Symplectic connections are (locally) Levi-Civita connections
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I was wondering... Is every symplectic connection $nabla$
on some symplectic manifold $(M,ω)$ the Levi-Civita connection of some metric $g$ on $M$? What about the local statement?
riemannian-geometry sg.symplectic-geometry connections
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up vote
7
down vote
favorite
I was wondering... Is every symplectic connection $nabla$
on some symplectic manifold $(M,ω)$ the Levi-Civita connection of some metric $g$ on $M$? What about the local statement?
riemannian-geometry sg.symplectic-geometry connections
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I was wondering... Is every symplectic connection $nabla$
on some symplectic manifold $(M,ω)$ the Levi-Civita connection of some metric $g$ on $M$? What about the local statement?
riemannian-geometry sg.symplectic-geometry connections
I was wondering... Is every symplectic connection $nabla$
on some symplectic manifold $(M,ω)$ the Levi-Civita connection of some metric $g$ on $M$? What about the local statement?
riemannian-geometry sg.symplectic-geometry connections
riemannian-geometry sg.symplectic-geometry connections
asked Nov 26 at 21:56
Valentino
835
835
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2 Answers
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10
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For any Riemannian metric $g$ on a symplectic manifold $(M, omega)$ there exists (canonical) almost complex structure $J$ making these threetensors compatible (any one can be defined by other 2). compatible with $omega$, which means that $omega(_, J_)$ defines a Riemannian metric $widetildeg$. For such a compatible triple and the Levi-Civita connection $widetildenabla$ the equation $widetildenabla omega = 0$ is equivalent to $J$ being integrable thus making $(M,widetildeg,omega, J)$ into Kähler manifold. There is plenty of symplectic manifolds which are not Kähler. So if there is metric $g$ for a non-Kähler Fedosov manifold $(M, omega, nabla)$ whose Levi-Civita connection is $nabla$ then it cannot be compatible with $omega.$
Oh, the question was so dumb! The answer for the local query is affirmative and is an immediate consequence of Darboux's theorem. Thank you, Vít Tuček!
– Valentino
Nov 26 at 23:39
Havin an easy answer doesn't make a question dumb. I think it's a good question and it's answer requires some nontrivial theorems.
– Vít Tuček
Nov 27 at 11:32
Wait... I was so excited that I didn't realize a lack of understanding of mine: why is it true that any Riemannian metric is the metric of some compatible triple?
– Valentino
Nov 27 at 13:40
See here: math.stackexchange.com/questions/2084282/…
– Paul Bryan
Nov 30 at 4:55
@Valentino I don't know! The answer is wrong as it stands as the metric $g$ is in general NOT the compatible metric for $J$ and $omega$. I will revise the answer.
– Vít Tuček
Nov 30 at 10:45
|
show 1 more comment
up vote
1
down vote
accepted
I think I've got the answer, and it is "no", at least for the global question.
When I started to try to understand the problem, I realized that to get some structure that is invariant by the connection we can (have to) fix the structure at some point and use parallel transport through picewise smooth paths to extend it. The resulting structure will be well-defined iff the initially fixed structure is invariant by the holonomy group at the point. The idea is then to construct some connection whose holonomy group at some point preserves some symplectic structure on the tangent space of the point but does not preserve any metric on such tanget space. To get the second feature, we may try to construct the connection in such a way that its holonomy group is unbounded. After some tries, I got the following:
On $mathbbR^2$, we consider the connection $nabla$ given by beginalign* nabla_fracpartialpartial xfracpartialpartial x & = yfracpartialpartial x & nabla_fracpartialpartial xfracpartialpartial y & = 0 \
nabla_fracpartialpartial yfracpartialpartial y & = xfracpartialpartial y & nabla_fracpartialpartial yfracpartialpartial x & = 0.
