Connected linear Lie groups $G$ generated by $exp (mathfrak{g})$.
0
$begingroup$
Let $G$ be a connected linear Lie group with Lie algebra $mathfrak{g}$.
I understand that any open neighborhood of the identity of $G$ generates it, but, why does $exp(mathfrak{g})$ also generate it? Is $exp(mathfrak{g})$ open?
exponential-function lie-groups lie-algebras
asked Jan 16 at 22:15
fer6268fer6268
1294
$endgroup$
add a comment |
0
$begingroup$
Let $G$ be a connected linear Lie group with Lie algebra $mathfrak{g}$.
I understand that any open neighborhood of the identity of $G$ generates it, but, why does $exp(mathfrak{g})$ also generate it? Is $exp(mathfrak{g})$ open?
exponential-function lie-groups lie-algebras
asked Jan 16 at 22:15
fer6268fer6268
1294
$endgroup$
$begingroup$
$exp$ is a local diffeomorphism at $0$, so it is open around $0$, hence for a neighbourhood $V$ of $0$, $exp(V)$ is a neighbourhood of $1$, which thus generates $G$; therefore so does $exp(mathfrak{g})$
$endgroup$
– Max
Jan 16 at 23:03
$begingroup$
@Max I wrote a comment exactly like yours at first but then I deleted it. I think the statement given by the OP is wrong. See my answer and the comments below it. What do you think?
$endgroup$
– stressed out
Jan 16 at 23:09
$begingroup$
@stressedout : you only proved that $exp (mathfrak{g})$ is not $G$, but that doesn't prove anything, the subgroup generated by $exp(mathfrak{g})$ can (and will in many cases) be strictly larger. Note that $exp(mathfrak{g})$ is not necessarily a subgroup !
$endgroup$
– Max
Jan 16 at 23:12
1
$begingroup$
@stressedout : read the first answer to the post. It begins with "This is not true in general"
$endgroup$
– Max
Jan 16 at 23:14
$begingroup$
@Max You're right. I misread his post. Sorry.
$endgroup$
– stressed out
Jan 16 at 23:19
add a comment |
0
0
0
$begingroup$
Let $G$ be a connected linear Lie group with Lie algebra $mathfrak{g}$.
I understand that any open neighborhood of the identity of $G$ generates it, but, why does $exp(mathfrak{g})$ also generate it? Is $exp(mathfrak{g})$ open?
exponential-function lie-groups lie-algebras
asked Jan 16 at 22:15
fer6268fer6268
1294
$endgroup$
Let $G$ be a connected linear Lie group with Lie algebra $mathfrak{g}$.
I understand that any open neighborhood of the identity of $G$ generates it, but, why does $exp(mathfrak{g})$ also generate it? Is $exp(mathfrak{g})$ open?
exponential-function lie-groups lie-algebras
exponential-function lie-groups lie-algebras
asked Jan 16 at 22:15
fer6268fer6268
1294
asked Jan 16 at 22:15
fer6268fer6268
1294
edited Jan 19 at 6:09
fer6268
asked Jan 16 at 22:15
fer6268fer6268
1294
asked Jan 16 at 22:15
fer6268fer6268
1294
asked Jan 16 at 22:15
fer6268fer6268
1294
1294
$begingroup$
$exp$ is a local diffeomorphism at $0$, so it is open around $0$, hence for a neighbourhood $V$ of $0$, $exp(V)$ is a neighbourhood of $1$, which thus generates $G$; therefore so does $exp(mathfrak{g})$
$endgroup$
– Max
Jan 16 at 23:03
$begingroup$
@Max I wrote a comment exactly like yours at first but then I deleted it. I think the statement given by the OP is wrong. See my answer and the comments below it. What do you think?
$endgroup$
– stressed out
Jan 16 at 23:09
$begingroup$
@stressedout : you only proved that $exp (mathfrak{g})$ is not $G$, but that doesn't prove anything, the subgroup generated by $exp(mathfrak{g})$ can (and will in many cases) be strictly larger. Note that $exp(mathfrak{g})$ is not necessarily a subgroup !
