Is the projection from $mathbb R^n$ locally isometric to the flat torus?
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I’m reading a problem asking to introduce a Riemannian metric on $T^n$ in such a way that the projection $pi(x_1,...,x_n)=(exp(ix_1),...,exp(ix_n))$ from $mathbb R^n$ to $T^n$ is a local isometry. Does the flat metric apply here?
By definition, flat torus is defined as the product of n copies of the Riemannian manifold $S^1$ with its metric induced by $mathbb R$.
differential-geometry riemannian-geometry smooth-manifolds
asked Dec 13 '18 at 15:12
User12239User12239
453216
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0
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I’m reading a problem asking to introduce a Riemannian metric on $T^n$ in such a way that the projection $pi(x_1,...,x_n)=(exp(ix_1),...,exp(ix_n))$ from $mathbb R^n$ to $T^n$ is a local isometry. Does the flat metric apply here?
By definition, flat torus is defined as the product of n copies of the Riemannian manifold $S^1$ with its metric induced by $mathbb R$.
differential-geometry riemannian-geometry smooth-manifolds
asked Dec 13 '18 at 15:12
User12239User12239
453216
$endgroup$
$begingroup$
Maybe you ought to clarify what the projection $pi$ you refer to is?
$endgroup$
– T_M
Dec 13 '18 at 15:16
$begingroup$
I editted my question @T_M
$endgroup$
– User12239
Dec 13 '18 at 15:19
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isn't the flat metric exactly defined via this property? related to this: math.stackexchange.com/questions/450110/…
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– Balou
Dec 13 '18 at 15:19
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I’m reading DoCarmo’s book and the definition there is different @Balou
$endgroup$
– User12239
Dec 13 '18 at 15:23
$begingroup$
maybe you should state then how "the flat metric" is defined in DoCarmo
$endgroup$
– Balou
Dec 13 '18 at 15:29
|
show 2 more comments
0
0
0
$begingroup$
I’m reading a problem asking to introduce a Riemannian metric on $T^n$ in such a way that the projection $pi(x_1,...,x_n)=(exp(ix_1),...,exp(ix_n))$ from $mathbb R^n$ to $T^n$ is a local isometry. Does the flat metric apply here?
By definition, flat torus is defined as the product of n copies of the Riemannian manifold $S^1$ with its metric induced by $mathbb R$.
differential-geometry riemannian-geometry smooth-manifolds
asked Dec 13 '18 at 15:12
User12239User12239
453216
$endgroup$
I’m reading a problem asking to introduce a Riemannian metric on $T^n$ in such a way that the projection $pi(x_1,...,x_n)=(exp(ix_1),...,exp(ix_n))$ from $mathbb R^n$ to $T^n$ is a local isometry. Does the flat metric apply here?
By definition, flat torus is defined as the product of n copies of the Riemannian manifold $S^1$ with its metric induced by $mathbb R$.
differential-geometry riemannian-geometry smooth-manifolds
differential-geometry riemannian-geometry smooth-manifolds
asked Dec 13 '18 at 15:12
User12239User12239
453216
asked Dec 13 '18 at 15:12
User12239User12239
453216
edited Dec 13 '18 at 15:38
User12239
asked Dec 13 '18 at 15:12
User12239User12239
453216
asked Dec 13 '18 at 15:12
User12239User12239
453216
asked Dec 13 '18 at 15:12
User12239User12239
453216
453216
$begingroup$
Maybe you ought to clarify what the projection $pi$ you refer to is?
$endgroup$
– T_M
Dec 13 '18 at 15:16
$begingroup$
I editted my question @T_M
$endgroup$
– User12239
Dec 13 '18 at 15:19
$begingroup$
isn't the flat metric exactly defined via this property? related to this: math.stackexchange.com/questions/450110/…
$endgroup$
– Balou
Dec 13 '18 at 15:19
$begingroup$
I’m reading DoCarmo’s book and the definition there is different @Balou
$endgroup$
– User12239
Dec 13 '18 at 15:23
$begingroup$
maybe you should state then how "the flat metric" is defined in DoCarmo
$endgroup$
– Balou
Dec 13 '18 at 15:29
|
show 2 more comments
$begingroup$
Maybe you ought to clarify what the projection $pi$ you refer to is?
$endgroup$
– T_M
Dec 13 '18 at 15:16
$begingroup$
I editted my question @T_M
$endgroup$
– User12239
Dec 13 '18 at 15:19
$begingroup$
isn't the flat metric exactly defined via this property? related to this: math.stackexchange.com/questions/450110/…
$endgroup$
– Balou
Dec 13 '18 at 15:19
$begingroup$
I’m reading DoCarmo’s book and the definition there is different @Balou
$endgroup$
– User12239
Dec 13 '18 at 15:23
$begingroup$
maybe you should state then how "the flat metric" is defined in DoCarmo
$endgroup$
– Balou
Dec 13 '18 at 15:29
$begingroup$
Maybe you ought to clarify what the projection $pi$ you refer to is?
$endgroup$
– T_M
Dec 13 '18 at 15:16
$begingroup$
Maybe you ought to clarify what the projection $pi$ you refer to is?
