Timeline for Norms of commutators
Current License: CC BY-SA 4.0
33 events
| when toggle format | what | by | license | comment | |
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| Aug 6, 2022 at 15:37 | history | edited | YCor | CC BY-SA 4.0 |
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| Aug 11, 2017 at 15:13 | answer | added | Suvrit | timeline score: 7 | |
| Feb 7, 2012 at 16:08 | answer | added | Bill Johnson | timeline score: 40 | |
| Jan 1, 2011 at 18:01 | answer | added | Suvrit | timeline score: 8 | |
| Jan 1, 2011 at 17:49 | comment | added | Suvrit | @Bill: I have a few references for you; let me post them below in the answer field. | |
| Jan 1, 2011 at 15:47 | comment | added | Bill Johnson | @Suvrit: No, I did not know that. Do you have a reference? | |
| Jan 1, 2011 at 13:20 | comment | added | Suvrit | I wanted to know if you are already aware of the result that for the Frobenius norm, the ratio $\|BC-CA\|/ (\|B\|\|C\|)$ for randomly chosen $B$ and $C$, tightly concentrates around a number that goes to zero as $n\to \infty$. Thus, it suggests that $\lambda(n) \to \infty$ as $n \to \infty$, right? | |
| Nov 19, 2010 at 15:18 | history | edited | Denis Serre |
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| Nov 3, 2010 at 21:01 | comment | added | Denis Serre | See Mark Sapir's answer mathoverflow.net/questions/44715 to a related question of mine, concerning the case where $A$ is nilpotent. We may choose a factor $B=A$, but the price to pay may be to high. | |
| Oct 2, 2010 at 14:43 | comment | added | Fiktor | @ Bill Johnson: Do you mean $\|C\|\leq\sqrt{2}\|A\|$ in "Added June 10"? Otherwise you (or Gideon Schechtman) should know the answer to mathoverflow.net/questions/40801/… . | |
| Sep 28, 2010 at 15:54 | comment | added | Bill Johnson | Off-line more has happened--the latest upper bound is a power of $\log n$ (sixth power, I think), resulting from combined efforts with N. Ozawa and G. Schechtman. I thought this thread had died and so did not post. The proofs are a bit beyond what should go on MO, but eventually we'll write what we can do and I'll then post a link here. | |
| Sep 28, 2010 at 14:23 | comment | added | Denis Serre | I did some calculations and found the bounds mentionned in several previous comments, namely $O(n)$ (resp. $O(n^{3/2}$) in the complex (resp. real) case. But what amazes me is that these bounds apply for both the Forbenius norm and the operator norm, and this for different reasons... | |
| Jun 18, 2010 at 16:22 | history | bounty ended | Bill Johnson | ||
| Jun 16, 2010 at 16:07 | history | edited | Bill Johnson |
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| Jun 12, 2010 at 5:30 | comment | added | Bill Johnson | We have not thought about that case, Mark. | |
| Jun 11, 2010 at 19:09 | comment | added | Mark Meckes | Can you handle as well the sum of powers of the shift (i.e. strictly upper triangular $A$ with all 1s above the diagonal)? | |
| Jun 11, 2010 at 18:50 | comment | added | Bill Johnson | No, Mark--at least not just from that condition, because the shift is not a problem. | |
| Jun 11, 2010 at 18:42 | comment | added | Mark Meckes | Here's a random quick thought. Schechtman's argument suggests that for a lower bound on $\lambda$ you should investigate $A$ which is far from normal. There exist matrices $A$ with $\| [A,A^*] \| = \| A \|^2$; can one bound below $\| B\| \| C \|$ for such $A$? | |
| Jun 11, 2010 at 15:48 | history | bounty started | Bill Johnson | ||
| Jun 10, 2010 at 15:45 | history | edited | Bill Johnson | CC BY-SA 2.5 |
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| Jun 8, 2010 at 17:26 | comment | added | Bill Johnson | I don't see how, Mark. The problem reduces to considering a real matrix $A$ with zeros on the diagonal, but then if you use a diagonal real matrix $B$ when writing $A=BC-CB$ you run into problems and get only $\|B\|\cdot\|C\|\le n^{3/2}$. | |
| Jun 8, 2010 at 14:47 | comment | added | Mark Meckes | Bill: Can you get $\lambda(n)\le n$ for real matrices? | |
| Jun 8, 2010 at 13:14 | comment | added | Steve Flammia | @gowers, the Pauli matrices show that a lower bound of λ=1/2 is tight, at least when n is even. | |
| Jun 8, 2010 at 11:55 | history | edited | Bill Johnson |
Added a tag to push thread to the top.
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| Jun 8, 2010 at 11:52 | comment | added | Bill Johnson | Schechtman showed that $\lambda(n) \le n$. WLOG (conjugate with an appropriate unitary) $C$ has zero diagonal and choose $A$ diagonal so that the magnitude of the difference of any two diagonal entries is at least one and the magnitude of every diagonal entry is less than $(n/2)^{1/2}$ (or a bit larger if $n$ is not a square). When you solve $AB-BA =C$ you see that the norm of $B$ is at most $n^{1/2} \|C\|$. | |
| Jun 8, 2010 at 10:27 | comment | added | Bill Johnson | No lower bound. Schechtman and I convinced ourselves that the argument in eecs.berkeley.edu/~wkahan/MathH110/trace0.pdf gives the upper bound $\lambda(n)\le n^{3/2}$. Same upper bound multiplied by a constant less than one for the Hilbert-Schmidt norm. It is easy to compute $\lambda(2) = 1/2$. It seems clear that $\lambda(n)\to \infty$, not that I have any idea how to prove it. We assumed that the actual growth rate was well known even if we could not find it in the books we checked nor by Googling. | |
| Jun 8, 2010 at 7:39 | comment | added | gowers | Do you have any lower bound better than λ=1/2 or any upper bound at all? I'd just like to get a feel for what the "obvious" bounds are that one should try to beat. | |
| Jun 7, 2010 at 17:05 | comment | added | Mark Meckes | @Gil: did you mean X and U have the same norm as B, and Y and Z have the same norm as C? | |
| Jun 7, 2010 at 16:53 | comment | added | Bill Johnson | I mean the operator norm, $\|A\|= \max \{\|Ax\|: \|x\|=1\}$ with $\|x\|$ the Euclidean norm. However, I do not know the answer if you use the Frobenius (Hilbert-Schmidt) norm. @Gil: I do not understand your comment. | |
| Jun 7, 2010 at 16:37 | comment | added | Sunni | I guess it is Frobenius norm first. For Frobenius norm it is true that $\sqrt{2} \|B\|\cdot \|C\| \ge \|A\|$ for all complex matrices $B, C$. | |
| Jun 7, 2010 at 16:09 | comment | added | Gil Kalai | Dear Bill, great question. Wild uneducated guess: isn't it the case that the norm of A can be essentially as large as the norm of XY-ZU where X and Z have the same notm as B and Y and U have the same norm as C? | |
| Jun 7, 2010 at 14:01 | comment | added | Wadim Zudilin | Do you mean the max norm? | |
| Jun 7, 2010 at 12:53 | history | asked | Bill Johnson | CC BY-SA 2.5 |