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Faithfull representation of $mathfrak h rtimes V$.
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Let $V$ be a finite dimensional vector space and $mathfrak h subset mathfrak {gl} (V)$ a Lie Algebra. Consider the semidirect product $mathfrak h rtimes V$
$$[(H_1,v_1),(H_2,v_2)] = ([H_1,H_2], H_1 v_2 - H_2 v_1). $$
I'm stuck in this question
Question: Let $mathfrak{h} subset mathfrak {gl} (V)$ a Lie Algebra of linear transformations and consider the semidirect product $ mathfrak g = mathfrak h rtimes V $. Find a faithful representation of $mathfrak h rtimes V$.
Does anyone have any ideas?
abstract-algebra representation-theory lie-algebras
asked yesterday
Matheus Manzatto
1,2541523
add a comment |
up vote
4
down vote
favorite
Let $V$ be a finite dimensional vector space and $mathfrak h subset mathfrak {gl} (V)$ a Lie Algebra. Consider the semidirect product $mathfrak h rtimes V$
$$[(H_1,v_1),(H_2,v_2)] = ([H_1,H_2], H_1 v_2 - H_2 v_1). $$
I'm stuck in this question
Question: Let $mathfrak{h} subset mathfrak {gl} (V)$ a Lie Algebra of linear transformations and consider the semidirect product $ mathfrak g = mathfrak h rtimes V $. Find a faithful representation of $mathfrak h rtimes V$.
Does anyone have any ideas?
abstract-algebra representation-theory lie-algebras
asked yesterday
Matheus Manzatto
1,2541523
1
Is $V$ finite dimensional at least? (I asked just in case. I don't have a solution even for finite dimensional $V$ yet.)
– Zvi
yesterday
I do think so. I will edit my question.
– Matheus Manzatto
yesterday
2
If $V$ is finite dim, then Ado's theorem guarantees that a finite dim faithful representation exists. I just don't know how to construct it. The easy case is when $bigcap_{Hinmathfrak{h}} ker H={0}$. Then, the Lie algebra $mathfrak{h} rtimes V$ has trivial center, and therefore the adjoint representation is faithful.
– Zvi
yesterday
Yes. This kind of statement always confuses me, I think that "find" means "explicit construct". But I agree with you that Ado's theorem guarantees that this representation exists.
– Matheus Manzatto
yesterday
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $V$ be a finite dimensional vector space and $mathfrak h subset mathfrak {gl} (V)$ a Lie Algebra. Consider the semidirect product $mathfrak h rtimes V$
$$[(H_1,v_1),(H_2,v_2)] = ([H_1,H_2], H_1 v_2 - H_2 v_1). $$
I'm stuck in this question
Question: Let $mathfrak{h} subset mathfrak {gl} (V)$ a Lie Algebra of linear transformations and consider the semidirect product $ mathfrak g = mathfrak h rtimes V $. Find a faithful representation of $mathfrak h rtimes V$.
Does anyone have any ideas?
abstract-algebra representation-theory lie-algebras
asked yesterday
Matheus Manzatto
1,2541523
Let $V$ be a finite dimensional vector space and $mathfrak h subset mathfrak {gl} (V)$ a Lie Algebra. Consider the semidirect product $mathfrak h rtimes V$
$$[(H_1,v_1),(H_2,v_2)] = ([H_1,H_2], H_1 v_2 - H_2 v_1). $$
I'm stuck in this question
Question: Let $mathfrak{h} subset mathfrak {gl} (V)$ a Lie Algebra of linear transformations and consider the semidirect product $ mathfrak g = mathfrak h rtimes V $. Find a faithful representation of $mathfrak h rtimes V$.
Does anyone have any ideas?
abstract-algebra representation-theory lie-algebras
abstract-algebra representation-theory lie-algebras
asked yesterday
Matheus Manzatto
1,2541523
asked yesterday
Matheus Manzatto
1,2541523
edited yesterday
asked yesterday
Matheus Manzatto
1,2541523
asked yesterday
Matheus Manzatto
1,2541523
asked yesterday
Matheus Manzatto
1,2541523
1,2541523
1
Is $V$ finite dimensional at least? (I asked just in case. I don't have a solution even for finite dimensional $V$ yet.)
– Zvi
yesterday
I do think so. I will edit my question.
– Matheus Manzatto
yesterday
2
If $V$ is finite dim, then Ado's theorem guarantees that a finite dim faithful representation exists. I just don't know how to construct it. The easy case is when $bigcap_{Hinmathfrak{h}} ker H={0}$. Then, the Lie algebra $mathfrak{h} rtimes V$ has trivial center, and therefore the adjoint representation is faithful.
