Mathematics

quallenjäger

6:33 PM

Is the concatenation between paths always defined?

I don't see why the set of Paths should not form a monoid

bolbteppa

Langlands gets 2018 Abel prize theguardian.com/science/alexs-adventures-in-numberland/2018/mar/…

GFauxPas

@AkivaWeinberger @quallenjäger how many times do you concatenate paths before it becomes a marathon

bolbteppa

ELI5 version of this madness:

"If we have a Galois group, then we can look at its representations. Galois Representations are really important, but they are never done in isolation. Wiles did a whole bunch of work to show that for ever Galois Representation of a certain type, you get a modular form that "matches" it. This means that you can use what you know about modular forms to conclude something about these kinds of Galois representations (eg, that there is no Galois representation of our type with a certain signature needed by a counterexample to FLT).

What is class field theory

quallenjäger

A lot

GFauxPas

@quallenjäger loops form a group as long as you pick a common point of origin for all loops

quallenjäger

6:35 PM

If I consider only paths?

Do they have a monoid structure?

GFauxPas

but for paths to be concatenated you have you deal with the possibility that the end of the first path might not be the beginning of the next path

so you'd need a way to define that

quallenjäger

I don't require my paths to be continuous

I mean, more is the problem with the associativity. Concatenation is not really associative because of different parametrization

But beside that, I don't see why the concatenation should not be defined for some paths

6

Q: Is there a monoid structure on the set of paths of a graph?

Given a graph G, and the set of paths in G called PathG. Is there a monoid structure on PathG? Will concatenation be the multiplication formula? even if it's not defined for some paths? What about the identity?

My question is related to this. I don't understand the answer.

anon

@quallenjäger how can you concatenate paths if the destination of one is not the beginning of the other?

quallenjäger

I thought it might have a jump at this point.

anon

that question is talking about paths as a sequence of edges, not a function from [0,1] into a topological space

(and in any case, the second usage of the term "path" would have to be continuous anyway, as semi notes)

quallenjäger

6:40 PM

Ah I see.

I am dealing with a set of general paths, where I only require it to be bounded variation

Does it have a monoid structure under equivalence of reparametrization?

anon

if you're allowing your "paths" to have jumps, then you're not actually talking about paths.

do you think it has an identity element?

quallenjäger

I don't know actually. I thought I could define a stationary point as an identity element.

But I need the reparametrization equivalence to achieve the associativity right?

This cannot be relaxed?

anon

it's certainly not associative without reparametrization. are you sure a stationary point is an identity?

Semiclassical

(As an aside, iirc one can do stuff involving homotopy of unbased paths. But then the space of paths forms a groupoid not a group.)

anon

yes, fundamental groupoid

quallenjäger

6:48 PM

I am not sure, I thought I can define the endpoint of the paths to be the unit element. Then under reparametrization, it will give me the original paths.

anon

"the endpoint of the paths" - wait, so they all have the same endpoint?

quallenjäger

Does it need to be unique? the identity element?

Ah ok I see.

Semiclassical

@anon kk. A pity I only know that at the level of a slogan

anon

you can't have a nonunique two-sided identity

(what do you think would happen if you were to combine distinct identity elements together?)

quallenjäger

Ok then It will only have a semi group structure I think.

anon

6:52 PM

yes

quallenjäger

@anon are you familiar with the iterated integral as a homomorphism to the semi group of the paths, which I have mentioned before.

Tobias Kildetoft

@quallenjäger If you want to stay in the realm of associative algebras (rather than move up to categories), then you just define the composition of paths to be zero if the start and end points do not match

usually you do this with quivers and get path algebras

anon

I think Tobias is talking about quiver algebras,

quallenjäger

What are quivers?

anon

basically graphs

Tobias Kildetoft

6:54 PM

just directed graphs

anon

we already found out quallen is talking about functions from [0,1] to a topological space, rather than graphs in discrete math

Tobias Kildetoft

@anon ahh, woops

but the linked question is about graphs in the sense I meant

anon

yes

quallenjäger

I am currently working on a paper by Chen, where he has developed a representation on the group of continuous paths (this can be constructed by introducing equivalence classes) into the tensor algebra. I was wondering, if the same applies to paths with bounded variation only. That's why I am asking. I need somehow a certain group structure on the set of paths with bounded variation.

Do you have an idea why it is useful to represent the group of paths into some algebraic object?

Is it because we can deal with concatenation of the paths more easily?

Tobias Kildetoft

What does it mean for a path to have bounded variation?

quallenjäger

7:01 PM

I consider a paths as a function from $[0,1]$

the function has bounded variation.

i.e. finite length

Tobias Kildetoft

Where does it go? I mean, this is not a term that applies to just arbitrary paths

quallenjäger

into $\Bbb R$

sorry

GFauxPas

whats a groupoid missing to be a group

?

Tobias Kildetoft

@GFauxPas You mean what it has that is too much

GFauxPas

idk I've never heard of it before

Tobias Kildetoft

7:03 PM

unless you mean groupoid in the older sense which is now usually called a magma

the old term is used for any set with a binary operation

the newer use is a category where all morphisms are isomorphisms

anon

If you think of a group as the set of functions on one set, a groupoid could be a set of functions between many sets, which includes all the functions' inverses and all the sets' identity functions.

GFauxPas

hm

thanks

anon

The 15-puzzle is easier to think of as a groupoid than a group, honestly

Tobias Kildetoft

@anon why? I find it more natural to consider it as a group action

anon

there are more states than just "lower right corner missing a square"

Tobias Kildetoft

7:06 PM

ahh, true

anon

each individual shift is a transition between the different states

quallenjäger

Is a polynomial function on the ring of polynomials always linear?

anon

one usually finishes the phrase "polynomial function on []" with a vector space or a variety

the ring of polynomials is itself an infinite-dimensional vector space. are you thinking of it that way?

quallenjäger

Yes

i.e. polynomials of polynomials.

anon

then replace the phrase "ring of polynomials" with "infinite-dimensional vector space." is a polynomial function on an infinite-dimensional vector space always linear?

quallenjäger

7:11 PM

That's sounds more correct :D Thanks.

anon

by "correct" do you mean accurate portrayal of your question, or do you mean more likely to have an answer of "yes"?

quallenjäger

I mean thats the formulation of the question sounds more correct.

I don't know the answer

anon

well, is a polynomial function on a finite-dimensional vector space always linear?

quallenjäger

yes