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Is there a natural category in which the morphisms are derivative operators?
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I'm studying general relativity. As I currently understand the theory, there's a part where we have a (differentiable) manifold $M$, and define a vector field on $M$ to be a function $v : (M to Re) to (M to Re)$ satisfying:
- (linearity) $v(alpha f + beta g) = alpha v(f) + beta v(g)$
- (Leibniz) $v(f cdot g) = f cdot v(g) + g cdot v(f)$
we might also need to further restrict our attention to $v$ that are continuous. We can then show that the set of functions $v$ that satisfy these properties form a vector space, at which point we can generalize from vector fields to tensor fields. (Then, given a metric on $M$ we can derive a natural notion of differentiation on tensor fields, at which point we're ready to state some properties that the spacetime metric and the stress-energy tensor obey.)
My question is, is there a (natural) category in which vector fields are just endomorphisms on $(M to Re)$? For example, condition (1) above arises automatically if we require that $v$ be an endomorphism of $(M to Re)$ in $Re$-Vect; is there a well-known category such that conditions (1) and (2) arise automatically if we require $v$ to be an endomorphism of $(M to Re)$?
Obviously I could simply define a category where the objects are vector spaces and the morphisms are linear maps that happen to satisfy the Leibniz rule [EDIT: This is wrong, as pointed out by Eric below -- given two $v$ that satisfy the property above, their composition does not in general satisfy the Leibniz property]; my question is, is this a well-known category (or, is there a simple variation on my question that allows me to see vector fields in a categorical light)?
As an example of the genre of question that I'm asking, recall that a vector space over a field $K$ with vectors $V$ can be viewed either as a function $* : K to V to V$ satisfying a handful of axioms, or simply as a ring homomorphism between $K$ and the ring of group endomorphisms on $V$. In other words, we can either specify a vector space as a function obeying a bunch of axioms, or we can choose a morphism between the right objects in the right category (in this case, any $phi : K to_textRing (V to_textGroup V)$) at which point the axioms come free. In the case of vector spaces, I could have defined a category in which all morphisms are vector spaces, but I probably wouldn't have noticed that I was trying to ask for a ring homomorphism between the scalar field and the ring of endomorphisms of the vector group. In my question here about a category where morphisms correspond to derivative operators, I'm hoping for an answer analogous to "you're looking for a ring homomorphism between $K$ and $V to_textGroup V$".
linear-algebra vector-spaces category-theory general-relativity
edited Jul 18 at 8:07
Pece
7,92211040
asked Jul 17 at 19:32
Nate
848818
add a comment |Â
up vote
5
down vote
favorite
I'm studying general relativity. As I currently understand the theory, there's a part where we have a (differentiable) manifold $M$, and define a vector field on $M$ to be a function $v : (M to Re) to (M to Re)$ satisfying:
- (linearity) $v(alpha f + beta g) = alpha v(f) + beta v(g)$
- (Leibniz) $v(f cdot g) = f cdot v(g) + g cdot v(f)$
we might also need to further restrict our attention to $v$ that are continuous. We can then show that the set of functions $v$ that satisfy these properties form a vector space, at which point we can generalize from vector fields to tensor fields. (Then, given a metric on $M$ we can derive a natural notion of differentiation on tensor fields, at which point we're ready to state some properties that the spacetime metric and the stress-energy tensor obey.)
My question is, is there a (natural) category in which vector fields are just endomorphisms on $(M to Re)$? For example, condition (1) above arises automatically if we require that $v$ be an endomorphism of $(M to Re)$ in $Re$-Vect; is there a well-known category such that conditions (1) and (2) arise automatically if we require $v$ to be an endomorphism of $(M to Re)$?
Obviously I could simply define a category where the objects are vector spaces and the morphisms are linear maps that happen to satisfy the Leibniz rule [EDIT: This is wrong, as pointed out by Eric below -- given two $v$ that satisfy the property above, their composition does not in general satisfy the Leibniz property]; my question is, is this a well-known category (or, is there a simple variation on my question that allows me to see vector fields in a categorical light)?
