@BalarkaSen the normal bundle only needs a metric to show it is a subbundle, right?

Yes, to get the embedding of the normal space in the ambient tangent space.

But otherwise as a topological object it doesn't need a metric, right?

Nope.

Nope as in that is right, or nope you are an idiot anakhro and u need 2 lern 2 math

(or both. :P)

It's right; you can define quotient of vector bundles without using a metric, like you can define quotient of vector spaces without an inner product.

For a submanifold $N \subset M$, the normal bundle is just $TM|_N/TN$

I have oddly only ever read it as being defined with a metric.

Which is an odd way.

If you have a metric on $M$, there is a natural embedding of $TM|_N/TN$ as the orthocomplement of $TN$ in $TM|_N$, that's all.

Well, "normal" makes more geometric sense if you have a metric. But eg look in Hirsch or something

He certainly defines normal bundle without using a metric.

Does choice of metric matter at all, or only geometric realization?

I don't know what that question means.

You can define the normal bundle without a metric. You cannot embed it in a canonical way inside $TM|_N$ without a metric.

As I understand it, choosing a metric is like choosing a basis, and gentlemen try to get away without doing it as it entails a choice

So the embedding would depend on the metric.

This is a linear algebra problem, @anakhro. Given a subspace $W \subset V$, the normal space of $W$ to $V$ is $V/W$. There is no canonical embedding $V/W \to V$ if you don't have an inner product on $V$. That's all.

Yes.

But if I have a metric $g$ and a metric $\tilde g$, the relation between the embeddings is purely visual?

I mean, then they are different embeddings.

ya the problem is I don't know anything about Riemannian geometry. Is there like an isotopy between any two embeddings then that arises from a connection between the metrics, orrrr....

For the linear algebra question, they are related by an invertible linear transformation of $V$, by the change of basis matrix the inner products are related by

isometric isotopy, maybe?

There is no Riemannian geometry in this, just linear algebra

O K

$\nu$ for a normal bundle

Wonder if that is just a play on _n_u or some better reason.

gonna start using $\tau$ for tangent bundles

Too small, looks funny.

@BalarkaSen a long time ago we talked about framings of knots as homotopy classes of trivialisations.

of normal bundles, that is.

The homotopy would be more precisely a homotopy in the complement $M\setminus K$, $M$ the ambient space and $K$ the knot, right?

I have forgotten what the context is, can you remind me?

knots

framings

I don't think there was more context.

Framing numbers?

Don't worry about it, I think I have it anyway.

k

@BalarkaSen what have you been up to?

Learning simplicial sets

Nice!

Anything interesting so far?

Also, listened to any good music lately?

I have been writing down whatever I'm thinking/stumbling into here, mostly it's abstract garbage but some things are nice

Not sure what kind of things I can prove with this machinery yet

But it's interesting

@anakhro I have been relistening to Persefone's "Spiritual Migration"

It's quite fun

Nice, nice, nice. Have you ever heard of the Physics House Band?

Nope

Experimental rock band, pretty cool stuff sometimes.

I'll check it out

Hello!

Let's talk about functions and relations

My professor said that any function has inverse, this inverse is always a relation but it can be a function too, if the original function is bijective

Is that true? Thanks!

@manooooh Any function is a relation. Any relation has an inverse relation.

The inverse relation of a function is a function by definition when a function is bijective.

The conditions of "surjective" and "injective" are basically the conditions "total" and "single valued" for the inverse relation.

Keep in mind the finely corrected statement that "any function (seen as a relation) has an inverse relation".

Saying "any function has an inverse" is ambiguous.

And 9 times out of 10, people will tell you that you are wrong as it colloquially would sound like you are saying any function has an inverse function.

@anakhro but I am not saying that, the people need to read carefully, I said "any function has inverse, this inverse is always a relation" which is the same as "any function has an inverse relation"

@anakhro ok, thanks!!

The statement "any function has inverse" is ambiguous.

"inverse" is not something.

"Inverse relation" is something. "Inverse function" is something different. etc.

@manooooh When people say "$f$ has an inverse", they usually mean "$f$ has an inverse function".

If you don't want to be misunderstood, you should say "$f$ has an inverse relation".

@user76284 but you have to read carefully, if not we can interpet anything

But thanks! I will try to be more careful

lol

It's not really a matter of reading carefully, it's just the way the phrase "$f$ has an inverse" is interpreted by most people.

