on the composition of derivations

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fbli
Posts: 1
Joined: Wed Oct 28, 2015 7:21 am

on the composition of derivations

Postby fbli » Wed Oct 28, 2015 7:24 am

This question arose while reading the paper [*On the composition of derivations*, by J.Krempa and J.Matczuk](http://w[/code]ww.researchgate.net/publication/226595021_On_the_composition_of_derivations).<br /><br />Let R denote an associative ring For convenience, we will take $chaR=\infty$,f the additive group of R is torsion-free.Lett $d_{1} ,...,d_{n}$ be derivations of $R$ and $U=\{1,...,n\}$.  For any subset of $V$ of $U$ we will put<br />$d_{v}(x)=d^{\epsilon_{1}}_{1}...d^{\epsilon_{n}}_{n}(x)$, where <br />$$\epsilon_{i}= \begin{array}{cc} <br /> \Big\{ & <br />    \begin{array}{cc}<br />1 & i\in V \\<br />0&   i\notin V. <br />\end{array}<br />\end{array}$$<br />In particular, we get $d_{U}(x)=d_{1}...d_{n}(x)$.<br />Further, by writing $V=V_{1}\sqcup...\sqcup V_{r}$ we will understand that $V$ is a sum of disjoint subset $V_{i}$.<br />It is easy to compute by induction on cardinality of $V$ contained in U that for any $x,y \in R$, <br /><br />  $$(2) \qquad d_{v}(xy)=\sum\limits_{V_{1}\sqcup V_{2}=V} d_{v_{1}}(x)d_{v_{2}}(y).$$<br />By substituting $d$ for all $ d_{i}$'s in (2) we easily obtained the Leibniz's formula $d^{r}(xy)=\sum\limits_{i=0}^r \binom{r}{i} \cdot d^{i}(x)d^{r-i}(y)$.<br />For any $k>1$ and $V \subset U, P_{k}(V)$ will stand for the family of all sequences<br />of sets $V_{1}, ..., V_{k}$ where all $V_{i}$'s are not empty and $V=V_{1}\sqcup...\sqcup V_{k}.$<br />In the above notation, it is claimed in the paper that (2) takes form of<br />  $$(3) \qquad d_{v}(xy)=d_{v}(x)y+xd_{v}(y)+ \sum\limits_{V_{1},V_{2}<br />  \in P_{2}(V)} d_{v_{1}}(x)d_{v_{2}}(y).$$<br /><br />My question is: how can we deduce (3) from (2)?$$$$I am trying to understand the concept  for (2).If just for x concept easy. if we can  take V={1,2,3,} and $V_{1}$={1},$V_{2}$={2,3}  then $d_{v}(x)=d_{1}d_{2}d_{3}(x) ,d_{V_{1}}(x)=d_{1}(x),d_{V_{2}}(x)=d_{2}d_{3}(x)$.  So, $d_{v}(x)=d_{1}d_{2}d_{3}(x)= d_{V_{1}\sqcup V_{2}=V}(x)=d_{V_{1}}           (x).d_{V_{2}}(x)=d_{1}(x)d_{2}d_{3}(x)=d_{1}d_{2}d_{3}(x)$.   But if we take x and y,then we can obtain  $d_{v}(xy)=d_{1}d_{2}d_{3}(xy)=d_{V_{1}\sqcup V_{2}=V}(xy)=d_{V_{1}}(xy).d_{V_{2}}(xy)=d_{1}(xy)d_{2}d_{3}(xy)=d_{1}d_{2}d_{3}(xy)$.  But how they do reduce (3) from (2)?.Maybe we must understand   meaning of $\sum\limits_{V_{1}\sqcup V_{2}=V}$ and $\sum\limits_{V_{1},V_{2}<br />  \in P_{2}(V)}$. this equation $d_{1}d_{2}d_{3}(xy)=d_{1}(xy)d_{2}d_{3}(xy)=d_{1}d_{2}d_{3}(xy)$ is it true i dont know.]