As an example of the genre of question that I'm asking, recall that a vector space over a field $K$ with vectors $V$ can be viewed either as a function $* : K to V to V$ satisfying a handful of axioms, or simply as a ring homomorphism between $K$ and the ring of group endomorphisms on $V$. In other words, we can either specify a vector space as a function obeying a bunch of axioms, or we can choose a morphism between the right objects in the right category (in this case, any $phi : K to_textRing (V to_textGroup V)$) at which point the axioms come free. In the case of vector spaces, I could have defined a category in which all morphisms are vector spaces, but I probably wouldn't have noticed that I was trying to ask for a ring homomorphism between the scalar field and the ring of endomorphisms of the vector group. In my question here about a category where morphisms correspond to derivative operators, I'm hoping for an answer analogous to "you're looking for a ring homomorphism between $K$ and $V to_textGroup V$".
linear-algebra vector-spaces category-theory general-relativity
edited Jul 18 at 8:07
Pece
7,92211040
asked Jul 17 at 19:32
Nate
848818
2
A composition of derivations is usually not a derivation, so this is not possible without generalizing what kind of morphisms you allow.
– Eric Wofsey
Jul 17 at 19:36
Oh, oops, well, that sinks that idea. Thanks! (If your comment was a top level answer I'd accept it.)
– Nate
Jul 17 at 19:44
However, augmented derivations may rescue the idea.
– Eric Towers
Jul 17 at 19:53
I feel like it's not actually the "correct" answer, though, since you can consider higher-order differential operators which will be closed under composition. I don't know their theory well enough to write a good answer, though.
– Eric Wofsey
Jul 17 at 20:13
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm studying general relativity. As I currently understand the theory, there's a part where we have a (differentiable) manifold $M$, and define a vector field on $M$ to be a function $v : (M to Re) to (M to Re)$ satisfying:
- (linearity) $v(alpha f + beta g) = alpha v(f) + beta v(g)$
- (Leibniz) $v(f cdot g) = f cdot v(g) + g cdot v(f)$
we might also need to further restrict our attention to $v$ that are continuous. We can then show that the set of functions $v$ that satisfy these properties form a vector space, at which point we can generalize from vector fields to tensor fields. (Then, given a metric on $M$ we can derive a natural notion of differentiation on tensor fields, at which point we're ready to state some properties that the spacetime metric and the stress-energy tensor obey.)
My question is, is there a (natural) category in which vector fields are just endomorphisms on $(M to Re)$? For example, condition (1) above arises automatically if we require that $v$ be an endomorphism of $(M to Re)$ in $Re$-Vect; is there a well-known category such that conditions (1) and (2) arise automatically if we require $v$ to be an endomorphism of $(M to Re)$?
Obviously I could simply define a category where the objects are vector spaces and the morphisms are linear maps that happen to satisfy the Leibniz rule [EDIT: This is wrong, as pointed out by Eric below -- given two $v$ that satisfy the property above, their composition does not in general satisfy the Leibniz property]; my question is, is this a well-known category (or, is there a simple variation on my question that allows me to see vector fields in a categorical light)?
As an example of the genre of question that I'm asking, recall that a vector space over a field $K$ with vectors $V$ can be viewed either as a function $* : K to V to V$ satisfying a handful of axioms, or simply as a ring homomorphism between $K$ and the ring of group endomorphisms on $V$. In other words, we can either specify a vector space as a function obeying a bunch of axioms, or we can choose a morphism between the right objects in the right category (in this case, any $phi : K to_textRing (V to_textGroup V)$) at which point the axioms come free. In the case of vector spaces, I could have defined a category in which all morphisms are vector spaces, but I probably wouldn't have noticed that I was trying to ask for a ring homomorphism between the scalar field and the ring of endomorphisms of the vector group. In my question here about a category where morphisms correspond to derivative operators, I'm hoping for an answer analogous to "you're looking for a ring homomorphism between $K$ and $V to_textGroup V$".
linear-algebra vector-spaces category-theory general-relativity
edited Jul 18 at 8:07
Pece
7,92211040
asked Jul 17 at 19:32
Nate
848818
I'm studying general relativity. As I currently understand the theory, there's a part where we have a (differentiable) manifold $M$, and define a vector field on $M$ to be a function $v : (M to Re) to (M to Re)$ satisfying:
- (linearity) $v(alpha f + beta g) = alpha v(f) + beta v(g)$
- (Leibniz) $v(f cdot g) = f cdot v(g) + g cdot v(f)$
we might also need to further restrict our attention to $v$ that are continuous. We can then show that the set of functions $v$ that satisfy these properties form a vector space, at which point we can generalize from vector fields to tensor fields. (Then, given a metric on $M$ we can derive a natural notion of differentiation on tensor fields, at which point we're ready to state some properties that the spacetime metric and the stress-energy tensor obey.)
