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1. • CommentRowNumber1.
   • CommentAuthorEric
   • CommentTimeAug 11th 2010
   • (edited Aug 11th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexHi, Over in the discussion on [semidirect product](http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1668&Focus=14840#Comment_14840), I cooked up something that looks like it might have general applicability. As such, it is almost certainly not "new". So one of the purposes for writing this here is to ask for references. If you've seen something like this, I'd love to read up on it, so please let me know. Of course, the whole thing could just be bogus :) Another purpose is to seek help ironing out any details I've left out. The idea is that given any $(k-1)$-[[k-morphism|morphisms]] $f,g:x\to y$ and $k$-morphism $$\alpha:f\to g,$$ we can define a $(k-1)$-endomorphism $$\partial\alpha:y\to y$$ such that $$\partial\alpha\circ g = f.$$ I think the "=" here is important. This makes sense even when the category is not a groupoid. Now, a $D$-valued differential form is just a functor $$F:C\to D.$$ If $C$ and $D$ are $(\infty,p)$-categories, we could call $F:C\to D$ a $D$-valued $p$-form. We can define the exterior derivative of a $D$-valued form in the obvious way such that Stokes' theory is satisfied, i.e. Stokes theorem is taken to define $d$. That is, we define $$d F(\alpha) = F(\partial\alpha)$$ so we have $$F(\partial\alpha\circ g) = F(\partial\alpha)\circ F(g) = d F(\alpha)\circ F(g) = F(f).$$ Note, this version of Stokes' theorem (I believe) is defined for any category. Also note that I've used bigons for the sake of communicating the idea, but the language I used to cook this up is that of $k$-maps (to avoid a name clash with $k$-morphisms) defined [here](http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1668&Focus=14840#Comment_14840).

   Hi,

   Over in the discussion on semidirect product, I cooked up something that looks like it might have general applicability. As such, it is almost certainly not “new”. So one of the purposes for writing this here is to ask for references. If you’ve seen something like this, I’d love to read up on it, so please let me know. Of course, the whole thing could just be bogus :)

   Another purpose is to seek help ironing out any details I’ve left out.

   The idea is that given any $(k-1)$-morphisms $f,g:x\to y$ and $k$-morphism

   $$\alpha:f\to g,$$

   we can define a $(k-1)$-endomorphism

   $$\partial\alpha:y\to y$$

   such that

   $$\partial\alpha\circ g = f.$$

   I think the “=” here is important.

   This makes sense even when the category is not a groupoid.

   Now, a $D$-valued differential form is just a functor

   $$F:C\to D.$$

   If $C$ and $D$ are $(\infty,p)$-categories, we could call $F:C\to D$ a $D$-valued $p$-form.

   We can define the exterior derivative of a $D$-valued form in the obvious way such that Stokes’ theory is satisfied, i.e. Stokes theorem is taken to define $d$. That is, we define

   $$d F(\alpha) = F(\partial\alpha)$$

   so we have

   $$F(\partial\alpha\circ g) = F(\partial\alpha)\circ F(g) = d F(\alpha)\circ F(g) = F(f).$$

   Note, this version of Stokes’ theorem (I believe) is defined for any category.

   Also note that I’ve used bigons for the sake of communicating the idea, but the language I used to cook this up is that of $k$-maps (to avoid a name clash with $k$-morphisms) defined here.

2. • CommentRowNumber2.
   • CommentAuthorTobyBartels
   • CommentTimeAug 11th 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItex>[…] given any […] we can […] That's not true in general, but maybe you want to focus attention on (higher) categories equipped with an operation $\partial$ such that this is true? Even so, I'd be inclined to think that this could only work if $\alpha$ is an automorphism (invertible, or at least weakly invertible). In that case, I believe that you should be able to replace $C$ with an equivalent category such that the statement is always true (when $\alpha$ is an automorphism). Also, I think that you should have both a target boundary and a source boundary; the target boundary is $\tau \alpha\colon y \to y$, while the source boundary is $\sigma \alpha\colon x \to x$. The boundary as we normally think of it is something that exists only in additive situations, where we can write $\partial \alpha = \tau \alpha - \sigma \alpha$. (And now that I write this down, I see that this really only makes sense when $x = y$.) That is my intuition, at least.

   […] given any […] we can […]

   That’s not true in general, but maybe you want to focus attention on (higher) categories equipped with an operation $\partial$ such that this is true? Even so, I’d be inclined to think that this could only work if $\alpha$ is an automorphism (invertible, or at least weakly invertible). In that case, I believe that you should be able to replace $C$ with an equivalent category such that the statement is always true (when $\alpha$ is an automorphism).

   Also, I think that you should have both a target boundary and a source boundary; the target boundary is $\tau \alpha\colon y \to y$, while the source boundary is $\sigma \alpha\colon x \to x$. The boundary as we normally think of it is something that exists only in additive situations, where we can write $\partial \alpha = \tau \alpha - \sigma \alpha$. (And now that I write this down, I see that this really only makes sense when $x = y$.) That is my intuition, at least.

