The idea behind Delzant construction of a toric manifold from a convex polytope
$begingroup$
I am trying to understand how to visualize a symplectic toric manifold from its moment polytope, following chapter 29.4 in "Lectures on Symplectic Geometry" by Ana Cannas da Silva: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf.
The explanation on page 188 is what interests me most (and what I understand the least). Here is a copy of that part of the text:
We can visualize $(M_Δ, ω_Δ, mathbb{T}^n, mu)$
from $Δ$ as follows. First take the product $mathbb T^n times Δ$. Let $p$ lie in the interior of $mathbb T^n times Δ$. The tangent space at $p$ is $mathbb R^n × (mathbb R^n)^∗$. Define $ω_p$ by:
$$ω_p(v, ξ) = ξ(v) = −ω_p(ξ, v),quad text{and}quad ω_p(v, v′) = ω(ξ, ξ′) = 0$$
for all $v, v′ in mathbb R^n$ and $ξ, ξ′ in (mathbb R^n)^∗$. Then $ω$ is a closed nondegenerate $2$-form on the interior of $mathbb T^n times Δ$.
At the corner there are directions missing in $(mathbb R^n)^∗$, so $ω$ is
a degenerate pairing. Hence, we need to eliminate the corresponding directions in
$mathbb R^n$. To do this, we collapse the orbits corresponding to subgroups of $mathbb T^n$ generated
by directions orthogonal to the annihilator of that face.
My questions are:
Is the idea of this construction just to build any symplectic toric manifold so that $Δ$ is the orbit space? If this is the case, I suppose we could conclude that this manifold we built must be $M_Δ$ by uniqueness in the Delzant correspondence?
Does this construction have anything to do with the construction of $M_Δ$ from the proof of the theorem in chapters 29.1-29.3? That construction is briefly described in this question Delzant theorem for polyhedra, and I understand it step by step, but I don't see whether it has anything to do with the visualisation above.
I also don't understand the boldface sentences in the text:
The definition of $omega$? How can we plug in two vectors from $mathbb R^n$, when the second argument of the function must be from the dual? I.e. why is it skew symmetric?
There are directions missing at the corners? Can you help me visualise this in the case $n=1$ and $Δ=[-1,1]$?
How do we see what are the orbits that we need to collapse? And why is the quotient a manifold? Is there some argument here that I'm missing? Normally, one has to be careful to obtain a manifold by passing to the quotient. In the actual proof mentioned in $2)$, symplectic reduction is used to justify that the quotient there is indeed a symplectic manifold. Is this where a connection to that construction comes in?
Answers to any of the questions would be much appreciated!
differential-geometry symplectic-geometry polytopes toric-geometry
edited Dec 10 '18 at 13:12
s.harp
8,67012250
asked Dec 9 '18 at 15:03
Layer CakeLayer Cake
292111
$endgroup$
add a comment |
$begingroup$
I am trying to understand how to visualize a symplectic toric manifold from its moment polytope, following chapter 29.4 in "Lectures on Symplectic Geometry" by Ana Cannas da Silva: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf.
The explanation on page 188 is what interests me most (and what I understand the least). Here is a copy of that part of the text:
We can visualize $(M_Δ, ω_Δ, mathbb{T}^n, mu)$
from $Δ$ as follows. First take the product $mathbb T^n times Δ$. Let $p$ lie in the interior of $mathbb T^n times Δ$. The tangent space at $p$ is $mathbb R^n × (mathbb R^n)^∗$. Define $ω_p$ by:
$$ω_p(v, ξ) = ξ(v) = −ω_p(ξ, v),quad text{and}quad ω_p(v, v′) = ω(ξ, ξ′) = 0$$
for all $v, v′ in mathbb R^n$ and $ξ, ξ′ in (mathbb R^n)^∗$. Then $ω$ is a closed nondegenerate $2$-form on the interior of $mathbb T^n times Δ$.
At the corner there are directions missing in $(mathbb R^n)^∗$, so $ω$ is
a degenerate pairing. Hence, we need to eliminate the corresponding directions in
$mathbb R^n$. To do this, we collapse the orbits corresponding to subgroups of $mathbb T^n$ generated
by directions orthogonal to the annihilator of that face.
My questions are:
Is the idea of this construction just to build any symplectic toric manifold so that $Δ$ is the orbit space? If this is the case, I suppose we could conclude that this manifold we built must be $M_Δ$ by uniqueness in the Delzant correspondence?
Does this construction have anything to do with the construction of $M_Δ$ from the proof of the theorem in chapters 29.1-29.3? That construction is briefly described in this question Delzant theorem for polyhedra, and I understand it step by step, but I don't see whether it has anything to do with the visualisation above.
