Mixing
Definition of smooth manifold using sheaves.
up vote
9
down vote
favorite
While defining differential manifolds using the concept of sheaves wikipedia gives the following definition.
A differentiable manifold (of class $C_k$) consists of a pair $(M, mathcal{O}_M)$ where $M$ is a topological space, and $mathcal{O}_M$ is a sheaf of local $mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,mathcal{O}_M)$ is locally isomorphic to $(mathbb{R}^n, mathcal{O})$.
[$mathcal{O}(U)=C^k(U,mathbb{R})$ is the structure sheaf on $mathbb{R}^n$.]
In one of my courses I have been asked to verify whether the above definition is equivalent to the standard definition using atlases, but in that the condition of "locally" ringed spaces is missing, that is I am supposed to prove that $M$ is a smooth manifold if and only if there is a sheaf $mathcal{O}_M$ of local $mathbb{R}$-algebras defined on $M$, such that the ringed space $(M,mathcal{O}_M)$ is locally isomorphic to $(mathbb{R}^n, mathcal{O})$ where $mathcal{O}(U)=C^{infty}(U,mathbb{R})$ is the structure sheaf on $mathbb{R}^n$.
So I was wondering if the condition of every stalk being a local ring (locally ringed space) is necessary in the case of smooth manifolds.
differential-geometry manifolds smooth-manifolds sheaf-theory ringed-spaces
edited 20 hours ago
Eric Wofsey
175k12202326
asked Nov 12 '16 at 2:57
HarshCurious
1729
add a comment |
up vote
9
down vote
favorite
While defining differential manifolds using the concept of sheaves wikipedia gives the following definition.
A differentiable manifold (of class $C_k$) consists of a pair $(M, mathcal{O}_M)$ where $M$ is a topological space, and $mathcal{O}_M$ is a sheaf of local $mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,mathcal{O}_M)$ is locally isomorphic to $(mathbb{R}^n, mathcal{O})$.
[$mathcal{O}(U)=C^k(U,mathbb{R})$ is the structure sheaf on $mathbb{R}^n$.]
In one of my courses I have been asked to verify whether the above definition is equivalent to the standard definition using atlases, but in that the condition of "locally" ringed spaces is missing, that is I am supposed to prove that $M$ is a smooth manifold if and only if there is a sheaf $mathcal{O}_M$ of local $mathbb{R}$-algebras defined on $M$, such that the ringed space $(M,mathcal{O}_M)$ is locally isomorphic to $(mathbb{R}^n, mathcal{O})$ where $mathcal{O}(U)=C^{infty}(U,mathbb{R})$ is the structure sheaf on $mathbb{R}^n$.
So I was wondering if the condition of every stalk being a local ring (locally ringed space) is necessary in the case of smooth manifolds.
differential-geometry manifolds smooth-manifolds sheaf-theory ringed-spaces
edited 20 hours ago
Eric Wofsey
175k12202326
asked Nov 12 '16 at 2:57
HarshCurious
1729
add a comment |
up vote
9
down vote
favorite
up vote
9
down vote
favorite
While defining differential manifolds using the concept of sheaves wikipedia gives the following definition.
A differentiable manifold (of class $C_k$) consists of a pair $(M, mathcal{O}_M)$ where $M$ is a topological space, and $mathcal{O}_M$ is a sheaf of local $mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,mathcal{O}_M)$ is locally isomorphic to $(mathbb{R}^n, mathcal{O})$.
[$mathcal{O}(U)=C^k(U,mathbb{R})$ is the structure sheaf on $mathbb{R}^n$.]
In one of my courses I have been asked to verify whether the above definition is equivalent to the standard definition using atlases, but in that the condition of "locally" ringed spaces is missing, that is I am supposed to prove that $M$ is a smooth manifold if and only if there is a sheaf $mathcal{O}_M$ of local $mathbb{R}$-algebras defined on $M$, such that the ringed space $(M,mathcal{O}_M)$ is locally isomorphic to $(mathbb{R}^n, mathcal{O})$ where $mathcal{O}(U)=C^{infty}(U,mathbb{R})$ is the structure sheaf on $mathbb{R}^n$.
So I was wondering if the condition of every stalk being a local ring (locally ringed space) is necessary in the case of smooth manifolds.
differential-geometry manifolds smooth-manifolds sheaf-theory ringed-spaces
edited 20 hours ago
Eric Wofsey
175k12202326
asked Nov 12 '16 at 2:57
HarshCurious
1729
While defining differential manifolds using the concept of sheaves wikipedia gives the following definition.
