Set of invertible elements in Banach algebra
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Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.
I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) subseteq G_0(A)$.
$B(e;1) subseteq G_0(A)$ follows from $$(e-x)^{-1} = sum_{n=0}^{infty} x^n $$
But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.
abstract-algebra functional-analysis
asked Dec 6 '18 at 23:08
ig97ig97
155
$endgroup$
add a comment |
$begingroup$
Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.
I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) subseteq G_0(A)$.
$B(e;1) subseteq G_0(A)$ follows from $$(e-x)^{-1} = sum_{n=0}^{infty} x^n $$
But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.
abstract-algebra functional-analysis
asked Dec 6 '18 at 23:08
ig97ig97
155
$endgroup$
3
$begingroup$
Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
$endgroup$
– KCd
Dec 6 '18 at 23:13
add a comment |
$begingroup$
Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.
I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) subseteq G_0(A)$.
$B(e;1) subseteq G_0(A)$ follows from $$(e-x)^{-1} = sum_{n=0}^{infty} x^n $$
But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.
abstract-algebra functional-analysis
asked Dec 6 '18 at 23:08
ig97ig97
155
$endgroup$
Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.
I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) subseteq G_0(A)$.
$B(e;1) subseteq G_0(A)$ follows from $$(e-x)^{-1} = sum_{n=0}^{infty} x^n $$
But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.
abstract-algebra functional-analysis
abstract-algebra functional-analysis
asked Dec 6 '18 at 23:08
ig97ig97
155
asked Dec 6 '18 at 23:08
ig97ig97
155
edited Dec 6 '18 at 23:26
ig97
asked Dec 6 '18 at 23:08
ig97ig97
155
asked Dec 6 '18 at 23:08
ig97ig97
155
asked Dec 6 '18 at 23:08
ig97ig97
155
155
3
$begingroup$
Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
$endgroup$
– KCd
Dec 6 '18 at 23:13
add a comment |
3
$begingroup$
Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
$endgroup$
– KCd
Dec 6 '18 at 23:13
3
3
$begingroup$
Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
$endgroup$
– KCd
Dec 6 '18 at 23:13
$begingroup$
Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
$endgroup$
– KCd
Dec 6 '18 at 23:13
add a comment |
1 Answer
1
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This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.
answered Dec 6 '18 at 23:22
Tsemo AristideTsemo Aristide
59.1k11445
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$begingroup$
This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.
answered Dec 6 '18 at 23:22
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
add a comment |
$begingroup$
This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.
answered Dec 6 '18 at 23:22
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
add a comment |
$begingroup$
This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.
answered Dec 6 '18 at 23:22
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.
answered Dec 6 '18 at 23:22
Tsemo AristideTsemo Aristide
59.1k11445
answered Dec 6 '18 at 23:22
Tsemo AristideTsemo Aristide
59.1k11445
answered Dec 6 '18 at 23:22
Tsemo AristideTsemo Aristide
59.1k11445
answered Dec 6 '18 at 23:22
Tsemo AristideTsemo Aristide
59.1k11445
59.1k11445
add a comment |
add a comment |
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Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
$endgroup$
– KCd
Dec 6 '18 at 23:13