One dimensional image of the adjoint action
$begingroup$
Let $K$ be an algebraically closed field of characteristic zero, and let $mathfrak{g}$ be a finite dimensional, nilpotent Lie algebra over $K$.
My question is, can we find an element $xinmathfrak{g}$ such that the image of $ad(x)$ is one dimensional over $K$? In other words can we find $xinmathfrak{g}$ such that $x$ is not central and there exists $zinmathfrak{g}$ such that for every $yinmathfrak{g}$, $[x,y]=alpha z$ for some $alphain K$.
This is not true if $K$ is not algebraically closed, but I haven't been able to find an example where we cannot find such an element $x$ after passing to a finite extension. Does anyone know whether this is true, or have a counterexample?
linear-algebra field-theory lie-algebras
asked Dec 23 '18 at 15:15
AdJoint-repAdJoint-rep
26718
$endgroup$
add a comment |
$begingroup$
Let $K$ be an algebraically closed field of characteristic zero, and let $mathfrak{g}$ be a finite dimensional, nilpotent Lie algebra over $K$.
My question is, can we find an element $xinmathfrak{g}$ such that the image of $ad(x)$ is one dimensional over $K$? In other words can we find $xinmathfrak{g}$ such that $x$ is not central and there exists $zinmathfrak{g}$ such that for every $yinmathfrak{g}$, $[x,y]=alpha z$ for some $alphain K$.
This is not true if $K$ is not algebraically closed, but I haven't been able to find an example where we cannot find such an element $x$ after passing to a finite extension. Does anyone know whether this is true, or have a counterexample?
linear-algebra field-theory lie-algebras
asked Dec 23 '18 at 15:15
AdJoint-repAdJoint-rep
26718
$endgroup$
add a comment |
$begingroup$
Let $K$ be an algebraically closed field of characteristic zero, and let $mathfrak{g}$ be a finite dimensional, nilpotent Lie algebra over $K$.
My question is, can we find an element $xinmathfrak{g}$ such that the image of $ad(x)$ is one dimensional over $K$? In other words can we find $xinmathfrak{g}$ such that $x$ is not central and there exists $zinmathfrak{g}$ such that for every $yinmathfrak{g}$, $[x,y]=alpha z$ for some $alphain K$.
This is not true if $K$ is not algebraically closed, but I haven't been able to find an example where we cannot find such an element $x$ after passing to a finite extension. Does anyone know whether this is true, or have a counterexample?
linear-algebra field-theory lie-algebras
asked Dec 23 '18 at 15:15
AdJoint-repAdJoint-rep
26718
$endgroup$
Let $K$ be an algebraically closed field of characteristic zero, and let $mathfrak{g}$ be a finite dimensional, nilpotent Lie algebra over $K$.
My question is, can we find an element $xinmathfrak{g}$ such that the image of $ad(x)$ is one dimensional over $K$? In other words can we find $xinmathfrak{g}$ such that $x$ is not central and there exists $zinmathfrak{g}$ such that for every $yinmathfrak{g}$, $[x,y]=alpha z$ for some $alphain K$.
This is not true if $K$ is not algebraically closed, but I haven't been able to find an example where we cannot find such an element $x$ after passing to a finite extension. Does anyone know whether this is true, or have a counterexample?
linear-algebra field-theory lie-algebras
linear-algebra field-theory lie-algebras
asked Dec 23 '18 at 15:15
AdJoint-repAdJoint-rep
26718
asked Dec 23 '18 at 15:15
AdJoint-repAdJoint-rep
26718
asked Dec 23 '18 at 15:15
AdJoint-repAdJoint-rep
26718
asked Dec 23 '18 at 15:15
AdJoint-repAdJoint-rep
26718
asked Dec 23 '18 at 15:15
AdJoint-repAdJoint-rep
26718
26718
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
$Vect(e_1,e_2,e_3,e_4,e_5,e_6)$. $[e_1,e_2]=e_4, [e_1,e_3]=e_5, [e_2,e_3]=e_6$ the other brackets are zero. The center is $Vect(e_4,e_5,e_6)$, $dim(Im(ad_{e_i})=3, i=1,2,3$, $dim(Imad_{e_i}))=0, i=4,5,6$....
answered Dec 23 '18 at 15:22
Tsemo AristideTsemo Aristide
57.8k11445
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$begingroup$
Yes, that works fine, thanks!
