Isometric embedding of a genus g surface The Next CEO of Stack OverflowNash embedding theorem for 2D manifoldsDo symmetric spaces admit isometric embeddings as intersections of quadrics?altering curvature on a tessellation representation of a compact surfaceSymmetries vs. Bound in codimension of Nash isometric embeddingNon-trivial isometric embedding of the standard sphere into $mathbbR^n$?Hilbert's Theorem relevance to positive curvatureIsometry group of a compact hyperbolic surfaceCompact surface with arbitrarily large eigenvalueIsometric embedding for manifolds with conical singularities?Isometric embedding of regular simplex into Riemannian manifoldQuantitative upper bound on mean curvature of an isometric embedding
Isometric embedding of a genus g surface
The Next CEO of Stack OverflowNash embedding theorem for 2D manifoldsDo symmetric spaces admit isometric embeddings as intersections of quadrics?altering curvature on a tessellation representation of a compact surfaceSymmetries vs. Bound in codimension of Nash isometric embeddingNon-trivial isometric embedding of the standard sphere into $mathbbR^n$?Hilbert's Theorem relevance to positive curvatureIsometry group of a compact hyperbolic surfaceCompact surface with arbitrarily large eigenvalueIsometric embedding for manifolds with conical singularities?Isometric embedding of regular simplex into Riemannian manifoldQuantitative upper bound on mean curvature of an isometric embedding
$begingroup$
Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $mathbbR^4?$
dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings
edited Mar 20 at 8:57
Sean Lawton
3,97622248
asked Mar 20 at 8:49
GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA
264
$endgroup$
add a comment |
$begingroup$
Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $mathbbR^4?$
dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings
edited Mar 20 at 8:57
Sean Lawton
3,97622248
asked Mar 20 at 8:49
GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA
264
$endgroup$
$begingroup$
Is there any example or counter example?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:53
4
$begingroup$
Relevant: math.stackexchange.com/questions/1528046/…
$endgroup$
– Aknazar Kazhymurat
Mar 20 at 8:57
$begingroup$
So, is this an open problem?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:59
3
$begingroup$
You might be interested in this question and its answers.
$endgroup$
– Michael Albanese
Mar 20 at 12:00
add a comment |
$begingroup$
Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $mathbbR^4?$
dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings
edited Mar 20 at 8:57
Sean Lawton
3,97622248
asked Mar 20 at 8:49
GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA
264
$endgroup$
Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $mathbbR^4?$
dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings
dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings
edited Mar 20 at 8:57
Sean Lawton
3,97622248
asked Mar 20 at 8:49
GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA
264
edited Mar 20 at 8:57
Sean Lawton
3,97622248
asked Mar 20 at 8:49
GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA
264
edited Mar 20 at 8:57
Sean Lawton
3,97622248
edited Mar 20 at 8:57
Sean Lawton
3,97622248
edited Mar 20 at 8:57
Sean Lawton
3,97622248
3,97622248
asked Mar 20 at 8:49
GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA
264
asked Mar 20 at 8:49
GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA
264
asked Mar 20 at 8:49
GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA
264
264
$begingroup$
Is there any example or counter example?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:53
4
$begingroup$
Relevant: math.stackexchange.com/questions/1528046/…
$endgroup$
– Aknazar Kazhymurat
Mar 20 at 8:57
$begingroup$
So, is this an open problem?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:59
3
$begingroup$
You might be interested in this question and its answers.
$endgroup$
– Michael Albanese
Mar 20 at 12:00
add a comment |
$begingroup$
Is there any example or counter example?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:53
4
$begingroup$
Relevant: math.stackexchange.com/questions/1528046/…
$endgroup$
– Aknazar Kazhymurat
Mar 20 at 8:57
$begingroup$
So, is this an open problem?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:59
3
$begingroup$
You might be interested in this question and its answers.
$endgroup$
– Michael Albanese
Mar 20 at 12:00
$begingroup$
Is there any example or counter example?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:53
$begingroup$
Is there any example or counter example?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:53
4
4
$begingroup$
Relevant: math.stackexchange.com/questions/1528046/…
$endgroup$
– Aknazar Kazhymurat
Mar 20 at 8:57
$begingroup$
Relevant: math.stackexchange.com/questions/1528046/…
$endgroup$
– Aknazar Kazhymurat
Mar 20 at 8:57
$begingroup$
So, is this an open problem?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:59
$begingroup$
So, is this an open problem?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:59
3
3
$begingroup$
You might be interested in this question and its answers.
$endgroup$
– Michael Albanese
Mar 20 at 12:00
$begingroup$
You might be interested in this question and its answers.
$endgroup$
– Michael Albanese
Mar 20 at 12:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbbR^4$).
Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbbR^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbbR^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.
I believe these facts and references may be found in:
Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.
answered Mar 20 at 9:08
Sean LawtonSean Lawton
3,97622248
$endgroup$
1
$begingroup$
There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
$endgroup$
– John Pardon
Mar 20 at 14:30
$begingroup$
@JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
$endgroup$
– Sean Lawton
Mar 20 at 14:33
1
$begingroup$
That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
$endgroup$
– John Pardon
Mar 20 at 14:38
1
$begingroup$
There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
$endgroup$
– John Pardon
Mar 20 at 14:45
1
$begingroup$
@JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
$endgroup$
– Sean Lawton
Mar 20 at 14:48
|
show 1 more comment
$begingroup$
I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^infty$ isometric embedding $V to mathbbR^5$. See Gromov's Partial Differential Relations, pages 298 - 303.
answered Mar 20 at 11:38
Michael AlbaneseMichael Albanese
7,88155393
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbbR^4$).
Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbbR^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbbR^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.