endalign*
The curvature $R$ of $nabla$ at $(0,0)$ is given by $R(fracpartialpartial x, fracpartialpartial y) = beginbmatrix
-1 & 0 \
0 & 1
endbmatrix in GL(mathbbR^2) = GL(T_(0,0)mathbbR^2)$, and all its covariant derivatives are of the form $beginbmatrix
-c & 0 \
0 & c
endbmatrix$ at such point. Because of that and of the connectedness of $Hol(nabla, (0,0))$ (simply-connectedness of $mathbbR^2$), Ambrose-Singer's theorem implies that the canonical symplectic form of $T_(0,0)mathbbR^2 = mathbbR^2$ is invariant by $Hol(nabla, (0,0))$. On the other hand, $Hol(nabla, (0,0))$ contains the exponential of $tR(fracpartialpartial x, fracpartialpartial y)$ for every $t in mathbbR$, and is therefore an unbounded subset of $GL(T_(0,0)mathbbR^2)$. As such, it doesn't fix any metric on $T_(0,0)mathbbR^2$.
$ textitedit:$ using Levi-Civita's formula, it is easy to conclude that the connection $nabla$ defined above is not the Levi-Civita connection of any metric, even locally.
With that in mind, Darboux's theorem shows us that for every symplectic manifold there exist some open subset $U$ of the manifold and some symplectic connection $nabla$ defined on $U$ such that $nabla$ is not the Levi-Civita connection of any metric on $U$.
By using partitions of unity, we conclude, then, that $textbfevery symplectic manifold admits$ $textbfa symplectic connection that is not a$ $textbfLevi-Civita connection.$
New contributor
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
For any Riemannian metric $g$ on a symplectic manifold $(M, omega)$ there exists (canonical) almost complex structure $J$ making these threetensors compatible (any one can be defined by other 2). compatible with $omega$, which means that $omega(_, J_)$ defines a Riemannian metric $widetildeg$. For such a compatible triple and the Levi-Civita connection $widetildenabla$ the equation $widetildenabla omega = 0$ is equivalent to $J$ being integrable thus making $(M,widetildeg,omega, J)$ into Kähler manifold. There is plenty of symplectic manifolds which are not Kähler. So if there is metric $g$ for a non-Kähler Fedosov manifold $(M, omega, nabla)$ whose Levi-Civita connection is $nabla$ then it cannot be compatible with $omega.$
Oh, the question was so dumb! The answer for the local query is affirmative and is an immediate consequence of Darboux's theorem. Thank you, Vít Tuček!
– Valentino
Nov 26 at 23:39
Havin an easy answer doesn't make a question dumb. I think it's a good question and it's answer requires some nontrivial theorems.
– Vít Tuček
Nov 27 at 11:32
Wait... I was so excited that I didn't realize a lack of understanding of mine: why is it true that any Riemannian metric is the metric of some compatible triple?
– Valentino
Nov 27 at 13:40
See here: math.stackexchange.com/questions/2084282/…
– Paul Bryan
Nov 30 at 4:55
@Valentino I don't know! The answer is wrong as it stands as the metric $g$ is in general NOT the compatible metric for $J$ and $omega$. I will revise the answer.
– Vít Tuček
Nov 30 at 10:45
|
show 1 more comment
up vote
10
down vote
For any Riemannian metric $g$ on a symplectic manifold $(M, omega)$ there exists (canonical) almost complex structure $J$ making these threetensors compatible (any one can be defined by other 2). compatible with $omega$, which means that $omega(_, J_)$ defines a Riemannian metric $widetildeg$. For such a compatible triple and the Levi-Civita connection $widetildenabla$ the equation $widetildenabla omega = 0$ is equivalent to $J$ being integrable thus making $(M,widetildeg,omega, J)$ into Kähler manifold. There is plenty of symplectic manifolds which are not Kähler. So if there is metric $g$ for a non-Kähler Fedosov manifold $(M, omega, nabla)$ whose Levi-Civita connection is $nabla$ then it cannot be compatible with $omega.$
Oh, the question was so dumb! The answer for the local query is affirmative and is an immediate consequence of Darboux's theorem. Thank you, Vít Tuček!
– Valentino
Nov 26 at 23:39
Havin an easy answer doesn't make a question dumb. I think it's a good question and it's answer requires some nontrivial theorems.
