Mixing
Derivative Operator as a Functor?
Clash Royale CLAN TAG#URR8PPP
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Whilst I was trying to think of a proof for the chain rule for Fréchet derivatives, I realized it looks very similar to the naturality axiom for functors. (Except for the need to specify points for the derivative.)
$$
D(f circ g)_p = (Df)_g(p)circ (Dg)_p sim F(f circ g) = F(f) circ F(g)
$$
Is this just a coincidence? Or is this hinting at some deeper meaning of derivatives?
differential-geometry category-theory
asked Jul 20 at 9:29
user577413
838
 |Â
show 3 more comments
up vote
3
down vote
favorite
Whilst I was trying to think of a proof for the chain rule for Fréchet derivatives, I realized it looks very similar to the naturality axiom for functors. (Except for the need to specify points for the derivative.)
$$
D(f circ g)_p = (Df)_g(p)circ (Dg)_p sim F(f circ g) = F(f) circ F(g)
$$
Is this just a coincidence? Or is this hinting at some deeper meaning of derivatives?
differential-geometry category-theory
asked Jul 20 at 9:29
user577413
838
1
You might find ncatlab.org/nlab/show/chain+rule illuminating.
– Ittay Weiss
Jul 20 at 9:34
Yes, I get the gist of it, but I’m afraid I have not reached manifolds yet in my reading
– user577413
Jul 20 at 9:51
So, basically your observation is certainly onto something. That something is investigated in category theoretic circles, as the nLab page indicates, and is a good starting point if you're interested in digging in deeper.
– Ittay Weiss
Jul 20 at 9:59
1
Let $mathttMan$ be the category of smooth manifolds and smooth maps between manifolds. Then the assignment $M mapsto TM$ taking a manifold $M$ to its tangent bundle $TM$ can be extended to a functor $mathttMan to mathttMan$. How does it act on arrows? It takes a smooth map $f : M to N$ to its derivative $Df$, which naturally defines a smooth map $Df : TM to TM$. Indeed, the chain rule really does become $D(f circ g) = Df circ Dg$ as maps between tangent bundles.
– Branimir Ćaćić
Jul 20 at 11:41
1
@MusaAl-hassy In this context, the derivative of $f : mathbbR^m to mathbbR^n$ at a point $x$ is its best approximation by a linear transformation $Df(x) : mathbbR^m to mathbbR^n$ at that point, in the sense that $|f(x+h)-f(x)-Df(x)(h)| = o(|h|)$; from this perspective, the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity itself, i.e., for every $x in mathbbR^n$, $DI(x) = I$, which means, in the language of vector bundles, that the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity $TmathbbR^n to TmathbbR^n$.
– Branimir Ćaćić
Jul 20 at 15:28
 |Â
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Whilst I was trying to think of a proof for the chain rule for Fréchet derivatives, I realized it looks very similar to the naturality axiom for functors. (Except for the need to specify points for the derivative.)
$$
D(f circ g)_p = (Df)_g(p)circ (Dg)_p sim F(f circ g) = F(f) circ F(g)
$$
Is this just a coincidence? Or is this hinting at some deeper meaning of derivatives?
differential-geometry category-theory
asked Jul 20 at 9:29
user577413
838
Whilst I was trying to think of a proof for the chain rule for Fréchet derivatives, I realized it looks very similar to the naturality axiom for functors. (Except for the need to specify points for the derivative.)
$$
D(f circ g)_p = (Df)_g(p)circ (Dg)_p sim F(f circ g) = F(f) circ F(g)
$$
Is this just a coincidence? Or is this hinting at some deeper meaning of derivatives?
differential-geometry category-theory
asked Jul 20 at 9:29
user577413
838
asked Jul 20 at 9:29
user577413
838
asked Jul 20 at 9:29
user577413
838
asked Jul 20 at 9:29
user577413
838
838
1
You might find ncatlab.org/nlab/show/chain+rule illuminating.
– Ittay Weiss
Jul 20 at 9:34
Yes, I get the gist of it, but I’m afraid I have not reached manifolds yet in my reading
– user577413
Jul 20 at 9:51
So, basically your observation is certainly onto something. That something is investigated in category theoretic circles, as the nLab page indicates, and is a good starting point if you're interested in digging in deeper.
