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Quantum functional analysis
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Can one explain some philosophy behind "quantum functional analysis" (or "quantized functional analysis") which was initiated and developed by such researchers as: Ruan Z.-J., Pisier J., Effros E.G., Haagerup U., et al....
The main notion of this subject is a quantum space: $(E, cdot)$ -- some normed space, equipped with "quantum norms" $|cdot|_n, , nin mathbbN$ (in turn, these quantum norms $cdot$ are not the norms on $E$ but on matrices $M_n(E)$). It appears that classical bounded operators in $(E, |cdot|_1)$ may have good ("quantum brothers") properties in whole quantum space, and as far as I understood they are called "completely bounded" operators. So quantum functional analysis is a theory about "such" operators, about matrices of operators (which are themselves operators) and from the book A.J. Helemsky I've read that it is a beautiful theory, "much simpler" than the classical functional analysis.
Also there should be a relationship with quantum mechanics, since
there is a notion of "quantization" of normed spaces (but I don't know anything more than that).
I was wondering, can one here post some beautiful results of this theory? (for example, problem of Halmos about polynomially contracting operators was negatively resolved by Pisier using techniques from this theory). Among these beautiful results those ones which speak about classical analysis are extremely beautiful. Are there many extremely beautiful results?
fa.functional-analysis ho.history-overview mathematical-philosophy banach-algebras
asked Aug 31 at 8:54
Fedor Goncharov
312112
add a comment |Â
up vote
10
down vote
favorite
Can one explain some philosophy behind "quantum functional analysis" (or "quantized functional analysis") which was initiated and developed by such researchers as: Ruan Z.-J., Pisier J., Effros E.G., Haagerup U., et al....
The main notion of this subject is a quantum space: $(E, cdot)$ -- some normed space, equipped with "quantum norms" $|cdot|_n, , nin mathbbN$ (in turn, these quantum norms $cdot$ are not the norms on $E$ but on matrices $M_n(E)$). It appears that classical bounded operators in $(E, |cdot|_1)$ may have good ("quantum brothers") properties in whole quantum space, and as far as I understood they are called "completely bounded" operators. So quantum functional analysis is a theory about "such" operators, about matrices of operators (which are themselves operators) and from the book A.J. Helemsky I've read that it is a beautiful theory, "much simpler" than the classical functional analysis.
Also there should be a relationship with quantum mechanics, since
there is a notion of "quantization" of normed spaces (but I don't know anything more than that).
I was wondering, can one here post some beautiful results of this theory? (for example, problem of Halmos about polynomially contracting operators was negatively resolved by Pisier using techniques from this theory). Among these beautiful results those ones which speak about classical analysis are extremely beautiful. Are there many extremely beautiful results?
fa.functional-analysis ho.history-overview mathematical-philosophy banach-algebras
asked Aug 31 at 8:54
Fedor Goncharov
312112
2
It may be worth pointing out that the $n$-th level "quantum norm" $|cdot|_n$ does not live on $E$ itself, but rather on the matrix space $M_n(E)$.
– Tobias Fritz
Aug 31 at 12:57
Yes, I added it in the text.
– Fedor Goncharov
Aug 31 at 14:13
2
With all due respect to Prof. Helemskii, I have very rarely seen this term "quantum normed space" or "quantum Banach space" used outside his school/seminar.
– Yemon Choi
Aug 31 at 16:10
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Can one explain some philosophy behind "quantum functional analysis" (or "quantized functional analysis") which was initiated and developed by such researchers as: Ruan Z.-J., Pisier J., Effros E.G., Haagerup U., et al....
The main notion of this subject is a quantum space: $(E, cdot)$ -- some normed space, equipped with "quantum norms" $|cdot|_n, , nin mathbbN$ (in turn, these quantum norms $cdot$ are not the norms on $E$ but on matrices $M_n(E)$). It appears that classical bounded operators in $(E, |cdot|_1)$ may have good ("quantum brothers") properties in whole quantum space, and as far as I understood they are called "completely bounded" operators. So quantum functional analysis is a theory about "such" operators, about matrices of operators (which are themselves operators) and from the book A.J. Helemsky I've read that it is a beautiful theory, "much simpler" than the classical functional analysis.
Also there should be a relationship with quantum mechanics, since
there is a notion of "quantization" of normed spaces (but I don't know anything more than that).