$endgroup$
– Max
Jan 16 at 23:12
1
$begingroup$
@stressedout : read the first answer to the post. It begins with "This is not true in general"
$endgroup$
– Max
Jan 16 at 23:14
$begingroup$
@Max You're right. I misread his post. Sorry.
$endgroup$
– stressed out
Jan 16 at 23:19
add a comment |
$begingroup$
$exp$ is a local diffeomorphism at $0$, so it is open around $0$, hence for a neighbourhood $V$ of $0$, $exp(V)$ is a neighbourhood of $1$, which thus generates $G$; therefore so does $exp(mathfrak{g})$
$endgroup$
– Max
Jan 16 at 23:03
$begingroup$
@Max I wrote a comment exactly like yours at first but then I deleted it. I think the statement given by the OP is wrong. See my answer and the comments below it. What do you think?
$endgroup$
– stressed out
Jan 16 at 23:09
$begingroup$
@stressedout : you only proved that $exp (mathfrak{g})$ is not $G$, but that doesn't prove anything, the subgroup generated by $exp(mathfrak{g})$ can (and will in many cases) be strictly larger. Note that $exp(mathfrak{g})$ is not necessarily a subgroup !
$endgroup$
– Max
Jan 16 at 23:12
1
$begingroup$
@stressedout : read the first answer to the post. It begins with "This is not true in general"
$endgroup$
– Max
Jan 16 at 23:14
$begingroup$
@Max You're right. I misread his post. Sorry.
$endgroup$
– stressed out
Jan 16 at 23:19
$begingroup$
$exp$ is a local diffeomorphism at $0$, so it is open around $0$, hence for a neighbourhood $V$ of $0$, $exp(V)$ is a neighbourhood of $1$, which thus generates $G$; therefore so does $exp(mathfrak{g})$
$endgroup$
– Max
Jan 16 at 23:03
$begingroup$
$exp$ is a local diffeomorphism at $0$, so it is open around $0$, hence for a neighbourhood $V$ of $0$, $exp(V)$ is a neighbourhood of $1$, which thus generates $G$; therefore so does $exp(mathfrak{g})$
$endgroup$
– Max
Jan 16 at 23:03
$begingroup$
@Max I wrote a comment exactly like yours at first but then I deleted it. I think the statement given by the OP is wrong. See my answer and the comments below it. What do you think?
$endgroup$
– stressed out
Jan 16 at 23:09
$begingroup$
@Max I wrote a comment exactly like yours at first but then I deleted it. I think the statement given by the OP is wrong. See my answer and the comments below it. What do you think?
$endgroup$
– stressed out
Jan 16 at 23:09
$begingroup$
@stressedout : you only proved that $exp (mathfrak{g})$ is not $G$, but that doesn't prove anything, the subgroup generated by $exp(mathfrak{g})$ can (and will in many cases) be strictly larger. Note that $exp(mathfrak{g})$ is not necessarily a subgroup !
$endgroup$
– Max
Jan 16 at 23:12
$begingroup$
@stressedout : you only proved that $exp (mathfrak{g})$ is not $G$, but that doesn't prove anything, the subgroup generated by $exp(mathfrak{g})$ can (and will in many cases) be strictly larger. Note that $exp(mathfrak{g})$ is not necessarily a subgroup !
$endgroup$
– Max
Jan 16 at 23:12
1
1
$begingroup$
@stressedout : read the first answer to the post. It begins with "This is not true in general"
$endgroup$
– Max
Jan 16 at 23:14
$begingroup$
@stressedout : read the first answer to the post. It begins with "This is not true in general"
$endgroup$
– Max
Jan 16 at 23:14
$begingroup$
@Max You're right. I misread his post. Sorry.
$endgroup$
– stressed out
Jan 16 at 23:19
$begingroup$
@Max You're right. I misread his post. Sorry.