$endgroup$
– T_M
Dec 13 '18 at 15:16
$begingroup$
I editted my question @T_M
$endgroup$
– User12239
Dec 13 '18 at 15:19
$begingroup$
I editted my question @T_M
$endgroup$
– User12239
Dec 13 '18 at 15:19
$begingroup$
isn't the flat metric exactly defined via this property? related to this: math.stackexchange.com/questions/450110/…
$endgroup$
– Balou
Dec 13 '18 at 15:19
$begingroup$
isn't the flat metric exactly defined via this property? related to this: math.stackexchange.com/questions/450110/…
$endgroup$
– Balou
Dec 13 '18 at 15:19
$begingroup$
I’m reading DoCarmo’s book and the definition there is different @Balou
$endgroup$
– User12239
Dec 13 '18 at 15:23
$begingroup$
I’m reading DoCarmo’s book and the definition there is different @Balou
$endgroup$
– User12239
Dec 13 '18 at 15:23
$begingroup$
maybe you should state then how "the flat metric" is defined in DoCarmo
$endgroup$
– Balou
Dec 13 '18 at 15:29
$begingroup$
maybe you should state then how "the flat metric" is defined in DoCarmo
$endgroup$
– Balou
Dec 13 '18 at 15:29
|
show 2 more comments
1 Answer
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Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.
answered Dec 13 '18 at 15:15
T_MT_M
1,12827
$endgroup$
$begingroup$
Why does DoCarmo following this question want to prove that the torus with the metric introduced is isometric to the flat torus? Is it not redundant?
$endgroup$
– User12239
Dec 13 '18 at 15:17
$begingroup$
I imagine it's just building familiarity with the rigorous definitions. Just because you know it's true, it certainly doesn't mean that its a pointless exercise
$endgroup$
– T_M
Dec 13 '18 at 19:08
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
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1
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Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.
answered Dec 13 '18 at 15:15
T_MT_M
1,12827
$endgroup$
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Why does DoCarmo following this question want to prove that the torus with the metric introduced is isometric to the flat torus? Is it not redundant?
$endgroup$
– User12239
Dec 13 '18 at 15:17
$begingroup$
I imagine it's just building familiarity with the rigorous definitions. Just because you know it's true, it certainly doesn't mean that its a pointless exercise
$endgroup$
– T_M
Dec 13 '18 at 19:08
add a comment |
1
$begingroup$
Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.
answered Dec 13 '18 at 15:15
T_MT_M
1,12827
$endgroup$
$begingroup$
Why does DoCarmo following this question want to prove that the torus with the metric introduced is isometric to the flat torus? Is it not redundant?
$endgroup$
– User12239
Dec 13 '18 at 15:17
$begingroup$
I imagine it's just building familiarity with the rigorous definitions. Just because you know it's true, it certainly doesn't mean that its a pointless exercise
$endgroup$
– T_M
Dec 13 '18 at 19:08
add a comment |
1
1
1
$begingroup$
Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.
answered Dec 13 '18 at 15:15
T_MT_M
1,12827
$endgroup$
Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.
answered Dec 13 '18 at 15:15
T_MT_M
1,12827
answered Dec 13 '18 at 15:15
T_MT_M
1,12827
answered Dec 13 '18 at 15:15
T_MT_M
1,12827
answered Dec 13 '18 at 15:15
T_MT_M
1,12827
1,12827
$begingroup$
Why does DoCarmo following this question want to prove that the torus with the metric introduced is isometric to the flat torus? Is it not redundant?
$endgroup$
– User12239
Dec 13 '18 at 15:17
$begingroup$
I imagine it's just building familiarity with the rigorous definitions. Just because you know it's true, it certainly doesn't mean that its a pointless exercise
$endgroup$
– T_M
Dec 13 '18 at 19:08
add a comment |
$begingroup$
Why does DoCarmo following this question want to prove that the torus with the metric introduced is isometric to the flat torus? Is it not redundant?
$endgroup$
– User12239
Dec 13 '18 at 15:17
$begingroup$
I imagine it's just building familiarity with the rigorous definitions. Just because you know it's true, it certainly doesn't mean that its a pointless exercise
$endgroup$
– T_M
Dec 13 '18 at 19:08
$begingroup$
Why does DoCarmo following this question want to prove that the torus with the metric introduced is isometric to the flat torus? Is it not redundant?
$endgroup$
– User12239
Dec 13 '18 at 15:17
$begingroup$
Why does DoCarmo following this question want to prove that the torus with the metric introduced is isometric to the flat torus? Is it not redundant?
$endgroup$
– User12239
Dec 13 '18 at 15:17
$begingroup$
I imagine it's just building familiarity with the rigorous definitions. Just because you know it's true, it certainly doesn't mean that its a pointless exercise
$endgroup$
– T_M
Dec 13 '18 at 19:08
$begingroup$
I imagine it's just building familiarity with the rigorous definitions. Just because you know it's true, it certainly doesn't mean that its a pointless exercise
$endgroup$
– T_M
Dec 13 '18 at 19:08
add a comment |
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$begingroup$
Maybe you ought to clarify what the projection $pi$ you refer to is?
$endgroup$
– T_M
Dec 13 '18 at 15:16
$begingroup$
I editted my question @T_M
$endgroup$
– User12239
Dec 13 '18 at 15:19
$begingroup$
isn't the flat metric exactly defined via this property? related to this: math.stackexchange.com/questions/450110/…
$endgroup$
– Balou
Dec 13 '18 at 15:19
$begingroup$
I’m reading DoCarmo’s book and the definition there is different @Balou
$endgroup$
– User12239
Dec 13 '18 at 15:23
$begingroup$
maybe you should state then how "the flat metric" is defined in DoCarmo
$endgroup$
– Balou
Dec 13 '18 at 15:29