My question is, is there a (natural) category in which vector fields are just endomorphisms on $(M to Re)$? For example, condition (1) above arises automatically if we require that $v$ be an endomorphism of $(M to Re)$ in $Re$-Vect; is there a well-known category such that conditions (1) and (2) arise automatically if we require $v$ to be an endomorphism of $(M to Re)$?
Obviously I could simply define a category where the objects are vector spaces and the morphisms are linear maps that happen to satisfy the Leibniz rule [EDIT: This is wrong, as pointed out by Eric below -- given two $v$ that satisfy the property above, their composition does not in general satisfy the Leibniz property]; my question is, is this a well-known category (or, is there a simple variation on my question that allows me to see vector fields in a categorical light)?
As an example of the genre of question that I'm asking, recall that a vector space over a field $K$ with vectors $V$ can be viewed either as a function $* : K to V to V$ satisfying a handful of axioms, or simply as a ring homomorphism between $K$ and the ring of group endomorphisms on $V$. In other words, we can either specify a vector space as a function obeying a bunch of axioms, or we can choose a morphism between the right objects in the right category (in this case, any $phi : K to_textRing (V to_textGroup V)$) at which point the axioms come free. In the case of vector spaces, I could have defined a category in which all morphisms are vector spaces, but I probably wouldn't have noticed that I was trying to ask for a ring homomorphism between the scalar field and the ring of endomorphisms of the vector group. In my question here about a category where morphisms correspond to derivative operators, I'm hoping for an answer analogous to "you're looking for a ring homomorphism between $K$ and $V to_textGroup V$".
linear-algebra vector-spaces category-theory general-relativity
edited Jul 18 at 8:07
Pece
7,92211040
asked Jul 17 at 19:32
Nate
848818
edited Jul 18 at 8:07
Pece
7,92211040
edited Jul 18 at 8:07
Pece
7,92211040
edited Jul 18 at 8:07
Pece
7,92211040
7,92211040
asked Jul 17 at 19:32
Nate
848818
asked Jul 17 at 19:32
Nate
848818
asked Jul 17 at 19:32
Nate
848818
848818
2
A composition of derivations is usually not a derivation, so this is not possible without generalizing what kind of morphisms you allow.
– Eric Wofsey
Jul 17 at 19:36
Oh, oops, well, that sinks that idea. Thanks! (If your comment was a top level answer I'd accept it.)
– Nate
Jul 17 at 19:44
However, augmented derivations may rescue the idea.
– Eric Towers
Jul 17 at 19:53
I feel like it's not actually the "correct" answer, though, since you can consider higher-order differential operators which will be closed under composition. I don't know their theory well enough to write a good answer, though.
– Eric Wofsey
Jul 17 at 20:13
add a comment |Â
2
A composition of derivations is usually not a derivation, so this is not possible without generalizing what kind of morphisms you allow.
– Eric Wofsey
Jul 17 at 19:36
Oh, oops, well, that sinks that idea. Thanks! (If your comment was a top level answer I'd accept it.)
– Nate
Jul 17 at 19:44
However, augmented derivations may rescue the idea.
– Eric Towers
Jul 17 at 19:53
I feel like it's not actually the "correct" answer, though, since you can consider higher-order differential operators which will be closed under composition. I don't know their theory well enough to write a good answer, though.
– Eric Wofsey
Jul 17 at 20:13
2
2
A composition of derivations is usually not a derivation, so this is not possible without generalizing what kind of morphisms you allow.
– Eric Wofsey
Jul 17 at 19:36
A composition of derivations is usually not a derivation, so this is not possible without generalizing what kind of morphisms you allow.
– Eric Wofsey
Jul 17 at 19:36
Oh, oops, well, that sinks that idea. Thanks! (If your comment was a top level answer I'd accept it.)
– Nate
Jul 17 at 19:44
Oh, oops, well, that sinks that idea. Thanks! (If your comment was a top level answer I'd accept it.)