$endgroup$
– stressed out
Jan 16 at 23:19
add a comment |
1 Answer
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$exp(cdot): mathfrak{g}to G$ is a local diffeomorphism. Hence, it is a local homeomorphism and it's open. This means that $exp(cdot)$ maps an open neighborhood $0in V subseteq mathfrak{g}$ to an open neighborhood around the identity of $G$. Now, $H=langle exp(V)rangle$ is a connected component of $G$ containing the identity. Since $G$ is connected, this subgroup must be $G$.
answered Jan 16 at 22:45
stressed outstressed out
4,9391634
$endgroup$
$begingroup$
By generating I mean that $G = bigcup_{k=1}^{infty}exp(mathfrak{g})^{k}$
$endgroup$
– fer6268
Jan 16 at 22:59
$begingroup$
@ stressed out - I agree with what you just proved, it happens that in the book of JACQUES FARAUT, Analysis on Lie Groups, page 43, Corollary 3.3.5., says that the connected component $G_{0}$ of the identity is generated by $exp(frak{g})$, so we can deduce that if $G$ is connected then it would be generated by $exp(frak{g})$. Is it probable that Fauraut is wrong with the statement $G_{0} = bigcup_{k=1}^{infty}{exp(frak{g})}^k$?
$endgroup$
– fer6268
Jan 17 at 0:09
$begingroup$
@fer6268 Unfortunately, I don't have access to that book right now. But I don't think that he's wrong. Saying that $langle exp(mathfrak{g}) rangle = G_0$ means that $G_0$ must contain all elements of $exp(mathfrak{g})$ and their powers because it's a group under multiplication. In fact, if you remember the previous example where I made a mistake, you need to be able to take square roots. So, you definitely need $exp(mathfrak{g})^2$ to be in $G_0$ and so forth. So, $G_0 = langle exp(mathfrak{g}) rangle = cup_{k=1}^{infty} exp(mathfrak{g})^k$ and I see nothing wrong with it.
$endgroup$
– stressed out
Jan 17 at 6:11
$begingroup$
@fer6268 I checked the book you mentioned. Check the proposition 1.1.1 for more information. But my last comment is also true. Proposition 1.1.1 will tell you why $langle exp(mathfrak{g}) rangle$ is the identity component $G_0$. But the algebraic reason I gave you also tells you why it is equal to the union of the powers of $exp(mathfrak{g})$ and there's no contradiction. Do you still have a question?
$endgroup$
– stressed out
Jan 19 at 5:36
$begingroup$
@ stressed out Thank you for reviewing the statement, and thank you for your algebraic explanations, which helped me understand why the statement is correct. What causes me confusion is that the proposition 1.1.1 says that the CONNECTED NEIGHBORS of identity generate $G_{0}$, and I am not very convinced that $exp (mathfrak{g})$ is a connected neighborhood of identity.
$endgroup$
– fer6268
Jan 19 at 5:48
|
show 3 more comments
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
$exp(cdot): mathfrak{g}to G$ is a local diffeomorphism. Hence, it is a local homeomorphism and it's open. This means that $exp(cdot)$ maps an open neighborhood $0in V subseteq mathfrak{g}$ to an open neighborhood around the identity of $G$. Now, $H=langle exp(V)rangle$ is a connected component of $G$ containing the identity. Since $G$ is connected, this subgroup must be $G$.
answered Jan 16 at 22:45
stressed outstressed out
4,9391634
$endgroup$
$begingroup$
By generating I mean that $G = bigcup_{k=1}^{infty}exp(mathfrak{g})^{k}$
$endgroup$
– fer6268
Jan 16 at 22:59
$begingroup$
@ stressed out - I agree with what you just proved, it happens that in the book of JACQUES FARAUT, Analysis on Lie Groups, page 43, Corollary 3.3.5., says that the connected component $G_{0}$ of the identity is generated by $exp(frak{g})$, so we can deduce that if $G$ is connected then it would be generated by $exp(frak{g})$. Is it probable that Fauraut is wrong with the statement $G_{0} = bigcup_{k=1}^{infty}{exp(frak{g})}^k$?