I also don't understand the boldface sentences in the text:
The definition of $omega$? How can we plug in two vectors from $mathbb R^n$, when the second argument of the function must be from the dual? I.e. why is it skew symmetric?
There are directions missing at the corners? Can you help me visualise this in the case $n=1$ and $Δ=[-1,1]$?
How do we see what are the orbits that we need to collapse? And why is the quotient a manifold? Is there some argument here that I'm missing? Normally, one has to be careful to obtain a manifold by passing to the quotient. In the actual proof mentioned in $2)$, symplectic reduction is used to justify that the quotient there is indeed a symplectic manifold. Is this where a connection to that construction comes in?
Answers to any of the questions would be much appreciated!
differential-geometry symplectic-geometry polytopes toric-geometry
edited Dec 10 '18 at 13:12
s.harp
8,67012250
asked Dec 9 '18 at 15:03
Layer CakeLayer Cake
292111
$endgroup$
3
$begingroup$
Q1.1 Yes, for maximal effective Hamiltonian toric actions on a symplectic manifolds, the moment map polytope turns out to be in bijection with the orbit space. Q1.2 Yes, but the Delzant correspondence is established by understanding this 'visualization' of symplectic toric manifold. Q3. If must view $mathbb{R}^n times Delta subset mathbb{R}^n times mathbb{R^n}^* cong T^*mathbb{R}^n$ equipped with its standard symplectic form $omega((v_1, xi_1), (v_2, xi_2)) = xi_1(v_2) - xi_2(v_1)$.
$endgroup$
– Jordan Payette
Dec 11 '18 at 21:54
2
$begingroup$
Q4. I'd prefer to speak of the boundary $mathbb{R}^n times partial Delta$ instead of corners. The boundary of the image of the moment map exists because, on this boundary, the inward-outward direction(s) is (are) no longer in the image of the differential of the moment map; these are the missing directions. As for questions 2 and 5, you are on the right track, but they would require a detailed answer (and some time...) and not mere comments.
$endgroup$
– Jordan Payette
Dec 11 '18 at 22:01
add a comment |
$begingroup$
I am trying to understand how to visualize a symplectic toric manifold from its moment polytope, following chapter 29.4 in "Lectures on Symplectic Geometry" by Ana Cannas da Silva: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf.
The explanation on page 188 is what interests me most (and what I understand the least). Here is a copy of that part of the text:
We can visualize $(M_Δ, ω_Δ, mathbb{T}^n, mu)$
from $Δ$ as follows. First take the product $mathbb T^n times Δ$. Let $p$ lie in the interior of $mathbb T^n times Δ$. The tangent space at $p$ is $mathbb R^n × (mathbb R^n)^∗$. Define $ω_p$ by:
$$ω_p(v, ξ) = ξ(v) = −ω_p(ξ, v),quad text{and}quad ω_p(v, v′) = ω(ξ, ξ′) = 0$$
for all $v, v′ in mathbb R^n$ and $ξ, ξ′ in (mathbb R^n)^∗$. Then $ω$ is a closed nondegenerate $2$-form on the interior of $mathbb T^n times Δ$.
At the corner there are directions missing in $(mathbb R^n)^∗$, so $ω$ is
a degenerate pairing. Hence, we need to eliminate the corresponding directions in
$mathbb R^n$. To do this, we collapse the orbits corresponding to subgroups of $mathbb T^n$ generated
by directions orthogonal to the annihilator of that face.
My questions are:
Is the idea of this construction just to build any symplectic toric manifold so that $Δ$ is the orbit space? If this is the case, I suppose we could conclude that this manifold we built must be $M_Δ$ by uniqueness in the Delzant correspondence?
Does this construction have anything to do with the construction of $M_Δ$ from the proof of the theorem in chapters 29.1-29.3? That construction is briefly described in this question Delzant theorem for polyhedra, and I understand it step by step, but I don't see whether it has anything to do with the visualisation above.
I also don't understand the boldface sentences in the text:
The definition of $omega$? How can we plug in two vectors from $mathbb R^n$, when the second argument of the function must be from the dual? I.e. why is it skew symmetric?
There are directions missing at the corners? Can you help me visualise this in the case $n=1$ and $Δ=[-1,1]$?
How do we see what are the orbits that we need to collapse? And why is the quotient a manifold? Is there some argument here that I'm missing? Normally, one has to be careful to obtain a manifold by passing to the quotient. In the actual proof mentioned in $2)$, symplectic reduction is used to justify that the quotient there is indeed a symplectic manifold. Is this where a connection to that construction comes in?