– Zvi
yesterday
Yes. This kind of statement always confuses me, I think that "find" means "explicit construct". But I agree with you that Ado's theorem guarantees that this representation exists.
– Matheus Manzatto
yesterday
add a comment |
1
Is $V$ finite dimensional at least? (I asked just in case. I don't have a solution even for finite dimensional $V$ yet.)
– Zvi
yesterday
I do think so. I will edit my question.
– Matheus Manzatto
yesterday
2
If $V$ is finite dim, then Ado's theorem guarantees that a finite dim faithful representation exists. I just don't know how to construct it. The easy case is when $bigcap_{Hinmathfrak{h}} ker H={0}$. Then, the Lie algebra $mathfrak{h} rtimes V$ has trivial center, and therefore the adjoint representation is faithful.
– Zvi
yesterday
Yes. This kind of statement always confuses me, I think that "find" means "explicit construct". But I agree with you that Ado's theorem guarantees that this representation exists.
– Matheus Manzatto
yesterday
1
1
Is $V$ finite dimensional at least? (I asked just in case. I don't have a solution even for finite dimensional $V$ yet.)
– Zvi
yesterday
Is $V$ finite dimensional at least? (I asked just in case. I don't have a solution even for finite dimensional $V$ yet.)
– Zvi
yesterday
I do think so. I will edit my question.
– Matheus Manzatto
yesterday
I do think so. I will edit my question.
– Matheus Manzatto
yesterday
2
2
If $V$ is finite dim, then Ado's theorem guarantees that a finite dim faithful representation exists. I just don't know how to construct it. The easy case is when $bigcap_{Hinmathfrak{h}} ker H={0}$. Then, the Lie algebra $mathfrak{h} rtimes V$ has trivial center, and therefore the adjoint representation is faithful.
– Zvi
yesterday
If $V$ is finite dim, then Ado's theorem guarantees that a finite dim faithful representation exists. I just don't know how to construct it. The easy case is when $bigcap_{Hinmathfrak{h}} ker H={0}$. Then, the Lie algebra $mathfrak{h} rtimes V$ has trivial center, and therefore the adjoint representation is faithful.
– Zvi
yesterday
Yes. This kind of statement always confuses me, I think that "find" means "explicit construct". But I agree with you that Ado's theorem guarantees that this representation exists.
– Matheus Manzatto
yesterday
Yes. This kind of statement always confuses me, I think that "find" means "explicit construct". But I agree with you that Ado's theorem guarantees that this representation exists.
– Matheus Manzatto
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
An explicit representation is given by $rho:mathfrak{g}to mathfrak{gl}(mathbb Rtimes V)$,
$$rho(H,v)=begin{pmatrix}0 &0_{1times n} \ 0_{ntimes 1} & Hend{pmatrix}+begin{pmatrix}0 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}.$$
(here I think on $V$ as $mathbb{R}^n$, and use matrices).
The verification is straightforward.
More important then this is how to get there. Suppose $H$ is a Lie subgroup of $GL(V)$ that integrates $mathfrak{h}$. If you think on the natural semidirect product of the groups $H$ and $V$, $Hrtimes V$, then $Hrtimes V$ acts on $V$ by combining multiplication and translation:
$$(h,v)x=hx+v.$$
You can even define the semidirect product through this action. This action, however, is not linear. You can fix this situation by considering a slightly extended action on $Vtimes mathbb{R}$:
$$ (h,v)(x,alpha)=hx+alpha v.$$
(note that the last action imitates the first one when you restrict to ${1}times V$.) The last action (magically) happens to be linear. Taking $V=mathbb{R}^n$, its explicit representation is
$$tilderho(h,v)=begin{pmatrix}1 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}begin{pmatrix}1 &0_{1times n} \ 0_{ntimes 1} & hend{pmatrix}.$$
You just need to differentiate $tilde rho$.
answered 19 hours ago
Llohann
70329
3
I think your construction works over any base field $mathbb{F}$ even when $V$ is infinite dim. Define the $(mathfrak{h}rtimes V)$-action on $mathbb{F}times V$ by $(H,v)cdot (t,w)=(0,tv+Hw)$ for any $(H,v)in mathfrak{h}rtimes V$ and for any $(t,w) in mathbb{F}times V$. Therefore, there is no need to assume that $V$ is finite dim at all.
– Zvi
18 hours ago
Nice approach thx!
– Matheus Manzatto
17 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
An explicit representation is given by $rho:mathfrak{g}to mathfrak{gl}(mathbb Rtimes V)$,
$$rho(H,v)=begin{pmatrix}0 &0_{1times n} \ 0_{ntimes 1} & Hend{pmatrix}+begin{pmatrix}0 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}.$$
(here I think on $V$ as $mathbb{R}^n$, and use matrices).
