Mixing
The identity cannot be a commutator in a Banach algebra?
$begingroup$
The Wikipedia article on Banach algebras claims, without a proof or reference, that there does not exist a (unital) Banach algebra $B$ and elements $x, y in B$ such that $xy - yx = 1$. This is surprising to me, but maybe the proof is straightforward; anyone have a proof and/or a reference?
More generally, I would have naively thought that I could embed any ring into a Banach algebra. I guess there are actually serious restrictions to doing this; are these issues discussed anywhere?
functional-analysis banach-algebras
asked Jul 29 '11 at 4:22
Qiaochu YuanQiaochu Yuan
279k32590935
$endgroup$
|
show 1 more comment
$begingroup$
The Wikipedia article on Banach algebras claims, without a proof or reference, that there does not exist a (unital) Banach algebra $B$ and elements $x, y in B$ such that $xy - yx = 1$. This is surprising to me, but maybe the proof is straightforward; anyone have a proof and/or a reference?
More generally, I would have naively thought that I could embed any ring into a Banach algebra. I guess there are actually serious restrictions to doing this; are these issues discussed anywhere?
functional-analysis banach-algebras
asked Jul 29 '11 at 4:22
Qiaochu YuanQiaochu Yuan
279k32590935
$endgroup$
5
$begingroup$
You may find of interest Halmos's exposition on commutators here.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 4:43
$begingroup$
By "ring" here, did you intend "algebra over $mathbb{R}$"? I suppose there is a subtlety in the question. I think the axiom of choice tells us that every linear space over $mathbb{R}$ admits a (subadditive) norm. But we see below that not every algebra over $mathbb{R}$ admits a submultiplicative norm, since there are real algebras which contains solutions of $xy - yx = I$.
$endgroup$
– Geoff Robinson
Jul 29 '11 at 11:59
$begingroup$
@Geoff: ah. Let's suppose I mean "finitely generated ring" (over $mathbb{Z}$).
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 13:27
$begingroup$
A small comment on your last question: if you take the algebra ${mathbb C}^{mathbb N}$ with pointwise product, then I think it cannot be given a submultiplicative norm. On the other hand, it is an old result of Allan that there is an embedding (no continuity) of ${mathbb C}[[X]]$ into the unitization of some radical Banach algebra.
$endgroup$
– user16299
Jan 7 '12 at 19:42
$begingroup$
Since nobody explicitly mentioned this, maybe I'll point it out in the comments here: In QM, the canonical commutation relation is $xy-yx= i hbar$ so, up to rescaling, the theorem under discussion is telling us, in particular, that the canonical commutation relation cannot be satisfied by two bounded operators. Hence, this particular part of QM necessarily involves unbounded operators.
$endgroup$
– Mike F
Feb 8 '16 at 6:39
|
show 1 more comment
$begingroup$
The Wikipedia article on Banach algebras claims, without a proof or reference, that there does not exist a (unital) Banach algebra $B$ and elements $x, y in B$ such that $xy - yx = 1$. This is surprising to me, but maybe the proof is straightforward; anyone have a proof and/or a reference?
More generally, I would have naively thought that I could embed any ring into a Banach algebra. I guess there are actually serious restrictions to doing this; are these issues discussed anywhere?
functional-analysis banach-algebras
asked Jul 29 '11 at 4:22
Qiaochu YuanQiaochu Yuan
279k32590935
$endgroup$
The Wikipedia article on Banach algebras claims, without a proof or reference, that there does not exist a (unital) Banach algebra $B$ and elements $x, y in B$ such that $xy - yx = 1$. This is surprising to me, but maybe the proof is straightforward; anyone have a proof and/or a reference?
More generally, I would have naively thought that I could embed any ring into a Banach algebra. I guess there are actually serious restrictions to doing this; are these issues discussed anywhere?
functional-analysis banach-algebras
functional-analysis banach-algebras
asked Jul 29 '11 at 4:22
Qiaochu YuanQiaochu Yuan
279k32590935
asked Jul 29 '11 at 4:22
Qiaochu YuanQiaochu Yuan
279k32590935
edited Jan 7 '12 at 19:39
user16299
asked Jul 29 '11 at 4:22
Qiaochu YuanQiaochu Yuan
279k32590935
asked Jul 29 '11 at 4:22
Qiaochu YuanQiaochu Yuan
279k32590935
asked Jul 29 '11 at 4:22
Qiaochu YuanQiaochu Yuan
279k32590935
279k32590935
5
$begingroup$
You may find of interest Halmos's exposition on commutators here.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 4:43
$begingroup$
By "ring" here, did you intend "algebra over $mathbb{R}$"? I suppose there is a subtlety in the question. I think the axiom of choice tells us that every linear space over $mathbb{R}$ admits a (subadditive) norm. But we see below that not every algebra over $mathbb{R}$ admits a submultiplicative norm, since there are real algebras which contains solutions of $xy - yx = I$.
$endgroup$
– Geoff Robinson
Jul 29 '11 at 11:59
$begingroup$
@Geoff: ah. Let's suppose I mean "finitely generated ring" (over $mathbb{Z}$).
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 13:27
$begingroup$
A small comment on your last question: if you take the algebra ${mathbb C}^{mathbb N}$ with pointwise product, then I think it cannot be given a submultiplicative norm. On the other hand, it is an old result of Allan that there is an embedding (no continuity) of ${mathbb C}[[X]]$ into the unitization of some radical Banach algebra.
$endgroup$
– user16299
Jan 7 '12 at 19:42
$begingroup$
Since nobody explicitly mentioned this, maybe I'll point it out in the comments here: In QM, the canonical commutation relation is $xy-yx= i hbar$ so, up to rescaling, the theorem under discussion is telling us, in particular, that the canonical commutation relation cannot be satisfied by two bounded operators. Hence, this particular part of QM necessarily involves unbounded operators.
$endgroup$
– Mike F
Feb 8 '16 at 6:39
|
show 1 more comment
5
$begingroup$
You may find of interest Halmos's exposition on commutators here.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 4:43
$begingroup$
By "ring" here, did you intend "algebra over $mathbb{R}$"? I suppose there is a subtlety in the question. I think the axiom of choice tells us that every linear space over $mathbb{R}$ admits a (subadditive) norm. But we see below that not every algebra over $mathbb{R}$ admits a submultiplicative norm, since there are real algebras which contains solutions of $xy - yx = I$.
$endgroup$
– Geoff Robinson
Jul 29 '11 at 11:59
$begingroup$
@Geoff: ah. Let's suppose I mean "finitely generated ring" (over $mathbb{Z}$).
