone can help me.
operator-theory operator-algebras c-star-algebras
asked Dec 17 '18 at 10:47
mathrookiemathrookie
914512
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add a comment |
$begingroup$
I want to find an element $x=(x_n)in prod_nM_n(Bbb C)$ such that $lim operatorname{tr}_n(x_n)=0$ but $lim operatorname{tr}_n(x_n^*x_n)not to 0$,where $tr$ is the unique tracial state on $M_n(Bbb C)$.But I cannot think of an example for a while.I'll appreciate it anyone can help me.
operator-theory operator-algebras c-star-algebras
asked Dec 17 '18 at 10:47
mathrookiemathrookie
914512
$endgroup$
add a comment |
$begingroup$
I want to find an element $x=(x_n)in prod_nM_n(Bbb C)$ such that $lim operatorname{tr}_n(x_n)=0$ but $lim operatorname{tr}_n(x_n^*x_n)not to 0$,where $tr$ is the unique tracial state on $M_n(Bbb C)$.But I cannot think of an example for a while.I'll appreciate it anyone can help me.
operator-theory operator-algebras c-star-algebras
asked Dec 17 '18 at 10:47
mathrookiemathrookie
914512
$endgroup$
I want to find an element $x=(x_n)in prod_nM_n(Bbb C)$ such that $lim operatorname{tr}_n(x_n)=0$ but $lim operatorname{tr}_n(x_n^*x_n)not to 0$,where $tr$ is the unique tracial state on $M_n(Bbb C)$.But I cannot think of an example for a while.I'll appreciate it anyone can help me.
operator-theory operator-algebras c-star-algebras
operator-theory operator-algebras c-star-algebras
asked Dec 17 '18 at 10:47
mathrookiemathrookie
914512
asked Dec 17 '18 at 10:47
mathrookiemathrookie
914512
edited Dec 17 '18 at 16:13
user593746
asked Dec 17 '18 at 10:47
mathrookiemathrookie
914512
asked Dec 17 '18 at 10:47
mathrookiemathrookie
914512
asked Dec 17 '18 at 10:47
mathrookiemathrookie
914512
914512
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the bounded sequence $(x_n)_{n=1}^infty$ given by
$$
begin{cases}
x_{2n} := 1_n oplus (-1_n), \
x_{2n+1} := 1_{2n+1}.
end{cases}
$$
Then $mathrm{tr}_n(x_n) = 0$ for every even natural nuber $n$. Now, let $omega$ be a free ultrafilter such that $omega$ contains the set of all even natural numbers. Then you easily check that
$$
lim_omega mathrm{tr}_n(x_n) = 0,
$$
but
$$
lim_omega mathrm{tr}_n(x_n^*x_n) = 1.
$$
$endgroup$
$begingroup$
I don't understand your question.
$endgroup$
– user42761
Dec 17 '18 at 18:31
add a comment |
$begingroup$
You can take $x_n$ to be the shift, that is $x_n=sum_{j=1}^n E_{j,j+1}$. Then $operatorname{tr}(x_n)=0$ for all $n$, while $operatorname{tr}(x_n^*x_n)=1-tfrac1n$. Then $$lim_{ntoinfty}operatorname{tr}(x_n^*x_n)=1,$$
while $$operatorname{tr}(x_n)=0$$ for all $n$.
answered Dec 17 '18 at 14:37
Martin ArgeramiMartin Argerami
128k1184184
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the bounded sequence $(x_n)_{n=1}^infty$ given by
$$
begin{cases}
x_{2n} := 1_n oplus (-1_n), \
x_{2n+1} := 1_{2n+1}.
end{cases}
$$
Then $mathrm{tr}_n(x_n) = 0$ for every even natural nuber $n$. Now, let $omega$ be a free ultrafilter such that $omega$ contains the set of all even natural numbers. Then you easily check that
$$
lim_omega mathrm{tr}_n(x_n) = 0,
$$
but
$$
lim_omega mathrm{tr}_n(x_n^*x_n) = 1.
$$
$endgroup$
$begingroup$
I don't understand your question.
$endgroup$
– user42761
Dec 17 '18 at 18:31
add a comment |
$begingroup$
Consider the bounded sequence $(x_n)_{n=1}^infty$ given by
$$
begin{cases}
x_{2n} := 1_n oplus (-1_n), \
x_{2n+1} := 1_{2n+1}.
end{cases}
$$
Then $mathrm{tr}_n(x_n) = 0$ for every even natural nuber $n$. Now, let $omega$ be a free ultrafilter such that $omega$ contains the set of all even natural numbers. Then you easily check that
$$
lim_omega mathrm{tr}_n(x_n) = 0,
$$
but
$$
lim_omega mathrm{tr}_n(x_n^*x_n) = 1.
$$
$endgroup$
$begingroup$
I don't understand your question.
