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• CommentRowNumber1.
• CommentAuthorporton
• CommentTimeNov 22nd 2018

I have defined “categories with restricted identities”.

See my blog

Basically, I invented a way to turn a category into a semigroup (abstract away objects). This seems a small but novel discovery.

I ask the moderators permission to add this to nLab.

• CommentRowNumber2.
• CommentAuthorDavidRoberts
• CommentTimeNov 22nd 2018
• (edited Nov 22nd 2018)

The concept is not defined at that blog page, but in an unspecified location in a 75-page pdf document to which you link. Browsing the contents I didn’t manage to see any reference to ’categories with restricted identities’. As to the novelty, I suspect the construction is related the one discussed in this n-category café post, but with linear categories replaced by ordinary categories, so that algebras are replaced by semigroups.

• CommentRowNumber3.
• CommentAuthorporton
• CommentTimeNov 22nd 2018

It is currently chapter 2, page 6, where ’category with restricted identities’ is defined.

The current draft is this file.

And no, my construction is unrelated to vector spaces. And there is (seemingly) no zero element in my construction, unlike $0$ in your n-category café post.

• CommentRowNumber4.
• CommentAuthorDavidRoberts
• CommentTimeNov 22nd 2018

I didn’t say it was related to vector spaces :-) I thought it might be analogous in a way that category theory makes precise. I see that you are pointing to Definition #1956, and that it relies on knowing what a ’filtrator’ is (not defined in that document, and I’m not going to hunt it down). Moreover the objects of the category you start with are equipped with the structure of a lattice, and this lattice is part of the data of the filtrator. You have not given a construction of a semigroup from a category, but from a category whose objects have a lot of extra structure. I for one don’t find this all that novel, considering that I can give a different such construction off that top of my head without relying on all that extra structure:

Take a (small, say) category $C$, with set of arrows $mor(C)$. Then $mor(C) \amalg \{\bot\}$ is a semigroup, with multiplication $(f,g) \mapsto f\circ g$ if this is defined, and $(f,g) \mapsto \bot$ otherwise.

It doesn’t seem useful to the nLab project to include this, I’m sorry to say.

• CommentRowNumber5.
• CommentAuthorHarry Gindi
• CommentTimeNov 23rd 2018
Going to be completely impolitic here and inform you that Porton is a notorious crank from ##math on freenode irc.

It's not worth humoring him, and he doesn't really seem to get offended if you tell this to his face.
• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeNov 23rd 2018
• (edited Nov 23rd 2018)

Harry, I think there was a period where you weren’t following the nForum closely, because we’ve had many dealings with Victor Porton here, and I think many of us have formed our own opinion by now.

I do think your assessment is not completely fair. I spent some time deciphering his funcoids as he calls them, and there’s a there there: it’s related to syntopogenous space, and I even have a page on my web called ’topogeny’ where I translate some of that stuff into what I think is more familiar language. I do agree that he does not listen well to what others say to him, as partly borne out in this thread and also at length in the history of my own interactions with him. This, combined with the heavy and obscurantist and overly lengthy style which does not make much contact with the literature, virtually guarantees that he will continue to work in isolation.

I guess you can call that ’crank’ if you want, but insofar as he is actually working with definitions and theorems and is honestly trying to get at the truth as a mathematician would, in his idiosyncratic way, I wouldn’t use that word myself.

However, based on my own interactions with him, I am inclined to recuse myself from weighing in directly on this thread. I’ll let David R. speak in my stead.

• CommentRowNumber7.
• CommentAuthorHarry Gindi
• CommentTimeNov 23rd 2018
Suit yourself =)
• CommentRowNumber8.
• CommentAuthorporton
• CommentTimeNov 23rd 2018
• (edited Nov 23rd 2018)

@DavidRoberts

No, it is dissimilar to your construction. Particularly semigroup multiplication $f\circ g$ is defined (an may be not the bottom element) even if the morphisms $f$ and $g$ are no composable.

Regarding my terminology: a filtrator $(\mathfrak{A},\mathfrak{Z})$ is just a pair of a poset $\mathfrak{A}$ and its subset $\mathfrak{Z}$.

• CommentRowNumber9.
• CommentAuthorporton
• CommentTimeNov 23rd 2018

@Todd_Trimble

Dear Todd,

I am not quite in isolation: I have 168 (not counting Twitter, Facebook, etc.) followers of my math blog.

However, I have no contributions to my research (not counting people helping me through Q/A sites), even despite the LaTeX files are available at GitHub as open source.

Can you give a specific advises how to not be in isolation?

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeNov 23rd 2018

Specific advice. Okay, I’ll just suggest one thing, and then that will be the end of what I have to say. Don’t ask me any more questions.

How about writing, for a change, normal papers in a normal language that topologists (or members of some other established community) already know?

I’ll even adapt this advice to your situation, and suggest you think about the following question. Do you have any result from your theory that you can state in a single sentence to a topologist, using language he already knows, and that is not already known? Imagine the “elevator pitch”: you are in an elevator for 90 seconds with a famous topologist you really want to impress. What do you say?

You do not get to tell him about your private terminology, and you don’t get to refer him to your massive screeds. He doesn’t care. (Don’t answer this question here.)

Even better: write the proof of your result, explain your insights, in normal language he already understands.

The point is: there is no reason in the world why anybody should force themselves to read hundreds of pages of your idiosyncratic stuff, if you refuse to speak to them in their native mathematical language. They are not going to accept that you deserve the Abel Prize, just because you think so.