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 13:27
$begingroup$
A small comment on your last question: if you take the algebra ${mathbb C}^{mathbb N}$ with pointwise product, then I think it cannot be given a submultiplicative norm. On the other hand, it is an old result of Allan that there is an embedding (no continuity) of ${mathbb C}[[X]]$ into the unitization of some radical Banach algebra.
$endgroup$
– user16299
Jan 7 '12 at 19:42
$begingroup$
Since nobody explicitly mentioned this, maybe I'll point it out in the comments here: In QM, the canonical commutation relation is $xy-yx= i hbar$ so, up to rescaling, the theorem under discussion is telling us, in particular, that the canonical commutation relation cannot be satisfied by two bounded operators. Hence, this particular part of QM necessarily involves unbounded operators.
$endgroup$
– Mike F
Feb 8 '16 at 6:39
5
5
$begingroup$
You may find of interest Halmos's exposition on commutators here.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 4:43
$begingroup$
You may find of interest Halmos's exposition on commutators here.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 4:43
$begingroup$
By "ring" here, did you intend "algebra over $mathbb{R}$"? I suppose there is a subtlety in the question. I think the axiom of choice tells us that every linear space over $mathbb{R}$ admits a (subadditive) norm. But we see below that not every algebra over $mathbb{R}$ admits a submultiplicative norm, since there are real algebras which contains solutions of $xy - yx = I$.
$endgroup$
– Geoff Robinson
Jul 29 '11 at 11:59
$begingroup$
By "ring" here, did you intend "algebra over $mathbb{R}$"? I suppose there is a subtlety in the question. I think the axiom of choice tells us that every linear space over $mathbb{R}$ admits a (subadditive) norm. But we see below that not every algebra over $mathbb{R}$ admits a submultiplicative norm, since there are real algebras which contains solutions of $xy - yx = I$.
$endgroup$
– Geoff Robinson
Jul 29 '11 at 11:59
$begingroup$
@Geoff: ah. Let's suppose I mean "finitely generated ring" (over $mathbb{Z}$).
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 13:27
$begingroup$
@Geoff: ah. Let's suppose I mean "finitely generated ring" (over $mathbb{Z}$).
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 13:27
$begingroup$
A small comment on your last question: if you take the algebra ${mathbb C}^{mathbb N}$ with pointwise product, then I think it cannot be given a submultiplicative norm. On the other hand, it is an old result of Allan that there is an embedding (no continuity) of ${mathbb C}[[X]]$ into the unitization of some radical Banach algebra.
$endgroup$
– user16299
Jan 7 '12 at 19:42
$begingroup$
A small comment on your last question: if you take the algebra ${mathbb C}^{mathbb N}$ with pointwise product, then I think it cannot be given a submultiplicative norm. On the other hand, it is an old result of Allan that there is an embedding (no continuity) of ${mathbb C}[[X]]$ into the unitization of some radical Banach algebra.
$endgroup$
– user16299
Jan 7 '12 at 19:42
$begingroup$
Since nobody explicitly mentioned this, maybe I'll point it out in the comments here: In QM, the canonical commutation relation is $xy-yx= i hbar$ so, up to rescaling, the theorem under discussion is telling us, in particular, that the canonical commutation relation cannot be satisfied by two bounded operators. Hence, this particular part of QM necessarily involves unbounded operators.
$endgroup$
– Mike F
Feb 8 '16 at 6:39
$begingroup$
Since nobody explicitly mentioned this, maybe I'll point it out in the comments here: In QM, the canonical commutation relation is $xy-yx= i hbar$ so, up to rescaling, the theorem under discussion is telling us, in particular, that the canonical commutation relation cannot be satisfied by two bounded operators. Hence, this particular part of QM necessarily involves unbounded operators.
$endgroup$
– Mike F
Feb 8 '16 at 6:39
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Here's a sketch of a proof. Let $sigma(x)$ denote the spectrum of $x$. Then $sigma(xy)cup{0} = sigma(yx)cup{0}$. On the other hand, $sigma(1+yx)=1+sigma(yx)$. If $xy=1+yx$, then the previous two sentences, along with the fact that the spectrum of each element of a Banach algebra is nonempty, imply that $sigma(xy)$ is unbounded. But every element of a Banach algebra has bounded spectrum.
(I don't remember where I first learned this proof, nor do I have a reference for it off-hand, but I did not come up with it myself.)
The proof that $sigma(xy)cup{0}=sigma(yx)cup{0}$ reduces to showing that $1-xy$ is invertible if and only if $1-yx$ is invertible, a problem that was the subject of a MathOverflow question.
There's a proof using derivations in section 2.2 of Sakai's book, Operator algebras in dynamical systems: the theory of unbounded derivations in $C^*$-algebras. A bounded derivation on a Banach algebra $A$ is a bounded linear map $delta$ on $A$ such that $delta(ab)=delta(a)b+adelta(b)$ for all $a$ and $b$ in $A$. Theorem 2.2.1 on page 18 shows that if $delta$ is a bounded derivation on $A$, and if $a$ is an element of $A$ such that $delta^2(a)=0$, then $limlimits_{ntoinfty}|delta(a)^n|^{1/n}=0$. The proof uses induction with a neat computation to show that $delta^2(a)=0$ implies that $n!delta(a)^n=delta^n(a^n)$, and then the result follows from boundedness of $delta$ and the fact that $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$.
Corollary 2.2.2 concludes that the identity is not a commutator. If $ab-ba=1$, then the bounded derivation $delta_a:Ato A$ defined by $delta_a(x)=ax-xa$ satisfies $delta_a^2(b)=delta_a(1)=0$. By the preceding theorem, this implies that $1=limlimits_{ntoinfty}|1^n|^{1/n}=limlimits_{ntoinfty}|delta_a(b)^n|^{1/n}=0$.
(Completeness is not used in this approach. An element $x$ of $A$ satisfies $limlimits_{ntoinfty}|x^n|^{1/n}=0$ if and only if $sigma(x)={0}$, and such an $x$ is called a generalized nilpotent. Incidentally, this also gives an approach to answering the example problem in the MathOverflow question Linear Algebra Problems? The remainder of Section 2.2 has a number of interesting results on bounded derivations and commutators of bounded operators.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Jul 29 '11 at 4:37
Jonas MeyerJonas Meyer
40.7k6147256
$endgroup$
$begingroup$
Oh, of course. Because $lambda - xy$ is invertible if and only if $lambda - yx$ is... thanks!
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 4:42
1
$begingroup$
For what it's worth: I know that I learned about this argument it from Appendix A, A.1, p. 409 in Pedersen's $C^{ast}$-algebras and their automorphism groups (so this would be a reference). I would be surprised if it weren't older.
