Vector space is to manifold as convex cone is to?
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In convex analysis, a convex cone can be viewed as being like a one-sided version of a subspace. (And the polar cone is analogous to the orthogonal complement. It's a nice analogy.)
A smooth manifold has a tangent space at every point. Is there an analogous type of mathematical structure which is sort of like a smooth manifold, but which has a convex cone at every point? (Or can be approximated in some sense by a convex cone at every point?)
If so then what is the name of this type of object?
convex-analysis
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add a comment |
$begingroup$
In convex analysis, a convex cone can be viewed as being like a one-sided version of a subspace. (And the polar cone is analogous to the orthogonal complement. It's a nice analogy.)
A smooth manifold has a tangent space at every point. Is there an analogous type of mathematical structure which is sort of like a smooth manifold, but which has a convex cone at every point? (Or can be approximated in some sense by a convex cone at every point?)
If so then what is the name of this type of object?
convex-analysis
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Timelike spaces?
$endgroup$
– mma
May 31 '17 at 20:13
add a comment |
$begingroup$
In convex analysis, a convex cone can be viewed as being like a one-sided version of a subspace. (And the polar cone is analogous to the orthogonal complement. It's a nice analogy.)
A smooth manifold has a tangent space at every point. Is there an analogous type of mathematical structure which is sort of like a smooth manifold, but which has a convex cone at every point? (Or can be approximated in some sense by a convex cone at every point?)
If so then what is the name of this type of object?
convex-analysis
$endgroup$
In convex analysis, a convex cone can be viewed as being like a one-sided version of a subspace. (And the polar cone is analogous to the orthogonal complement. It's a nice analogy.)
A smooth manifold has a tangent space at every point. Is there an analogous type of mathematical structure which is sort of like a smooth manifold, but which has a convex cone at every point? (Or can be approximated in some sense by a convex cone at every point?)
If so then what is the name of this type of object?
convex-analysis
convex-analysis
edited Jan 23 '17 at 19:13
eternalGoldenBraid
asked Jan 23 '17 at 10:36
eternalGoldenBraideternalGoldenBraid
727314
727314
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Timelike spaces?
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– mma
May 31 '17 at 20:13
add a comment |
$begingroup$
Timelike spaces?
$endgroup$
– mma
May 31 '17 at 20:13
$begingroup$
Timelike spaces?
$endgroup$
– mma
May 31 '17 at 20:13
$begingroup$
Timelike spaces?
$endgroup$
– mma
May 31 '17 at 20:13
add a comment |
1 Answer
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My opinion is that the answer is contained in your first words: convex sets.
- Convex sets have tangent and normal cones at any point.
- The tangent cone to a convex set is a good local approximation $quad T_C(x) := text{cl } mathbb{R}_+(C-x)$
- These tangent and normal cones are convex, and in duality by polarity.
This being said, you could argue that it is not really satisfying because "convex sets" and "manifolds" are two separate class of sets (even though there is some overlap between the two).
Then, you might want to go into nonsmooth variational analysis, a field where people study nonconvex nonsmooth sets, their first-order approximations, etc.
For instance there is the interesting class of prox-regular sets (for simplicity I consider the ambient space to be $mathbb{R}^N$):
- A set $S$ is said to be prox regular if the metric projection $x mapsto text{proj}_C(x)$ is uniquely defined in a neighbourhood of $S$.
- Examples of prox-regular sets are convex sets and manifolds of class $C^2$ (but not only).
- There are various possible notions of tangent/normal cones for these sets in the litterature. The interesting point is that they all reduce to the classic tangent/normal cones when $S$ is convex, or to the classic tangent/normal spaces when $S$ is a manifold. Generally people consider the so-called Limiting normal cone [1, Theorem 1.6]
$$ N^{L}_S(x) := underset{y to x}{text{Limsup}} mathbb{R}_+(y - text{proj}_S(y)), $$
and take its polar as a tangent cone: $T^L_S(x) := N^L_S(x)^0$. - By definition, the Limiting tangent space is convex, but unfortunately, the Limiting normal cone is not convex in general. We also lose a perfect duality between tangent and normal cones : $N^L_S(x) subsetneq T^L_S(x)^0$.
