Quotient by a subring and not an idealRing germs of $C^infty$ functions on the real lineTangent space as the dual of an ideal quotientAlgebraic definition of Tangent space of manifold.Where can I find this result?Zariski cotangent space, as defined in Arapura's “Algebraic Geometry over the Complex Numbers”Questions about tangent and cotangent bundle on schemesDerivatives and the cotangent spacetom Dieck's universal definition of a tangent spaceLocally Analytic NullstellensatzIdentification of the tangent space of a manifold and the tangent vectors to curves
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Quotient by a subring and not an ideal
Ring germs of $C^infty$ functions on the real lineTangent space as the dual of an ideal quotientAlgebraic definition of Tangent space of manifold.Where can I find this result?Zariski cotangent space, as defined in Arapura's “Algebraic Geometry over the Complex Numbers”Questions about tangent and cotangent bundle on schemesDerivatives and the cotangent spacetom Dieck's universal definition of a tangent spaceLocally Analytic NullstellensatzIdentification of the tangent space of a manifold and the tangent vectors to curves
$begingroup$
I'm working towards understanding the Zariski tangent space of a $C^k$ manifold, using this pdf.
The author defines $mathcalO^(k)_M,p$ as the set of germs of $C^k$ functions at $p$, which forms a ring as well as an algebra over $mathbbR$ (and itself, of course) with operations defined pointwise.
They further define $mathcalS^(k)_M,p$ as the set of germs of $C^k$ functions stationary at $p$, i.e.
$mathbff in mathcalS^(k)_M,p iff forall(f in mathbff).(f circ varphi^-1)'(varphi(p))=0$, for some (and any) chart $varphi$ at $p$.
This is clearly a subring (and subalgebra) of $mathcalO^(k)_M,p$, but not an ideal (as the author points out).
The author then asserts that $mathcalO^(k)_M,p / mathcalS^(k)_M,p$ is (1) a vector space that is (2) isomorphic to the space of linear derivations on $mathcalO^(k)_M,p$ that vanish on $mathcalS^(k)_M,p$, in turn (3) isomorphic to $T_p^*M$ the (usual) cotangent space. However, I do not understand the quotient $mathcalO^(k)_M,p / mathcalS^(k)_M,p$, since $mathcalS^(k)_M,p$ is not an ideal. Is this a quotient of algebras/vector spaces? What is the implied equivalence relation?
algebraic-geometry algebraic-topology manifolds tangent-spaces
asked Mar 21 at 19:20
terrygarciaterrygarcia
1509
$endgroup$
add a comment |
$begingroup$
I'm working towards understanding the Zariski tangent space of a $C^k$ manifold, using this pdf.
The author defines $mathcalO^(k)_M,p$ as the set of germs of $C^k$ functions at $p$, which forms a ring as well as an algebra over $mathbbR$ (and itself, of course) with operations defined pointwise.
They further define $mathcalS^(k)_M,p$ as the set of germs of $C^k$ functions stationary at $p$, i.e.
$mathbff in mathcalS^(k)_M,p iff forall(f in mathbff).(f circ varphi^-1)'(varphi(p))=0$, for some (and any) chart $varphi$ at $p$.
This is clearly a subring (and subalgebra) of $mathcalO^(k)_M,p$, but not an ideal (as the author points out).
The author then asserts that $mathcalO^(k)_M,p / mathcalS^(k)_M,p$ is (1) a vector space that is (2) isomorphic to the space of linear derivations on $mathcalO^(k)_M,p$ that vanish on $mathcalS^(k)_M,p$, in turn (3) isomorphic to $T_p^*M$ the (usual) cotangent space. However, I do not understand the quotient $mathcalO^(k)_M,p / mathcalS^(k)_M,p$, since $mathcalS^(k)_M,p$ is not an ideal. Is this a quotient of algebras/vector spaces? What is the implied equivalence relation?
algebraic-geometry algebraic-topology manifolds tangent-spaces
asked Mar 21 at 19:20
terrygarciaterrygarcia
1509
$endgroup$
add a comment |
$begingroup$
I'm working towards understanding the Zariski tangent space of a $C^k$ manifold, using this pdf.