$endgroup$
– t.b.
Jul 29 '11 at 8:32
1
$begingroup$
@Theo: Thanks for the reference. I meant the proof that the identity is not a commutator. I learned about $sigma(xy)cup{0}=sigma(yx)cup{0}$ from Arveson's A short course on spectral theory, Exercises 3 and 4 on page 7. I like the exercise because it gives the seemingly nonsense series method to discover the proof, which is the subject of Bill Dubuque's question linked above.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 8:42
2
$begingroup$
I see. Yes, that's really nice. The funny thing is that this nonsense series is the only way for me to remember the relevant identity. If I really needed a reference for the fact that the identity isn't a commutator, I'd probably start looking in the context of "no-go theorems" in quantum mechanics. This must go way back (von Neumann?). Added Ah, I see only now that Halmos refers to Wintnter [160] and Wielandt [158], but I can't see the detailed references.
$endgroup$
– t.b.
Jul 29 '11 at 8:47
2
$begingroup$
Sorry to pursue this further: Is there such a thing as a detailed and reliable early history of Banach algebras/operator algebras? What I have in mind is something comparable to Dieudonnés History of Functional Analysis but with more focus on the algebra aspect? Of course one could just say: read the Gel'fand school (and what I've seen from that is well worth it). Dixmier has comprehensive references in both his big books, but the history parts are rather terse.
$endgroup$
– t.b.
Jul 29 '11 at 12:16
|
show 9 more comments
$begingroup$
There is a Theorem of Wielandt which asserts that if $A$ is any normed algebra, complete or not, we can't express $I = 1_{A}$ in the form $xy - yx$. The proof is given in Rudin's book, but it is so beautiful that I give it here. Suppose that $xy -yx = I$. I claim that $xy^{n} - y^{n}x = ny^{n-1}$ for all $n in mathbb{N}$. We have the case $n = 1.$
Suppose that $xy^k - y^kx = ky^{k-1}$ for some $k$. Then $$xy^{k+1} - y^{k+1}x =
(xy^{k} - y^{k}x)y +y^{k}(xy-yx) =ky^{k-1}y +y^{k}.I = (k+1)y^{k},$$ so the claim is established by induction. Note that $y^n neq 0$ for any $n$, since otherwise there is a smallest value of $n$ with $y^n = 0$, leading to $0 = xy^n - y^nx = ny^{n-1}$, contrary to the choice of $n$.
But now, for any $n$, we have $$n|y^{n-1} | = |xy^{n} -y^{n}x| leq 2|x|. |y| .
|y^{n-1} | .$$ Since $y^{n-1} neq 0$, as remarked above, we have $ 2 |x| . |y|
geq n$, a contradiction, as $n$ is arbitrary.
answered Jul 29 '11 at 6:57
Geoff RobinsonGeoff Robinson
20.6k13043
$endgroup$
2
$begingroup$
That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 7:12
$begingroup$
That (and more) is mentioned in Halmos's exposition that I already linked to in a comment.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 15:08
2
$begingroup$
@ Bill: not everyone can see that link (me, for example)
$endgroup$
– Geoff Robinson
Jul 29 '11 at 17:25
2
$begingroup$
This is a very cool argument, but a little mysterious...
$endgroup$
– Igor Rivin
Dec 4 '13 at 4:10
$begingroup$
that's the exact answer I'm looking for.
$endgroup$
– Tomas
Dec 4 '13 at 4:16
add a comment |
$begingroup$
$$
xy-yx = 1 \
(x-alpha 1)y-y(x-alpha 1)=1 \
y(alpha 1-x)-(alpha 1-x)y = -1 \
(alpha 1-x)^{-1}y-y(alpha 1-x)^{-1}=frac{d}{dalpha}(alpha 1-x)^{-1}.
$$
Using the functional calculus for a function $F$ that is holomorphic on an open neighborhood of $sigma(x)$ leads to a bounded differentiation:
$$
F(x)y-yF(x)=-F'(x) \
|F'(x)| le 2|y||F(x)|.
$$
If you set $F(alpha)=e^{talpha}$ for large enough real $t$, then the above yields the contradiction $|t| le 2|y|$.
answered Jan 16 at 9:31
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
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$begingroup$
Here's a sketch of a proof. Let $sigma(x)$ denote the spectrum of $x$. Then $sigma(xy)cup{0} = sigma(yx)cup{0}$. On the other hand, $sigma(1+yx)=1+sigma(yx)$. If $xy=1+yx$, then the previous two sentences, along with the fact that the spectrum of each element of a Banach algebra is nonempty, imply that $sigma(xy)$ is unbounded. But every element of a Banach algebra has bounded spectrum.
(I don't remember where I first learned this proof, nor do I have a reference for it off-hand, but I did not come up with it myself.)
The proof that $sigma(xy)cup{0}=sigma(yx)cup{0}$ reduces to showing that $1-xy$ is invertible if and only if $1-yx$ is invertible, a problem that was the subject of a MathOverflow question.
There's a proof using derivations in section 2.2 of Sakai's book, Operator algebras in dynamical systems: the theory of unbounded derivations in $C^*$-algebras. A bounded derivation on a Banach algebra $A$ is a bounded linear map $delta$ on $A$ such that $delta(ab)=delta(a)b+adelta(b)$ for all $a$ and $b$ in $A$. Theorem 2.2.1 on page 18 shows that if $delta$ is a bounded derivation on $A$, and if $a$ is an element of $A$ such that $delta^2(a)=0$, then $limlimits_{ntoinfty}|delta(a)^n|^{1/n}=0$. The proof uses induction with a neat computation to show that $delta^2(a)=0$ implies that $n!delta(a)^n=delta^n(a^n)$, and then the result follows from boundedness of $delta$ and the fact that $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$.
Corollary 2.2.2 concludes that the identity is not a commutator. If $ab-ba=1$, then the bounded derivation $delta_a:Ato A$ defined by $delta_a(x)=ax-xa$ satisfies $delta_a^2(b)=delta_a(1)=0$. By the preceding theorem, this implies that $1=limlimits_{ntoinfty}|1^n|^{1/n}=limlimits_{ntoinfty}|delta_a(b)^n|^{1/n}=0$.
(Completeness is not used in this approach. An element $x$ of $A$ satisfies $limlimits_{ntoinfty}|x^n|^{1/n}=0$ if and only if $sigma(x)={0}$, and such an $x$ is called a generalized nilpotent. Incidentally, this also gives an approach to answering the example problem in the MathOverflow question Linear Algebra Problems? The remainder of Section 2.2 has a number of interesting results on bounded derivations and commutators of bounded operators.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Jul 29 '11 at 4:37
Jonas MeyerJonas Meyer
40.7k6147256
$endgroup$
$begingroup$
Oh, of course. Because $lambda - xy$ is invertible if and only if $lambda - yx$ is... thanks!
