Timeline for Is $H$ closed in $G$?
Current License: CC BY-SA 4.0
18 events
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| Oct 15, 2019 at 23:10 | history | edited | YCor | CC BY-SA 4.0 |
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| Oct 15, 2019 at 23:05 | history | edited | Born to be proud | CC BY-SA 4.0 |
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| Oct 7, 2019 at 21:54 | comment | added | Francois Ziegler | Oh! great, that completely cleans it up. And I guess (by e.g. your posted answer) we learn that such an $H$ always has discrete induced structure. | |
| Oct 7, 2019 at 21:38 | comment | added | YCor | @FrancoisZiegler Take in $\mathbf{R}\times\mathbf{R}$, $N=\mathbf{R}\times\{0\}$ and $H$ the graph of a non-continuous group isomorphism $\mathbf{R}\to\mathbf{R}$. | |
| Oct 7, 2019 at 21:31 | comment | added | Francois Ziegler | @YCor Given the closed normal $N\subset G$, I’d say the natural generality is $H=$ arbitrary subgroup with its induced Lie group structure (Bourbaki Chap. III §4 nº5) (and still $NH=G$, $N\cap H=\{e\}$). In that setting I wonder if $\sigma$-compactness is even needed to exclude non-closed $H$...? | |
| Oct 7, 2019 at 19:07 | vote | accept | Born to be proud | ||
| Oct 7, 2019 at 14:34 | answer | added | YCor | timeline score: 7 | |
| Oct 7, 2019 at 14:26 | comment | added | YCor | Oh indeed, I didn't see the condition $H\cap N=\{e\}$. Indeed in this case the answer is yes. | |
| Oct 7, 2019 at 12:49 | comment | added | Vincent | @Ycor I really like your example of a non-closed Lie subgroup - it is easy to visualize and insightful. However I don't believe it satisfies the condition $H \cap N = \{e\}$ from the original post. Doesn't the line intersect the circle infinitely many times? | |
| Oct 7, 2019 at 12:29 | comment | added | Born to be proud | @YCor math.stackexchange.com/questions/3384118/… | |
| Oct 7, 2019 at 11:54 | history | edited | Born to be proud | CC BY-SA 4.0 |
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| Oct 7, 2019 at 11:52 | comment | added | Born to be proud | @YCor Every smooth manifold is assumed to be second countable. | |
| Oct 7, 2019 at 7:30 | review | Close votes | |||
| Oct 13, 2019 at 3:32 | |||||
| Oct 7, 2019 at 7:12 | comment | added | YCor | In any case, take $G$ to be the 2-torus, $N$ a closed circle subgroup, $H$ a dense line. Then $NH=G$ and $H$ is not closed. This is the simplest example of a non-closed connected immersed Lie subgroup... | |
| Oct 7, 2019 at 7:10 | comment | added | YCor | Hence, according to the definition, every subgroup $H$ of every Lie group $G$ is a Lie subgroup (being the image in $G$ of $H$ endowed with the discrete topology). Since your question does not refer to the additional structure on $H$, it's just the same as assuming that $H$ is an arbitrary subgroup. | |
| Oct 6, 2019 at 21:29 | history | edited | Born to be proud | CC BY-SA 4.0 |
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| Oct 6, 2019 at 20:57 | comment | added | YCor | Can you define "Lie subgroup"? Some authors mean by this a closed subgroup. Do you mean the image of connected Lie group by a continuous homomorphism? or something weaker without connectedness? | |
| Oct 6, 2019 at 20:37 | history | asked | Born to be proud | CC BY-SA 4.0 |