Mixing
Bracket of Malcev Closures
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I'm trying to understand why I don't need one assumption in an exercise from Vinberg, Onishchik book on Lie groups, and so I'm asking for help.
The exercise is the following:
Let $mathfraka$ and $mathfrak b$ be subalgebras of a Lie algebra of a Lie group such that $[mathfrak a, mathfrak b] subseteq mathfrak a cap mathfrak b$. Let $mathfrak a^M$ and $mathfrak b^M$ be their Malcev closures (i.e., the smallest Lie subalgebras containing $mathfrak a$ and $mathfrak b$ respectively and such that the corresponding Lie groups are closed). Prove that $[mathfrak a ^M, mathfrak b ^M] = [mathfrak a, mathfrak b]$ (the subalgebra $[mathfrak a, mathfrak b]$ is generated by brackets of the form $[xi, eta]$ where $xi in mathfrak a$ and $eta in mathfrak b$).
Obviously, $[mathfrak a^M, mathfrak b^M] supseteq [mathfrak a, mathfrak b]$. We can rewrite the first bracket as $[mathfrak a^M, mathfrak b^M] = [bigcap_alphaI_alpha, bigcap_beta J_beta]$, where the intersections are from the definition of the Malcev closure; now, if $[xi, beta]$ is a generator of the latter subalgebra, then for all $alpha$ and $beta$ we see that $xi in I_alpha$ and $eta in J_beta$, which means that $[xi,eta] in [I_alpha,J_beta]$. Therefore, $[xi,eta] in bigcap_alpha,beta[I_alpha,J_beta]$. But the last intersection is equal to $[mathfrak a, mathfrak b]$ since $I_alpha supseteq mathfrak a$ and $J_beta supseteq mathfrak b$ for every $alpha$ and $beta$. Thus $[mathfrak a^M,mathfrak b^M] = [mathfrak a,mathfrak b]$.
So, I've not used that $[mathfrak a,mathfrak b] subseteq mathfrak a cap mathfrak b]$.
lie-algebras
asked Jul 26 at 16:45
Grabovsky
134
add a comment |Â
up vote
0
down vote
favorite
I'm trying to understand why I don't need one assumption in an exercise from Vinberg, Onishchik book on Lie groups, and so I'm asking for help.
The exercise is the following:
Let $mathfraka$ and $mathfrak b$ be subalgebras of a Lie algebra of a Lie group such that $[mathfrak a, mathfrak b] subseteq mathfrak a cap mathfrak b$. Let $mathfrak a^M$ and $mathfrak b^M$ be their Malcev closures (i.e., the smallest Lie subalgebras containing $mathfrak a$ and $mathfrak b$ respectively and such that the corresponding Lie groups are closed). Prove that $[mathfrak a ^M, mathfrak b ^M] = [mathfrak a, mathfrak b]$ (the subalgebra $[mathfrak a, mathfrak b]$ is generated by brackets of the form $[xi, eta]$ where $xi in mathfrak a$ and $eta in mathfrak b$).
Obviously, $[mathfrak a^M, mathfrak b^M] supseteq [mathfrak a, mathfrak b]$. We can rewrite the first bracket as $[mathfrak a^M, mathfrak b^M] = [bigcap_alphaI_alpha, bigcap_beta J_beta]$, where the intersections are from the definition of the Malcev closure; now, if $[xi, beta]$ is a generator of the latter subalgebra, then for all $alpha$ and $beta$ we see that $xi in I_alpha$ and $eta in J_beta$, which means that $[xi,eta] in [I_alpha,J_beta]$. Therefore, $[xi,eta] in bigcap_alpha,beta[I_alpha,J_beta]$. But the last intersection is equal to $[mathfrak a, mathfrak b]$ since $I_alpha supseteq mathfrak a$ and $J_beta supseteq mathfrak b$ for every $alpha$ and $beta$. Thus $[mathfrak a^M,mathfrak b^M] = [mathfrak a,mathfrak b]$.
So, I've not used that $[mathfrak a,mathfrak b] subseteq mathfrak a cap mathfrak b]$.
lie-algebras
asked Jul 26 at 16:45
Grabovsky
134
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to understand why I don't need one assumption in an exercise from Vinberg, Onishchik book on Lie groups, and so I'm asking for help.