$endgroup$
– fer6268
Jan 17 at 0:09
$begingroup$
@fer6268 Unfortunately, I don't have access to that book right now. But I don't think that he's wrong. Saying that $langle exp(mathfrak{g}) rangle = G_0$ means that $G_0$ must contain all elements of $exp(mathfrak{g})$ and their powers because it's a group under multiplication. In fact, if you remember the previous example where I made a mistake, you need to be able to take square roots. So, you definitely need $exp(mathfrak{g})^2$ to be in $G_0$ and so forth. So, $G_0 = langle exp(mathfrak{g}) rangle = cup_{k=1}^{infty} exp(mathfrak{g})^k$ and I see nothing wrong with it.
$endgroup$
– stressed out
Jan 17 at 6:11
$begingroup$
@fer6268 I checked the book you mentioned. Check the proposition 1.1.1 for more information. But my last comment is also true. Proposition 1.1.1 will tell you why $langle exp(mathfrak{g}) rangle$ is the identity component $G_0$. But the algebraic reason I gave you also tells you why it is equal to the union of the powers of $exp(mathfrak{g})$ and there's no contradiction. Do you still have a question?
$endgroup$
– stressed out
Jan 19 at 5:36
$begingroup$
@ stressed out Thank you for reviewing the statement, and thank you for your algebraic explanations, which helped me understand why the statement is correct. What causes me confusion is that the proposition 1.1.1 says that the CONNECTED NEIGHBORS of identity generate $G_{0}$, and I am not very convinced that $exp (mathfrak{g})$ is a connected neighborhood of identity.
$endgroup$
– fer6268
Jan 19 at 5:48
|
show 3 more comments
0
$begingroup$
$exp(cdot): mathfrak{g}to G$ is a local diffeomorphism. Hence, it is a local homeomorphism and it's open. This means that $exp(cdot)$ maps an open neighborhood $0in V subseteq mathfrak{g}$ to an open neighborhood around the identity of $G$. Now, $H=langle exp(V)rangle$ is a connected component of $G$ containing the identity. Since $G$ is connected, this subgroup must be $G$.
answered Jan 16 at 22:45
stressed outstressed out
4,9391634
$endgroup$
$begingroup$
By generating I mean that $G = bigcup_{k=1}^{infty}exp(mathfrak{g})^{k}$
$endgroup$
– fer6268
Jan 16 at 22:59
$begingroup$
@ stressed out - I agree with what you just proved, it happens that in the book of JACQUES FARAUT, Analysis on Lie Groups, page 43, Corollary 3.3.5., says that the connected component $G_{0}$ of the identity is generated by $exp(frak{g})$, so we can deduce that if $G$ is connected then it would be generated by $exp(frak{g})$. Is it probable that Fauraut is wrong with the statement $G_{0} = bigcup_{k=1}^{infty}{exp(frak{g})}^k$?
$endgroup$
– fer6268
Jan 17 at 0:09
$begingroup$
@fer6268 Unfortunately, I don't have access to that book right now. But I don't think that he's wrong. Saying that $langle exp(mathfrak{g}) rangle = G_0$ means that $G_0$ must contain all elements of $exp(mathfrak{g})$ and their powers because it's a group under multiplication. In fact, if you remember the previous example where I made a mistake, you need to be able to take square roots. So, you definitely need $exp(mathfrak{g})^2$ to be in $G_0$ and so forth. So, $G_0 = langle exp(mathfrak{g}) rangle = cup_{k=1}^{infty} exp(mathfrak{g})^k$ and I see nothing wrong with it.
$endgroup$
– stressed out
Jan 17 at 6:11
$begingroup$
@fer6268 I checked the book you mentioned. Check the proposition 1.1.1 for more information. But my last comment is also true. Proposition 1.1.1 will tell you why $langle exp(mathfrak{g}) rangle$ is the identity component $G_0$. But the algebraic reason I gave you also tells you why it is equal to the union of the powers of $exp(mathfrak{g})$ and there's no contradiction. Do you still have a question?