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 4:42
1
$begingroup$
For what it's worth: I know that I learned about this argument it from Appendix A, A.1, p. 409 in Pedersen's $C^{ast}$-algebras and their automorphism groups (so this would be a reference). I would be surprised if it weren't older.
$endgroup$
– t.b.
Jul 29 '11 at 8:32
1
$begingroup$
@Theo: Thanks for the reference. I meant the proof that the identity is not a commutator. I learned about $sigma(xy)cup{0}=sigma(yx)cup{0}$ from Arveson's A short course on spectral theory, Exercises 3 and 4 on page 7. I like the exercise because it gives the seemingly nonsense series method to discover the proof, which is the subject of Bill Dubuque's question linked above.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 8:42
2
$begingroup$
I see. Yes, that's really nice. The funny thing is that this nonsense series is the only way for me to remember the relevant identity. If I really needed a reference for the fact that the identity isn't a commutator, I'd probably start looking in the context of "no-go theorems" in quantum mechanics. This must go way back (von Neumann?). Added Ah, I see only now that Halmos refers to Wintnter [160] and Wielandt [158], but I can't see the detailed references.
$endgroup$
– t.b.
Jul 29 '11 at 8:47
2
$begingroup$
Sorry to pursue this further: Is there such a thing as a detailed and reliable early history of Banach algebras/operator algebras? What I have in mind is something comparable to Dieudonnés History of Functional Analysis but with more focus on the algebra aspect? Of course one could just say: read the Gel'fand school (and what I've seen from that is well worth it). Dixmier has comprehensive references in both his big books, but the history parts are rather terse.
$endgroup$
– t.b.
Jul 29 '11 at 12:16
|
show 9 more comments
$begingroup$
Here's a sketch of a proof. Let $sigma(x)$ denote the spectrum of $x$. Then $sigma(xy)cup{0} = sigma(yx)cup{0}$. On the other hand, $sigma(1+yx)=1+sigma(yx)$. If $xy=1+yx$, then the previous two sentences, along with the fact that the spectrum of each element of a Banach algebra is nonempty, imply that $sigma(xy)$ is unbounded. But every element of a Banach algebra has bounded spectrum.
(I don't remember where I first learned this proof, nor do I have a reference for it off-hand, but I did not come up with it myself.)
The proof that $sigma(xy)cup{0}=sigma(yx)cup{0}$ reduces to showing that $1-xy$ is invertible if and only if $1-yx$ is invertible, a problem that was the subject of a MathOverflow question.
There's a proof using derivations in section 2.2 of Sakai's book, Operator algebras in dynamical systems: the theory of unbounded derivations in $C^*$-algebras. A bounded derivation on a Banach algebra $A$ is a bounded linear map $delta$ on $A$ such that $delta(ab)=delta(a)b+adelta(b)$ for all $a$ and $b$ in $A$. Theorem 2.2.1 on page 18 shows that if $delta$ is a bounded derivation on $A$, and if $a$ is an element of $A$ such that $delta^2(a)=0$, then $limlimits_{ntoinfty}|delta(a)^n|^{1/n}=0$. The proof uses induction with a neat computation to show that $delta^2(a)=0$ implies that $n!delta(a)^n=delta^n(a^n)$, and then the result follows from boundedness of $delta$ and the fact that $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$.
Corollary 2.2.2 concludes that the identity is not a commutator. If $ab-ba=1$, then the bounded derivation $delta_a:Ato A$ defined by $delta_a(x)=ax-xa$ satisfies $delta_a^2(b)=delta_a(1)=0$. By the preceding theorem, this implies that $1=limlimits_{ntoinfty}|1^n|^{1/n}=limlimits_{ntoinfty}|delta_a(b)^n|^{1/n}=0$.
(Completeness is not used in this approach. An element $x$ of $A$ satisfies $limlimits_{ntoinfty}|x^n|^{1/n}=0$ if and only if $sigma(x)={0}$, and such an $x$ is called a generalized nilpotent. Incidentally, this also gives an approach to answering the example problem in the MathOverflow question Linear Algebra Problems? The remainder of Section 2.2 has a number of interesting results on bounded derivations and commutators of bounded operators.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Jul 29 '11 at 4:37
Jonas MeyerJonas Meyer
40.7k6147256
$endgroup$
$begingroup$
Oh, of course. Because $lambda - xy$ is invertible if and only if $lambda - yx$ is... thanks!
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 4:42
1
$begingroup$
For what it's worth: I know that I learned about this argument it from Appendix A, A.1, p. 409 in Pedersen's $C^{ast}$-algebras and their automorphism groups (so this would be a reference). I would be surprised if it weren't older.
$endgroup$
– t.b.
Jul 29 '11 at 8:32
1
$begingroup$
@Theo: Thanks for the reference. I meant the proof that the identity is not a commutator. I learned about $sigma(xy)cup{0}=sigma(yx)cup{0}$ from Arveson's A short course on spectral theory, Exercises 3 and 4 on page 7. I like the exercise because it gives the seemingly nonsense series method to discover the proof, which is the subject of Bill Dubuque's question linked above.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 8:42
2
$begingroup$
I see. Yes, that's really nice. The funny thing is that this nonsense series is the only way for me to remember the relevant identity. If I really needed a reference for the fact that the identity isn't a commutator, I'd probably start looking in the context of "no-go theorems" in quantum mechanics. This must go way back (von Neumann?). Added Ah, I see only now that Halmos refers to Wintnter [160] and Wielandt [158], but I can't see the detailed references.
$endgroup$
– t.b.
Jul 29 '11 at 8:47
2
$begingroup$
Sorry to pursue this further: Is there such a thing as a detailed and reliable early history of Banach algebras/operator algebras? What I have in mind is something comparable to Dieudonnés History of Functional Analysis but with more focus on the algebra aspect? Of course one could just say: read the Gel'fand school (and what I've seen from that is well worth it). Dixmier has comprehensive references in both his big books, but the history parts are rather terse.
$endgroup$
– t.b.
Jul 29 '11 at 12:16
|
show 9 more comments
$begingroup$
Here's a sketch of a proof. Let $sigma(x)$ denote the spectrum of $x$. Then $sigma(xy)cup{0} = sigma(yx)cup{0}$. On the other hand, $sigma(1+yx)=1+sigma(yx)$. If $xy=1+yx$, then the previous two sentences, along with the fact that the spectrum of each element of a Banach algebra is nonempty, imply that $sigma(xy)$ is unbounded. But every element of a Banach algebra has bounded spectrum.