A differentiable manifold (of class $C_k$) consists of a pair $(M, mathcal{O}_M)$ where $M$ is a topological space, and $mathcal{O}_M$ is a sheaf of local $mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,mathcal{O}_M)$ is locally isomorphic to $(mathbb{R}^n, mathcal{O})$.
[$mathcal{O}(U)=C^k(U,mathbb{R})$ is the structure sheaf on $mathbb{R}^n$.]
In one of my courses I have been asked to verify whether the above definition is equivalent to the standard definition using atlases, but in that the condition of "locally" ringed spaces is missing, that is I am supposed to prove that $M$ is a smooth manifold if and only if there is a sheaf $mathcal{O}_M$ of local $mathbb{R}$-algebras defined on $M$, such that the ringed space $(M,mathcal{O}_M)$ is locally isomorphic to $(mathbb{R}^n, mathcal{O})$ where $mathcal{O}(U)=C^{infty}(U,mathbb{R})$ is the structure sheaf on $mathbb{R}^n$.
So I was wondering if the condition of every stalk being a local ring (locally ringed space) is necessary in the case of smooth manifolds.
differential-geometry manifolds smooth-manifolds sheaf-theory ringed-spaces
differential-geometry manifolds smooth-manifolds sheaf-theory ringed-spaces
edited 20 hours ago
Eric Wofsey
175k12202326
asked Nov 12 '16 at 2:57
HarshCurious
1729
edited 20 hours ago
Eric Wofsey
175k12202326
asked Nov 12 '16 at 2:57
HarshCurious
1729
edited 20 hours ago
Eric Wofsey
175k12202326
edited 20 hours ago
Eric Wofsey
175k12202326
edited 20 hours ago
Eric Wofsey
175k12202326
175k12202326
asked Nov 12 '16 at 2:57
HarshCurious
1729
asked Nov 12 '16 at 2:57
HarshCurious
1729
asked Nov 12 '16 at 2:57
HarshCurious
1729
1729
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Isomorphisms of locally ringed spaces are the same as isomorphisms of ringed spaces (which happen to be locally ringed), so you do not have to specify that $(M,mathcal{O}_M)$ is locally ringed. That is, if $(M,mathcal{O}_M)$ is a ringed space which is locally isomorphic to $(mathbb{R}^n, mathcal{O})$, then the stalks of $mathcal{O}_M$ are isomorphic to the corresponding stalks of $mathcal{O}$, and thus are local rings. The local isomorphisms from $(M,mathcal{O}_M)$ to $(mathbb{R}^n, mathcal{O})$ are then automatically isomorphisms of locally ringed spaces (basically, because the condition for a morphism to be "local" involves only the ringed space structure, and so is satisfied by any isomorphism of ringed spaces).
So you can define a manifold as a ringed space that is locally isomorphic to $(mathbb{R}^n, mathcal{O})$. However, it is better to define it as a locally ringed space because smooth maps between manifolds correspond to morphisms of locally ringed spaces, not morphisms of ringed spaces. I don't know a counterexample off the top of my head (it seems rather complicated to construct one), but there isn't any reason to expect that a morphism of ringed spaces between two manifolds is the same thing as a smooth map. The local condition on morphisms says that pullback respects evaluation of smooth functions at points: that is, if $varphi:(M,mathcal{O}_M)to(N,mathcal{O}_N)$ is a morphism, $finmathcal{O}_N(U)$, and $pin varphi^{-1}(U)$, then $(varphi^*f)(p)=f(varphi(p))$. In this way, the pullback map $varphi^*$ on the sheaves is completely determined by the map of sets $Mto N$. If you drop this condition (i.e., you talk only about morphisms of ringed spaces instead of morphisms of locally ringed spaces), it seems very difficult to control what the pullback map $varphi^*$ can possibly look like.