– Nate
Jul 17 at 19:44
However, augmented derivations may rescue the idea.
– Eric Towers
Jul 17 at 19:53
However, augmented derivations may rescue the idea.
– Eric Towers
Jul 17 at 19:53
I feel like it's not actually the "correct" answer, though, since you can consider higher-order differential operators which will be closed under composition. I don't know their theory well enough to write a good answer, though.
– Eric Wofsey
Jul 17 at 20:13
I feel like it's not actually the "correct" answer, though, since you can consider higher-order differential operators which will be closed under composition. I don't know their theory well enough to write a good answer, though.
– Eric Wofsey
Jul 17 at 20:13
add a comment |Â
1 Answer
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oldest
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Instead of talking about manifolds let's talk for a bit about vector fields on affine schemes. For the sake of minimizing technical detail my definition will be that the category $textAff$ of affine schemes is just the opposite of the category $textCRing$ of commutative rings, and if $R$ is a commutative ring then $textSpec R$ will just be my name for $R$ but regarded as an object in the opposite category.
Definition: A vector field on $textSpec R$ is a derivation $D : R to R$; that is, a linear function satisfying $D(ab) = a D(b) + D(a) b$ for all $a, b in R$.
The intuition here, as in the manifold case, is that derivations are "infinitesimal automorphisms" of $R$, or equivalently of $textSpec R$. The lovely benefit of switching to working with affine schemes is that it is possible to make this intuition completely precise as follows.
Exercise: A derivation $D$ on a commutative ring $R$ is the same thing as an $epsilon$-linear homomorphism $R[epsilon]/epsilon^2 to R[epsilon]/epsilon^2$ which reduces to the identity $bmod epsilon$.
In particular, the Leibniz rule is equivalent to the assertion that the map $I + epsilon D$ preserves multiplication, and composition of homomorphisms as above corresponds to addition of derivations. See this blog post for more on what can be done from this perspective, including a clean conceptual proof that the commutator of two derivations is a derivation.
Geometrically the above definition corresponds to asking for a map from $textSpec R times textSpec mathbbZ[epsilon]/epsilon^2$ to $textSpec R$; you can think of $textSpec mathbbZ[epsilon]/epsilon^2$ as the "walking tangent vector" and of this definition as asking for a tangent vector to the identity in the "space" (it would be an affine scheme if the category of affine schemes were cartesian closed, which it's not; however it is a presheaf on the category of affine schemes, just not a representable one) of endomorphisms of $textSpec R$.
More generally you can contemplate homomorphisms $R to S[epsilon]/epsilon^2$ for two rings $R$ to $S$, or equivalently $epsilon$-linear homomorphisms $R[epsilon]/epsilon^2 to S[epsilon]/epsilon^2$; this works out to a map of the form $f + epsilon g$ where $f : R to S$ is a homomorphism and $g : R to S$ is a linear map satisfying
$$g(ab) = f(a) g(b) + g(a) f(b).$$
This is a useful generalization of the notion of a derivation; for example it can be used to define Zariski tangent spaces. It can be thought of as describing a tangent vector to $f$ in the "space" of maps from $R$ to $S$, or equivalently from $textSpec S$ to $textSpec R$. The corresponding object in differential geometry is a pair consisting of a map $f : M to N$ of smooth manifolds and a section of the pullback $f^ast(TN)$ of the tangent bundle of $N$ to $M$.
You can try to make this sort of thing work out for smooth manifolds by passing to a somewhat more complicated category that includes smooth manifolds as well as various "infinitesimal" spaces; see smooth algebra for some details.
On the more general topic of categorical approaches to tensor fields, you might enjoy skimming Kolar, Michor, and Slovak's Natural Operations in Differential Geometry.
answered Jul 17 at 23:53
Qiaochu Yuan
269k32564900
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Instead of talking about manifolds let's talk for a bit about vector fields on affine schemes. For the sake of minimizing technical detail my definition will be that the category $textAff$ of affine schemes is just the opposite of the category $textCRing$ of commutative rings, and if $R$ is a commutative ring then $textSpec R$ will just be my name for $R$ but regarded as an object in the opposite category.
Definition: A vector field on $textSpec R$ is a derivation $D : R to R$; that is, a linear function satisfying $D(ab) = a D(b) + D(a) b$ for all $a, b in R$.