The author defines $mathcalO^(k)_M,p$ as the set of germs of $C^k$ functions at $p$, which forms a ring as well as an algebra over $mathbbR$ (and itself, of course) with operations defined pointwise.
They further define $mathcalS^(k)_M,p$ as the set of germs of $C^k$ functions stationary at $p$, i.e.
$mathbff in mathcalS^(k)_M,p iff forall(f in mathbff).(f circ varphi^-1)'(varphi(p))=0$, for some (and any) chart $varphi$ at $p$.
This is clearly a subring (and subalgebra) of $mathcalO^(k)_M,p$, but not an ideal (as the author points out).
The author then asserts that $mathcalO^(k)_M,p / mathcalS^(k)_M,p$ is (1) a vector space that is (2) isomorphic to the space of linear derivations on $mathcalO^(k)_M,p$ that vanish on $mathcalS^(k)_M,p$, in turn (3) isomorphic to $T_p^*M$ the (usual) cotangent space. However, I do not understand the quotient $mathcalO^(k)_M,p / mathcalS^(k)_M,p$, since $mathcalS^(k)_M,p$ is not an ideal. Is this a quotient of algebras/vector spaces? What is the implied equivalence relation?
algebraic-geometry algebraic-topology manifolds tangent-spaces
asked Mar 21 at 19:20
terrygarciaterrygarcia
1509
$endgroup$
I'm working towards understanding the Zariski tangent space of a $C^k$ manifold, using this pdf.
The author defines $mathcalO^(k)_M,p$ as the set of germs of $C^k$ functions at $p$, which forms a ring as well as an algebra over $mathbbR$ (and itself, of course) with operations defined pointwise.
They further define $mathcalS^(k)_M,p$ as the set of germs of $C^k$ functions stationary at $p$, i.e.
$mathbff in mathcalS^(k)_M,p iff forall(f in mathbff).(f circ varphi^-1)'(varphi(p))=0$, for some (and any) chart $varphi$ at $p$.
This is clearly a subring (and subalgebra) of $mathcalO^(k)_M,p$, but not an ideal (as the author points out).
The author then asserts that $mathcalO^(k)_M,p / mathcalS^(k)_M,p$ is (1) a vector space that is (2) isomorphic to the space of linear derivations on $mathcalO^(k)_M,p$ that vanish on $mathcalS^(k)_M,p$, in turn (3) isomorphic to $T_p^*M$ the (usual) cotangent space. However, I do not understand the quotient $mathcalO^(k)_M,p / mathcalS^(k)_M,p$, since $mathcalS^(k)_M,p$ is not an ideal. Is this a quotient of algebras/vector spaces? What is the implied equivalence relation?
algebraic-geometry algebraic-topology manifolds tangent-spaces
algebraic-geometry algebraic-topology manifolds tangent-spaces
asked Mar 21 at 19:20
terrygarciaterrygarcia
1509
asked Mar 21 at 19:20
terrygarciaterrygarcia
1509
asked Mar 21 at 19:20
terrygarciaterrygarcia
1509
asked Mar 21 at 19:20
terrygarciaterrygarcia
1509
asked Mar 21 at 19:20
terrygarciaterrygarcia
1509
1509
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since it is a subalgebra, it is also a vector subspace. Do you know how to take the quotient by a vector subspace? It preserves the vector space structure, but the ring structure disappears.
You're doing the same thing when you take the quotient of any ring by an ideal: primarily it is a quotient of an abelian group by a subgroup. The ring structure is a "bonus" that you get when the subgroup is an ideal.
answered Mar 21 at 19:30
Matt SamuelMatt Samuel
39.1k63770
$endgroup$
$begingroup$
I see, so then the relation is the usual $a sim b iff a - b in mathcalS^(k)_M,p$ and the quotient is $mathcalO^(k)_M,p / sim$?
$endgroup$
– terrygarcia
Mar 21 at 19:40
1
$begingroup$
@terry Yes, that's right.
$endgroup$
– Matt Samuel
Mar 21 at 19:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since it is a subalgebra, it is also a vector subspace. Do you know how to take the quotient by a vector subspace? It preserves the vector space structure, but the ring structure disappears.