– Vít Tuček
Nov 27 at 11:32
Wait... I was so excited that I didn't realize a lack of understanding of mine: why is it true that any Riemannian metric is the metric of some compatible triple?
– Valentino
Nov 27 at 13:40
See here: math.stackexchange.com/questions/2084282/…
– Paul Bryan
Nov 30 at 4:55
@Valentino I don't know! The answer is wrong as it stands as the metric $g$ is in general NOT the compatible metric for $J$ and $omega$. I will revise the answer.
– Vít Tuček
Nov 30 at 10:45
|
show 1 more comment
up vote
10
down vote
up vote
10
down vote
For any Riemannian metric $g$ on a symplectic manifold $(M, omega)$ there exists (canonical) almost complex structure $J$ making these threetensors compatible (any one can be defined by other 2). compatible with $omega$, which means that $omega(_, J_)$ defines a Riemannian metric $widetildeg$. For such a compatible triple and the Levi-Civita connection $widetildenabla$ the equation $widetildenabla omega = 0$ is equivalent to $J$ being integrable thus making $(M,widetildeg,omega, J)$ into Kähler manifold. There is plenty of symplectic manifolds which are not Kähler. So if there is metric $g$ for a non-Kähler Fedosov manifold $(M, omega, nabla)$ whose Levi-Civita connection is $nabla$ then it cannot be compatible with $omega.$
For any Riemannian metric $g$ on a symplectic manifold $(M, omega)$ there exists (canonical) almost complex structure $J$ making these threetensors compatible (any one can be defined by other 2). compatible with $omega$, which means that $omega(_, J_)$ defines a Riemannian metric $widetildeg$. For such a compatible triple and the Levi-Civita connection $widetildenabla$ the equation $widetildenabla omega = 0$ is equivalent to $J$ being integrable thus making $(M,widetildeg,omega, J)$ into Kähler manifold. There is plenty of symplectic manifolds which are not Kähler. So if there is metric $g$ for a non-Kähler Fedosov manifold $(M, omega, nabla)$ whose Levi-Civita connection is $nabla$ then it cannot be compatible with $omega.$
edited 14 hours ago
answered Nov 26 at 22:31
Vít Tuček
4,92911748
4,92911748
Oh, the question was so dumb! The answer for the local query is affirmative and is an immediate consequence of Darboux's theorem. Thank you, Vít Tuček!
– Valentino
Nov 26 at 23:39
Havin an easy answer doesn't make a question dumb. I think it's a good question and it's answer requires some nontrivial theorems.
– Vít Tuček
Nov 27 at 11:32
Wait... I was so excited that I didn't realize a lack of understanding of mine: why is it true that any Riemannian metric is the metric of some compatible triple?
– Valentino
Nov 27 at 13:40
See here: math.stackexchange.com/questions/2084282/…
– Paul Bryan
Nov 30 at 4:55
@Valentino I don't know! The answer is wrong as it stands as the metric $g$ is in general NOT the compatible metric for $J$ and $omega$. I will revise the answer.
– Vít Tuček
Nov 30 at 10:45
|
show 1 more comment
Oh, the question was so dumb! The answer for the local query is affirmative and is an immediate consequence of Darboux's theorem. Thank you, Vít Tuček!
– Valentino
Nov 26 at 23:39
Havin an easy answer doesn't make a question dumb. I think it's a good question and it's answer requires some nontrivial theorems.
– Vít Tuček
Nov 27 at 11:32
Wait... I was so excited that I didn't realize a lack of understanding of mine: why is it true that any Riemannian metric is the metric of some compatible triple?
– Valentino
Nov 27 at 13:40
See here: math.stackexchange.com/questions/2084282/…
– Paul Bryan
Nov 30 at 4:55
@Valentino I don't know! The answer is wrong as it stands as the metric $g$ is in general NOT the compatible metric for $J$ and $omega$. I will revise the answer.
– Vít Tuček
Nov 30 at 10:45
Oh, the question was so dumb! The answer for the local query is affirmative and is an immediate consequence of Darboux's theorem. Thank you, Vít Tuček!