– Ittay Weiss
Jul 20 at 9:59
1
Let $mathttMan$ be the category of smooth manifolds and smooth maps between manifolds. Then the assignment $M mapsto TM$ taking a manifold $M$ to its tangent bundle $TM$ can be extended to a functor $mathttMan to mathttMan$. How does it act on arrows? It takes a smooth map $f : M to N$ to its derivative $Df$, which naturally defines a smooth map $Df : TM to TM$. Indeed, the chain rule really does become $D(f circ g) = Df circ Dg$ as maps between tangent bundles.
– Branimir Ćaćić
Jul 20 at 11:41
1
@MusaAl-hassy In this context, the derivative of $f : mathbbR^m to mathbbR^n$ at a point $x$ is its best approximation by a linear transformation $Df(x) : mathbbR^m to mathbbR^n$ at that point, in the sense that $|f(x+h)-f(x)-Df(x)(h)| = o(|h|)$; from this perspective, the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity itself, i.e., for every $x in mathbbR^n$, $DI(x) = I$, which means, in the language of vector bundles, that the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity $TmathbbR^n to TmathbbR^n$.
– Branimir Ćaćić
Jul 20 at 15:28
 |Â
show 3 more comments
1
You might find ncatlab.org/nlab/show/chain+rule illuminating.
– Ittay Weiss
Jul 20 at 9:34
Yes, I get the gist of it, but I’m afraid I have not reached manifolds yet in my reading
– user577413
Jul 20 at 9:51
So, basically your observation is certainly onto something. That something is investigated in category theoretic circles, as the nLab page indicates, and is a good starting point if you're interested in digging in deeper.
– Ittay Weiss
Jul 20 at 9:59
1
Let $mathttMan$ be the category of smooth manifolds and smooth maps between manifolds. Then the assignment $M mapsto TM$ taking a manifold $M$ to its tangent bundle $TM$ can be extended to a functor $mathttMan to mathttMan$. How does it act on arrows? It takes a smooth map $f : M to N$ to its derivative $Df$, which naturally defines a smooth map $Df : TM to TM$. Indeed, the chain rule really does become $D(f circ g) = Df circ Dg$ as maps between tangent bundles.
– Branimir Ćaćić
Jul 20 at 11:41
1
@MusaAl-hassy In this context, the derivative of $f : mathbbR^m to mathbbR^n$ at a point $x$ is its best approximation by a linear transformation $Df(x) : mathbbR^m to mathbbR^n$ at that point, in the sense that $|f(x+h)-f(x)-Df(x)(h)| = o(|h|)$; from this perspective, the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity itself, i.e., for every $x in mathbbR^n$, $DI(x) = I$, which means, in the language of vector bundles, that the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity $TmathbbR^n to TmathbbR^n$.
– Branimir Ćaćić
Jul 20 at 15:28
1
1
You might find ncatlab.org/nlab/show/chain+rule illuminating.
– Ittay Weiss
Jul 20 at 9:34
You might find ncatlab.org/nlab/show/chain+rule illuminating.
– Ittay Weiss
Jul 20 at 9:34
Yes, I get the gist of it, but I’m afraid I have not reached manifolds yet in my reading
– user577413
Jul 20 at 9:51
Yes, I get the gist of it, but I’m afraid I have not reached manifolds yet in my reading
– user577413
Jul 20 at 9:51
So, basically your observation is certainly onto something. That something is investigated in category theoretic circles, as the nLab page indicates, and is a good starting point if you're interested in digging in deeper.
– Ittay Weiss
Jul 20 at 9:59
So, basically your observation is certainly onto something. That something is investigated in category theoretic circles, as the nLab page indicates, and is a good starting point if you're interested in digging in deeper.
– Ittay Weiss
Jul 20 at 9:59
1
1
Let $mathttMan$ be the category of smooth manifolds and smooth maps between manifolds. Then the assignment $M mapsto TM$ taking a manifold $M$ to its tangent bundle $TM$ can be extended to a functor $mathttMan to mathttMan$. How does it act on arrows? It takes a smooth map $f : M to N$ to its derivative $Df$, which naturally defines a smooth map $Df : TM to TM$. Indeed, the chain rule really does become $D(f circ g) = Df circ Dg$ as maps between tangent bundles.
– Branimir Ćaćić
Jul 20 at 11:41
Let $mathttMan$ be the category of smooth manifolds and smooth maps between manifolds. Then the assignment $M mapsto TM$ taking a manifold $M$ to its tangent bundle $TM$ can be extended to a functor $mathttMan to mathttMan$. How does it act on arrows? It takes a smooth map $f : M to N$ to its derivative $Df$, which naturally defines a smooth map $Df : TM to TM$. Indeed, the chain rule really does become $D(f circ g) = Df circ Dg$ as maps between tangent bundles.