Answers to any of the questions would be much appreciated!
differential-geometry symplectic-geometry polytopes toric-geometry
edited Dec 10 '18 at 13:12
s.harp
8,67012250
asked Dec 9 '18 at 15:03
Layer CakeLayer Cake
292111
$endgroup$
I am trying to understand how to visualize a symplectic toric manifold from its moment polytope, following chapter 29.4 in "Lectures on Symplectic Geometry" by Ana Cannas da Silva: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf.
The explanation on page 188 is what interests me most (and what I understand the least). Here is a copy of that part of the text:
We can visualize $(M_Δ, ω_Δ, mathbb{T}^n, mu)$
from $Δ$ as follows. First take the product $mathbb T^n times Δ$. Let $p$ lie in the interior of $mathbb T^n times Δ$. The tangent space at $p$ is $mathbb R^n × (mathbb R^n)^∗$. Define $ω_p$ by:
$$ω_p(v, ξ) = ξ(v) = −ω_p(ξ, v),quad text{and}quad ω_p(v, v′) = ω(ξ, ξ′) = 0$$
for all $v, v′ in mathbb R^n$ and $ξ, ξ′ in (mathbb R^n)^∗$. Then $ω$ is a closed nondegenerate $2$-form on the interior of $mathbb T^n times Δ$.
At the corner there are directions missing in $(mathbb R^n)^∗$, so $ω$ is
a degenerate pairing. Hence, we need to eliminate the corresponding directions in
$mathbb R^n$. To do this, we collapse the orbits corresponding to subgroups of $mathbb T^n$ generated
by directions orthogonal to the annihilator of that face.
My questions are:
Is the idea of this construction just to build any symplectic toric manifold so that $Δ$ is the orbit space? If this is the case, I suppose we could conclude that this manifold we built must be $M_Δ$ by uniqueness in the Delzant correspondence?
Does this construction have anything to do with the construction of $M_Δ$ from the proof of the theorem in chapters 29.1-29.3? That construction is briefly described in this question Delzant theorem for polyhedra, and I understand it step by step, but I don't see whether it has anything to do with the visualisation above.
I also don't understand the boldface sentences in the text:
The definition of $omega$? How can we plug in two vectors from $mathbb R^n$, when the second argument of the function must be from the dual? I.e. why is it skew symmetric?
There are directions missing at the corners? Can you help me visualise this in the case $n=1$ and $Δ=[-1,1]$?
How do we see what are the orbits that we need to collapse? And why is the quotient a manifold? Is there some argument here that I'm missing? Normally, one has to be careful to obtain a manifold by passing to the quotient. In the actual proof mentioned in $2)$, symplectic reduction is used to justify that the quotient there is indeed a symplectic manifold. Is this where a connection to that construction comes in?
Answers to any of the questions would be much appreciated!
differential-geometry symplectic-geometry polytopes toric-geometry
differential-geometry symplectic-geometry polytopes toric-geometry
edited Dec 10 '18 at 13:12
s.harp
8,67012250
asked Dec 9 '18 at 15:03
Layer CakeLayer Cake
292111
edited Dec 10 '18 at 13:12
s.harp
8,67012250
asked Dec 9 '18 at 15:03
Layer CakeLayer Cake
292111
edited Dec 10 '18 at 13:12
s.harp
8,67012250
edited Dec 10 '18 at 13:12
s.harp
8,67012250
edited Dec 10 '18 at 13:12
s.harp
8,67012250
8,67012250
asked Dec 9 '18 at 15:03
Layer CakeLayer Cake
292111
asked Dec 9 '18 at 15:03
Layer CakeLayer Cake
292111
asked Dec 9 '18 at 15:03
Layer CakeLayer Cake
292111
292111
3
$begingroup$
Q1.1 Yes, for maximal effective Hamiltonian toric actions on a symplectic manifolds, the moment map polytope turns out to be in bijection with the orbit space. Q1.2 Yes, but the Delzant correspondence is established by understanding this 'visualization' of symplectic toric manifold. Q3. If must view $mathbb{R}^n times Delta subset mathbb{R}^n times mathbb{R^n}^* cong T^*mathbb{R}^n$ equipped with its standard symplectic form $omega((v_1, xi_1), (v_2, xi_2)) = xi_1(v_2) - xi_2(v_1)$.
$endgroup$
– Jordan Payette
Dec 11 '18 at 21:54
2
$begingroup$
Q4. I'd prefer to speak of the boundary $mathbb{R}^n times partial Delta$ instead of corners. The boundary of the image of the moment map exists because, on this boundary, the inward-outward direction(s) is (are) no longer in the image of the differential of the moment map; these are the missing directions. As for questions 2 and 5, you are on the right track, but they would require a detailed answer (and some time...) and not mere comments.