The exercise is the following:
Let $mathfraka$ and $mathfrak b$ be subalgebras of a Lie algebra of a Lie group such that $[mathfrak a, mathfrak b] subseteq mathfrak a cap mathfrak b$. Let $mathfrak a^M$ and $mathfrak b^M$ be their Malcev closures (i.e., the smallest Lie subalgebras containing $mathfrak a$ and $mathfrak b$ respectively and such that the corresponding Lie groups are closed). Prove that $[mathfrak a ^M, mathfrak b ^M] = [mathfrak a, mathfrak b]$ (the subalgebra $[mathfrak a, mathfrak b]$ is generated by brackets of the form $[xi, eta]$ where $xi in mathfrak a$ and $eta in mathfrak b$).
Obviously, $[mathfrak a^M, mathfrak b^M] supseteq [mathfrak a, mathfrak b]$. We can rewrite the first bracket as $[mathfrak a^M, mathfrak b^M] = [bigcap_alphaI_alpha, bigcap_beta J_beta]$, where the intersections are from the definition of the Malcev closure; now, if $[xi, beta]$ is a generator of the latter subalgebra, then for all $alpha$ and $beta$ we see that $xi in I_alpha$ and $eta in J_beta$, which means that $[xi,eta] in [I_alpha,J_beta]$. Therefore, $[xi,eta] in bigcap_alpha,beta[I_alpha,J_beta]$. But the last intersection is equal to $[mathfrak a, mathfrak b]$ since $I_alpha supseteq mathfrak a$ and $J_beta supseteq mathfrak b$ for every $alpha$ and $beta$. Thus $[mathfrak a^M,mathfrak b^M] = [mathfrak a,mathfrak b]$.
So, I've not used that $[mathfrak a,mathfrak b] subseteq mathfrak a cap mathfrak b]$.
lie-algebras
asked Jul 26 at 16:45
Grabovsky
134
I'm trying to understand why I don't need one assumption in an exercise from Vinberg, Onishchik book on Lie groups, and so I'm asking for help.
The exercise is the following:
Let $mathfraka$ and $mathfrak b$ be subalgebras of a Lie algebra of a Lie group such that $[mathfrak a, mathfrak b] subseteq mathfrak a cap mathfrak b$. Let $mathfrak a^M$ and $mathfrak b^M$ be their Malcev closures (i.e., the smallest Lie subalgebras containing $mathfrak a$ and $mathfrak b$ respectively and such that the corresponding Lie groups are closed). Prove that $[mathfrak a ^M, mathfrak b ^M] = [mathfrak a, mathfrak b]$ (the subalgebra $[mathfrak a, mathfrak b]$ is generated by brackets of the form $[xi, eta]$ where $xi in mathfrak a$ and $eta in mathfrak b$).
Obviously, $[mathfrak a^M, mathfrak b^M] supseteq [mathfrak a, mathfrak b]$. We can rewrite the first bracket as $[mathfrak a^M, mathfrak b^M] = [bigcap_alphaI_alpha, bigcap_beta J_beta]$, where the intersections are from the definition of the Malcev closure; now, if $[xi, beta]$ is a generator of the latter subalgebra, then for all $alpha$ and $beta$ we see that $xi in I_alpha$ and $eta in J_beta$, which means that $[xi,eta] in [I_alpha,J_beta]$. Therefore, $[xi,eta] in bigcap_alpha,beta[I_alpha,J_beta]$. But the last intersection is equal to $[mathfrak a, mathfrak b]$ since $I_alpha supseteq mathfrak a$ and $J_beta supseteq mathfrak b$ for every $alpha$ and $beta$. Thus $[mathfrak a^M,mathfrak b^M] = [mathfrak a,mathfrak b]$.
So, I've not used that $[mathfrak a,mathfrak b] subseteq mathfrak a cap mathfrak b]$.
lie-algebras
asked Jul 26 at 16:45
Grabovsky
134
asked Jul 26 at 16:45
Grabovsky
134
asked Jul 26 at 16:45
Grabovsky
134
asked Jul 26 at 16:45
Grabovsky
134
134
add a comment |Â
add a comment |Â
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