– Valentino
Nov 26 at 23:39
Oh, the question was so dumb! The answer for the local query is affirmative and is an immediate consequence of Darboux's theorem. Thank you, Vít Tuček!
– Valentino
Nov 26 at 23:39
Havin an easy answer doesn't make a question dumb. I think it's a good question and it's answer requires some nontrivial theorems.
– Vít Tuček
Nov 27 at 11:32
Havin an easy answer doesn't make a question dumb. I think it's a good question and it's answer requires some nontrivial theorems.
– Vít Tuček
Nov 27 at 11:32
Wait... I was so excited that I didn't realize a lack of understanding of mine: why is it true that any Riemannian metric is the metric of some compatible triple?
– Valentino
Nov 27 at 13:40
Wait... I was so excited that I didn't realize a lack of understanding of mine: why is it true that any Riemannian metric is the metric of some compatible triple?
– Valentino
Nov 27 at 13:40
See here: math.stackexchange.com/questions/2084282/…
– Paul Bryan
Nov 30 at 4:55
See here: math.stackexchange.com/questions/2084282/…
– Paul Bryan
Nov 30 at 4:55
@Valentino I don't know! The answer is wrong as it stands as the metric $g$ is in general NOT the compatible metric for $J$ and $omega$. I will revise the answer.
– Vít Tuček
Nov 30 at 10:45
@Valentino I don't know! The answer is wrong as it stands as the metric $g$ is in general NOT the compatible metric for $J$ and $omega$. I will revise the answer.
– Vít Tuček
Nov 30 at 10:45
|
show 1 more comment
up vote
1
down vote
accepted
I think I've got the answer, and it is "no", at least for the global question.
When I started to try to understand the problem, I realized that to get some structure that is invariant by the connection we can (have to) fix the structure at some point and use parallel transport through picewise smooth paths to extend it. The resulting structure will be well-defined iff the initially fixed structure is invariant by the holonomy group at the point. The idea is then to construct some connection whose holonomy group at some point preserves some symplectic structure on the tangent space of the point but does not preserve any metric on such tanget space. To get the second feature, we may try to construct the connection in such a way that its holonomy group is unbounded. After some tries, I got the following:
On $mathbbR^2$, we consider the connection $nabla$ given by beginalign* nabla_fracpartialpartial xfracpartialpartial x & = yfracpartialpartial x & nabla_fracpartialpartial xfracpartialpartial y & = 0 \
nabla_fracpartialpartial yfracpartialpartial y & = xfracpartialpartial y & nabla_fracpartialpartial yfracpartialpartial x & = 0.
endalign*
The curvature $R$ of $nabla$ at $(0,0)$ is given by $R(fracpartialpartial x, fracpartialpartial y) = beginbmatrix
-1 & 0 \
0 & 1
endbmatrix in GL(mathbbR^2) = GL(T_(0,0)mathbbR^2)$, and all its covariant derivatives are of the form $beginbmatrix
-c & 0 \
0 & c
endbmatrix$ at such point. Because of that and of the connectedness of $Hol(nabla, (0,0))$ (simply-connectedness of $mathbbR^2$), Ambrose-Singer's theorem implies that the canonical symplectic form of $T_(0,0)mathbbR^2 = mathbbR^2$ is invariant by $Hol(nabla, (0,0))$. On the other hand, $Hol(nabla, (0,0))$ contains the exponential of $tR(fracpartialpartial x, fracpartialpartial y)$ for every $t in mathbbR$, and is therefore an unbounded subset of $GL(T_(0,0)mathbbR^2)$. As such, it doesn't fix any metric on $T_(0,0)mathbbR^2$.
$ textitedit:$ using Levi-Civita's formula, it is easy to conclude that the connection $nabla$ defined above is not the Levi-Civita connection of any metric, even locally.
With that in mind, Darboux's theorem shows us that for every symplectic manifold there exist some open subset $U$ of the manifold and some symplectic connection $nabla$ defined on $U$ such that $nabla$ is not the Levi-Civita connection of any metric on $U$.