I believe these facts and references may be found in:
Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.
answered Mar 20 at 9:08
Sean LawtonSean Lawton
3,97622248
$endgroup$
1
$begingroup$
There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
$endgroup$
– John Pardon
Mar 20 at 14:30
$begingroup$
@JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
$endgroup$
– Sean Lawton
Mar 20 at 14:33
1
$begingroup$
That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
$endgroup$
– John Pardon
Mar 20 at 14:38
1
$begingroup$
There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
$endgroup$
– John Pardon
Mar 20 at 14:45
1
$begingroup$
@JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
$endgroup$
– Sean Lawton
Mar 20 at 14:48
|
show 1 more comment
$begingroup$
The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbbR^4$).
Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbbR^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbbR^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.
I believe these facts and references may be found in:
Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.
answered Mar 20 at 9:08
Sean LawtonSean Lawton
3,97622248
$endgroup$
1
$begingroup$
There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
$endgroup$
– John Pardon
Mar 20 at 14:30
$begingroup$
@JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
$endgroup$
– Sean Lawton
Mar 20 at 14:33
1
$begingroup$
That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
$endgroup$
– John Pardon
Mar 20 at 14:38
1
$begingroup$
There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
$endgroup$
– John Pardon
Mar 20 at 14:45
1
$begingroup$
@JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
$endgroup$
– Sean Lawton
Mar 20 at 14:48
|
show 1 more comment
$begingroup$
The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbbR^4$).
Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbbR^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbbR^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.
I believe these facts and references may be found in:
Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.
answered Mar 20 at 9:08
Sean LawtonSean Lawton
3,97622248
$endgroup$
The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbbR^4$).
Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbbR^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbbR^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.
I believe these facts and references may be found in:
Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.
answered Mar 20 at 9:08
Sean LawtonSean Lawton
3,97622248
edited Mar 20 at 14:42
answered Mar 20 at 9:08
Sean LawtonSean Lawton
3,97622248
answered Mar 20 at 9:08
Sean LawtonSean Lawton
3,97622248
answered Mar 20 at 9:08
Sean LawtonSean Lawton
3,97622248
3,97622248
1
$begingroup$
There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
$endgroup$
– John Pardon
Mar 20 at 14:30
$begingroup$
@JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
$endgroup$
– Sean Lawton
Mar 20 at 14:33
1
$begingroup$
That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
$endgroup$
– John Pardon
Mar 20 at 14:38
1
$begingroup$
There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
$endgroup$
– John Pardon
Mar 20 at 14:45
1
$begingroup$
@JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
$endgroup$
– Sean Lawton
Mar 20 at 14:48
|
show 1 more comment
1
$begingroup$
There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
$endgroup$
– John Pardon
Mar 20 at 14:30
$begingroup$
@JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
$endgroup$
– Sean Lawton
Mar 20 at 14:33
1
$begingroup$
That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
$endgroup$
– John Pardon
Mar 20 at 14:38
1
$begingroup$
There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
$endgroup$
– John Pardon
Mar 20 at 14:45
1
$begingroup$
@JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
$endgroup$
– Sean Lawton
Mar 20 at 14:48
1
1
$begingroup$
There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
$endgroup$
– John Pardon
Mar 20 at 14:30
$begingroup$
There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
$endgroup$
– John Pardon
Mar 20 at 14:30
$begingroup$
@JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
$endgroup$
– Sean Lawton
Mar 20 at 14:33
$begingroup$
@JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
$endgroup$
– Sean Lawton
Mar 20 at 14:33
1
1
$begingroup$
That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
$endgroup$
– John Pardon
Mar 20 at 14:38
$begingroup$
That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
$endgroup$
– John Pardon
Mar 20 at 14:38
1
1
$begingroup$
There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
$endgroup$
– John Pardon
Mar 20 at 14:45
$begingroup$
There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
$endgroup$
– John Pardon
Mar 20 at 14:45
1
1
$begingroup$
@JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
$endgroup$
– Sean Lawton
Mar 20 at 14:48
$begingroup$
@JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
$endgroup$
– Sean Lawton
Mar 20 at 14:48
|
show 1 more comment
$begingroup$
I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^infty$ isometric embedding $V to mathbbR^5$. See Gromov's Partial Differential Relations, pages 298 - 303.
answered Mar 20 at 11:38
Michael AlbaneseMichael Albanese
7,88155393
$endgroup$
add a comment |
$begingroup$
I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^infty$ isometric embedding $V to mathbbR^5$. See Gromov's Partial Differential Relations, pages 298 - 303.
answered Mar 20 at 11:38
Michael AlbaneseMichael Albanese
7,88155393
$endgroup$
add a comment |
$begingroup$
I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^infty$ isometric embedding $V to mathbbR^5$. See Gromov's Partial Differential Relations, pages 298 - 303.
answered Mar 20 at 11:38
Michael AlbaneseMichael Albanese
7,88155393
$endgroup$
I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^infty$ isometric embedding $V to mathbbR^5$. See Gromov's Partial Differential Relations, pages 298 - 303.
answered Mar 20 at 11:38
Michael AlbaneseMichael Albanese
7,88155393
edited Mar 20 at 11:50
answered Mar 20 at 11:38
Michael AlbaneseMichael Albanese
7,88155393
answered Mar 20 at 11:38
Michael AlbaneseMichael Albanese
7,88155393
answered Mar 20 at 11:38
Michael AlbaneseMichael Albanese
7,88155393
7,88155393
add a comment |
add a comment |
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Is there any example or counter example?
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– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:53
4
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Relevant: math.stackexchange.com/questions/1528046/…
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– Aknazar Kazhymurat
Mar 20 at 8:57
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So, is this an open problem?
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– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:59
3
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You might be interested in this question and its answers.
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– Michael Albanese
Mar 20 at 12:00