$endgroup$
– Jordan Payette
Dec 11 '18 at 22:01
add a comment |
3
$begingroup$
Q1.1 Yes, for maximal effective Hamiltonian toric actions on a symplectic manifolds, the moment map polytope turns out to be in bijection with the orbit space. Q1.2 Yes, but the Delzant correspondence is established by understanding this 'visualization' of symplectic toric manifold. Q3. If must view $mathbb{R}^n times Delta subset mathbb{R}^n times mathbb{R^n}^* cong T^*mathbb{R}^n$ equipped with its standard symplectic form $omega((v_1, xi_1), (v_2, xi_2)) = xi_1(v_2) - xi_2(v_1)$.
$endgroup$
– Jordan Payette
Dec 11 '18 at 21:54
2
$begingroup$
Q4. I'd prefer to speak of the boundary $mathbb{R}^n times partial Delta$ instead of corners. The boundary of the image of the moment map exists because, on this boundary, the inward-outward direction(s) is (are) no longer in the image of the differential of the moment map; these are the missing directions. As for questions 2 and 5, you are on the right track, but they would require a detailed answer (and some time...) and not mere comments.
$endgroup$
– Jordan Payette
Dec 11 '18 at 22:01
3
3
$begingroup$
Q1.1 Yes, for maximal effective Hamiltonian toric actions on a symplectic manifolds, the moment map polytope turns out to be in bijection with the orbit space. Q1.2 Yes, but the Delzant correspondence is established by understanding this 'visualization' of symplectic toric manifold. Q3. If must view $mathbb{R}^n times Delta subset mathbb{R}^n times mathbb{R^n}^* cong T^*mathbb{R}^n$ equipped with its standard symplectic form $omega((v_1, xi_1), (v_2, xi_2)) = xi_1(v_2) - xi_2(v_1)$.
$endgroup$
– Jordan Payette
Dec 11 '18 at 21:54
$begingroup$
Q1.1 Yes, for maximal effective Hamiltonian toric actions on a symplectic manifolds, the moment map polytope turns out to be in bijection with the orbit space. Q1.2 Yes, but the Delzant correspondence is established by understanding this 'visualization' of symplectic toric manifold. Q3. If must view $mathbb{R}^n times Delta subset mathbb{R}^n times mathbb{R^n}^* cong T^*mathbb{R}^n$ equipped with its standard symplectic form $omega((v_1, xi_1), (v_2, xi_2)) = xi_1(v_2) - xi_2(v_1)$.
$endgroup$
– Jordan Payette
Dec 11 '18 at 21:54
2
2
$begingroup$
Q4. I'd prefer to speak of the boundary $mathbb{R}^n times partial Delta$ instead of corners. The boundary of the image of the moment map exists because, on this boundary, the inward-outward direction(s) is (are) no longer in the image of the differential of the moment map; these are the missing directions. As for questions 2 and 5, you are on the right track, but they would require a detailed answer (and some time...) and not mere comments.
$endgroup$
– Jordan Payette
Dec 11 '18 at 22:01
$begingroup$
Q4. I'd prefer to speak of the boundary $mathbb{R}^n times partial Delta$ instead of corners. The boundary of the image of the moment map exists because, on this boundary, the inward-outward direction(s) is (are) no longer in the image of the differential of the moment map; these are the missing directions. As for questions 2 and 5, you are on the right track, but they would require a detailed answer (and some time...) and not mere comments.
$endgroup$
– Jordan Payette
Dec 11 '18 at 22:01
add a comment |
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$begingroup$
Q1.1 Yes, for maximal effective Hamiltonian toric actions on a symplectic manifolds, the moment map polytope turns out to be in bijection with the orbit space. Q1.2 Yes, but the Delzant correspondence is established by understanding this 'visualization' of symplectic toric manifold. Q3. If must view $mathbb{R}^n times Delta subset mathbb{R}^n times mathbb{R^n}^* cong T^*mathbb{R}^n$ equipped with its standard symplectic form $omega((v_1, xi_1), (v_2, xi_2)) = xi_1(v_2) - xi_2(v_1)$.
$endgroup$
– Jordan Payette
Dec 11 '18 at 21:54
2
$begingroup$
Q4. I'd prefer to speak of the boundary $mathbb{R}^n times partial Delta$ instead of corners. The boundary of the image of the moment map exists because, on this boundary, the inward-outward direction(s) is (are) no longer in the image of the differential of the moment map; these are the missing directions. As for questions 2 and 5, you are on the right track, but they would require a detailed answer (and some time...) and not mere comments.
$endgroup$
– Jordan Payette
Dec 11 '18 at 22:01