Timeline for Is always the ratio (number of commuting pairs of elements in a ring or a Lie algebra)/(the size of the ring or the Lie algebra) integer?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 11, 2016 at 10:44 | history | edited | Alireza Abdollahi | CC BY-SA 3.0 |
added 73 characters in body
|
| Jan 11, 2016 at 10:40 | comment | added | Frieder Ladisch | Well, for (1) the answer is the same as for groups, simply because $ 1+R $ is a group, and $(1+x)(1+y) = (1+y)(1+x)$ iff $ xy=yx $. | |
| Jan 11, 2016 at 9:45 | history | edited | Alireza Abdollahi | CC BY-SA 3.0 |
added 5 characters in body
|
| Jan 11, 2016 at 6:36 | comment | added | Alireza Abdollahi | \begin{align*} |\{(x,y)\in L\times L \;|\; [x,y]=0\}|&=&\\ &\sum_{x\in L} |C_L(x)|= \sum_{i=1}^{n} |C_L(x_i)||[x_i]|\\ &=\sum_{i=1}^n |C_L(x_i)||L:C_L(x)|=\sum_{i=1}^n |L|=n|L|. \end{align*} Hence the ratio is equal to $n$. This completes the proof. | |
| Jan 11, 2016 at 6:36 | comment | added | Alireza Abdollahi | (2) $\sim$ is transitive. Let $x\sim y$ and $y\sim z$. Then there exist $t,s\in L$ such that $x+[x,t]=y$ and $y+[y,s]=z$. It follows that \begin{align*}z=y+[y,s]&=&\\&x+[x,t]+[x+[x,t],s]=\\ &x+[x,t]+[x,s]+[x,t,s]=x+[x,t+s]. \end{align*} Thus $z=x+[x,t+s]$ and so $x\sim z$. Thus, $L$ is partitioned by $[x]$'s. Suppose that $[x_1],\dots,[x_n]$ are all orbits of the equivalence relation $\sim$ on $L$. Now note that is any Lie ring, we have $|L:C_L(x)|=|[x]|$. Therefore, | |
| Jan 11, 2016 at 6:35 | comment | added | Alireza Abdollahi | $$x \sim y \Leftrightarrow \exists\; t\in L \;\;\;\; x+[x,t]=y.$$ We now prove that $\sim$ is reflexive, symmetric and transitive. (1) $x+[x,0]=x+0=x$ ($\sim$ is reflexive). \item If $x\sim y$, then there exists $t\in L$ such that $x+[x,t]=y$. Thus $[y,t]=[x+[x,t],t]=[x,t]+[x,t,t]$ which implies $[y,t]=[x,t]$. So, $x=y-[x,t]=y-[y,t]=y+[y,-t]$. Thus $y\sim x$. Hence $\sim$ is symmetric. | |
| Jan 11, 2016 at 6:32 | comment | added | Alireza Abdollahi | Let $L$ be a Lie ring such that $[L,L,L]=0$. Then $c(L)$ is an integer. For amt $x\in L$, define $[x]=\{x+[x,y] \;|\; y\in L\}$. The latter set is an orbit of the equivalence relation $\sim$ on $L$ which is defined as follows: | |
| Jan 10, 2016 at 20:05 | comment | added | YCor | Anyway I initially understood from your last line that your claim is that there are counterexamples in the case of finite (nilpotent??) groups, while you mean the opposite. | |
| Jan 10, 2016 at 19:50 | comment | added | Alireza Abdollahi | @YCor. I used "always" in the title so I have to write in positive. I wrote the question in the body in negative as my guess is that an example is elusive. | |
| Jan 10, 2016 at 19:47 | comment | added | Alireza Abdollahi | @DavidSpeyer. I have a proof for a finite nilpotent Lie ring of nilpotent class $2$. I will publish it here tommorow, as it is now in my computer at university. | |
| Jan 10, 2016 at 17:19 | comment | added | David E Speyer | @AlirezaAbdollahi I also got the nilpotency $\geq 3$ result. Proof for the curious: Let $L$ be a Lie algebra with nilpotency $2$. Let $Z$ be the center. Let $\ell=\dim L$ and $z=\dim Z$. We write $\bar{\ }$ for reduction from $L$ to $L/Z$. For $g \in L$, we have $[g,h]=0$ iff $\bar{h}$ is in $\mathrm{Ker}([\bar{g}, \ ] : L/Z \to Z)$. The dimension of the kernel is $\geq (\ell-z)-z=\ell-2z$. For each possible $\bar{g} \in L/Z$, this contributes $q^{\ell-2z} q^{2z}=|L|$ pairs $(g,h)$, since there are $q^z$ ways each to lift $\bar{g}$ and $\bar{h}$ to $L$. Sum over $\bar{g}$ to conclude. | |
| Jan 10, 2016 at 14:05 | comment | added | YCor | The title is written is the positive while the question is in the negative, which makes "a negative answer" quite ambiguous. It's more natural to ask everything in the positive. | |
| Jan 10, 2016 at 13:10 | comment | added | Alireza Abdollahi | @DavidSpeyer. In the meanwhile, the nilpotency class of a possible counterexample should be at least $3$. I have proved the latter. | |
| Jan 10, 2016 at 13:05 | comment | added | Alireza Abdollahi | @DavidSpeyer. Yes. By Lazard's correspondence there is a bijective map $\mathcal{L}$ from an arbitrary nilpotent finite dimensional Lie algebra $L$ over a finite field of characteristic $p$ of nilpotent class less than $p$ to a finite $p$-group $G$ (of the same nilpotency class as $L$) such that $[x, y]^\mathcal{L} = [x^{\mathcal{L}}, y^{\mathcal{L}}]$ for all $x, y \in L$, where the right hand side is the usual group commutator $[a, b] = a^{-1}b^{-1}ab$. This implies that the answer to the question is negative for all such Lie algebras $L$. | |
| Jan 10, 2016 at 12:56 | review | Close votes | |||
| Jan 10, 2016 at 19:22 | |||||
| Jan 10, 2016 at 12:48 | comment | added | David E Speyer | Counter examples for nilpotent Lie algebras will involve "small primes". If $p>n$, then $\exp$ is a well defined bijection from $N_n(\mathbb{F}_p)$ to $U_n(\mathbb{F}_p)$ (where $N_n$ and $U_n$ are upper triangular matrices with $0$'s and $1$'s on the diagonal respectively). The map $\exp$ takes sub-Lie-algebras to sub-groups and commuting pairs to commuting pairs. | |
| Jan 10, 2016 at 12:45 | history | edited | Alireza Abdollahi | CC BY-SA 3.0 |
added 77 characters in body
|
| Jan 10, 2016 at 12:44 | comment | added | Alireza Abdollahi | @YCor. I think I have missed another hypothesis that the Lie algebra should be nilpotent. | |
| Jan 10, 2016 at 12:37 | comment | added | YCor | What have you tried? in a Lie algebra of dimension $n$ over a field of cardinal $q$ such that every commuting pair is collinear, the number of commuting pairs is $q^{n+1}+q^n-q$, so the ratio is $q+1-1/q$. There are such examples with $n=2$ or $n=3$ (so, the smallest possible non-abelian Lie algebra is a counterexample!). | |
| Jan 10, 2016 at 12:12 | history | edited | Alireza Abdollahi | CC BY-SA 3.0 |
edited title
|
| Jan 10, 2016 at 11:58 | history | asked | Alireza Abdollahi | CC BY-SA 3.0 |