The intuition here, as in the manifold case, is that derivations are "infinitesimal automorphisms" of $R$, or equivalently of $textSpec R$. The lovely benefit of switching to working with affine schemes is that it is possible to make this intuition completely precise as follows.
Exercise: A derivation $D$ on a commutative ring $R$ is the same thing as an $epsilon$-linear homomorphism $R[epsilon]/epsilon^2 to R[epsilon]/epsilon^2$ which reduces to the identity $bmod epsilon$.
In particular, the Leibniz rule is equivalent to the assertion that the map $I + epsilon D$ preserves multiplication, and composition of homomorphisms as above corresponds to addition of derivations. See this blog post for more on what can be done from this perspective, including a clean conceptual proof that the commutator of two derivations is a derivation.
Geometrically the above definition corresponds to asking for a map from $textSpec R times textSpec mathbbZ[epsilon]/epsilon^2$ to $textSpec R$; you can think of $textSpec mathbbZ[epsilon]/epsilon^2$ as the "walking tangent vector" and of this definition as asking for a tangent vector to the identity in the "space" (it would be an affine scheme if the category of affine schemes were cartesian closed, which it's not; however it is a presheaf on the category of affine schemes, just not a representable one) of endomorphisms of $textSpec R$.
More generally you can contemplate homomorphisms $R to S[epsilon]/epsilon^2$ for two rings $R$ to $S$, or equivalently $epsilon$-linear homomorphisms $R[epsilon]/epsilon^2 to S[epsilon]/epsilon^2$; this works out to a map of the form $f + epsilon g$ where $f : R to S$ is a homomorphism and $g : R to S$ is a linear map satisfying
$$g(ab) = f(a) g(b) + g(a) f(b).$$
This is a useful generalization of the notion of a derivation; for example it can be used to define Zariski tangent spaces. It can be thought of as describing a tangent vector to $f$ in the "space" of maps from $R$ to $S$, or equivalently from $textSpec S$ to $textSpec R$. The corresponding object in differential geometry is a pair consisting of a map $f : M to N$ of smooth manifolds and a section of the pullback $f^ast(TN)$ of the tangent bundle of $N$ to $M$.
You can try to make this sort of thing work out for smooth manifolds by passing to a somewhat more complicated category that includes smooth manifolds as well as various "infinitesimal" spaces; see smooth algebra for some details.
On the more general topic of categorical approaches to tensor fields, you might enjoy skimming Kolar, Michor, and Slovak's Natural Operations in Differential Geometry.
answered Jul 17 at 23:53
Qiaochu Yuan
269k32564900
add a comment |Â
up vote
5
down vote
accepted
Instead of talking about manifolds let's talk for a bit about vector fields on affine schemes. For the sake of minimizing technical detail my definition will be that the category $textAff$ of affine schemes is just the opposite of the category $textCRing$ of commutative rings, and if $R$ is a commutative ring then $textSpec R$ will just be my name for $R$ but regarded as an object in the opposite category.
Definition: A vector field on $textSpec R$ is a derivation $D : R to R$; that is, a linear function satisfying $D(ab) = a D(b) + D(a) b$ for all $a, b in R$.
The intuition here, as in the manifold case, is that derivations are "infinitesimal automorphisms" of $R$, or equivalently of $textSpec R$. The lovely benefit of switching to working with affine schemes is that it is possible to make this intuition completely precise as follows.
Exercise: A derivation $D$ on a commutative ring $R$ is the same thing as an $epsilon$-linear homomorphism $R[epsilon]/epsilon^2 to R[epsilon]/epsilon^2$ which reduces to the identity $bmod epsilon$.
In particular, the Leibniz rule is equivalent to the assertion that the map $I + epsilon D$ preserves multiplication, and composition of homomorphisms as above corresponds to addition of derivations. See this blog post for more on what can be done from this perspective, including a clean conceptual proof that the commutator of two derivations is a derivation.
Geometrically the above definition corresponds to asking for a map from $textSpec R times textSpec mathbbZ[epsilon]/epsilon^2$ to $textSpec R$; you can think of $textSpec mathbbZ[epsilon]/epsilon^2$ as the "walking tangent vector" and of this definition as asking for a tangent vector to the identity in the "space" (it would be an affine scheme if the category of affine schemes were cartesian closed, which it's not; however it is a presheaf on the category of affine schemes, just not a representable one) of endomorphisms of $textSpec R$.