By using partitions of unity, we conclude, then, that $textbfevery symplectic manifold admits$ $textbfa symplectic connection that is not a$ $textbfLevi-Civita connection.$
New contributor
add a comment |
up vote
1
down vote
accepted
I think I've got the answer, and it is "no", at least for the global question.
When I started to try to understand the problem, I realized that to get some structure that is invariant by the connection we can (have to) fix the structure at some point and use parallel transport through picewise smooth paths to extend it. The resulting structure will be well-defined iff the initially fixed structure is invariant by the holonomy group at the point. The idea is then to construct some connection whose holonomy group at some point preserves some symplectic structure on the tangent space of the point but does not preserve any metric on such tanget space. To get the second feature, we may try to construct the connection in such a way that its holonomy group is unbounded. After some tries, I got the following:
On $mathbbR^2$, we consider the connection $nabla$ given by beginalign* nabla_fracpartialpartial xfracpartialpartial x & = yfracpartialpartial x & nabla_fracpartialpartial xfracpartialpartial y & = 0 \
nabla_fracpartialpartial yfracpartialpartial y & = xfracpartialpartial y & nabla_fracpartialpartial yfracpartialpartial x & = 0.
endalign*
The curvature $R$ of $nabla$ at $(0,0)$ is given by $R(fracpartialpartial x, fracpartialpartial y) = beginbmatrix
-1 & 0 \
0 & 1
endbmatrix in GL(mathbbR^2) = GL(T_(0,0)mathbbR^2)$, and all its covariant derivatives are of the form $beginbmatrix
-c & 0 \
0 & c
endbmatrix$ at such point. Because of that and of the connectedness of $Hol(nabla, (0,0))$ (simply-connectedness of $mathbbR^2$), Ambrose-Singer's theorem implies that the canonical symplectic form of $T_(0,0)mathbbR^2 = mathbbR^2$ is invariant by $Hol(nabla, (0,0))$. On the other hand, $Hol(nabla, (0,0))$ contains the exponential of $tR(fracpartialpartial x, fracpartialpartial y)$ for every $t in mathbbR$, and is therefore an unbounded subset of $GL(T_(0,0)mathbbR^2)$. As such, it doesn't fix any metric on $T_(0,0)mathbbR^2$.
$ textitedit:$ using Levi-Civita's formula, it is easy to conclude that the connection $nabla$ defined above is not the Levi-Civita connection of any metric, even locally.
With that in mind, Darboux's theorem shows us that for every symplectic manifold there exist some open subset $U$ of the manifold and some symplectic connection $nabla$ defined on $U$ such that $nabla$ is not the Levi-Civita connection of any metric on $U$.
By using partitions of unity, we conclude, then, that $textbfevery symplectic manifold admits$ $textbfa symplectic connection that is not a$ $textbfLevi-Civita connection.$
New contributor
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think I've got the answer, and it is "no", at least for the global question.
When I started to try to understand the problem, I realized that to get some structure that is invariant by the connection we can (have to) fix the structure at some point and use parallel transport through picewise smooth paths to extend it. The resulting structure will be well-defined iff the initially fixed structure is invariant by the holonomy group at the point. The idea is then to construct some connection whose holonomy group at some point preserves some symplectic structure on the tangent space of the point but does not preserve any metric on such tanget space. To get the second feature, we may try to construct the connection in such a way that its holonomy group is unbounded. After some tries, I got the following:
On $mathbbR^2$, we consider the connection $nabla$ given by beginalign* nabla_fracpartialpartial xfracpartialpartial x & = yfracpartialpartial x & nabla_fracpartialpartial xfracpartialpartial y & = 0 \
nabla_fracpartialpartial yfracpartialpartial y & = xfracpartialpartial y & nabla_fracpartialpartial yfracpartialpartial x & = 0.