I was wondering, can one here post some beautiful results of this theory? (for example, problem of Halmos about polynomially contracting operators was negatively resolved by Pisier using techniques from this theory). Among these beautiful results those ones which speak about classical analysis are extremely beautiful. Are there many extremely beautiful results?
fa.functional-analysis ho.history-overview mathematical-philosophy banach-algebras
asked Aug 31 at 8:54
Fedor Goncharov
312112
Can one explain some philosophy behind "quantum functional analysis" (or "quantized functional analysis") which was initiated and developed by such researchers as: Ruan Z.-J., Pisier J., Effros E.G., Haagerup U., et al....
The main notion of this subject is a quantum space: $(E, cdot)$ -- some normed space, equipped with "quantum norms" $|cdot|_n, , nin mathbbN$ (in turn, these quantum norms $cdot$ are not the norms on $E$ but on matrices $M_n(E)$). It appears that classical bounded operators in $(E, |cdot|_1)$ may have good ("quantum brothers") properties in whole quantum space, and as far as I understood they are called "completely bounded" operators. So quantum functional analysis is a theory about "such" operators, about matrices of operators (which are themselves operators) and from the book A.J. Helemsky I've read that it is a beautiful theory, "much simpler" than the classical functional analysis.
Also there should be a relationship with quantum mechanics, since
there is a notion of "quantization" of normed spaces (but I don't know anything more than that).
I was wondering, can one here post some beautiful results of this theory? (for example, problem of Halmos about polynomially contracting operators was negatively resolved by Pisier using techniques from this theory). Among these beautiful results those ones which speak about classical analysis are extremely beautiful. Are there many extremely beautiful results?
fa.functional-analysis ho.history-overview mathematical-philosophy banach-algebras
asked Aug 31 at 8:54
Fedor Goncharov
312112
edited Sep 4 at 9:40
asked Aug 31 at 8:54
Fedor Goncharov
312112
asked Aug 31 at 8:54
Fedor Goncharov
312112
asked Aug 31 at 8:54
Fedor Goncharov
312112
312112
2
It may be worth pointing out that the $n$-th level "quantum norm" $|cdot|_n$ does not live on $E$ itself, but rather on the matrix space $M_n(E)$.
– Tobias Fritz
Aug 31 at 12:57
Yes, I added it in the text.
– Fedor Goncharov
Aug 31 at 14:13
2
With all due respect to Prof. Helemskii, I have very rarely seen this term "quantum normed space" or "quantum Banach space" used outside his school/seminar.
– Yemon Choi
Aug 31 at 16:10
add a comment |Â
2
It may be worth pointing out that the $n$-th level "quantum norm" $|cdot|_n$ does not live on $E$ itself, but rather on the matrix space $M_n(E)$.
– Tobias Fritz
Aug 31 at 12:57
Yes, I added it in the text.
– Fedor Goncharov
Aug 31 at 14:13
2
With all due respect to Prof. Helemskii, I have very rarely seen this term "quantum normed space" or "quantum Banach space" used outside his school/seminar.
– Yemon Choi
Aug 31 at 16:10
2
2
It may be worth pointing out that the $n$-th level "quantum norm" $|cdot|_n$ does not live on $E$ itself, but rather on the matrix space $M_n(E)$.
– Tobias Fritz
Aug 31 at 12:57
It may be worth pointing out that the $n$-th level "quantum norm" $|cdot|_n$ does not live on $E$ itself, but rather on the matrix space $M_n(E)$.
– Tobias Fritz
Aug 31 at 12:57
Yes, I added it in the text.
– Fedor Goncharov
Aug 31 at 14:13
Yes, I added it in the text.
– Fedor Goncharov
Aug 31 at 14:13
2
2
With all due respect to Prof. Helemskii, I have very rarely seen this term "quantum normed space" or "quantum Banach space" used outside his school/seminar.
– Yemon Choi
Aug 31 at 16:10
With all due respect to Prof. Helemskii, I have very rarely seen this term "quantum normed space" or "quantum Banach space" used outside his school/seminar.
– Yemon Choi
Aug 31 at 16:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
15
down vote
accepted
Okay, I'll take this one. First let me say that the English term is "completely bounded" (or "complete isometry", etc.).
About the term "quantum". The general principle is that analyzing some aspect of a physical system typically involves very different kinds of mathematical structures, depending on whether the system is classical or quantum. Thus if you study a classical system using topological techniques, or measure theory, or Lie groups, or graphs, the analogous analysis of a quantum system would probably involve C*-algebras, or von Neumann algebras, or quantum groups, or operator systems. (For instance, here's a paper of mine that talks about how operator systems arise in quantum error correction in the same way that graphs arise in classical error correction.)