So to sum up: you don't need to say "locally ringed" instead of "ringed" when talking about the objects of the category of manifolds, but you (probably) do need to say it when talking about the morphisms (though I don't know a counterexample that would prove this).
answered Feb 8 '17 at 7:25
Eric Wofsey
175k12202326
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Isomorphisms of locally ringed spaces are the same as isomorphisms of ringed spaces (which happen to be locally ringed), so you do not have to specify that $(M,mathcal{O}_M)$ is locally ringed. That is, if $(M,mathcal{O}_M)$ is a ringed space which is locally isomorphic to $(mathbb{R}^n, mathcal{O})$, then the stalks of $mathcal{O}_M$ are isomorphic to the corresponding stalks of $mathcal{O}$, and thus are local rings. The local isomorphisms from $(M,mathcal{O}_M)$ to $(mathbb{R}^n, mathcal{O})$ are then automatically isomorphisms of locally ringed spaces (basically, because the condition for a morphism to be "local" involves only the ringed space structure, and so is satisfied by any isomorphism of ringed spaces).
So you can define a manifold as a ringed space that is locally isomorphic to $(mathbb{R}^n, mathcal{O})$. However, it is better to define it as a locally ringed space because smooth maps between manifolds correspond to morphisms of locally ringed spaces, not morphisms of ringed spaces. I don't know a counterexample off the top of my head (it seems rather complicated to construct one), but there isn't any reason to expect that a morphism of ringed spaces between two manifolds is the same thing as a smooth map. The local condition on morphisms says that pullback respects evaluation of smooth functions at points: that is, if $varphi:(M,mathcal{O}_M)to(N,mathcal{O}_N)$ is a morphism, $finmathcal{O}_N(U)$, and $pin varphi^{-1}(U)$, then $(varphi^*f)(p)=f(varphi(p))$. In this way, the pullback map $varphi^*$ on the sheaves is completely determined by the map of sets $Mto N$. If you drop this condition (i.e., you talk only about morphisms of ringed spaces instead of morphisms of locally ringed spaces), it seems very difficult to control what the pullback map $varphi^*$ can possibly look like.
So to sum up: you don't need to say "locally ringed" instead of "ringed" when talking about the objects of the category of manifolds, but you (probably) do need to say it when talking about the morphisms (though I don't know a counterexample that would prove this).
answered Feb 8 '17 at 7:25
Eric Wofsey
175k12202326
add a comment |
up vote
5
down vote
accepted
Isomorphisms of locally ringed spaces are the same as isomorphisms of ringed spaces (which happen to be locally ringed), so you do not have to specify that $(M,mathcal{O}_M)$ is locally ringed. That is, if $(M,mathcal{O}_M)$ is a ringed space which is locally isomorphic to $(mathbb{R}^n, mathcal{O})$, then the stalks of $mathcal{O}_M$ are isomorphic to the corresponding stalks of $mathcal{O}$, and thus are local rings. The local isomorphisms from $(M,mathcal{O}_M)$ to $(mathbb{R}^n, mathcal{O})$ are then automatically isomorphisms of locally ringed spaces (basically, because the condition for a morphism to be "local" involves only the ringed space structure, and so is satisfied by any isomorphism of ringed spaces).
So you can define a manifold as a ringed space that is locally isomorphic to $(mathbb{R}^n, mathcal{O})$. However, it is better to define it as a locally ringed space because smooth maps between manifolds correspond to morphisms of locally ringed spaces, not morphisms of ringed spaces. I don't know a counterexample off the top of my head (it seems rather complicated to construct one), but there isn't any reason to expect that a morphism of ringed spaces between two manifolds is the same thing as a smooth map. The local condition on morphisms says that pullback respects evaluation of smooth functions at points: that is, if $varphi:(M,mathcal{O}_M)to(N,mathcal{O}_N)$ is a morphism, $finmathcal{O}_N(U)$, and $pin varphi^{-1}(U)$, then $(varphi^*f)(p)=f(varphi(p))$. In this way, the pullback map $varphi^*$ on the sheaves is completely determined by the map of sets $Mto N$. If you drop this condition (i.e., you talk only about morphisms of ringed spaces instead of morphisms of locally ringed spaces), it seems very difficult to control what the pullback map $varphi^*$ can possibly look like.