(I don't remember where I first learned this proof, nor do I have a reference for it off-hand, but I did not come up with it myself.)
The proof that $sigma(xy)cup{0}=sigma(yx)cup{0}$ reduces to showing that $1-xy$ is invertible if and only if $1-yx$ is invertible, a problem that was the subject of a MathOverflow question.
There's a proof using derivations in section 2.2 of Sakai's book, Operator algebras in dynamical systems: the theory of unbounded derivations in $C^*$-algebras. A bounded derivation on a Banach algebra $A$ is a bounded linear map $delta$ on $A$ such that $delta(ab)=delta(a)b+adelta(b)$ for all $a$ and $b$ in $A$. Theorem 2.2.1 on page 18 shows that if $delta$ is a bounded derivation on $A$, and if $a$ is an element of $A$ such that $delta^2(a)=0$, then $limlimits_{ntoinfty}|delta(a)^n|^{1/n}=0$. The proof uses induction with a neat computation to show that $delta^2(a)=0$ implies that $n!delta(a)^n=delta^n(a^n)$, and then the result follows from boundedness of $delta$ and the fact that $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$.
Corollary 2.2.2 concludes that the identity is not a commutator. If $ab-ba=1$, then the bounded derivation $delta_a:Ato A$ defined by $delta_a(x)=ax-xa$ satisfies $delta_a^2(b)=delta_a(1)=0$. By the preceding theorem, this implies that $1=limlimits_{ntoinfty}|1^n|^{1/n}=limlimits_{ntoinfty}|delta_a(b)^n|^{1/n}=0$.
(Completeness is not used in this approach. An element $x$ of $A$ satisfies $limlimits_{ntoinfty}|x^n|^{1/n}=0$ if and only if $sigma(x)={0}$, and such an $x$ is called a generalized nilpotent. Incidentally, this also gives an approach to answering the example problem in the MathOverflow question Linear Algebra Problems? The remainder of Section 2.2 has a number of interesting results on bounded derivations and commutators of bounded operators.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Jul 29 '11 at 4:37
Jonas MeyerJonas Meyer
40.7k6147256
$endgroup$
Here's a sketch of a proof. Let $sigma(x)$ denote the spectrum of $x$. Then $sigma(xy)cup{0} = sigma(yx)cup{0}$. On the other hand, $sigma(1+yx)=1+sigma(yx)$. If $xy=1+yx$, then the previous two sentences, along with the fact that the spectrum of each element of a Banach algebra is nonempty, imply that $sigma(xy)$ is unbounded. But every element of a Banach algebra has bounded spectrum.
(I don't remember where I first learned this proof, nor do I have a reference for it off-hand, but I did not come up with it myself.)
The proof that $sigma(xy)cup{0}=sigma(yx)cup{0}$ reduces to showing that $1-xy$ is invertible if and only if $1-yx$ is invertible, a problem that was the subject of a MathOverflow question.
There's a proof using derivations in section 2.2 of Sakai's book, Operator algebras in dynamical systems: the theory of unbounded derivations in $C^*$-algebras. A bounded derivation on a Banach algebra $A$ is a bounded linear map $delta$ on $A$ such that $delta(ab)=delta(a)b+adelta(b)$ for all $a$ and $b$ in $A$. Theorem 2.2.1 on page 18 shows that if $delta$ is a bounded derivation on $A$, and if $a$ is an element of $A$ such that $delta^2(a)=0$, then $limlimits_{ntoinfty}|delta(a)^n|^{1/n}=0$. The proof uses induction with a neat computation to show that $delta^2(a)=0$ implies that $n!delta(a)^n=delta^n(a^n)$, and then the result follows from boundedness of $delta$ and the fact that $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$.
Corollary 2.2.2 concludes that the identity is not a commutator. If $ab-ba=1$, then the bounded derivation $delta_a:Ato A$ defined by $delta_a(x)=ax-xa$ satisfies $delta_a^2(b)=delta_a(1)=0$. By the preceding theorem, this implies that $1=limlimits_{ntoinfty}|1^n|^{1/n}=limlimits_{ntoinfty}|delta_a(b)^n|^{1/n}=0$.
(Completeness is not used in this approach. An element $x$ of $A$ satisfies $limlimits_{ntoinfty}|x^n|^{1/n}=0$ if and only if $sigma(x)={0}$, and such an $x$ is called a generalized nilpotent. Incidentally, this also gives an approach to answering the example problem in the MathOverflow question Linear Algebra Problems? The remainder of Section 2.2 has a number of interesting results on bounded derivations and commutators of bounded operators.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Jul 29 '11 at 4:37
Jonas MeyerJonas Meyer
40.7k6147256
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Jul 29 '11 at 4:37
Jonas MeyerJonas Meyer
40.7k6147256
answered Jul 29 '11 at 4:37
Jonas MeyerJonas Meyer
40.7k6147256
answered Jul 29 '11 at 4:37
Jonas MeyerJonas Meyer
40.7k6147256
40.7k6147256
$begingroup$
Oh, of course. Because $lambda - xy$ is invertible if and only if $lambda - yx$ is... thanks!
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 4:42
1
$begingroup$
For what it's worth: I know that I learned about this argument it from Appendix A, A.1, p. 409 in Pedersen's $C^{ast}$-algebras and their automorphism groups (so this would be a reference). I would be surprised if it weren't older.
$endgroup$
– t.b.
Jul 29 '11 at 8:32
1
$begingroup$
@Theo: Thanks for the reference. I meant the proof that the identity is not a commutator. I learned about $sigma(xy)cup{0}=sigma(yx)cup{0}$ from Arveson's A short course on spectral theory, Exercises 3 and 4 on page 7. I like the exercise because it gives the seemingly nonsense series method to discover the proof, which is the subject of Bill Dubuque's question linked above.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 8:42
2
$begingroup$
I see. Yes, that's really nice. The funny thing is that this nonsense series is the only way for me to remember the relevant identity. If I really needed a reference for the fact that the identity isn't a commutator, I'd probably start looking in the context of "no-go theorems" in quantum mechanics. This must go way back (von Neumann?). Added Ah, I see only now that Halmos refers to Wintnter [160] and Wielandt [158], but I can't see the detailed references.
$endgroup$
– t.b.
Jul 29 '11 at 8:47
2
$begingroup$
Sorry to pursue this further: Is there such a thing as a detailed and reliable early history of Banach algebras/operator algebras? What I have in mind is something comparable to Dieudonnés History of Functional Analysis but with more focus on the algebra aspect? Of course one could just say: read the Gel'fand school (and what I've seen from that is well worth it). Dixmier has comprehensive references in both his big books, but the history parts are rather terse.
