Question regarding $C^*$ algebra generated by a product
1
$begingroup$
Let $V=M_{1,infty}$,the row Hilbert space. Suppose $W$ denotes the $C^*-$ algebra generated by $V^*V={x^*y : x,y in V }$
Is it true that $W= K(l_2)$, space of compact operators on $l_2$?
I can see that $V^*V$ is nothing but $M_infty$ but I cannot see the exact claim. Please help.
functional-analysis c-star-algebras
asked Jan 3 at 6:00
Math LoverMath Lover
1,034315
$endgroup$
|
show 1 more comment
1
$begingroup$
Let $V=M_{1,infty}$,the row Hilbert space. Suppose $W$ denotes the $C^*-$ algebra generated by $V^*V={x^*y : x,y in V }$
Is it true that $W= K(l_2)$, space of compact operators on $l_2$?
I can see that $V^*V$ is nothing but $M_infty$ but I cannot see the exact claim. Please help.
functional-analysis c-star-algebras
asked Jan 3 at 6:00
Math LoverMath Lover
1,034315
$endgroup$
$begingroup$
Where does your notation come from? If I translate this to the terminology I am familiar with, it seems to be asking whether or not the closed linear span of the finite-rank operators on a Hilbert space is the compact operators on the Hilbert space. Is this what you are asking?
$endgroup$
– Aweygan
Jan 3 at 6:37
$begingroup$
@Aweygan The notation is taken from a paper by Ruan and Kaur. I know whatever you have written but how does that help? Are you saying that $V^*V= l_2$?
$endgroup$
– Math Lover
Jan 3 at 6:46
$begingroup$
I'm confused by the notation $M_{1,infty}$, what you refer to as a "row Hilbert space", and the product $x^*y$ in a Hilbert space.
$endgroup$
– Aweygan
Jan 3 at 7:02
$begingroup$
If you could provide a PDF link to the paper you are reading that would help. Otherwise, I feel like the notation is to nonstandard to allow anyone to provide specific advice to help you.
$endgroup$
– Aweygan
Jan 3 at 7:06
$begingroup$
@Aweygan: "row Hilbert (operator) space" is standard operator-space terminology. Same with "column". But your intuition is exactly what it is.
$endgroup$
– Martin Argerami
Jan 3 at 7:11
|
show 1 more comment
1
1
1
$begingroup$
Let $V=M_{1,infty}$,the row Hilbert space. Suppose $W$ denotes the $C^*-$ algebra generated by $V^*V={x^*y : x,y in V }$
Is it true that $W= K(l_2)$, space of compact operators on $l_2$?
I can see that $V^*V$ is nothing but $M_infty$ but I cannot see the exact claim. Please help.
functional-analysis c-star-algebras
asked Jan 3 at 6:00
Math LoverMath Lover
1,034315
$endgroup$
Let $V=M_{1,infty}$,the row Hilbert space. Suppose $W$ denotes the $C^*-$ algebra generated by $V^*V={x^*y : x,y in V }$
Is it true that $W= K(l_2)$, space of compact operators on $l_2$?
I can see that $V^*V$ is nothing but $M_infty$ but I cannot see the exact claim. Please help.
functional-analysis c-star-algebras
functional-analysis c-star-algebras
asked Jan 3 at 6:00
Math LoverMath Lover
1,034315
asked Jan 3 at 6:00
Math LoverMath Lover
1,034315
asked Jan 3 at 6:00
Math LoverMath Lover
1,034315
asked Jan 3 at 6:00
Math LoverMath Lover
1,034315
asked Jan 3 at 6:00
Math LoverMath Lover
1,034315
1,034315
$begingroup$
Where does your notation come from? If I translate this to the terminology I am familiar with, it seems to be asking whether or not the closed linear span of the finite-rank operators on a Hilbert space is the compact operators on the Hilbert space. Is this what you are asking?
$endgroup$
– Aweygan
Jan 3 at 6:37
$begingroup$
@Aweygan The notation is taken from a paper by Ruan and Kaur. I know whatever you have written but how does that help? Are you saying that $V^*V= l_2$?
