Closed subgroup of $GL({cal A})$
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I'm trying to prove that for any algebra ${cal A}$, $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}}$ is a closed subgroup of $GL({cal A})$.
I can prove that it is a subgroup just showing that $acirc a^{'-1}$ is in $Aut({cal A})$ which is direct computation. Nevertherless, I don't know how to show that thi group is closed. I've been thinking about proving that it is the isotropy group of $GL({cal A})$, but I don't get it. I tried to impose that $[x, y]$ is the fixed point for the isotropy subgroup, but I just get:
$$a([x, y]) = a(x)a(y) - a(y)a(x) = xy - yx$$
But this means nothing because $a in Aut({cal A})$ is not imposing that $[x, y]$ is fixed.
How do I solve it?
group-theory lie-groups lie-algebras group-isomorphism
asked Nov 19 at 3:36
Vicky
1387
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I'm trying to prove that for any algebra ${cal A}$, $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}}$ is a closed subgroup of $GL({cal A})$.
I can prove that it is a subgroup just showing that $acirc a^{'-1}$ is in $Aut({cal A})$ which is direct computation. Nevertherless, I don't know how to show that thi group is closed. I've been thinking about proving that it is the isotropy group of $GL({cal A})$, but I don't get it. I tried to impose that $[x, y]$ is the fixed point for the isotropy subgroup, but I just get:
$$a([x, y]) = a(x)a(y) - a(y)a(x) = xy - yx$$
But this means nothing because $a in Aut({cal A})$ is not imposing that $[x, y]$ is fixed.
How do I solve it?
group-theory lie-groups lie-algebras group-isomorphism
asked Nov 19 at 3:36
Vicky
1387
First $ain GL(A)$, not $End(A)$. Otherwise you'd have $0in Aut(A)$. For the question, try to exhibit $Aut(A)$ as a zero set of polynomials
– leibnewtz
Nov 19 at 3:44
$a in End({cal A})$ because $a$ is a linear map $a: {cal A} rightarrow {cal A}$, and I don't know what you mean by exhibit the automorphism group as a set of zero polynomials
– Vicky
Nov 19 at 3:48
You need yo specify that $a$ is invertible if that's what you want. A polynomial on a manifold $X$ is a smooth function. So the preimage of zero is closed
– leibnewtz
Nov 19 at 4:22
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to prove that for any algebra ${cal A}$, $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}}$ is a closed subgroup of $GL({cal A})$.
I can prove that it is a subgroup just showing that $acirc a^{'-1}$ is in $Aut({cal A})$ which is direct computation. Nevertherless, I don't know how to show that thi group is closed. I've been thinking about proving that it is the isotropy group of $GL({cal A})$, but I don't get it. I tried to impose that $[x, y]$ is the fixed point for the isotropy subgroup, but I just get:
$$a([x, y]) = a(x)a(y) - a(y)a(x) = xy - yx$$
But this means nothing because $a in Aut({cal A})$ is not imposing that $[x, y]$ is fixed.
How do I solve it?
group-theory lie-groups lie-algebras group-isomorphism
asked Nov 19 at 3:36
Vicky
1387
I'm trying to prove that for any algebra ${cal A}$, $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}}$ is a closed subgroup of $GL({cal A})$.
I can prove that it is a subgroup just showing that $acirc a^{'-1}$ is in $Aut({cal A})$ which is direct computation. Nevertherless, I don't know how to show that thi group is closed. I've been thinking about proving that it is the isotropy group of $GL({cal A})$, but I don't get it. I tried to impose that $[x, y]$ is the fixed point for the isotropy subgroup, but I just get:
$$a([x, y]) = a(x)a(y) - a(y)a(x) = xy - yx$$
But this means nothing because $a in Aut({cal A})$ is not imposing that $[x, y]$ is fixed.
How do I solve it?
group-theory lie-groups lie-algebras group-isomorphism
group-theory lie-groups lie-algebras group-isomorphism
asked Nov 19 at 3:36
Vicky
1387
asked Nov 19 at 3:36
Vicky
1387
edited Nov 19 at 4:17
asked Nov 19 at 3:36
Vicky
1387
asked Nov 19 at 3:36
Vicky
1387
asked Nov 19 at 3:36
Vicky
1387
1387
First $ain GL(A)$, not $End(A)$. Otherwise you'd have $0in Aut(A)$. For the question, try to exhibit $Aut(A)$ as a zero set of polynomials
– leibnewtz
Nov 19 at 3:44
$a in End({cal A})$ because $a$ is a linear map $a: {cal A} rightarrow {cal A}$, and I don't know what you mean by exhibit the automorphism group as a set of zero polynomials
– Vicky
Nov 19 at 3:48
You need yo specify that $a$ is invertible if that's what you want. A polynomial on a manifold $X$ is a smooth function. So the preimage of zero is closed
– leibnewtz
Nov 19 at 4:22
add a comment |
First $ain GL(A)$, not $End(A)$. Otherwise you'd have $0in Aut(A)$. For the question, try to exhibit $Aut(A)$ as a zero set of polynomials
– leibnewtz
Nov 19 at 3:44
$a in End({cal A})$ because $a$ is a linear map $a: {cal A} rightarrow {cal A}$, and I don't know what you mean by exhibit the automorphism group as a set of zero polynomials
– Vicky
Nov 19 at 3:48
You need yo specify that $a$ is invertible if that's what you want. A polynomial on a manifold $X$ is a smooth function. So the preimage of zero is closed
– leibnewtz
Nov 19 at 4:22
First $ain GL(A)$, not $End(A)$. Otherwise you'd have $0in Aut(A)$. For the question, try to exhibit $Aut(A)$ as a zero set of polynomials
– leibnewtz
Nov 19 at 3:44
First $ain GL(A)$, not $End(A)$. Otherwise you'd have $0in Aut(A)$. For the question, try to exhibit $Aut(A)$ as a zero set of polynomials
– leibnewtz
Nov 19 at 3:44
$a in End({cal A})$ because $a$ is a linear map $a: {cal A} rightarrow {cal A}$, and I don't know what you mean by exhibit the automorphism group as a set of zero polynomials
– Vicky
Nov 19 at 3:48
$a in End({cal A})$ because $a$ is a linear map $a: {cal A} rightarrow {cal A}$, and I don't know what you mean by exhibit the automorphism group as a set of zero polynomials
– Vicky
Nov 19 at 3:48
You need yo specify that $a$ is invertible if that's what you want. A polynomial on a manifold $X$ is a smooth function. So the preimage of zero is closed
– leibnewtz
Nov 19 at 4:22
You need yo specify that $a$ is invertible if that's what you want. A polynomial on a manifold $X$ is a smooth function. So the preimage of zero is closed
– leibnewtz
Nov 19 at 4:22
add a comment |
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First $ain GL(A)$, not $End(A)$. Otherwise you'd have $0in Aut(A)$. For the question, try to exhibit $Aut(A)$ as a zero set of polynomials
– leibnewtz
Nov 19 at 3:44
$a in End({cal A})$ because $a$ is a linear map $a: {cal A} rightarrow {cal A}$, and I don't know what you mean by exhibit the automorphism group as a set of zero polynomials
– Vicky
Nov 19 at 3:48
You need yo specify that $a$ is invertible if that's what you want. A polynomial on a manifold $X$ is a smooth function. So the preimage of zero is closed
– leibnewtz
Nov 19 at 4:22