$endgroup$
– t.b.
Jul 29 '11 at 12:16
|
show 9 more comments
$begingroup$
Oh, of course. Because $lambda - xy$ is invertible if and only if $lambda - yx$ is... thanks!
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 4:42
1
$begingroup$
For what it's worth: I know that I learned about this argument it from Appendix A, A.1, p. 409 in Pedersen's $C^{ast}$-algebras and their automorphism groups (so this would be a reference). I would be surprised if it weren't older.
$endgroup$
– t.b.
Jul 29 '11 at 8:32
1
$begingroup$
@Theo: Thanks for the reference. I meant the proof that the identity is not a commutator. I learned about $sigma(xy)cup{0}=sigma(yx)cup{0}$ from Arveson's A short course on spectral theory, Exercises 3 and 4 on page 7. I like the exercise because it gives the seemingly nonsense series method to discover the proof, which is the subject of Bill Dubuque's question linked above.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 8:42
2
$begingroup$
I see. Yes, that's really nice. The funny thing is that this nonsense series is the only way for me to remember the relevant identity. If I really needed a reference for the fact that the identity isn't a commutator, I'd probably start looking in the context of "no-go theorems" in quantum mechanics. This must go way back (von Neumann?). Added Ah, I see only now that Halmos refers to Wintnter [160] and Wielandt [158], but I can't see the detailed references.
$endgroup$
– t.b.
Jul 29 '11 at 8:47
2
$begingroup$
Sorry to pursue this further: Is there such a thing as a detailed and reliable early history of Banach algebras/operator algebras? What I have in mind is something comparable to Dieudonnés History of Functional Analysis but with more focus on the algebra aspect? Of course one could just say: read the Gel'fand school (and what I've seen from that is well worth it). Dixmier has comprehensive references in both his big books, but the history parts are rather terse.
$endgroup$
– t.b.
Jul 29 '11 at 12:16
$begingroup$
Oh, of course. Because $lambda - xy$ is invertible if and only if $lambda - yx$ is... thanks!
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 4:42
$begingroup$
Oh, of course. Because $lambda - xy$ is invertible if and only if $lambda - yx$ is... thanks!
$endgroup$
– Qiaochu Yuan
Jul 29 '11 at 4:42
1
1
$begingroup$
For what it's worth: I know that I learned about this argument it from Appendix A, A.1, p. 409 in Pedersen's $C^{ast}$-algebras and their automorphism groups (so this would be a reference). I would be surprised if it weren't older.
$endgroup$
– t.b.
Jul 29 '11 at 8:32
$begingroup$
For what it's worth: I know that I learned about this argument it from Appendix A, A.1, p. 409 in Pedersen's $C^{ast}$-algebras and their automorphism groups (so this would be a reference). I would be surprised if it weren't older.
$endgroup$
– t.b.
Jul 29 '11 at 8:32
1
1
$begingroup$
@Theo: Thanks for the reference. I meant the proof that the identity is not a commutator. I learned about $sigma(xy)cup{0}=sigma(yx)cup{0}$ from Arveson's A short course on spectral theory, Exercises 3 and 4 on page 7. I like the exercise because it gives the seemingly nonsense series method to discover the proof, which is the subject of Bill Dubuque's question linked above.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 8:42
$begingroup$
@Theo: Thanks for the reference. I meant the proof that the identity is not a commutator. I learned about $sigma(xy)cup{0}=sigma(yx)cup{0}$ from Arveson's A short course on spectral theory, Exercises 3 and 4 on page 7. I like the exercise because it gives the seemingly nonsense series method to discover the proof, which is the subject of Bill Dubuque's question linked above.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 8:42
2
2
$begingroup$
I see. Yes, that's really nice. The funny thing is that this nonsense series is the only way for me to remember the relevant identity. If I really needed a reference for the fact that the identity isn't a commutator, I'd probably start looking in the context of "no-go theorems" in quantum mechanics. This must go way back (von Neumann?). Added Ah, I see only now that Halmos refers to Wintnter [160] and Wielandt [158], but I can't see the detailed references.
$endgroup$
– t.b.
Jul 29 '11 at 8:47
$begingroup$
I see. Yes, that's really nice. The funny thing is that this nonsense series is the only way for me to remember the relevant identity. If I really needed a reference for the fact that the identity isn't a commutator, I'd probably start looking in the context of "no-go theorems" in quantum mechanics. This must go way back (von Neumann?). Added Ah, I see only now that Halmos refers to Wintnter [160] and Wielandt [158], but I can't see the detailed references.
$endgroup$
– t.b.
Jul 29 '11 at 8:47
2
2
$begingroup$
Sorry to pursue this further: Is there such a thing as a detailed and reliable early history of Banach algebras/operator algebras? What I have in mind is something comparable to Dieudonnés History of Functional Analysis but with more focus on the algebra aspect? Of course one could just say: read the Gel'fand school (and what I've seen from that is well worth it). Dixmier has comprehensive references in both his big books, but the history parts are rather terse.
$endgroup$
– t.b.
Jul 29 '11 at 12:16
$begingroup$
Sorry to pursue this further: Is there such a thing as a detailed and reliable early history of Banach algebras/operator algebras? What I have in mind is something comparable to Dieudonnés History of Functional Analysis but with more focus on the algebra aspect? Of course one could just say: read the Gel'fand school (and what I've seen from that is well worth it). Dixmier has comprehensive references in both his big books, but the history parts are rather terse.
$endgroup$
– t.b.
Jul 29 '11 at 12:16
|
show 9 more comments
$begingroup$
There is a Theorem of Wielandt which asserts that if $A$ is any normed algebra, complete or not, we can't express $I = 1_{A}$ in the form $xy - yx$. The proof is given in Rudin's book, but it is so beautiful that I give it here. Suppose that $xy -yx = I$. I claim that $xy^{n} - y^{n}x = ny^{n-1}$ for all $n in mathbb{N}$. We have the case $n = 1.$
Suppose that $xy^k - y^kx = ky^{k-1}$ for some $k$. Then $$xy^{k+1} - y^{k+1}x =
(xy^{k} - y^{k}x)y +y^{k}(xy-yx) =ky^{k-1}y +y^{k}.I = (k+1)y^{k},$$ so the claim is established by induction. Note that $y^n neq 0$ for any $n$, since otherwise there is a smallest value of $n$ with $y^n = 0$, leading to $0 = xy^n - y^nx = ny^{n-1}$, contrary to the choice of $n$.