$endgroup$
– Math Lover
Jan 3 at 6:46
$begingroup$
I'm confused by the notation $M_{1,infty}$, what you refer to as a "row Hilbert space", and the product $x^*y$ in a Hilbert space.
$endgroup$
– Aweygan
Jan 3 at 7:02
$begingroup$
If you could provide a PDF link to the paper you are reading that would help. Otherwise, I feel like the notation is to nonstandard to allow anyone to provide specific advice to help you.
$endgroup$
– Aweygan
Jan 3 at 7:06
$begingroup$
@Aweygan: "row Hilbert (operator) space" is standard operator-space terminology. Same with "column". But your intuition is exactly what it is.
$endgroup$
– Martin Argerami
Jan 3 at 7:11
|
show 1 more comment
$begingroup$
Where does your notation come from? If I translate this to the terminology I am familiar with, it seems to be asking whether or not the closed linear span of the finite-rank operators on a Hilbert space is the compact operators on the Hilbert space. Is this what you are asking?
$endgroup$
– Aweygan
Jan 3 at 6:37
$begingroup$
@Aweygan The notation is taken from a paper by Ruan and Kaur. I know whatever you have written but how does that help? Are you saying that $V^*V= l_2$?
$endgroup$
– Math Lover
Jan 3 at 6:46
$begingroup$
I'm confused by the notation $M_{1,infty}$, what you refer to as a "row Hilbert space", and the product $x^*y$ in a Hilbert space.
$endgroup$
– Aweygan
Jan 3 at 7:02
$begingroup$
If you could provide a PDF link to the paper you are reading that would help. Otherwise, I feel like the notation is to nonstandard to allow anyone to provide specific advice to help you.
$endgroup$
– Aweygan
Jan 3 at 7:06
$begingroup$
@Aweygan: "row Hilbert (operator) space" is standard operator-space terminology. Same with "column". But your intuition is exactly what it is.
$endgroup$
– Martin Argerami
Jan 3 at 7:11
$begingroup$
Where does your notation come from? If I translate this to the terminology I am familiar with, it seems to be asking whether or not the closed linear span of the finite-rank operators on a Hilbert space is the compact operators on the Hilbert space. Is this what you are asking?
$endgroup$
– Aweygan
Jan 3 at 6:37
$begingroup$
Where does your notation come from? If I translate this to the terminology I am familiar with, it seems to be asking whether or not the closed linear span of the finite-rank operators on a Hilbert space is the compact operators on the Hilbert space. Is this what you are asking?
$endgroup$
– Aweygan
Jan 3 at 6:37
$begingroup$
@Aweygan The notation is taken from a paper by Ruan and Kaur. I know whatever you have written but how does that help? Are you saying that $V^*V= l_2$?
$endgroup$
– Math Lover
Jan 3 at 6:46
$begingroup$
@Aweygan The notation is taken from a paper by Ruan and Kaur. I know whatever you have written but how does that help? Are you saying that $V^*V= l_2$?
$endgroup$
– Math Lover
Jan 3 at 6:46
$begingroup$
I'm confused by the notation $M_{1,infty}$, what you refer to as a "row Hilbert space", and the product $x^*y$ in a Hilbert space.
$endgroup$
– Aweygan
Jan 3 at 7:02
$begingroup$
I'm confused by the notation $M_{1,infty}$, what you refer to as a "row Hilbert space", and the product $x^*y$ in a Hilbert space.
$endgroup$
– Aweygan
Jan 3 at 7:02
$begingroup$
If you could provide a PDF link to the paper you are reading that would help. Otherwise, I feel like the notation is to nonstandard to allow anyone to provide specific advice to help you.
$endgroup$
– Aweygan
Jan 3 at 7:06
$begingroup$
If you could provide a PDF link to the paper you are reading that would help. Otherwise, I feel like the notation is to nonstandard to allow anyone to provide specific advice to help you.
$endgroup$
– Aweygan
Jan 3 at 7:06
$begingroup$
@Aweygan: "row Hilbert (operator) space" is standard operator-space terminology. Same with "column". But your intuition is exactly what it is.
$endgroup$
– Martin Argerami
Jan 3 at 7:11
$begingroup$
@Aweygan: "row Hilbert (operator) space" is standard operator-space terminology. Same with "column". But your intuition is exactly what it is.