$endgroup$
– user42761
Dec 17 '18 at 18:31
add a comment |
$begingroup$
Consider the bounded sequence $(x_n)_{n=1}^infty$ given by
$$
begin{cases}
x_{2n} := 1_n oplus (-1_n), \
x_{2n+1} := 1_{2n+1}.
end{cases}
$$
Then $mathrm{tr}_n(x_n) = 0$ for every even natural nuber $n$. Now, let $omega$ be a free ultrafilter such that $omega$ contains the set of all even natural numbers. Then you easily check that
$$
lim_omega mathrm{tr}_n(x_n) = 0,
$$
but
$$
lim_omega mathrm{tr}_n(x_n^*x_n) = 1.
$$
$endgroup$
Consider the bounded sequence $(x_n)_{n=1}^infty$ given by
$$
begin{cases}
x_{2n} := 1_n oplus (-1_n), \
x_{2n+1} := 1_{2n+1}.
end{cases}
$$
Then $mathrm{tr}_n(x_n) = 0$ for every even natural nuber $n$. Now, let $omega$ be a free ultrafilter such that $omega$ contains the set of all even natural numbers. Then you easily check that
$$
lim_omega mathrm{tr}_n(x_n) = 0,
$$
but
$$
lim_omega mathrm{tr}_n(x_n^*x_n) = 1.
$$
answered Dec 17 '18 at 11:20
user42761
$begingroup$
I don't understand your question.
$endgroup$
– user42761
Dec 17 '18 at 18:31
add a comment |
$begingroup$
I don't understand your question.
$endgroup$
– user42761
Dec 17 '18 at 18:31
$begingroup$
I don't understand your question.
$endgroup$
– user42761
Dec 17 '18 at 18:31
$begingroup$
I don't understand your question.
$endgroup$
– user42761
Dec 17 '18 at 18:31
add a comment |
$begingroup$
You can take $x_n$ to be the shift, that is $x_n=sum_{j=1}^n E_{j,j+1}$. Then $operatorname{tr}(x_n)=0$ for all $n$, while $operatorname{tr}(x_n^*x_n)=1-tfrac1n$. Then $$lim_{ntoinfty}operatorname{tr}(x_n^*x_n)=1,$$
while $$operatorname{tr}(x_n)=0$$ for all $n$.
answered Dec 17 '18 at 14:37
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
You can take $x_n$ to be the shift, that is $x_n=sum_{j=1}^n E_{j,j+1}$. Then $operatorname{tr}(x_n)=0$ for all $n$, while $operatorname{tr}(x_n^*x_n)=1-tfrac1n$. Then $$lim_{ntoinfty}operatorname{tr}(x_n^*x_n)=1,$$
while $$operatorname{tr}(x_n)=0$$ for all $n$.
answered Dec 17 '18 at 14:37
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
You can take $x_n$ to be the shift, that is $x_n=sum_{j=1}^n E_{j,j+1}$. Then $operatorname{tr}(x_n)=0$ for all $n$, while $operatorname{tr}(x_n^*x_n)=1-tfrac1n$. Then $$lim_{ntoinfty}operatorname{tr}(x_n^*x_n)=1,$$
while $$operatorname{tr}(x_n)=0$$ for all $n$.
answered Dec 17 '18 at 14:37
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
You can take $x_n$ to be the shift, that is $x_n=sum_{j=1}^n E_{j,j+1}$. Then $operatorname{tr}(x_n)=0$ for all $n$, while $operatorname{tr}(x_n^*x_n)=1-tfrac1n$. Then $$lim_{ntoinfty}operatorname{tr}(x_n^*x_n)=1,$$
while $$operatorname{tr}(x_n)=0$$ for all $n$.
answered Dec 17 '18 at 14:37
Martin ArgeramiMartin Argerami
128k1184184
answered Dec 17 '18 at 14:37
Martin ArgeramiMartin Argerami
128k1184184
answered Dec 17 '18 at 14:37
Martin ArgeramiMartin Argerami
128k1184184
answered Dec 17 '18 at 14:37
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
add a comment |
add a comment |
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