There is no simple rule for translating classical structures into quantum structures, but we have a pretty large dictionary and one can identify broad themes. My take is that operator spaces fit into this scheme as the quantum analog of relations on a set. See Section 2 of the paper linked above. I'm not aware of any really compelling direct connections with real-world quantum mechanics in this instance, but the word "quantum" may be appropriate just because the subject fits into the general scheme of mathematical quantization which does include many direct links.
Now, as to the "beautiful results" you ask for. Well, there's a lot of good stuff so the selection would be idiosyncratic. One general principle is that when you're studying things related to Hilbert space (single operators, C*-algebras, etc.) it is often the case that you get much nicer results if you assume the "complete" (at all matrix levels) version of your hypothesis. For example, if two C*-algebras are linearly isometric, are they isomorphic as C*-algebras? Not in general, but yes if they are completely linearly isometric, i.e., linearly isometric at all matrix levels. (I think this is folklore.)
The result of Pisier that you refer to was actually a counterexample, showing that a polynomially bounded operator need not be similar to a contraction. The positive result is due to Paulsen and says that any completely polynomially bounded operator is similar to a contraction.
Here's another nice result, this one due to Zhong-Jin Ruan, one of the giants of the subject. B. Johnson proved that a locally compact group $G$ is amenable if and only if the Banach algebra $L^1(G)$ is amenable. Ruan showed that this happens if and only if the Fourier algebra $A(G)$ is completely amenable.
A good place to learn more is the book Operator Spaces by Effros and Ruan.
References:
G. Pisier, A polynomially bounded operator on Hilbert space which is not similar to a contraction, J. Amer. Math. Soc. 10 (1997), 351–369.
V. Paulsen, Every completely polynomially bounded operator is similar to a contraction, J. Funct. Anal. 55 (1984), 1–17.
Z-J. Ruan, The operator amenability of A(G), Amer. J. Math. 117 (1995), 1449–1474.
answered Aug 31 at 15:11
Nik Weaver
17.9k142114
1
I have to say I prefer "completely amenable" to the horrible-but-standard "operator amenable", although the former does make me wonder if I should go ahead and define a notion of something being utterly amenable ...
– Yemon Choi
Aug 31 at 16:09
12
I once heard a talk by David Blecher where he started by explaining the meaning of "completely bounded", "completely positive", and so on. Then he put up a lemma and said "The proof of this is completely obvious." I think I was the only one who laughed, though.
– Nik Weaver
Aug 31 at 16:20
1
Thanks for remark about 'completely bounded operator' etc., and the example with amenable groups is very good. Thank you!
– Fedor Goncharov
Aug 31 at 16:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
Okay, I'll take this one. First let me say that the English term is "completely bounded" (or "complete isometry", etc.).
About the term "quantum". The general principle is that analyzing some aspect of a physical system typically involves very different kinds of mathematical structures, depending on whether the system is classical or quantum. Thus if you study a classical system using topological techniques, or measure theory, or Lie groups, or graphs, the analogous analysis of a quantum system would probably involve C*-algebras, or von Neumann algebras, or quantum groups, or operator systems. (For instance, here's a paper of mine that talks about how operator systems arise in quantum error correction in the same way that graphs arise in classical error correction.)
There is no simple rule for translating classical structures into quantum structures, but we have a pretty large dictionary and one can identify broad themes. My take is that operator spaces fit into this scheme as the quantum analog of relations on a set. See Section 2 of the paper linked above. I'm not aware of any really compelling direct connections with real-world quantum mechanics in this instance, but the word "quantum" may be appropriate just because the subject fits into the general scheme of mathematical quantization which does include many direct links.
Now, as to the "beautiful results" you ask for. Well, there's a lot of good stuff so the selection would be idiosyncratic. One general principle is that when you're studying things related to Hilbert space (single operators, C*-algebras, etc.) it is often the case that you get much nicer results if you assume the "complete" (at all matrix levels) version of your hypothesis. For example, if two C*-algebras are linearly isometric, are they isomorphic as C*-algebras? Not in general, but yes if they are completely linearly isometric, i.e., linearly isometric at all matrix levels. (I think this is folklore.)
The result of Pisier that you refer to was actually a counterexample, showing that a polynomially bounded operator need not be similar to a contraction. The positive result is due to Paulsen and says that any completely polynomially bounded operator is similar to a contraction.