$endgroup$
– Martin Argerami
Jan 3 at 7:11
|
show 1 more comment
1 Answer
1
active
oldest
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1
$begingroup$
Since the row Hilbert operator space is ${e_{1j}: jinmathbb N}subset B(ell^2(mathbb N))$, your C$^*$-algebra $W$ is generated by the matrix units $e_{kj}=e_{1k}^*e_{1j}$. So $W=K(ell^2(mathbb N))$, the compact operators.
answered Jan 3 at 7:12
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thanks. I am so dumb it was easy.
$endgroup$
– Math Lover
Jan 3 at 7:39
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
1
$begingroup$
Since the row Hilbert operator space is ${e_{1j}: jinmathbb N}subset B(ell^2(mathbb N))$, your C$^*$-algebra $W$ is generated by the matrix units $e_{kj}=e_{1k}^*e_{1j}$. So $W=K(ell^2(mathbb N))$, the compact operators.
answered Jan 3 at 7:12
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thanks. I am so dumb it was easy.
$endgroup$
– Math Lover
Jan 3 at 7:39
add a comment |
1
$begingroup$
Since the row Hilbert operator space is ${e_{1j}: jinmathbb N}subset B(ell^2(mathbb N))$, your C$^*$-algebra $W$ is generated by the matrix units $e_{kj}=e_{1k}^*e_{1j}$. So $W=K(ell^2(mathbb N))$, the compact operators.
answered Jan 3 at 7:12
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thanks. I am so dumb it was easy.
$endgroup$
– Math Lover
Jan 3 at 7:39
add a comment |
1
1
1
$begingroup$
Since the row Hilbert operator space is ${e_{1j}: jinmathbb N}subset B(ell^2(mathbb N))$, your C$^*$-algebra $W$ is generated by the matrix units $e_{kj}=e_{1k}^*e_{1j}$. So $W=K(ell^2(mathbb N))$, the compact operators.
answered Jan 3 at 7:12
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
Since the row Hilbert operator space is ${e_{1j}: jinmathbb N}subset B(ell^2(mathbb N))$, your C$^*$-algebra $W$ is generated by the matrix units $e_{kj}=e_{1k}^*e_{1j}$. So $W=K(ell^2(mathbb N))$, the compact operators.
answered Jan 3 at 7:12
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 3 at 7:12
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 3 at 7:12
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 3 at 7:12
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
Thanks. I am so dumb it was easy.
$endgroup$
– Math Lover
Jan 3 at 7:39
add a comment |
$begingroup$
Thanks. I am so dumb it was easy.
$endgroup$
– Math Lover
Jan 3 at 7:39
$begingroup$
Thanks. I am so dumb it was easy.
$endgroup$
– Math Lover
Jan 3 at 7:39
$begingroup$
Thanks. I am so dumb it was easy.
$endgroup$
– Math Lover
Jan 3 at 7:39
add a comment |
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$begingroup$
Where does your notation come from? If I translate this to the terminology I am familiar with, it seems to be asking whether or not the closed linear span of the finite-rank operators on a Hilbert space is the compact operators on the Hilbert space. Is this what you are asking?
$endgroup$
– Aweygan
Jan 3 at 6:37
$begingroup$
@Aweygan The notation is taken from a paper by Ruan and Kaur. I know whatever you have written but how does that help? Are you saying that $V^*V= l_2$?
$endgroup$
– Math Lover
Jan 3 at 6:46
$begingroup$
I'm confused by the notation $M_{1,infty}$, what you refer to as a "row Hilbert space", and the product $x^*y$ in a Hilbert space.
$endgroup$
– Aweygan
Jan 3 at 7:02
$begingroup$
If you could provide a PDF link to the paper you are reading that would help. Otherwise, I feel like the notation is to nonstandard to allow anyone to provide specific advice to help you.
$endgroup$
– Aweygan
Jan 3 at 7:06
$begingroup$
@Aweygan: "row Hilbert (operator) space" is standard operator-space terminology. Same with "column". But your intuition is exactly what it is.
$endgroup$
– Martin Argerami
Jan 3 at 7:11