[1] Mordukhovich, B. S. (2006). Variational analysis and generalized differentiation I: Basic theory (Vol. 330). Springer Science & Business Media.
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1 Answer
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1 Answer
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My opinion is that the answer is contained in your first words: convex sets.
- Convex sets have tangent and normal cones at any point.
- The tangent cone to a convex set is a good local approximation $quad T_C(x) := text{cl } mathbb{R}_+(C-x)$
- These tangent and normal cones are convex, and in duality by polarity.
This being said, you could argue that it is not really satisfying because "convex sets" and "manifolds" are two separate class of sets (even though there is some overlap between the two).
Then, you might want to go into nonsmooth variational analysis, a field where people study nonconvex nonsmooth sets, their first-order approximations, etc.
For instance there is the interesting class of prox-regular sets (for simplicity I consider the ambient space to be $mathbb{R}^N$):
- A set $S$ is said to be prox regular if the metric projection $x mapsto text{proj}_C(x)$ is uniquely defined in a neighbourhood of $S$.
- Examples of prox-regular sets are convex sets and manifolds of class $C^2$ (but not only).
- There are various possible notions of tangent/normal cones for these sets in the litterature. The interesting point is that they all reduce to the classic tangent/normal cones when $S$ is convex, or to the classic tangent/normal spaces when $S$ is a manifold. Generally people consider the so-called Limiting normal cone [1, Theorem 1.6]
$$ N^{L}_S(x) := underset{y to x}{text{Limsup}} mathbb{R}_+(y - text{proj}_S(y)), $$
and take its polar as a tangent cone: $T^L_S(x) := N^L_S(x)^0$. - By definition, the Limiting tangent space is convex, but unfortunately, the Limiting normal cone is not convex in general. We also lose a perfect duality between tangent and normal cones : $N^L_S(x) subsetneq T^L_S(x)^0$.
[1] Mordukhovich, B. S. (2006). Variational analysis and generalized differentiation I: Basic theory (Vol. 330). Springer Science & Business Media.
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add a comment |
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My opinion is that the answer is contained in your first words: convex sets.
- Convex sets have tangent and normal cones at any point.
- The tangent cone to a convex set is a good local approximation $quad T_C(x) := text{cl } mathbb{R}_+(C-x)$
- These tangent and normal cones are convex, and in duality by polarity.
This being said, you could argue that it is not really satisfying because "convex sets" and "manifolds" are two separate class of sets (even though there is some overlap between the two).
Then, you might want to go into nonsmooth variational analysis, a field where people study nonconvex nonsmooth sets, their first-order approximations, etc.
For instance there is the interesting class of prox-regular sets (for simplicity I consider the ambient space to be $mathbb{R}^N$):
- A set $S$ is said to be prox regular if the metric projection $x mapsto text{proj}_C(x)$ is uniquely defined in a neighbourhood of $S$.
- Examples of prox-regular sets are convex sets and manifolds of class $C^2$ (but not only).
- There are various possible notions of tangent/normal cones for these sets in the litterature. The interesting point is that they all reduce to the classic tangent/normal cones when $S$ is convex, or to the classic tangent/normal spaces when $S$ is a manifold. Generally people consider the so-called Limiting normal cone [1, Theorem 1.6]
$$ N^{L}_S(x) := underset{y to x}{text{Limsup}} mathbb{R}_+(y - text{proj}_S(y)), $$
and take its polar as a tangent cone: $T^L_S(x) := N^L_S(x)^0$. - By definition, the Limiting tangent space is convex, but unfortunately, the Limiting normal cone is not convex in general. We also lose a perfect duality between tangent and normal cones : $N^L_S(x) subsetneq T^L_S(x)^0$.
