Mixing
Irreducibility of the derived representation
$begingroup$
Let $G$ a linear Lie group with lie algebra $mathfrak{g}$.
If $(pi, mathcal{H})$ is a irreducible representation of $G$. Does the irreducibility of $pi$ imply the irreducibility of derived representation $dpi$?
$$dpi:mathfrak{g}longrightarrow End(mathcal{H}),,,
Xlongmapsto frac{d}{dt}Bigr|_{t=0}pi(exp(tX))$$
representation-theory lie-groups lie-algebras
asked Jan 18 at 20:57
fer6268fer6268
1294
$endgroup$
add a comment |
$begingroup$
Let $G$ a linear Lie group with lie algebra $mathfrak{g}$.
If $(pi, mathcal{H})$ is a irreducible representation of $G$. Does the irreducibility of $pi$ imply the irreducibility of derived representation $dpi$?
$$dpi:mathfrak{g}longrightarrow End(mathcal{H}),,,
Xlongmapsto frac{d}{dt}Bigr|_{t=0}pi(exp(tX))$$
representation-theory lie-groups lie-algebras
asked Jan 18 at 20:57
fer6268fer6268
1294
$endgroup$
$begingroup$
Rephrase this as a question.
$endgroup$
– J.F
Jan 18 at 21:04
add a comment |
$begingroup$
Let $G$ a linear Lie group with lie algebra $mathfrak{g}$.
If $(pi, mathcal{H})$ is a irreducible representation of $G$. Does the irreducibility of $pi$ imply the irreducibility of derived representation $dpi$?
$$dpi:mathfrak{g}longrightarrow End(mathcal{H}),,,
Xlongmapsto frac{d}{dt}Bigr|_{t=0}pi(exp(tX))$$
representation-theory lie-groups lie-algebras
asked Jan 18 at 20:57
fer6268fer6268
1294
$endgroup$
Let $G$ a linear Lie group with lie algebra $mathfrak{g}$.
If $(pi, mathcal{H})$ is a irreducible representation of $G$. Does the irreducibility of $pi$ imply the irreducibility of derived representation $dpi$?
$$dpi:mathfrak{g}longrightarrow End(mathcal{H}),,,
Xlongmapsto frac{d}{dt}Bigr|_{t=0}pi(exp(tX))$$
representation-theory lie-groups lie-algebras
representation-theory lie-groups lie-algebras
asked Jan 18 at 20:57
fer6268fer6268
1294
asked Jan 18 at 20:57
fer6268fer6268
1294
edited Jan 18 at 21:05
fer6268
asked Jan 18 at 20:57
fer6268fer6268
1294
asked Jan 18 at 20:57
fer6268fer6268
1294
asked Jan 18 at 20:57
fer6268fer6268
1294
1294
$begingroup$
Rephrase this as a question.
$endgroup$
– J.F
Jan 18 at 21:04
add a comment |
$begingroup$
Rephrase this as a question.
$endgroup$
– J.F
Jan 18 at 21:04
$begingroup$
Rephrase this as a question.
$endgroup$
– J.F
Jan 18 at 21:04
$begingroup$
Rephrase this as a question.
$endgroup$
– J.F
Jan 18 at 21:04
add a comment |
1 Answer
1
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$begingroup$
If all vectors in $pi$ are differentiable, then yes. It suffices to show that a closed subspace $V subset mathcal{H}$ which is invariant under $G$ is also invariant under $mathfrak{g}$. Let $v in V$ and let $X in mathfrak{g}$. Then by $G$-invariance of $W$, we have $t^{-1}(pi(exp{tX})v-v) in V$ for all $t in mathbb{R}^{times}$. Since $V$ is assumed to by closed, the limit as $t rightarrow 0$ also belongs to $V$.
If $G$ is conected and $mathcal{H}$ is finite dimensional a (necessarily closed) subsapce $W$ of $mathcal{H}$ which is $mathfrak{g}$-invariant, is also $G$-invaraint.
answered Jan 18 at 21:21
m.sm.s
1,324313
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
If all vectors in $pi$ are differentiable, then yes. It suffices to show that a closed subspace $V subset mathcal{H}$ which is invariant under $G$ is also invariant under $mathfrak{g}$. Let $v in V$ and let $X in mathfrak{g}$. Then by $G$-invariance of $W$, we have $t^{-1}(pi(exp{tX})v-v) in V$ for all $t in mathbb{R}^{times}$. Since $V$ is assumed to by closed, the limit as $t rightarrow 0$ also belongs to $V$.
If $G$ is conected and $mathcal{H}$ is finite dimensional a (necessarily closed) subsapce $W$ of $mathcal{H}$ which is $mathfrak{g}$-invariant, is also $G$-invaraint.
answered Jan 18 at 21:21
m.sm.s
1,324313
$endgroup$
add a comment |
$begingroup$
If all vectors in $pi$ are differentiable, then yes. It suffices to show that a closed subspace $V subset mathcal{H}$ which is invariant under $G$ is also invariant under $mathfrak{g}$. Let $v in V$ and let $X in mathfrak{g}$. Then by $G$-invariance of $W$, we have $t^{-1}(pi(exp{tX})v-v) in V$ for all $t in mathbb{R}^{times}$. Since $V$ is assumed to by closed, the limit as $t rightarrow 0$ also belongs to $V$.
If $G$ is conected and $mathcal{H}$ is finite dimensional a (necessarily closed) subsapce $W$ of $mathcal{H}$ which is $mathfrak{g}$-invariant, is also $G$-invaraint.
answered Jan 18 at 21:21
m.sm.s
1,324313
$endgroup$
add a comment |
$begingroup$
If all vectors in $pi$ are differentiable, then yes. It suffices to show that a closed subspace $V subset mathcal{H}$ which is invariant under $G$ is also invariant under $mathfrak{g}$. Let $v in V$ and let $X in mathfrak{g}$. Then by $G$-invariance of $W$, we have $t^{-1}(pi(exp{tX})v-v) in V$ for all $t in mathbb{R}^{times}$. Since $V$ is assumed to by closed, the limit as $t rightarrow 0$ also belongs to $V$.
If $G$ is conected and $mathcal{H}$ is finite dimensional a (necessarily closed) subsapce $W$ of $mathcal{H}$ which is $mathfrak{g}$-invariant, is also $G$-invaraint.
answered Jan 18 at 21:21
m.sm.s
1,324313
$endgroup$
If all vectors in $pi$ are differentiable, then yes. It suffices to show that a closed subspace $V subset mathcal{H}$ which is invariant under $G$ is also invariant under $mathfrak{g}$. Let $v in V$ and let $X in mathfrak{g}$. Then by $G$-invariance of $W$, we have $t^{-1}(pi(exp{tX})v-v) in V$ for all $t in mathbb{R}^{times}$. Since $V$ is assumed to by closed, the limit as $t rightarrow 0$ also belongs to $V$.
If $G$ is conected and $mathcal{H}$ is finite dimensional a (necessarily closed) subsapce $W$ of $mathcal{H}$ which is $mathfrak{g}$-invariant, is also $G$-invaraint.
answered Jan 18 at 21:21
m.sm.s
1,324313
answered Jan 18 at 21:21
m.sm.s
1,324313
answered Jan 18 at 21:21
m.sm.s
1,324313
answered Jan 18 at 21:21
m.sm.s
1,324313
1,324313
add a comment |
add a comment |
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$begingroup$
Rephrase this as a question.
$endgroup$
– J.F
Jan 18 at 21:04