$endgroup$
– stressed out
Jan 19 at 5:36
$begingroup$
@ stressed out Thank you for reviewing the statement, and thank you for your algebraic explanations, which helped me understand why the statement is correct. What causes me confusion is that the proposition 1.1.1 says that the CONNECTED NEIGHBORS of identity generate $G_{0}$, and I am not very convinced that $exp (mathfrak{g})$ is a connected neighborhood of identity.
$endgroup$
– fer6268
Jan 19 at 5:48
|
show 3 more comments
0
0
0
$begingroup$
$exp(cdot): mathfrak{g}to G$ is a local diffeomorphism. Hence, it is a local homeomorphism and it's open. This means that $exp(cdot)$ maps an open neighborhood $0in V subseteq mathfrak{g}$ to an open neighborhood around the identity of $G$. Now, $H=langle exp(V)rangle$ is a connected component of $G$ containing the identity. Since $G$ is connected, this subgroup must be $G$.
answered Jan 16 at 22:45
stressed outstressed out
4,9391634
$endgroup$
$exp(cdot): mathfrak{g}to G$ is a local diffeomorphism. Hence, it is a local homeomorphism and it's open. This means that $exp(cdot)$ maps an open neighborhood $0in V subseteq mathfrak{g}$ to an open neighborhood around the identity of $G$. Now, $H=langle exp(V)rangle$ is a connected component of $G$ containing the identity. Since $G$ is connected, this subgroup must be $G$.
answered Jan 16 at 22:45
stressed outstressed out
4,9391634
edited Jan 16 at 23:26
answered Jan 16 at 22:45
stressed outstressed out
4,9391634
answered Jan 16 at 22:45
stressed outstressed out
4,9391634
answered Jan 16 at 22:45
stressed outstressed out
4,9391634
4,9391634
$begingroup$
By generating I mean that $G = bigcup_{k=1}^{infty}exp(mathfrak{g})^{k}$
$endgroup$
– fer6268
Jan 16 at 22:59
$begingroup$
@ stressed out - I agree with what you just proved, it happens that in the book of JACQUES FARAUT, Analysis on Lie Groups, page 43, Corollary 3.3.5., says that the connected component $G_{0}$ of the identity is generated by $exp(frak{g})$, so we can deduce that if $G$ is connected then it would be generated by $exp(frak{g})$. Is it probable that Fauraut is wrong with the statement $G_{0} = bigcup_{k=1}^{infty}{exp(frak{g})}^k$?
$endgroup$
– fer6268
Jan 17 at 0:09
$begingroup$
@fer6268 Unfortunately, I don't have access to that book right now. But I don't think that he's wrong. Saying that $langle exp(mathfrak{g}) rangle = G_0$ means that $G_0$ must contain all elements of $exp(mathfrak{g})$ and their powers because it's a group under multiplication. In fact, if you remember the previous example where I made a mistake, you need to be able to take square roots. So, you definitely need $exp(mathfrak{g})^2$ to be in $G_0$ and so forth. So, $G_0 = langle exp(mathfrak{g}) rangle = cup_{k=1}^{infty} exp(mathfrak{g})^k$ and I see nothing wrong with it.
$endgroup$
– stressed out
Jan 17 at 6:11
$begingroup$
@fer6268 I checked the book you mentioned. Check the proposition 1.1.1 for more information. But my last comment is also true. Proposition 1.1.1 will tell you why $langle exp(mathfrak{g}) rangle$ is the identity component $G_0$. But the algebraic reason I gave you also tells you why it is equal to the union of the powers of $exp(mathfrak{g})$ and there's no contradiction. Do you still have a question?
$endgroup$
– stressed out
Jan 19 at 5:36
$begingroup$
@ stressed out Thank you for reviewing the statement, and thank you for your algebraic explanations, which helped me understand why the statement is correct. What causes me confusion is that the proposition 1.1.1 says that the CONNECTED NEIGHBORS of identity generate $G_{0}$, and I am not very convinced that $exp (mathfrak{g})$ is a connected neighborhood of identity.