$endgroup$
– AdJoint-rep
Dec 23 '18 at 16:55
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$Vect(e_1,e_2,e_3,e_4,e_5,e_6)$. $[e_1,e_2]=e_4, [e_1,e_3]=e_5, [e_2,e_3]=e_6$ the other brackets are zero. The center is $Vect(e_4,e_5,e_6)$, $dim(Im(ad_{e_i})=3, i=1,2,3$, $dim(Imad_{e_i}))=0, i=4,5,6$....
answered Dec 23 '18 at 15:22
Tsemo AristideTsemo Aristide
57.8k11445
$endgroup$
$begingroup$
Yes, that works fine, thanks!
$endgroup$
– AdJoint-rep
Dec 23 '18 at 16:55
add a comment |
$begingroup$
$Vect(e_1,e_2,e_3,e_4,e_5,e_6)$. $[e_1,e_2]=e_4, [e_1,e_3]=e_5, [e_2,e_3]=e_6$ the other brackets are zero. The center is $Vect(e_4,e_5,e_6)$, $dim(Im(ad_{e_i})=3, i=1,2,3$, $dim(Imad_{e_i}))=0, i=4,5,6$....
answered Dec 23 '18 at 15:22
Tsemo AristideTsemo Aristide
57.8k11445
$endgroup$
$begingroup$
Yes, that works fine, thanks!
$endgroup$
– AdJoint-rep
Dec 23 '18 at 16:55
add a comment |
$begingroup$
$Vect(e_1,e_2,e_3,e_4,e_5,e_6)$. $[e_1,e_2]=e_4, [e_1,e_3]=e_5, [e_2,e_3]=e_6$ the other brackets are zero. The center is $Vect(e_4,e_5,e_6)$, $dim(Im(ad_{e_i})=3, i=1,2,3$, $dim(Imad_{e_i}))=0, i=4,5,6$....
answered Dec 23 '18 at 15:22
Tsemo AristideTsemo Aristide
57.8k11445
$endgroup$
$Vect(e_1,e_2,e_3,e_4,e_5,e_6)$. $[e_1,e_2]=e_4, [e_1,e_3]=e_5, [e_2,e_3]=e_6$ the other brackets are zero. The center is $Vect(e_4,e_5,e_6)$, $dim(Im(ad_{e_i})=3, i=1,2,3$, $dim(Imad_{e_i}))=0, i=4,5,6$....
answered Dec 23 '18 at 15:22
Tsemo AristideTsemo Aristide
57.8k11445
edited Dec 23 '18 at 15:31
answered Dec 23 '18 at 15:22
Tsemo AristideTsemo Aristide
57.8k11445
answered Dec 23 '18 at 15:22
Tsemo AristideTsemo Aristide
57.8k11445
answered Dec 23 '18 at 15:22
Tsemo AristideTsemo Aristide
57.8k11445
57.8k11445
$begingroup$
Yes, that works fine, thanks!
$endgroup$
– AdJoint-rep
Dec 23 '18 at 16:55
add a comment |
$begingroup$
Yes, that works fine, thanks!
$endgroup$
– AdJoint-rep
Dec 23 '18 at 16:55
$begingroup$
Yes, that works fine, thanks!
$endgroup$
– AdJoint-rep
Dec 23 '18 at 16:55
$begingroup$
Yes, that works fine, thanks!
$endgroup$
– AdJoint-rep
Dec 23 '18 at 16:55
add a comment |
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