More generally you can contemplate homomorphisms $R to S[epsilon]/epsilon^2$ for two rings $R$ to $S$, or equivalently $epsilon$-linear homomorphisms $R[epsilon]/epsilon^2 to S[epsilon]/epsilon^2$; this works out to a map of the form $f + epsilon g$ where $f : R to S$ is a homomorphism and $g : R to S$ is a linear map satisfying
$$g(ab) = f(a) g(b) + g(a) f(b).$$
This is a useful generalization of the notion of a derivation; for example it can be used to define Zariski tangent spaces. It can be thought of as describing a tangent vector to $f$ in the "space" of maps from $R$ to $S$, or equivalently from $textSpec S$ to $textSpec R$. The corresponding object in differential geometry is a pair consisting of a map $f : M to N$ of smooth manifolds and a section of the pullback $f^ast(TN)$ of the tangent bundle of $N$ to $M$.
You can try to make this sort of thing work out for smooth manifolds by passing to a somewhat more complicated category that includes smooth manifolds as well as various "infinitesimal" spaces; see smooth algebra for some details.
On the more general topic of categorical approaches to tensor fields, you might enjoy skimming Kolar, Michor, and Slovak's Natural Operations in Differential Geometry.
answered Jul 17 at 23:53
Qiaochu Yuan
269k32564900
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Instead of talking about manifolds let's talk for a bit about vector fields on affine schemes. For the sake of minimizing technical detail my definition will be that the category $textAff$ of affine schemes is just the opposite of the category $textCRing$ of commutative rings, and if $R$ is a commutative ring then $textSpec R$ will just be my name for $R$ but regarded as an object in the opposite category.
Definition: A vector field on $textSpec R$ is a derivation $D : R to R$; that is, a linear function satisfying $D(ab) = a D(b) + D(a) b$ for all $a, b in R$.
The intuition here, as in the manifold case, is that derivations are "infinitesimal automorphisms" of $R$, or equivalently of $textSpec R$. The lovely benefit of switching to working with affine schemes is that it is possible to make this intuition completely precise as follows.
Exercise: A derivation $D$ on a commutative ring $R$ is the same thing as an $epsilon$-linear homomorphism $R[epsilon]/epsilon^2 to R[epsilon]/epsilon^2$ which reduces to the identity $bmod epsilon$.
In particular, the Leibniz rule is equivalent to the assertion that the map $I + epsilon D$ preserves multiplication, and composition of homomorphisms as above corresponds to addition of derivations. See this blog post for more on what can be done from this perspective, including a clean conceptual proof that the commutator of two derivations is a derivation.
Geometrically the above definition corresponds to asking for a map from $textSpec R times textSpec mathbbZ[epsilon]/epsilon^2$ to $textSpec R$; you can think of $textSpec mathbbZ[epsilon]/epsilon^2$ as the "walking tangent vector" and of this definition as asking for a tangent vector to the identity in the "space" (it would be an affine scheme if the category of affine schemes were cartesian closed, which it's not; however it is a presheaf on the category of affine schemes, just not a representable one) of endomorphisms of $textSpec R$.
More generally you can contemplate homomorphisms $R to S[epsilon]/epsilon^2$ for two rings $R$ to $S$, or equivalently $epsilon$-linear homomorphisms $R[epsilon]/epsilon^2 to S[epsilon]/epsilon^2$; this works out to a map of the form $f + epsilon g$ where $f : R to S$ is a homomorphism and $g : R to S$ is a linear map satisfying
$$g(ab) = f(a) g(b) + g(a) f(b).$$
This is a useful generalization of the notion of a derivation; for example it can be used to define Zariski tangent spaces. It can be thought of as describing a tangent vector to $f$ in the "space" of maps from $R$ to $S$, or equivalently from $textSpec S$ to $textSpec R$. The corresponding object in differential geometry is a pair consisting of a map $f : M to N$ of smooth manifolds and a section of the pullback $f^ast(TN)$ of the tangent bundle of $N$ to $M$.
You can try to make this sort of thing work out for smooth manifolds by passing to a somewhat more complicated category that includes smooth manifolds as well as various "infinitesimal" spaces; see smooth algebra for some details.