– Branimir Ćaćić
Jul 20 at 11:41
1
1
@MusaAl-hassy In this context, the derivative of $f : mathbbR^m to mathbbR^n$ at a point $x$ is its best approximation by a linear transformation $Df(x) : mathbbR^m to mathbbR^n$ at that point, in the sense that $|f(x+h)-f(x)-Df(x)(h)| = o(|h|)$; from this perspective, the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity itself, i.e., for every $x in mathbbR^n$, $DI(x) = I$, which means, in the language of vector bundles, that the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity $TmathbbR^n to TmathbbR^n$.
– Branimir Ćaćić
Jul 20 at 15:28
@MusaAl-hassy In this context, the derivative of $f : mathbbR^m to mathbbR^n$ at a point $x$ is its best approximation by a linear transformation $Df(x) : mathbbR^m to mathbbR^n$ at that point, in the sense that $|f(x+h)-f(x)-Df(x)(h)| = o(|h|)$; from this perspective, the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity itself, i.e., for every $x in mathbbR^n$, $DI(x) = I$, which means, in the language of vector bundles, that the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity $TmathbbR^n to TmathbbR^n$.
– Branimir Ćaćić
Jul 20 at 15:28
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
The way I prefer to say things is this. There is a functor from, say, the category of pointed smooth manifolds to the category of vector spaces, which takes a smooth map $f : (M, m) to (N, n)$ (so $m in M, n in N$ and $f(m) = n$) to the derivative $df_m : T_m(M) to T_n(N)$. The fact that this respects composition is precisely the chain rule. There are many variations on this.
answered Jul 20 at 18:58
Qiaochu Yuan
269k32564900
add a comment |Â
up vote
2
down vote
If you're not ready to think about manifolds, you can still think of the derivative as a functor on the category whose objects are $mathbbR^m$ and whose morphisms are smooth maps. The (total) derivative of $f:mathbbR^mto mathbbR^n$ is $Tf:mathbbR^2mto mathbbR^2n$, sending $(vec x,vec y)$ to $(f(vec x),Df_vec x(vec y))$. This is the specialization of the notion of "tangent bundle" mentioned in the comments to Euclidean spaces.
answered Jul 20 at 18:41
Kevin Carlson
29.2k23065
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The way I prefer to say things is this. There is a functor from, say, the category of pointed smooth manifolds to the category of vector spaces, which takes a smooth map $f : (M, m) to (N, n)$ (so $m in M, n in N$ and $f(m) = n$) to the derivative $df_m : T_m(M) to T_n(N)$. The fact that this respects composition is precisely the chain rule. There are many variations on this.
answered Jul 20 at 18:58
Qiaochu Yuan
269k32564900
add a comment |Â
up vote
4
down vote
accepted
The way I prefer to say things is this. There is a functor from, say, the category of pointed smooth manifolds to the category of vector spaces, which takes a smooth map $f : (M, m) to (N, n)$ (so $m in M, n in N$ and $f(m) = n$) to the derivative $df_m : T_m(M) to T_n(N)$. The fact that this respects composition is precisely the chain rule. There are many variations on this.
answered Jul 20 at 18:58
Qiaochu Yuan
269k32564900
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The way I prefer to say things is this. There is a functor from, say, the category of pointed smooth manifolds to the category of vector spaces, which takes a smooth map $f : (M, m) to (N, n)$ (so $m in M, n in N$ and $f(m) = n$) to the derivative $df_m : T_m(M) to T_n(N)$. The fact that this respects composition is precisely the chain rule. There are many variations on this.
answered Jul 20 at 18:58
Qiaochu Yuan
269k32564900
The way I prefer to say things is this. There is a functor from, say, the category of pointed smooth manifolds to the category of vector spaces, which takes a smooth map $f : (M, m) to (N, n)$ (so $m in M, n in N$ and $f(m) = n$) to the derivative $df_m : T_m(M) to T_n(N)$. The fact that this respects composition is precisely the chain rule. There are many variations on this.