endalign*
The curvature $R$ of $nabla$ at $(0,0)$ is given by $R(fracpartialpartial x, fracpartialpartial y) = beginbmatrix
-1 & 0 \
0 & 1
endbmatrix in GL(mathbbR^2) = GL(T_(0,0)mathbbR^2)$, and all its covariant derivatives are of the form $beginbmatrix
-c & 0 \
0 & c
endbmatrix$ at such point. Because of that and of the connectedness of $Hol(nabla, (0,0))$ (simply-connectedness of $mathbbR^2$), Ambrose-Singer's theorem implies that the canonical symplectic form of $T_(0,0)mathbbR^2 = mathbbR^2$ is invariant by $Hol(nabla, (0,0))$. On the other hand, $Hol(nabla, (0,0))$ contains the exponential of $tR(fracpartialpartial x, fracpartialpartial y)$ for every $t in mathbbR$, and is therefore an unbounded subset of $GL(T_(0,0)mathbbR^2)$. As such, it doesn't fix any metric on $T_(0,0)mathbbR^2$.
$ textitedit:$ using Levi-Civita's formula, it is easy to conclude that the connection $nabla$ defined above is not the Levi-Civita connection of any metric, even locally.
With that in mind, Darboux's theorem shows us that for every symplectic manifold there exist some open subset $U$ of the manifold and some symplectic connection $nabla$ defined on $U$ such that $nabla$ is not the Levi-Civita connection of any metric on $U$.
By using partitions of unity, we conclude, then, that $textbfevery symplectic manifold admits$ $textbfa symplectic connection that is not a$ $textbfLevi-Civita connection.$
New contributor
I think I've got the answer, and it is "no", at least for the global question.
When I started to try to understand the problem, I realized that to get some structure that is invariant by the connection we can (have to) fix the structure at some point and use parallel transport through picewise smooth paths to extend it. The resulting structure will be well-defined iff the initially fixed structure is invariant by the holonomy group at the point. The idea is then to construct some connection whose holonomy group at some point preserves some symplectic structure on the tangent space of the point but does not preserve any metric on such tanget space. To get the second feature, we may try to construct the connection in such a way that its holonomy group is unbounded. After some tries, I got the following:
On $mathbbR^2$, we consider the connection $nabla$ given by beginalign* nabla_fracpartialpartial xfracpartialpartial x & = yfracpartialpartial x & nabla_fracpartialpartial xfracpartialpartial y & = 0 \
nabla_fracpartialpartial yfracpartialpartial y & = xfracpartialpartial y & nabla_fracpartialpartial yfracpartialpartial x & = 0.
endalign*
The curvature $R$ of $nabla$ at $(0,0)$ is given by $R(fracpartialpartial x, fracpartialpartial y) = beginbmatrix
-1 & 0 \
0 & 1
endbmatrix in GL(mathbbR^2) = GL(T_(0,0)mathbbR^2)$, and all its covariant derivatives are of the form $beginbmatrix
-c & 0 \
0 & c
endbmatrix$ at such point. Because of that and of the connectedness of $Hol(nabla, (0,0))$ (simply-connectedness of $mathbbR^2$), Ambrose-Singer's theorem implies that the canonical symplectic form of $T_(0,0)mathbbR^2 = mathbbR^2$ is invariant by $Hol(nabla, (0,0))$. On the other hand, $Hol(nabla, (0,0))$ contains the exponential of $tR(fracpartialpartial x, fracpartialpartial y)$ for every $t in mathbbR$, and is therefore an unbounded subset of $GL(T_(0,0)mathbbR^2)$. As such, it doesn't fix any metric on $T_(0,0)mathbbR^2$.
$ textitedit:$ using Levi-Civita's formula, it is easy to conclude that the connection $nabla$ defined above is not the Levi-Civita connection of any metric, even locally.
With that in mind, Darboux's theorem shows us that for every symplectic manifold there exist some open subset $U$ of the manifold and some symplectic connection $nabla$ defined on $U$ such that $nabla$ is not the Levi-Civita connection of any metric on $U$.
By using partitions of unity, we conclude, then, that $textbfevery symplectic manifold admits$ $textbfa symplectic connection that is not a$ $textbfLevi-Civita connection.$
New contributor
edited Dec 1 at 16:33
New contributor
answered Dec 1 at 4:07
Valentino
835
835
New contributor
New contributor
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