3. • CommentRowNumber3.
   • CommentAuthorEric
   • CommentTimeAug 11th 2010
   • (edited Aug 11th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexHi Toby, Thanks for your feedback :) I'll have to think about the automorphism issue, but I thought I would try to clarify one point. Integration is actually more like "product integration" not "summation". In the case of groupoid, where the intuition is clear, we'd have $\partial\alpha\circ g = f$ as usual, but since it is a groupoid, this can be rewritten as $$\partial\alpha = f\circ g^{-1}.$$ This is opposed to the way we usually might think of it, i.e. as a difference $f - g$. All we need is composition. Not summation or subtraction. I'm not sure how/if this viewpoint might change your earlier comments.

   Hi Toby,

   Thanks for your feedback :)

   I’ll have to think about the automorphism issue, but I thought I would try to clarify one point.

   Integration is actually more like “product integration” not “summation”. In the case of groupoid, where the intuition is clear, we’d have $\partial\alpha\circ g = f$ as usual, but since it is a groupoid, this can be rewritten as

   $$\partial\alpha = f\circ g^{-1}.$$

   This is opposed to the way we usually might think of it, i.e. as a difference $f - g$. All we need is composition. Not summation or subtraction.

   I’m not sure how/if this viewpoint might change your earlier comments.

4. • CommentRowNumber4.
   • CommentAuthorEric
   • CommentTimeAug 11th 2010
   • (edited Aug 11th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexThis is a thought that just popped into my head: maybe we can think of $\partial\alpha$ as a (higher) [[coequalizer]] of $(k-1)$-morphisms $f,g:x\to y$. Or something... PS: Everything I write here should be prefixed with "(higher)", but I was being lazy and assuming all higher stuff exists. Trying to channel Urs. Everything is an $\infinity$-category :) Edit: In grad school, my friend and I had an expression: "Every idea is a good idea... for the first 5 minutes." Then the reality sets in :) This one took less than 5 minutes. I no longer want to think of $\partial\alpha$ as a coequalizer, but will leave this comment here for the record. The "spirit" of $\partial\alpha$ is similar to that of a coequalizer though. It is an endomorphism (as opposed to object with components) that is inserted in order to make diagrams commute.

   This is a thought that just popped into my head: maybe we can think of $\partial\alpha$ as a (higher) coequalizer of $(k-1)$-morphisms $f,g:x\to y$. Or something…

   PS: Everything I write here should be prefixed with “(higher)”, but I was being lazy and assuming all higher stuff exists. Trying to channel Urs. Everything is an $\infinity$-category :)

   Edit: In grad school, my friend and I had an expression: “Every idea is a good idea… for the first 5 minutes.” Then the reality sets in :)

   This one took less than 5 minutes. I no longer want to think of $\partial\alpha$ as a coequalizer, but will leave this comment here for the record.

   The “spirit” of $\partial\alpha$ is similar to that of a coequalizer though. It is an endomorphism (as opposed to object with components) that is inserted in order to make diagrams commute.

5. • CommentRowNumber5.
   • CommentAuthorTobyBartels
   • CommentTimeAug 12th 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexYou're right, what I wrote about subtraction was silly. The target boundary of $\alpha$ is simply $g$, and the source boundary is simply $f$, so it makes sense to define $\partial \alpha$ as $f \circ g^{-1}$. However, there are still two possible meanings of $\partial \alpha$; the other is $g \circ f^{-1}$. So that is what led to my comment, mistaken as that comment was.

   You’re right, what I wrote about subtraction was silly.

   The target boundary of $\alpha$ is simply $g$, and the source boundary is simply $f$, so it makes sense to define $\partial \alpha$ as $f \circ g^{-1}$.

   However, there are still two possible meanings of $\partial \alpha$; the other is $g \circ f^{-1}$. So that is what led to my comment, mistaken as that comment was.

6. • CommentRowNumber6.
   • CommentAuthorEric
   • CommentTimeAug 12th 2010
   • (edited Aug 12th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexYeah. There is a choice. I think I am going to switch to the second option you mention, i.e. $$\partial\alpha\circ f = g.$$ This way, the action of $\alpha$ on $f$ is post composition with the boundary, i.e. $$\alpha\cdot f = \partial\alpha\circ f = g.$$ Using this boundary operation, last night I demonstrated that if the boundary exists, it defines a calculus and using this calculus you can derive/confirm the interchange law algebraically in a straightforward manner. I suspect the proof goes both ways and the interchange law implies the existence of a boundary so that a boundary operation of 2-morphisms is inherent in the definition of a (strict?) 2-category. It may not be "new", but it is not devoid of utility either. I think with this, I am finally starting to understand 2-categories :) I'll put the proof on my personal web and provide a link shortly. Edit: Here is the link * [[ericforgy:Boundary of a 2-Morphism and the Interchange Law]]

   Yeah. There is a choice. I think I am going to switch to the second option you mention, i.e.

   $$\partial\alpha\circ f = g.$$

   This way, the action of $\alpha$ on $f$ is post composition with the boundary, i.e.

   $$\alpha\cdot f = \partial\alpha\circ f = g.$$

   Using this boundary operation, last night I demonstrated that if the boundary exists, it defines a calculus and using this calculus you can derive/confirm the interchange law algebraically in a straightforward manner. I suspect the proof goes both ways and the interchange law implies the existence of a boundary so that a boundary operation of 2-morphisms is inherent in the definition of a (strict?) 2-category.

   It may not be “new”, but it is not devoid of utility either. I think with this, I am finally starting to understand 2-categories :)

   I’ll put the proof on my personal web and provide a link shortly.

   Edit: Here is the link

   • Boundary of a 2-Morphism and the Interchange Law (ericforgy)