Here's another nice result, this one due to Zhong-Jin Ruan, one of the giants of the subject. B. Johnson proved that a locally compact group $G$ is amenable if and only if the Banach algebra $L^1(G)$ is amenable. Ruan showed that this happens if and only if the Fourier algebra $A(G)$ is completely amenable.
A good place to learn more is the book Operator Spaces by Effros and Ruan.
References:
G. Pisier, A polynomially bounded operator on Hilbert space which is not similar to a contraction, J. Amer. Math. Soc. 10 (1997), 351–369.
V. Paulsen, Every completely polynomially bounded operator is similar to a contraction, J. Funct. Anal. 55 (1984), 1–17.
Z-J. Ruan, The operator amenability of A(G), Amer. J. Math. 117 (1995), 1449–1474.
answered Aug 31 at 15:11
Nik Weaver
17.9k142114
1
I have to say I prefer "completely amenable" to the horrible-but-standard "operator amenable", although the former does make me wonder if I should go ahead and define a notion of something being utterly amenable ...
– Yemon Choi
Aug 31 at 16:09
12
I once heard a talk by David Blecher where he started by explaining the meaning of "completely bounded", "completely positive", and so on. Then he put up a lemma and said "The proof of this is completely obvious." I think I was the only one who laughed, though.
– Nik Weaver
Aug 31 at 16:20
1
Thanks for remark about 'completely bounded operator' etc., and the example with amenable groups is very good. Thank you!
– Fedor Goncharov
Aug 31 at 16:30
add a comment |Â
up vote
15
down vote
accepted
Okay, I'll take this one. First let me say that the English term is "completely bounded" (or "complete isometry", etc.).
About the term "quantum". The general principle is that analyzing some aspect of a physical system typically involves very different kinds of mathematical structures, depending on whether the system is classical or quantum. Thus if you study a classical system using topological techniques, or measure theory, or Lie groups, or graphs, the analogous analysis of a quantum system would probably involve C*-algebras, or von Neumann algebras, or quantum groups, or operator systems. (For instance, here's a paper of mine that talks about how operator systems arise in quantum error correction in the same way that graphs arise in classical error correction.)
There is no simple rule for translating classical structures into quantum structures, but we have a pretty large dictionary and one can identify broad themes. My take is that operator spaces fit into this scheme as the quantum analog of relations on a set. See Section 2 of the paper linked above. I'm not aware of any really compelling direct connections with real-world quantum mechanics in this instance, but the word "quantum" may be appropriate just because the subject fits into the general scheme of mathematical quantization which does include many direct links.
Now, as to the "beautiful results" you ask for. Well, there's a lot of good stuff so the selection would be idiosyncratic. One general principle is that when you're studying things related to Hilbert space (single operators, C*-algebras, etc.) it is often the case that you get much nicer results if you assume the "complete" (at all matrix levels) version of your hypothesis. For example, if two C*-algebras are linearly isometric, are they isomorphic as C*-algebras? Not in general, but yes if they are completely linearly isometric, i.e., linearly isometric at all matrix levels. (I think this is folklore.)
The result of Pisier that you refer to was actually a counterexample, showing that a polynomially bounded operator need not be similar to a contraction. The positive result is due to Paulsen and says that any completely polynomially bounded operator is similar to a contraction.
Here's another nice result, this one due to Zhong-Jin Ruan, one of the giants of the subject. B. Johnson proved that a locally compact group $G$ is amenable if and only if the Banach algebra $L^1(G)$ is amenable. Ruan showed that this happens if and only if the Fourier algebra $A(G)$ is completely amenable.
A good place to learn more is the book Operator Spaces by Effros and Ruan.
References:
G. Pisier, A polynomially bounded operator on Hilbert space which is not similar to a contraction, J. Amer. Math. Soc. 10 (1997), 351–369.
V. Paulsen, Every completely polynomially bounded operator is similar to a contraction, J. Funct. Anal. 55 (1984), 1–17.
Z-J. Ruan, The operator amenability of A(G), Amer. J. Math. 117 (1995), 1449–1474.
answered Aug 31 at 15:11
Nik Weaver
17.9k142114
1
I have to say I prefer "completely amenable" to the horrible-but-standard "operator amenable", although the former does make me wonder if I should go ahead and define a notion of something being utterly amenable ...
– Yemon Choi
Aug 31 at 16:09
12
I once heard a talk by David Blecher where he started by explaining the meaning of "completely bounded", "completely positive", and so on. Then he put up a lemma and said "The proof of this is completely obvious." I think I was the only one who laughed, though.