answered Jul 20 at 18:58
Qiaochu Yuan
269k32564900
answered Jul 20 at 18:58
Qiaochu Yuan
269k32564900
answered Jul 20 at 18:58
Qiaochu Yuan
269k32564900
answered Jul 20 at 18:58
Qiaochu Yuan
269k32564900
269k32564900
add a comment |Â
add a comment |Â
up vote
2
down vote
If you're not ready to think about manifolds, you can still think of the derivative as a functor on the category whose objects are $mathbbR^m$ and whose morphisms are smooth maps. The (total) derivative of $f:mathbbR^mto mathbbR^n$ is $Tf:mathbbR^2mto mathbbR^2n$, sending $(vec x,vec y)$ to $(f(vec x),Df_vec x(vec y))$. This is the specialization of the notion of "tangent bundle" mentioned in the comments to Euclidean spaces.
answered Jul 20 at 18:41
Kevin Carlson
29.2k23065
add a comment |Â
up vote
2
down vote
If you're not ready to think about manifolds, you can still think of the derivative as a functor on the category whose objects are $mathbbR^m$ and whose morphisms are smooth maps. The (total) derivative of $f:mathbbR^mto mathbbR^n$ is $Tf:mathbbR^2mto mathbbR^2n$, sending $(vec x,vec y)$ to $(f(vec x),Df_vec x(vec y))$. This is the specialization of the notion of "tangent bundle" mentioned in the comments to Euclidean spaces.
answered Jul 20 at 18:41
Kevin Carlson
29.2k23065
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you're not ready to think about manifolds, you can still think of the derivative as a functor on the category whose objects are $mathbbR^m$ and whose morphisms are smooth maps. The (total) derivative of $f:mathbbR^mto mathbbR^n$ is $Tf:mathbbR^2mto mathbbR^2n$, sending $(vec x,vec y)$ to $(f(vec x),Df_vec x(vec y))$. This is the specialization of the notion of "tangent bundle" mentioned in the comments to Euclidean spaces.
answered Jul 20 at 18:41
Kevin Carlson
29.2k23065
If you're not ready to think about manifolds, you can still think of the derivative as a functor on the category whose objects are $mathbbR^m$ and whose morphisms are smooth maps. The (total) derivative of $f:mathbbR^mto mathbbR^n$ is $Tf:mathbbR^2mto mathbbR^2n$, sending $(vec x,vec y)$ to $(f(vec x),Df_vec x(vec y))$. This is the specialization of the notion of "tangent bundle" mentioned in the comments to Euclidean spaces.
answered Jul 20 at 18:41
Kevin Carlson
29.2k23065
answered Jul 20 at 18:41
Kevin Carlson
29.2k23065
answered Jul 20 at 18:41
Kevin Carlson
29.2k23065
answered Jul 20 at 18:41
Kevin Carlson
29.2k23065
29.2k23065
add a comment |Â
add a comment |Â
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1
You might find ncatlab.org/nlab/show/chain+rule illuminating.
– Ittay Weiss
Jul 20 at 9:34
Yes, I get the gist of it, but I’m afraid I have not reached manifolds yet in my reading
– user577413
Jul 20 at 9:51
So, basically your observation is certainly onto something. That something is investigated in category theoretic circles, as the nLab page indicates, and is a good starting point if you're interested in digging in deeper.
– Ittay Weiss
Jul 20 at 9:59
1
Let $mathttMan$ be the category of smooth manifolds and smooth maps between manifolds. Then the assignment $M mapsto TM$ taking a manifold $M$ to its tangent bundle $TM$ can be extended to a functor $mathttMan to mathttMan$. How does it act on arrows? It takes a smooth map $f : M to N$ to its derivative $Df$, which naturally defines a smooth map $Df : TM to TM$. Indeed, the chain rule really does become $D(f circ g) = Df circ Dg$ as maps between tangent bundles.
– Branimir Ćaćić
Jul 20 at 11:41
1
@MusaAl-hassy In this context, the derivative of $f : mathbbR^m to mathbbR^n$ at a point $x$ is its best approximation by a linear transformation $Df(x) : mathbbR^m to mathbbR^n$ at that point, in the sense that $|f(x+h)-f(x)-Df(x)(h)| = o(|h|)$; from this perspective, the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity itself, i.e., for every $x in mathbbR^n$, $DI(x) = I$, which means, in the language of vector bundles, that the derivative of the identity $I : mathbbR^n to mathbbR^n$ is the identity $TmathbbR^n to TmathbbR^n$.
– Branimir Ćaćić
Jul 20 at 15:28