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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
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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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I am so dumb it was easy.
    $endgroup$
    – Math Lover
    Jan 3 at 7:39












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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes






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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I am so dumb it was easy.
    $endgroup$
    – Math Lover
    Jan 3 at 7:39
















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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I am so dumb it was easy.
    $endgroup$
    – Math Lover
    Jan 3 at 7:39














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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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