– Nik Weaver
Aug 31 at 16:20
1
Thanks for remark about 'completely bounded operator' etc., and the example with amenable groups is very good. Thank you!
– Fedor Goncharov
Aug 31 at 16:30
add a comment |Â
up vote
15
down vote
accepted
up vote
15
down vote
accepted
Okay, I'll take this one. First let me say that the English term is "completely bounded" (or "complete isometry", etc.).
About the term "quantum". The general principle is that analyzing some aspect of a physical system typically involves very different kinds of mathematical structures, depending on whether the system is classical or quantum. Thus if you study a classical system using topological techniques, or measure theory, or Lie groups, or graphs, the analogous analysis of a quantum system would probably involve C*-algebras, or von Neumann algebras, or quantum groups, or operator systems. (For instance, here's a paper of mine that talks about how operator systems arise in quantum error correction in the same way that graphs arise in classical error correction.)
There is no simple rule for translating classical structures into quantum structures, but we have a pretty large dictionary and one can identify broad themes. My take is that operator spaces fit into this scheme as the quantum analog of relations on a set. See Section 2 of the paper linked above. I'm not aware of any really compelling direct connections with real-world quantum mechanics in this instance, but the word "quantum" may be appropriate just because the subject fits into the general scheme of mathematical quantization which does include many direct links.
Now, as to the "beautiful results" you ask for. Well, there's a lot of good stuff so the selection would be idiosyncratic. One general principle is that when you're studying things related to Hilbert space (single operators, C*-algebras, etc.) it is often the case that you get much nicer results if you assume the "complete" (at all matrix levels) version of your hypothesis. For example, if two C*-algebras are linearly isometric, are they isomorphic as C*-algebras? Not in general, but yes if they are completely linearly isometric, i.e., linearly isometric at all matrix levels. (I think this is folklore.)
The result of Pisier that you refer to was actually a counterexample, showing that a polynomially bounded operator need not be similar to a contraction. The positive result is due to Paulsen and says that any completely polynomially bounded operator is similar to a contraction.
Here's another nice result, this one due to Zhong-Jin Ruan, one of the giants of the subject. B. Johnson proved that a locally compact group $G$ is amenable if and only if the Banach algebra $L^1(G)$ is amenable. Ruan showed that this happens if and only if the Fourier algebra $A(G)$ is completely amenable.
A good place to learn more is the book Operator Spaces by Effros and Ruan.
References:
G. Pisier, A polynomially bounded operator on Hilbert space which is not similar to a contraction, J. Amer. Math. Soc. 10 (1997), 351–369.
V. Paulsen, Every completely polynomially bounded operator is similar to a contraction, J. Funct. Anal. 55 (1984), 1–17.
Z-J. Ruan, The operator amenability of A(G), Amer. J. Math. 117 (1995), 1449–1474.
answered Aug 31 at 15:11
Nik Weaver
17.9k142114
Okay, I'll take this one. First let me say that the English term is "completely bounded" (or "complete isometry", etc.).
About the term "quantum". The general principle is that analyzing some aspect of a physical system typically involves very different kinds of mathematical structures, depending on whether the system is classical or quantum. Thus if you study a classical system using topological techniques, or measure theory, or Lie groups, or graphs, the analogous analysis of a quantum system would probably involve C*-algebras, or von Neumann algebras, or quantum groups, or operator systems. (For instance, here's a paper of mine that talks about how operator systems arise in quantum error correction in the same way that graphs arise in classical error correction.)
There is no simple rule for translating classical structures into quantum structures, but we have a pretty large dictionary and one can identify broad themes. My take is that operator spaces fit into this scheme as the quantum analog of relations on a set. See Section 2 of the paper linked above. I'm not aware of any really compelling direct connections with real-world quantum mechanics in this instance, but the word "quantum" may be appropriate just because the subject fits into the general scheme of mathematical quantization which does include many direct links.
Now, as to the "beautiful results" you ask for. Well, there's a lot of good stuff so the selection would be idiosyncratic. One general principle is that when you're studying things related to Hilbert space (single operators, C*-algebras, etc.) it is often the case that you get much nicer results if you assume the "complete" (at all matrix levels) version of your hypothesis. For example, if two C*-algebras are linearly isometric, are they isomorphic as C*-algebras? Not in general, but yes if they are completely linearly isometric, i.e., linearly isometric at all matrix levels. (I think this is folklore.)