But now, for any $n$, we have $$n|y^{n-1} | = |xy^{n} -y^{n}x| leq 2|x|. |y| .
|y^{n-1} | .$$ Since $y^{n-1} neq 0$, as remarked above, we have $ 2 |x| . |y|
geq n$, a contradiction, as $n$ is arbitrary.
answered Jul 29 '11 at 6:57
Geoff RobinsonGeoff Robinson
20.6k13043
$endgroup$
2
$begingroup$
That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 7:12
$begingroup$
That (and more) is mentioned in Halmos's exposition that I already linked to in a comment.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 15:08
2
$begingroup$
@ Bill: not everyone can see that link (me, for example)
$endgroup$
– Geoff Robinson
Jul 29 '11 at 17:25
2
$begingroup$
This is a very cool argument, but a little mysterious...
$endgroup$
– Igor Rivin
Dec 4 '13 at 4:10
$begingroup$
that's the exact answer I'm looking for.
$endgroup$
– Tomas
Dec 4 '13 at 4:16
add a comment |
$begingroup$
There is a Theorem of Wielandt which asserts that if $A$ is any normed algebra, complete or not, we can't express $I = 1_{A}$ in the form $xy - yx$. The proof is given in Rudin's book, but it is so beautiful that I give it here. Suppose that $xy -yx = I$. I claim that $xy^{n} - y^{n}x = ny^{n-1}$ for all $n in mathbb{N}$. We have the case $n = 1.$
Suppose that $xy^k - y^kx = ky^{k-1}$ for some $k$. Then $$xy^{k+1} - y^{k+1}x =
(xy^{k} - y^{k}x)y +y^{k}(xy-yx) =ky^{k-1}y +y^{k}.I = (k+1)y^{k},$$ so the claim is established by induction. Note that $y^n neq 0$ for any $n$, since otherwise there is a smallest value of $n$ with $y^n = 0$, leading to $0 = xy^n - y^nx = ny^{n-1}$, contrary to the choice of $n$.
But now, for any $n$, we have $$n|y^{n-1} | = |xy^{n} -y^{n}x| leq 2|x|. |y| .
|y^{n-1} | .$$ Since $y^{n-1} neq 0$, as remarked above, we have $ 2 |x| . |y|
geq n$, a contradiction, as $n$ is arbitrary.
answered Jul 29 '11 at 6:57
Geoff RobinsonGeoff Robinson
20.6k13043
$endgroup$
2
$begingroup$
That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 7:12
$begingroup$
That (and more) is mentioned in Halmos's exposition that I already linked to in a comment.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 15:08
2
$begingroup$
@ Bill: not everyone can see that link (me, for example)
$endgroup$
– Geoff Robinson
Jul 29 '11 at 17:25
2
$begingroup$
This is a very cool argument, but a little mysterious...
$endgroup$
– Igor Rivin
Dec 4 '13 at 4:10
$begingroup$
that's the exact answer I'm looking for.
$endgroup$
– Tomas
Dec 4 '13 at 4:16
add a comment |
$begingroup$
There is a Theorem of Wielandt which asserts that if $A$ is any normed algebra, complete or not, we can't express $I = 1_{A}$ in the form $xy - yx$. The proof is given in Rudin's book, but it is so beautiful that I give it here. Suppose that $xy -yx = I$. I claim that $xy^{n} - y^{n}x = ny^{n-1}$ for all $n in mathbb{N}$. We have the case $n = 1.$
Suppose that $xy^k - y^kx = ky^{k-1}$ for some $k$. Then $$xy^{k+1} - y^{k+1}x =
(xy^{k} - y^{k}x)y +y^{k}(xy-yx) =ky^{k-1}y +y^{k}.I = (k+1)y^{k},$$ so the claim is established by induction. Note that $y^n neq 0$ for any $n$, since otherwise there is a smallest value of $n$ with $y^n = 0$, leading to $0 = xy^n - y^nx = ny^{n-1}$, contrary to the choice of $n$.
But now, for any $n$, we have $$n|y^{n-1} | = |xy^{n} -y^{n}x| leq 2|x|. |y| .
|y^{n-1} | .$$ Since $y^{n-1} neq 0$, as remarked above, we have $ 2 |x| . |y|
geq n$, a contradiction, as $n$ is arbitrary.
answered Jul 29 '11 at 6:57
Geoff RobinsonGeoff Robinson
20.6k13043
$endgroup$
There is a Theorem of Wielandt which asserts that if $A$ is any normed algebra, complete or not, we can't express $I = 1_{A}$ in the form $xy - yx$. The proof is given in Rudin's book, but it is so beautiful that I give it here. Suppose that $xy -yx = I$. I claim that $xy^{n} - y^{n}x = ny^{n-1}$ for all $n in mathbb{N}$. We have the case $n = 1.$
Suppose that $xy^k - y^kx = ky^{k-1}$ for some $k$. Then $$xy^{k+1} - y^{k+1}x =
(xy^{k} - y^{k}x)y +y^{k}(xy-yx) =ky^{k-1}y +y^{k}.I = (k+1)y^{k},$$ so the claim is established by induction. Note that $y^n neq 0$ for any $n$, since otherwise there is a smallest value of $n$ with $y^n = 0$, leading to $0 = xy^n - y^nx = ny^{n-1}$, contrary to the choice of $n$.
But now, for any $n$, we have $$n|y^{n-1} | = |xy^{n} -y^{n}x| leq 2|x|. |y| .
|y^{n-1} | .$$ Since $y^{n-1} neq 0$, as remarked above, we have $ 2 |x| . |y|
geq n$, a contradiction, as $n$ is arbitrary.
answered Jul 29 '11 at 6:57
Geoff RobinsonGeoff Robinson
20.6k13043
edited Jan 6 '12 at 0:33
answered Jul 29 '11 at 6:57
Geoff RobinsonGeoff Robinson
20.6k13043
answered Jul 29 '11 at 6:57
Geoff RobinsonGeoff Robinson
20.6k13043
answered Jul 29 '11 at 6:57
Geoff RobinsonGeoff Robinson
20.6k13043
20.6k13043
2
$begingroup$
That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 7:12
$begingroup$
That (and more) is mentioned in Halmos's exposition that I already linked to in a comment.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 15:08
2
$begingroup$
@ Bill: not everyone can see that link (me, for example)
$endgroup$
– Geoff Robinson
Jul 29 '11 at 17:25
2
$begingroup$
This is a very cool argument, but a little mysterious...
$endgroup$
– Igor Rivin
Dec 4 '13 at 4:10
$begingroup$
that's the exact answer I'm looking for.