On the more general topic of categorical approaches to tensor fields, you might enjoy skimming Kolar, Michor, and Slovak's Natural Operations in Differential Geometry.
answered Jul 17 at 23:53
Qiaochu Yuan
269k32564900
Instead of talking about manifolds let's talk for a bit about vector fields on affine schemes. For the sake of minimizing technical detail my definition will be that the category $textAff$ of affine schemes is just the opposite of the category $textCRing$ of commutative rings, and if $R$ is a commutative ring then $textSpec R$ will just be my name for $R$ but regarded as an object in the opposite category.
Definition: A vector field on $textSpec R$ is a derivation $D : R to R$; that is, a linear function satisfying $D(ab) = a D(b) + D(a) b$ for all $a, b in R$.
The intuition here, as in the manifold case, is that derivations are "infinitesimal automorphisms" of $R$, or equivalently of $textSpec R$. The lovely benefit of switching to working with affine schemes is that it is possible to make this intuition completely precise as follows.
Exercise: A derivation $D$ on a commutative ring $R$ is the same thing as an $epsilon$-linear homomorphism $R[epsilon]/epsilon^2 to R[epsilon]/epsilon^2$ which reduces to the identity $bmod epsilon$.
In particular, the Leibniz rule is equivalent to the assertion that the map $I + epsilon D$ preserves multiplication, and composition of homomorphisms as above corresponds to addition of derivations. See this blog post for more on what can be done from this perspective, including a clean conceptual proof that the commutator of two derivations is a derivation.
Geometrically the above definition corresponds to asking for a map from $textSpec R times textSpec mathbbZ[epsilon]/epsilon^2$ to $textSpec R$; you can think of $textSpec mathbbZ[epsilon]/epsilon^2$ as the "walking tangent vector" and of this definition as asking for a tangent vector to the identity in the "space" (it would be an affine scheme if the category of affine schemes were cartesian closed, which it's not; however it is a presheaf on the category of affine schemes, just not a representable one) of endomorphisms of $textSpec R$.
More generally you can contemplate homomorphisms $R to S[epsilon]/epsilon^2$ for two rings $R$ to $S$, or equivalently $epsilon$-linear homomorphisms $R[epsilon]/epsilon^2 to S[epsilon]/epsilon^2$; this works out to a map of the form $f + epsilon g$ where $f : R to S$ is a homomorphism and $g : R to S$ is a linear map satisfying
$$g(ab) = f(a) g(b) + g(a) f(b).$$
This is a useful generalization of the notion of a derivation; for example it can be used to define Zariski tangent spaces. It can be thought of as describing a tangent vector to $f$ in the "space" of maps from $R$ to $S$, or equivalently from $textSpec S$ to $textSpec R$. The corresponding object in differential geometry is a pair consisting of a map $f : M to N$ of smooth manifolds and a section of the pullback $f^ast(TN)$ of the tangent bundle of $N$ to $M$.
You can try to make this sort of thing work out for smooth manifolds by passing to a somewhat more complicated category that includes smooth manifolds as well as various "infinitesimal" spaces; see smooth algebra for some details.
On the more general topic of categorical approaches to tensor fields, you might enjoy skimming Kolar, Michor, and Slovak's Natural Operations in Differential Geometry.
answered Jul 17 at 23:53
Qiaochu Yuan
269k32564900
edited Jul 18 at 6:51
answered Jul 17 at 23:53
Qiaochu Yuan
269k32564900
answered Jul 17 at 23:53
Qiaochu Yuan
269k32564900
answered Jul 17 at 23:53
Qiaochu Yuan
269k32564900
269k32564900
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2
A composition of derivations is usually not a derivation, so this is not possible without generalizing what kind of morphisms you allow.
– Eric Wofsey
Jul 17 at 19:36
Oh, oops, well, that sinks that idea. Thanks! (If your comment was a top level answer I'd accept it.)
– Nate
Jul 17 at 19:44
However, augmented derivations may rescue the idea.
– Eric Towers
Jul 17 at 19:53
I feel like it's not actually the "correct" answer, though, since you can consider higher-order differential operators which will be closed under composition. I don't know their theory well enough to write a good answer, though.
– Eric Wofsey
Jul 17 at 20:13