Geometry & Topology: Dank Ass Ganja

3:05 AM

Jan 20 at 7:56, by Martin Sleziak

Posting a dummy message to keep the room from getting frozen.

16 hours later…

6:59 PM

Hello all, I would like to know why the sets of germs at different points are disjoint

@MartinSleziak do you think you can help me with that doubt?

7:12 PM

What is a germ of a function at a point?

8:04 PM

From what I see in the Wikipedia article, germs are equivalence classes of functions. The equivalence relation is determined by the condition that the two functions coincide on some neighborhood of $x$.

Let us denote them by $[f]_x$, $[f]_y$ as in the Wikipedia article.

@MartinSleziak If you are answering me: it was a rhetorical question. I think that given suitable slight pushes, one should be able to answer the question by chasing definitions.

@Anoldmaninthesea. So you are basically asking to show that $x\ne y$ implies $[f]_x\ne [f]_y$, right?

Just think whether there is a neighborhood $U$ of $x$ and a function $g$ such that $g|_U=0$ and $g(y)\ne 0$.

@MikeMiller This is not really my area, but I suppose that this is the direction to go...?

I'm not sure whether this falls under "slight push", or whether this was too much.

I see now what you are thinking. I wouldn't have phrased it that way, but if you ask any two mathematicians to prove something, the arguments you get will probably look a little different.

Personally, my tendency is to say almost nothing (I like to be socratic). But I don't begrudge someone who has different taste.

Yes, I was assuming that we have sufficient separation axioms. (Response to the deleted message.)

Do you think that Hausdorff should be enough?

Hausdorff is certainly enough. I suspect this question arises in the context of manifolds, too.

(Which are essentially always assumed Hausdorff.)

8:14 PM

Hm, so my approach requires something stronger that Hausdorffness. (Probably normal Hausdorff. I have used a function separating the point $y$ and $\overline U$.)

In any case, I should probably leave it to you and An old man in the sea. We will see whether they respond.

Actually, let me take a step back. What we are proving is best left to a definition.

These sort of set-level things are unimportant, because one should never be thinking of $[f]_x$ and $[f]_y$ as meaning the same thing.

I would say that $[f]_x$ is equivalence classes of the following data: $(x, U, f)$, where $x$ is a fixed point, $x \in U$, ans $f: U \to \Bbb R$ is continuous. The equivalence relation will not mention $x$.

Therefore, these are defined to be disjoint.

Are there even situations where it is useful to look at germs at different points?

Not that I know of.

On the other hand, it is interesting to ask what conditions you need to avoid pushing this to a definition.

(It's just not necessary for the theory.)

8:31 PM

@MartinSleziak It seems that there are T3 spaces for which no two points can be separated by a function. So your T4 assumption seems necessary.

8:58 PM

@MartinSleziak why is disjointness equivalent to $\neq$ for the sets of germs? I'm thinking of $C^{\infty}_x$

$C^{\infty}_x(M)$

9:22 PM

@Anoldmaninthesea. You are dealing with two partitions (given by the equivalence relations).

If I have an equivalence class $[f]_x$ which is a germ at $x$, I want to show that it is not a germ at $y$.

I.e., I want to show that it is not equal to any equivalence class of the eq. relation $\sim_y$.

But since it contains $f$, the only possible class at $y$ which it could be equal is $[f]_y$.

So if I know that $[f]_x\ne [f]_y$ for any $f$, I am basically done.

If I take any element of the set of germs at $x$, it has the form $[f]_x$. But from what I wrote above, it is not equal to any of the germs at $y$.

BTW where did the question come from? Is this formulated in this way in some textbook?

@MartinSleziak my question comes from the answer to this question. math.stackexchange.com/questions/228840/…

I posted a comment there, but I didn't get a response

Q: Why are $C^\infty_p\neq C^\infty_q$ when $p\neq q$?

I see.

But how are we sure that we always have $C^\infty_p \neq C^\infty_q$ for $q\neq p$? — An old man in the sea. Nov 3 '18 at 12:50

Let $C^\infty_p$ be the set of all germs of $C^\infty(\mathbb{R}^n)$ at point $p$. Why do we always have $C^\infty_p\neq C^\infty_q$ when $p\neq q$? I need to understand this in order to understand how the tangent spaces 'as the' set of derivations are always disjoint for different points.

yes, but today I wrote a comment to the answer I had previously accepted. I think it's not correct. I left there a comment

It seems that this depends quite a lot in definition.

What I wrote here was assuming that $C^\infty_p$ is partition of a set of functions.

An answer to your question uses equivalence relations on pairs $(U,f)$.

Mike Miller suggested here in chat to use triples.

Maybe you could help me with the comment I left with a doubt about the answer«I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a V? The representatives of f are representatives, but at different points... » I