$endgroup$
– Tomas
Dec 4 '13 at 4:16
add a comment |
2
$begingroup$
That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 7:12
$begingroup$
That (and more) is mentioned in Halmos's exposition that I already linked to in a comment.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 15:08
2
$begingroup$
@ Bill: not everyone can see that link (me, for example)
$endgroup$
– Geoff Robinson
Jul 29 '11 at 17:25
2
$begingroup$
This is a very cool argument, but a little mysterious...
$endgroup$
– Igor Rivin
Dec 4 '13 at 4:10
$begingroup$
that's the exact answer I'm looking for.
$endgroup$
– Tomas
Dec 4 '13 at 4:16
2
2
$begingroup$
That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 7:12
$begingroup$
That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion.
$endgroup$
– Jonas Meyer
Jul 29 '11 at 7:12
$begingroup$
That (and more) is mentioned in Halmos's exposition that I already linked to in a comment.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 15:08
$begingroup$
That (and more) is mentioned in Halmos's exposition that I already linked to in a comment.
$endgroup$
– Bill Dubuque
Jul 29 '11 at 15:08
2
2
$begingroup$
@ Bill: not everyone can see that link (me, for example)
$endgroup$
– Geoff Robinson
Jul 29 '11 at 17:25
$begingroup$
@ Bill: not everyone can see that link (me, for example)
$endgroup$
– Geoff Robinson
Jul 29 '11 at 17:25
2
2
$begingroup$
This is a very cool argument, but a little mysterious...
$endgroup$
– Igor Rivin
Dec 4 '13 at 4:10
$begingroup$
This is a very cool argument, but a little mysterious...
$endgroup$
– Igor Rivin
Dec 4 '13 at 4:10
$begingroup$
that's the exact answer I'm looking for.
$endgroup$
– Tomas
Dec 4 '13 at 4:16
$begingroup$
that's the exact answer I'm looking for.
$endgroup$
– Tomas
Dec 4 '13 at 4:16
add a comment |
$begingroup$
$$
xy-yx = 1 \
(x-alpha 1)y-y(x-alpha 1)=1 \
y(alpha 1-x)-(alpha 1-x)y = -1 \
(alpha 1-x)^{-1}y-y(alpha 1-x)^{-1}=frac{d}{dalpha}(alpha 1-x)^{-1}.
$$
Using the functional calculus for a function $F$ that is holomorphic on an open neighborhood of $sigma(x)$ leads to a bounded differentiation:
$$
F(x)y-yF(x)=-F'(x) \
|F'(x)| le 2|y||F(x)|.
$$
If you set $F(alpha)=e^{talpha}$ for large enough real $t$, then the above yields the contradiction $|t| le 2|y|$.
answered Jan 16 at 9:31
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
$$
xy-yx = 1 \
(x-alpha 1)y-y(x-alpha 1)=1 \
y(alpha 1-x)-(alpha 1-x)y = -1 \
(alpha 1-x)^{-1}y-y(alpha 1-x)^{-1}=frac{d}{dalpha}(alpha 1-x)^{-1}.
$$
Using the functional calculus for a function $F$ that is holomorphic on an open neighborhood of $sigma(x)$ leads to a bounded differentiation:
$$
F(x)y-yF(x)=-F'(x) \
|F'(x)| le 2|y||F(x)|.
$$
If you set $F(alpha)=e^{talpha}$ for large enough real $t$, then the above yields the contradiction $|t| le 2|y|$.
answered Jan 16 at 9:31
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
$$
xy-yx = 1 \
(x-alpha 1)y-y(x-alpha 1)=1 \
y(alpha 1-x)-(alpha 1-x)y = -1 \
(alpha 1-x)^{-1}y-y(alpha 1-x)^{-1}=frac{d}{dalpha}(alpha 1-x)^{-1}.
$$
Using the functional calculus for a function $F$ that is holomorphic on an open neighborhood of $sigma(x)$ leads to a bounded differentiation:
$$
F(x)y-yF(x)=-F'(x) \
|F'(x)| le 2|y||F(x)|.
$$
If you set $F(alpha)=e^{talpha}$ for large enough real $t$, then the above yields the contradiction $|t| le 2|y|$.
answered Jan 16 at 9:31
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$$
xy-yx = 1 \
(x-alpha 1)y-y(x-alpha 1)=1 \
y(alpha 1-x)-(alpha 1-x)y = -1 \
(alpha 1-x)^{-1}y-y(alpha 1-x)^{-1}=frac{d}{dalpha}(alpha 1-x)^{-1}.
$$
Using the functional calculus for a function $F$ that is holomorphic on an open neighborhood of $sigma(x)$ leads to a bounded differentiation:
$$
F(x)y-yF(x)=-F'(x) \
|F'(x)| le 2|y||F(x)|.
$$
If you set $F(alpha)=e^{talpha}$ for large enough real $t$, then the above yields the contradiction $|t| le 2|y|$.
answered Jan 16 at 9:31
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 16 at 9:31
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 16 at 9:31
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 16 at 9:31
DisintegratingByPartsDisintegratingByParts
59.4k42580
59.4k42580
add a comment |
add a comment |
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5
$begingroup$
You may find of interest Halmos's exposition on commutators here.
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– Bill Dubuque
Jul 29 '11 at 4:43
$begingroup$
By "ring" here, did you intend "algebra over $mathbb{R}$"? I suppose there is a subtlety in the question. I think the axiom of choice tells us that every linear space over $mathbb{R}$ admits a (subadditive) norm. But we see below that not every algebra over $mathbb{R}$ admits a submultiplicative norm, since there are real algebras which contains solutions of $xy - yx = I$.
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– Geoff Robinson
Jul 29 '11 at 11:59
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@Geoff: ah. Let's suppose I mean "finitely generated ring" (over $mathbb{Z}$).
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– Qiaochu Yuan
Jul 29 '11 at 13:27
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A small comment on your last question: if you take the algebra ${mathbb C}^{mathbb N}$ with pointwise product, then I think it cannot be given a submultiplicative norm. On the other hand, it is an old result of Allan that there is an embedding (no continuity) of ${mathbb C}[[X]]$ into the unitization of some radical Banach algebra.
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– user16299
Jan 7 '12 at 19:42
$begingroup$
Since nobody explicitly mentioned this, maybe I'll point it out in the comments here: In QM, the canonical commutation relation is $xy-yx= i hbar$ so, up to rescaling, the theorem under discussion is telling us, in particular, that the canonical commutation relation cannot be satisfied by two bounded operators. Hence, this particular part of QM necessarily involves unbounded operators.
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– Mike F
Feb 8 '16 at 6:39