You're doing the same thing when you take the quotient of any ring by an ideal: primarily it is a quotient of an abelian group by a subgroup. The ring structure is a "bonus" that you get when the subgroup is an ideal.
answered Mar 21 at 19:30
Matt SamuelMatt Samuel
39.1k63770
$endgroup$
$begingroup$
I see, so then the relation is the usual $a sim b iff a - b in mathcalS^(k)_M,p$ and the quotient is $mathcalO^(k)_M,p / sim$?
$endgroup$
– terrygarcia
Mar 21 at 19:40
1
$begingroup$
@terry Yes, that's right.
$endgroup$
– Matt Samuel
Mar 21 at 19:41
add a comment |
$begingroup$
Since it is a subalgebra, it is also a vector subspace. Do you know how to take the quotient by a vector subspace? It preserves the vector space structure, but the ring structure disappears.
You're doing the same thing when you take the quotient of any ring by an ideal: primarily it is a quotient of an abelian group by a subgroup. The ring structure is a "bonus" that you get when the subgroup is an ideal.
answered Mar 21 at 19:30
Matt SamuelMatt Samuel
39.1k63770
$endgroup$
$begingroup$
I see, so then the relation is the usual $a sim b iff a - b in mathcalS^(k)_M,p$ and the quotient is $mathcalO^(k)_M,p / sim$?
$endgroup$
– terrygarcia
Mar 21 at 19:40
1
$begingroup$
@terry Yes, that's right.
$endgroup$
– Matt Samuel
Mar 21 at 19:41
add a comment |
$begingroup$
Since it is a subalgebra, it is also a vector subspace. Do you know how to take the quotient by a vector subspace? It preserves the vector space structure, but the ring structure disappears.
You're doing the same thing when you take the quotient of any ring by an ideal: primarily it is a quotient of an abelian group by a subgroup. The ring structure is a "bonus" that you get when the subgroup is an ideal.
answered Mar 21 at 19:30
Matt SamuelMatt Samuel
39.1k63770
$endgroup$
Since it is a subalgebra, it is also a vector subspace. Do you know how to take the quotient by a vector subspace? It preserves the vector space structure, but the ring structure disappears.
You're doing the same thing when you take the quotient of any ring by an ideal: primarily it is a quotient of an abelian group by a subgroup. The ring structure is a "bonus" that you get when the subgroup is an ideal.
answered Mar 21 at 19:30
Matt SamuelMatt Samuel
39.1k63770
answered Mar 21 at 19:30
Matt SamuelMatt Samuel
39.1k63770
answered Mar 21 at 19:30
Matt SamuelMatt Samuel
39.1k63770
answered Mar 21 at 19:30
Matt SamuelMatt Samuel
39.1k63770
39.1k63770
$begingroup$
I see, so then the relation is the usual $a sim b iff a - b in mathcalS^(k)_M,p$ and the quotient is $mathcalO^(k)_M,p / sim$?
$endgroup$
– terrygarcia
Mar 21 at 19:40
1
$begingroup$
@terry Yes, that's right.
$endgroup$
– Matt Samuel
Mar 21 at 19:41
add a comment |
$begingroup$
I see, so then the relation is the usual $a sim b iff a - b in mathcalS^(k)_M,p$ and the quotient is $mathcalO^(k)_M,p / sim$?
$endgroup$
– terrygarcia
Mar 21 at 19:40
1
$begingroup$
@terry Yes, that's right.
$endgroup$
– Matt Samuel
Mar 21 at 19:41
$begingroup$
I see, so then the relation is the usual $a sim b iff a - b in mathcalS^(k)_M,p$ and the quotient is $mathcalO^(k)_M,p / sim$?
$endgroup$
– terrygarcia
Mar 21 at 19:40
$begingroup$
I see, so then the relation is the usual $a sim b iff a - b in mathcalS^(k)_M,p$ and the quotient is $mathcalO^(k)_M,p / sim$?
$endgroup$
– terrygarcia
Mar 21 at 19:40
1
1
$begingroup$
@terry Yes, that's right.
$endgroup$
– Matt Samuel
Mar 21 at 19:41
$begingroup$
@terry Yes, that's right.
$endgroup$
– Matt Samuel
Mar 21 at 19:41
add a comment |
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