$endgroup$
– fer6268
Jan 19 at 5:48
|
show 3 more comments
$begingroup$
By generating I mean that $G = bigcup_{k=1}^{infty}exp(mathfrak{g})^{k}$
$endgroup$
– fer6268
Jan 16 at 22:59
$begingroup$
@ stressed out - I agree with what you just proved, it happens that in the book of JACQUES FARAUT, Analysis on Lie Groups, page 43, Corollary 3.3.5., says that the connected component $G_{0}$ of the identity is generated by $exp(frak{g})$, so we can deduce that if $G$ is connected then it would be generated by $exp(frak{g})$. Is it probable that Fauraut is wrong with the statement $G_{0} = bigcup_{k=1}^{infty}{exp(frak{g})}^k$?
$endgroup$
– fer6268
Jan 17 at 0:09
$begingroup$
@fer6268 Unfortunately, I don't have access to that book right now. But I don't think that he's wrong. Saying that $langle exp(mathfrak{g}) rangle = G_0$ means that $G_0$ must contain all elements of $exp(mathfrak{g})$ and their powers because it's a group under multiplication. In fact, if you remember the previous example where I made a mistake, you need to be able to take square roots. So, you definitely need $exp(mathfrak{g})^2$ to be in $G_0$ and so forth. So, $G_0 = langle exp(mathfrak{g}) rangle = cup_{k=1}^{infty} exp(mathfrak{g})^k$ and I see nothing wrong with it.
$endgroup$
– stressed out
Jan 17 at 6:11
$begingroup$
@fer6268 I checked the book you mentioned. Check the proposition 1.1.1 for more information. But my last comment is also true. Proposition 1.1.1 will tell you why $langle exp(mathfrak{g}) rangle$ is the identity component $G_0$. But the algebraic reason I gave you also tells you why it is equal to the union of the powers of $exp(mathfrak{g})$ and there's no contradiction. Do you still have a question?
$endgroup$
– stressed out
Jan 19 at 5:36
$begingroup$
@ stressed out Thank you for reviewing the statement, and thank you for your algebraic explanations, which helped me understand why the statement is correct. What causes me confusion is that the proposition 1.1.1 says that the CONNECTED NEIGHBORS of identity generate $G_{0}$, and I am not very convinced that $exp (mathfrak{g})$ is a connected neighborhood of identity.
$endgroup$
– fer6268
Jan 19 at 5:48
$begingroup$
By generating I mean that $G = bigcup_{k=1}^{infty}exp(mathfrak{g})^{k}$
$endgroup$
– fer6268
Jan 16 at 22:59
$begingroup$
By generating I mean that $G = bigcup_{k=1}^{infty}exp(mathfrak{g})^{k}$
$endgroup$
– fer6268
Jan 16 at 22:59
$begingroup$
@ stressed out - I agree with what you just proved, it happens that in the book of JACQUES FARAUT, Analysis on Lie Groups, page 43, Corollary 3.3.5., says that the connected component $G_{0}$ of the identity is generated by $exp(frak{g})$, so we can deduce that if $G$ is connected then it would be generated by $exp(frak{g})$. Is it probable that Fauraut is wrong with the statement $G_{0} = bigcup_{k=1}^{infty}{exp(frak{g})}^k$?
$endgroup$
– fer6268
Jan 17 at 0:09
$begingroup$
@ stressed out - I agree with what you just proved, it happens that in the book of JACQUES FARAUT, Analysis on Lie Groups, page 43, Corollary 3.3.5., says that the connected component $G_{0}$ of the identity is generated by $exp(frak{g})$, so we can deduce that if $G$ is connected then it would be generated by $exp(frak{g})$. Is it probable that Fauraut is wrong with the statement $G_{0} = bigcup_{k=1}^{infty}{exp(frak{g})}^k$?