The verification is straightforward.
More important then this is how to get there. Suppose $H$ is a Lie subgroup of $GL(V)$ that integrates $mathfrak{h}$. If you think on the natural semidirect product of the groups $H$ and $V$, $Hrtimes V$, then $Hrtimes V$ acts on $V$ by combining multiplication and translation:
$$(h,v)x=hx+v.$$
You can even define the semidirect product through this action. This action, however, is not linear. You can fix this situation by considering a slightly extended action on $Vtimes mathbb{R}$:
$$ (h,v)(x,alpha)=hx+alpha v.$$
(note that the last action imitates the first one when you restrict to ${1}times V$.) The last action (magically) happens to be linear. Taking $V=mathbb{R}^n$, its explicit representation is
$$tilderho(h,v)=begin{pmatrix}1 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}begin{pmatrix}1 &0_{1times n} \ 0_{ntimes 1} & hend{pmatrix}.$$
You just need to differentiate $tilde rho$.
answered 19 hours ago
Llohann
70329
3
I think your construction works over any base field $mathbb{F}$ even when $V$ is infinite dim. Define the $(mathfrak{h}rtimes V)$-action on $mathbb{F}times V$ by $(H,v)cdot (t,w)=(0,tv+Hw)$ for any $(H,v)in mathfrak{h}rtimes V$ and for any $(t,w) in mathbb{F}times V$. Therefore, there is no need to assume that $V$ is finite dim at all.
– Zvi
18 hours ago
Nice approach thx!
– Matheus Manzatto
17 hours ago
add a comment |
up vote
5
down vote
accepted
An explicit representation is given by $rho:mathfrak{g}to mathfrak{gl}(mathbb Rtimes V)$,
$$rho(H,v)=begin{pmatrix}0 &0_{1times n} \ 0_{ntimes 1} & Hend{pmatrix}+begin{pmatrix}0 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}.$$
(here I think on $V$ as $mathbb{R}^n$, and use matrices).
The verification is straightforward.
More important then this is how to get there. Suppose $H$ is a Lie subgroup of $GL(V)$ that integrates $mathfrak{h}$. If you think on the natural semidirect product of the groups $H$ and $V$, $Hrtimes V$, then $Hrtimes V$ acts on $V$ by combining multiplication and translation:
$$(h,v)x=hx+v.$$
You can even define the semidirect product through this action. This action, however, is not linear. You can fix this situation by considering a slightly extended action on $Vtimes mathbb{R}$:
$$ (h,v)(x,alpha)=hx+alpha v.$$
(note that the last action imitates the first one when you restrict to ${1}times V$.) The last action (magically) happens to be linear. Taking $V=mathbb{R}^n$, its explicit representation is
$$tilderho(h,v)=begin{pmatrix}1 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}begin{pmatrix}1 &0_{1times n} \ 0_{ntimes 1} & hend{pmatrix}.$$
You just need to differentiate $tilde rho$.
answered 19 hours ago
Llohann
70329
3
I think your construction works over any base field $mathbb{F}$ even when $V$ is infinite dim. Define the $(mathfrak{h}rtimes V)$-action on $mathbb{F}times V$ by $(H,v)cdot (t,w)=(0,tv+Hw)$ for any $(H,v)in mathfrak{h}rtimes V$ and for any $(t,w) in mathbb{F}times V$. Therefore, there is no need to assume that $V$ is finite dim at all.
– Zvi
18 hours ago
Nice approach thx!
– Matheus Manzatto
17 hours ago
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
An explicit representation is given by $rho:mathfrak{g}to mathfrak{gl}(mathbb Rtimes V)$,
$$rho(H,v)=begin{pmatrix}0 &0_{1times n} \ 0_{ntimes 1} & Hend{pmatrix}+begin{pmatrix}0 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}.$$
(here I think on $V$ as $mathbb{R}^n$, and use matrices).
The verification is straightforward.
More important then this is how to get there. Suppose $H$ is a Lie subgroup of $GL(V)$ that integrates $mathfrak{h}$. If you think on the natural semidirect product of the groups $H$ and $V$, $Hrtimes V$, then $Hrtimes V$ acts on $V$ by combining multiplication and translation:
$$(h,v)x=hx+v.$$
You can even define the semidirect product through this action. This action, however, is not linear. You can fix this situation by considering a slightly extended action on $Vtimes mathbb{R}$:
$$ (h,v)(x,alpha)=hx+alpha v.$$
(note that the last action imitates the first one when you restrict to ${1}times V$.) The last action (magically) happens to be linear. Taking $V=mathbb{R}^n$, its explicit representation is
$$tilderho(h,v)=begin{pmatrix}1 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}begin{pmatrix}1 &0_{1times n} \ 0_{ntimes 1} & hend{pmatrix}.$$
You just need to differentiate $tilde rho$.
answered 19 hours ago
Llohann
70329
An explicit representation is given by $rho:mathfrak{g}to mathfrak{gl}(mathbb Rtimes V)$,
$$rho(H,v)=begin{pmatrix}0 &0_{1times n} \ 0_{ntimes 1} & Hend{pmatrix}+begin{pmatrix}0 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}.$$
(here I think on $V$ as $mathbb{R}^n$, and use matrices).