The result of Pisier that you refer to was actually a counterexample, showing that a polynomially bounded operator need not be similar to a contraction. The positive result is due to Paulsen and says that any completely polynomially bounded operator is similar to a contraction.
Here's another nice result, this one due to Zhong-Jin Ruan, one of the giants of the subject. B. Johnson proved that a locally compact group $G$ is amenable if and only if the Banach algebra $L^1(G)$ is amenable. Ruan showed that this happens if and only if the Fourier algebra $A(G)$ is completely amenable.
A good place to learn more is the book Operator Spaces by Effros and Ruan.
References:
G. Pisier, A polynomially bounded operator on Hilbert space which is not similar to a contraction, J. Amer. Math. Soc. 10 (1997), 351–369.
V. Paulsen, Every completely polynomially bounded operator is similar to a contraction, J. Funct. Anal. 55 (1984), 1–17.
Z-J. Ruan, The operator amenability of A(G), Amer. J. Math. 117 (1995), 1449–1474.
answered Aug 31 at 15:11
Nik Weaver
17.9k142114
answered Aug 31 at 15:11
Nik Weaver
17.9k142114
answered Aug 31 at 15:11
Nik Weaver
17.9k142114
answered Aug 31 at 15:11
Nik Weaver
17.9k142114
17.9k142114
1
I have to say I prefer "completely amenable" to the horrible-but-standard "operator amenable", although the former does make me wonder if I should go ahead and define a notion of something being utterly amenable ...
– Yemon Choi
Aug 31 at 16:09
12
I once heard a talk by David Blecher where he started by explaining the meaning of "completely bounded", "completely positive", and so on. Then he put up a lemma and said "The proof of this is completely obvious." I think I was the only one who laughed, though.
– Nik Weaver
Aug 31 at 16:20
1
Thanks for remark about 'completely bounded operator' etc., and the example with amenable groups is very good. Thank you!
– Fedor Goncharov
Aug 31 at 16:30
add a comment |Â
1
I have to say I prefer "completely amenable" to the horrible-but-standard "operator amenable", although the former does make me wonder if I should go ahead and define a notion of something being utterly amenable ...
– Yemon Choi
Aug 31 at 16:09
12
I once heard a talk by David Blecher where he started by explaining the meaning of "completely bounded", "completely positive", and so on. Then he put up a lemma and said "The proof of this is completely obvious." I think I was the only one who laughed, though.
– Nik Weaver
Aug 31 at 16:20
1
Thanks for remark about 'completely bounded operator' etc., and the example with amenable groups is very good. Thank you!
– Fedor Goncharov
Aug 31 at 16:30
1
1
I have to say I prefer "completely amenable" to the horrible-but-standard "operator amenable", although the former does make me wonder if I should go ahead and define a notion of something being utterly amenable ...
– Yemon Choi
Aug 31 at 16:09
I have to say I prefer "completely amenable" to the horrible-but-standard "operator amenable", although the former does make me wonder if I should go ahead and define a notion of something being utterly amenable ...
– Yemon Choi
Aug 31 at 16:09
12
12
I once heard a talk by David Blecher where he started by explaining the meaning of "completely bounded", "completely positive", and so on. Then he put up a lemma and said "The proof of this is completely obvious." I think I was the only one who laughed, though.
– Nik Weaver
Aug 31 at 16:20
I once heard a talk by David Blecher where he started by explaining the meaning of "completely bounded", "completely positive", and so on. Then he put up a lemma and said "The proof of this is completely obvious." I think I was the only one who laughed, though.
– Nik Weaver
Aug 31 at 16:20
1
1
Thanks for remark about 'completely bounded operator' etc., and the example with amenable groups is very good. Thank you!
– Fedor Goncharov
Aug 31 at 16:30
Thanks for remark about 'completely bounded operator' etc., and the example with amenable groups is very good. Thank you!
– Fedor Goncharov
Aug 31 at 16:30
add a comment |Â
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2
It may be worth pointing out that the $n$-th level "quantum norm" $|cdot|_n$ does not live on $E$ itself, but rather on the matrix space $M_n(E)$.
– Tobias Fritz
Aug 31 at 12:57
Yes, I added it in the text.
– Fedor Goncharov
Aug 31 at 14:13
2
With all due respect to Prof. Helemskii, I have very rarely seen this term "quantum normed space" or "quantum Banach space" used outside his school/seminar.
– Yemon Choi
Aug 31 at 16:10