$endgroup$
– fer6268
Jan 17 at 0:09
$begingroup$
@fer6268 Unfortunately, I don't have access to that book right now. But I don't think that he's wrong. Saying that $langle exp(mathfrak{g}) rangle = G_0$ means that $G_0$ must contain all elements of $exp(mathfrak{g})$ and their powers because it's a group under multiplication. In fact, if you remember the previous example where I made a mistake, you need to be able to take square roots. So, you definitely need $exp(mathfrak{g})^2$ to be in $G_0$ and so forth. So, $G_0 = langle exp(mathfrak{g}) rangle = cup_{k=1}^{infty} exp(mathfrak{g})^k$ and I see nothing wrong with it.
$endgroup$
– stressed out
Jan 17 at 6:11
$begingroup$
@fer6268 Unfortunately, I don't have access to that book right now. But I don't think that he's wrong. Saying that $langle exp(mathfrak{g}) rangle = G_0$ means that $G_0$ must contain all elements of $exp(mathfrak{g})$ and their powers because it's a group under multiplication. In fact, if you remember the previous example where I made a mistake, you need to be able to take square roots. So, you definitely need $exp(mathfrak{g})^2$ to be in $G_0$ and so forth. So, $G_0 = langle exp(mathfrak{g}) rangle = cup_{k=1}^{infty} exp(mathfrak{g})^k$ and I see nothing wrong with it.
$endgroup$
– stressed out
Jan 17 at 6:11
$begingroup$
@fer6268 I checked the book you mentioned. Check the proposition 1.1.1 for more information. But my last comment is also true. Proposition 1.1.1 will tell you why $langle exp(mathfrak{g}) rangle$ is the identity component $G_0$. But the algebraic reason I gave you also tells you why it is equal to the union of the powers of $exp(mathfrak{g})$ and there's no contradiction. Do you still have a question?
$endgroup$
– stressed out
Jan 19 at 5:36
$begingroup$
@fer6268 I checked the book you mentioned. Check the proposition 1.1.1 for more information. But my last comment is also true. Proposition 1.1.1 will tell you why $langle exp(mathfrak{g}) rangle$ is the identity component $G_0$. But the algebraic reason I gave you also tells you why it is equal to the union of the powers of $exp(mathfrak{g})$ and there's no contradiction. Do you still have a question?
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– stressed out
Jan 19 at 5:36
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@ stressed out Thank you for reviewing the statement, and thank you for your algebraic explanations, which helped me understand why the statement is correct. What causes me confusion is that the proposition 1.1.1 says that the CONNECTED NEIGHBORS of identity generate $G_{0}$, and I am not very convinced that $exp (mathfrak{g})$ is a connected neighborhood of identity.
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– fer6268
Jan 19 at 5:48
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@ stressed out Thank you for reviewing the statement, and thank you for your algebraic explanations, which helped me understand why the statement is correct. What causes me confusion is that the proposition 1.1.1 says that the CONNECTED NEIGHBORS of identity generate $G_{0}$, and I am not very convinced that $exp (mathfrak{g})$ is a connected neighborhood of identity.
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– fer6268
Jan 19 at 5:48
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$exp$ is a local diffeomorphism at $0$, so it is open around $0$, hence for a neighbourhood $V$ of $0$, $exp(V)$ is a neighbourhood of $1$, which thus generates $G$; therefore so does $exp(mathfrak{g})$
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– Max
Jan 16 at 23:03
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@Max I wrote a comment exactly like yours at first but then I deleted it. I think the statement given by the OP is wrong. See my answer and the comments below it. What do you think?
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– stressed out
Jan 16 at 23:09
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@stressedout : you only proved that $exp (mathfrak{g})$ is not $G$, but that doesn't prove anything, the subgroup generated by $exp(mathfrak{g})$ can (and will in many cases) be strictly larger. Note that $exp(mathfrak{g})$ is not necessarily a subgroup !
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– Max
Jan 16 at 23:12
1
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@stressedout : read the first answer to the post. It begins with "This is not true in general"
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– Max
Jan 16 at 23:14
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@Max You're right. I misread his post. Sorry.
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– stressed out
Jan 16 at 23:19