The verification is straightforward.
More important then this is how to get there. Suppose $H$ is a Lie subgroup of $GL(V)$ that integrates $mathfrak{h}$. If you think on the natural semidirect product of the groups $H$ and $V$, $Hrtimes V$, then $Hrtimes V$ acts on $V$ by combining multiplication and translation:
$$(h,v)x=hx+v.$$
You can even define the semidirect product through this action. This action, however, is not linear. You can fix this situation by considering a slightly extended action on $Vtimes mathbb{R}$:
$$ (h,v)(x,alpha)=hx+alpha v.$$
(note that the last action imitates the first one when you restrict to ${1}times V$.) The last action (magically) happens to be linear. Taking $V=mathbb{R}^n$, its explicit representation is
$$tilderho(h,v)=begin{pmatrix}1 &0_{1times n} \ v & 0_{ntimes n}end{pmatrix}begin{pmatrix}1 &0_{1times n} \ 0_{ntimes 1} & hend{pmatrix}.$$
You just need to differentiate $tilde rho$.
answered 19 hours ago
Llohann
70329
answered 19 hours ago
Llohann
70329
answered 19 hours ago
Llohann
70329
answered 19 hours ago
Llohann
70329
70329
3
I think your construction works over any base field $mathbb{F}$ even when $V$ is infinite dim. Define the $(mathfrak{h}rtimes V)$-action on $mathbb{F}times V$ by $(H,v)cdot (t,w)=(0,tv+Hw)$ for any $(H,v)in mathfrak{h}rtimes V$ and for any $(t,w) in mathbb{F}times V$. Therefore, there is no need to assume that $V$ is finite dim at all.
– Zvi
18 hours ago
Nice approach thx!
– Matheus Manzatto
17 hours ago
add a comment |
3
I think your construction works over any base field $mathbb{F}$ even when $V$ is infinite dim. Define the $(mathfrak{h}rtimes V)$-action on $mathbb{F}times V$ by $(H,v)cdot (t,w)=(0,tv+Hw)$ for any $(H,v)in mathfrak{h}rtimes V$ and for any $(t,w) in mathbb{F}times V$. Therefore, there is no need to assume that $V$ is finite dim at all.
– Zvi
18 hours ago
Nice approach thx!
– Matheus Manzatto
17 hours ago
3
3
I think your construction works over any base field $mathbb{F}$ even when $V$ is infinite dim. Define the $(mathfrak{h}rtimes V)$-action on $mathbb{F}times V$ by $(H,v)cdot (t,w)=(0,tv+Hw)$ for any $(H,v)in mathfrak{h}rtimes V$ and for any $(t,w) in mathbb{F}times V$. Therefore, there is no need to assume that $V$ is finite dim at all.
– Zvi
18 hours ago
I think your construction works over any base field $mathbb{F}$ even when $V$ is infinite dim. Define the $(mathfrak{h}rtimes V)$-action on $mathbb{F}times V$ by $(H,v)cdot (t,w)=(0,tv+Hw)$ for any $(H,v)in mathfrak{h}rtimes V$ and for any $(t,w) in mathbb{F}times V$. Therefore, there is no need to assume that $V$ is finite dim at all.
– Zvi
18 hours ago
Nice approach thx!
– Matheus Manzatto
17 hours ago
Nice approach thx!
– Matheus Manzatto
17 hours ago
add a comment |
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Is $V$ finite dimensional at least? (I asked just in case. I don't have a solution even for finite dimensional $V$ yet.)
– Zvi
yesterday
I do think so. I will edit my question.
– Matheus Manzatto
yesterday
2
If $V$ is finite dim, then Ado's theorem guarantees that a finite dim faithful representation exists. I just don't know how to construct it. The easy case is when $bigcap_{Hinmathfrak{h}} ker H={0}$. Then, the Lie algebra $mathfrak{h} rtimes V$ has trivial center, and therefore the adjoint representation is faithful.
– Zvi
yesterday
Yes. This kind of statement always confuses me, I think that "find" means "explicit construct". But I agree with you that Ado's theorem guarantees that this representation exists.
– Matheus Manzatto
yesterday