So to sum up: you don't need to say "locally ringed" instead of "ringed" when talking about the objects of the category of manifolds, but you (probably) do need to say it when talking about the morphisms (though I don't know a counterexample that would prove this).
answered Feb 8 '17 at 7:25
Eric Wofsey
175k12202326
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Isomorphisms of locally ringed spaces are the same as isomorphisms of ringed spaces (which happen to be locally ringed), so you do not have to specify that $(M,mathcal{O}_M)$ is locally ringed. That is, if $(M,mathcal{O}_M)$ is a ringed space which is locally isomorphic to $(mathbb{R}^n, mathcal{O})$, then the stalks of $mathcal{O}_M$ are isomorphic to the corresponding stalks of $mathcal{O}$, and thus are local rings. The local isomorphisms from $(M,mathcal{O}_M)$ to $(mathbb{R}^n, mathcal{O})$ are then automatically isomorphisms of locally ringed spaces (basically, because the condition for a morphism to be "local" involves only the ringed space structure, and so is satisfied by any isomorphism of ringed spaces).
So you can define a manifold as a ringed space that is locally isomorphic to $(mathbb{R}^n, mathcal{O})$. However, it is better to define it as a locally ringed space because smooth maps between manifolds correspond to morphisms of locally ringed spaces, not morphisms of ringed spaces. I don't know a counterexample off the top of my head (it seems rather complicated to construct one), but there isn't any reason to expect that a morphism of ringed spaces between two manifolds is the same thing as a smooth map. The local condition on morphisms says that pullback respects evaluation of smooth functions at points: that is, if $varphi:(M,mathcal{O}_M)to(N,mathcal{O}_N)$ is a morphism, $finmathcal{O}_N(U)$, and $pin varphi^{-1}(U)$, then $(varphi^*f)(p)=f(varphi(p))$. In this way, the pullback map $varphi^*$ on the sheaves is completely determined by the map of sets $Mto N$. If you drop this condition (i.e., you talk only about morphisms of ringed spaces instead of morphisms of locally ringed spaces), it seems very difficult to control what the pullback map $varphi^*$ can possibly look like.
So to sum up: you don't need to say "locally ringed" instead of "ringed" when talking about the objects of the category of manifolds, but you (probably) do need to say it when talking about the morphisms (though I don't know a counterexample that would prove this).
answered Feb 8 '17 at 7:25
Eric Wofsey
175k12202326
Isomorphisms of locally ringed spaces are the same as isomorphisms of ringed spaces (which happen to be locally ringed), so you do not have to specify that $(M,mathcal{O}_M)$ is locally ringed. That is, if $(M,mathcal{O}_M)$ is a ringed space which is locally isomorphic to $(mathbb{R}^n, mathcal{O})$, then the stalks of $mathcal{O}_M$ are isomorphic to the corresponding stalks of $mathcal{O}$, and thus are local rings. The local isomorphisms from $(M,mathcal{O}_M)$ to $(mathbb{R}^n, mathcal{O})$ are then automatically isomorphisms of locally ringed spaces (basically, because the condition for a morphism to be "local" involves only the ringed space structure, and so is satisfied by any isomorphism of ringed spaces).
So you can define a manifold as a ringed space that is locally isomorphic to $(mathbb{R}^n, mathcal{O})$. However, it is better to define it as a locally ringed space because smooth maps between manifolds correspond to morphisms of locally ringed spaces, not morphisms of ringed spaces. I don't know a counterexample off the top of my head (it seems rather complicated to construct one), but there isn't any reason to expect that a morphism of ringed spaces between two manifolds is the same thing as a smooth map. The local condition on morphisms says that pullback respects evaluation of smooth functions at points: that is, if $varphi:(M,mathcal{O}_M)to(N,mathcal{O}_N)$ is a morphism, $finmathcal{O}_N(U)$, and $pin varphi^{-1}(U)$, then $(varphi^*f)(p)=f(varphi(p))$. In this way, the pullback map $varphi^*$ on the sheaves is completely determined by the map of sets $Mto N$. If you drop this condition (i.e., you talk only about morphisms of ringed spaces instead of morphisms of locally ringed spaces), it seems very difficult to control what the pullback map $varphi^*$ can possibly look like.
So to sum up: you don't need to say "locally ringed" instead of "ringed" when talking about the objects of the category of manifolds, but you (probably) do need to say it when talking about the morphisms (though I don't know a counterexample that would prove this).
answered Feb 8 '17 at 7:25
Eric Wofsey
175k12202326
edited Feb 8 '17 at 7:42
answered Feb 8 '17 at 7:25
Eric Wofsey
175k12202326
answered Feb 8 '17 at 7:25
Eric Wofsey
175k12202326
answered Feb 8 '17 at 7:25
Eric Wofsey
175k12202326
175k12202326
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2010035%2fdefinition-of-smooth-manifold-using-sheaves%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