GeraltofRivia
Posts: 13
Joined: Tue Oct 06, 2015 12:33 pm

Re: on the composition of derivations

Postby GeraltofRivia » Wed Oct 28, 2015 8:32 am

fbli wrote:This question arose while reading the paper [*On the composition of derivations*, by J.Krempa and J.Matczuk](http://w[/code]ww.researchgate.net/publication/226595021_On_the_composition_of_derivations).<br /><br />Let R denote an associative ring For convenience, we will take $chaR=\infty$,f the additive group of R is torsion-free.Lett $d_{1} ,...,d_{n}$ be derivations of $R$ and $U=\{1,...,n\}$.  For any subset of $V$ of $U$ we will put<br />$d_{v}(x)=d^{\epsilon_{1}}_{1}...d^{\epsilon_{n}}_{n}(x)$, where <br />$$\epsilon_{i}= \begin{array}{cc} <br /> \Big\{ & <br />    \begin{array}{cc}<br />1 & i\in V \\<br />0&   i\notin V. <br />\end{array}<br />\end{array}$$<br />In particular, we get $d_{U}(x)=d_{1}...d_{n}(x)$.<br />Further, by writing $V=V_{1}\sqcup...\sqcup V_{r}$ we will understand that $V$ is a sum of disjoint subset $V_{i}$.<br />It is easy to compute by induction on cardinality of $V$ contained in U that for any $x,y \in R$, <br /><br />  $$(2) \qquad d_{v}(xy)=\sum\limits_{V_{1}\sqcup V_{2}=V} d_{v_{1}}(x)d_{v_{2}}(y).$$<br />By substituting $d$ for all $ d_{i}$'s in (2) we easily obtained the Leibniz's formula $d^{r}(xy)=\sum\limits_{i=0}^r \binom{r}{i} \cdot d^{i}(x)d^{r-i}(y)$.<br />For any $k>1$ and $V \subset U, P_{k}(V)$ will stand for the family of all sequences<br />of sets $V_{1}, ..., V_{k}$ where all $V_{i}$'s are not empty and $V=V_{1}\sqcup...\sqcup V_{k}.$<br />In the above notation, it is claimed in the paper that (2) takes form of<br />  $$(3) \qquad d_{v}(xy)=d_{v}(x)y+xd_{v}(y)+ \sum\limits_{V_{1},V_{2}<br />  \in P_{2}(V)} d_{v_{1}}(x)d_{v_{2}}(y).$$<br /><br />My question is: how can we deduce (3) from (2)?$$$$I am trying to understand the concept  for (2).If just for x concept easy. if we can  take V={1,2,3,} and $V_{1}$={1},$V_{2}$={2,3}  then $d_{v}(x)=d_{1}d_{2}d_{3}(x) ,d_{V_{1}}(x)=d_{1}(x),d_{V_{2}}(x)=d_{2}d_{3}(x)$.  So, $d_{v}(x)=d_{1}d_{2}d_{3}(x)= d_{V_{1}\sqcup V_{2}=V}(x)=d_{V_{1}}           (x).d_{V_{2}}(x)=d_{1}(x)d_{2}d_{3}(x)=d_{1}d_{2}d_{3}(x)$.   But if we take x and y,then we can obtain  $d_{v}(xy)=d_{1}d_{2}d_{3}(xy)=d_{V_{1}\sqcup V_{2}=V}(xy)=d_{V_{1}}(xy).d_{V_{2}}(xy)=d_{1}(xy)d_{2}d_{3}(xy)=d_{1}d_{2}d_{3}(xy)$.  But how they do reduce (3) from (2)?.Maybe we must understand   meaning of $\sum\limits_{V_{1}\sqcup V_{2}=V}$ and $\sum\limits_{V_{1},V_{2}<br />  \in P_{2}(V)}$. this equation $d_{1}d_{2}d_{3}(xy)=d_{1}(xy)d_{2}d_{3}(xy)=d_{1}d_{2}d_{3}(xy)$ is it true i dont know.]

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