[1] Mordukhovich, B. S. (2006). Variational analysis and generalized differentiation I: Basic theory (Vol. 330). Springer Science & Business Media.
$endgroup$
add a comment |
$begingroup$
My opinion is that the answer is contained in your first words: convex sets.
- Convex sets have tangent and normal cones at any point.
- The tangent cone to a convex set is a good local approximation $quad T_C(x) := text{cl } mathbb{R}_+(C-x)$
- These tangent and normal cones are convex, and in duality by polarity.
This being said, you could argue that it is not really satisfying because "convex sets" and "manifolds" are two separate class of sets (even though there is some overlap between the two).
Then, you might want to go into nonsmooth variational analysis, a field where people study nonconvex nonsmooth sets, their first-order approximations, etc.
For instance there is the interesting class of prox-regular sets (for simplicity I consider the ambient space to be $mathbb{R}^N$):
- A set $S$ is said to be prox regular if the metric projection $x mapsto text{proj}_C(x)$ is uniquely defined in a neighbourhood of $S$.
- Examples of prox-regular sets are convex sets and manifolds of class $C^2$ (but not only).
- There are various possible notions of tangent/normal cones for these sets in the litterature. The interesting point is that they all reduce to the classic tangent/normal cones when $S$ is convex, or to the classic tangent/normal spaces when $S$ is a manifold. Generally people consider the so-called Limiting normal cone [1, Theorem 1.6]
$$ N^{L}_S(x) := underset{y to x}{text{Limsup}} mathbb{R}_+(y - text{proj}_S(y)), $$
and take its polar as a tangent cone: $T^L_S(x) := N^L_S(x)^0$. - By definition, the Limiting tangent space is convex, but unfortunately, the Limiting normal cone is not convex in general. We also lose a perfect duality between tangent and normal cones : $N^L_S(x) subsetneq T^L_S(x)^0$.
[1] Mordukhovich, B. S. (2006). Variational analysis and generalized differentiation I: Basic theory (Vol. 330). Springer Science & Business Media.
$endgroup$
My opinion is that the answer is contained in your first words: convex sets.
- Convex sets have tangent and normal cones at any point.
- The tangent cone to a convex set is a good local approximation $quad T_C(x) := text{cl } mathbb{R}_+(C-x)$
- These tangent and normal cones are convex, and in duality by polarity.
This being said, you could argue that it is not really satisfying because "convex sets" and "manifolds" are two separate class of sets (even though there is some overlap between the two).
Then, you might want to go into nonsmooth variational analysis, a field where people study nonconvex nonsmooth sets, their first-order approximations, etc.
For instance there is the interesting class of prox-regular sets (for simplicity I consider the ambient space to be $mathbb{R}^N$):
- A set $S$ is said to be prox regular if the metric projection $x mapsto text{proj}_C(x)$ is uniquely defined in a neighbourhood of $S$.
- Examples of prox-regular sets are convex sets and manifolds of class $C^2$ (but not only).
- There are various possible notions of tangent/normal cones for these sets in the litterature. The interesting point is that they all reduce to the classic tangent/normal cones when $S$ is convex, or to the classic tangent/normal spaces when $S$ is a manifold. Generally people consider the so-called Limiting normal cone [1, Theorem 1.6]
$$ N^{L}_S(x) := underset{y to x}{text{Limsup}} mathbb{R}_+(y - text{proj}_S(y)), $$
and take its polar as a tangent cone: $T^L_S(x) := N^L_S(x)^0$. - By definition, the Limiting tangent space is convex, but unfortunately, the Limiting normal cone is not convex in general. We also lose a perfect duality between tangent and normal cones : $N^L_S(x) subsetneq T^L_S(x)^0$.
[1] Mordukhovich, B. S. (2006). Variational analysis and generalized differentiation I: Basic theory (Vol. 330). Springer Science & Business Media.
answered Dec 4 '18 at 17:15
Guillaume GarrigosGuillaume Garrigos
1619
1619
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$begingroup$
Timelike spaces?
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– mma
May 31 '17 at 20:13