@user76284 I have not end the sentence in "$f$ has an inverse"...

You have ended a clause though, so it should be grammatically complete.

It's not complete though.

@anakhro sorry but a , is a pause, I can perfectly delete it and now you have no arguments to say that I ended the sentence there

I think you misunderstand. The issue is you say "any function has [an] inverse" (I assume you mean to have "an" there), which is not complete.

What is an "inverse"?

You don't ever specify.

Hello! Numerics question from non-mathematician : let Ax = y be over-determined real finite system of linear equations. The number of linear independent equations is of the same order as variables, each about 10^5. Now, what happens to me is I can find many "equally-good" approximate solutions. I don't know is this typical behavior or specific to my problem. For an example, for different initial guesses solver gives me different solutions,

its like the least-square solutions are kinda degenerated. This gives me the opportunity to setup additional constrains to pick up solutions that have properties that I like and still be almost "the best" least-square approximate solution. I LIKE this but it bothers me that I get free stuff by imposing additional constrains on already over-determined system.

I understand that question is about something that in nowhere near of mathematical rigor, but shows up consistantly. Anyone care to comment?

Over-determined doesn't mean that you have completely determined a system.

It just means that there are more equations than variables. Many of these equations might be superfluous.

Ok, I meant that there are more linearly independent equations than variables

or, I can prove that there is no exact solution.

uhh, that should not be possible.

A is just a matrix?

yes

Say A is mxn, m rows, n variables. Then if you have k rows that are linearly independent, then you must have k<=n

Since each row is from R^n.

And R^n has dimension n.

So any linearly independent subset of R^n has cardinality <=n.

Hm

Ok, let say we work in R^n. Each equation defines a line in that space, right?

Yes.

Inconsistent system is when I pick m lines that don't share a common point?

As far as I understand this, I can pick M lines that don't share common point but are also mutually different

Yes. That is, the system of the equations has no solution.

and this M can be arbitrarly high.

M>n in any case

That's not "linear independence" though.

So, this is what I thought that over-determined. incosistant system is - such M lines that don't intersect and my solver is looking for a point that is as close as possible in a least square sense

Inconsistent means there are NO solutions to the system. Over-determined is when there are more equations than variables.

Your system is known to be inconsistent?

yes

what did you mean by "Over-determined doesn't mean that you have completely determined a system."

Hmmm, this is kind of out of my knowledge.

yeah no problem

I meant that it doesn't mean that it gives a unique solution.

That is, it doesn't mean that adding information won't fine tune your solutions.

@Olexot sure

@StipeGalić This still holds true in your case, though.

Over-determined doesn't imply adding new information won't narrow down solutions.

yeah thats what bugs me

An extreme situation is the matrix A with 100 rows, 1 variable, each row being the equation 0 = 1.

I don't know how you really are approximating such a system, but like I mean, making it overdetermined doesn't change any information than leaving it as not over-determined.

So if your approximation depends on some other information, and you throw more information in there, then that's why you are getting something different.

In the non-approximation setting, inconsistency sort of makes the problem pointless/trivial (since you will always have 0 solutions for an inconsistent system, no matter how much information you give).

But it still holds without the inconsistency.

@StipeGalić hopefully this sort of makes sense.

hi chat

h i

it does

@StipeGalić good!

hullo

hi @ÍgjøgnumMeg

Heya @Ted

Quiet in here tonight

Yup. It is.

Hi professor @TedShifrin

hi @skull

reading some Pre-Gauss number theory, it's pretty dry

I haven't read original source materials in number theory, I confess

Well, I really mean pre-Gauss-style

It's not that the proofs aren't "nice", it's just that the proofs seem all to be using the same 2 or 3 techniques and it's not so fun to read atm

I guess fancy stuff like descent is post-Gauss.

Idk if that's sarcasm or not lol

How old is this material?

Well, actually, I don't know if ideas were there before.

it's a technical book on the history of Fermat's last theorem from the 70s I think

@Ted I see, Fermat came up with descent

but Gauss kinda formalised the language of congruences and canonized preceding works

So I'm confused. Fermat is way before Gauss.

Right, so I mean.. descent is really really cool

But only 2 or 3 of the proofs I've read so far have been using descent

the others have been a lot of coprimality arguments and parity

Modular arithmetic is amazingly powerful for elementary stuff.

Yeah exactly

in fact I struggle not to use the language

Why avoid the language?

Just because it's not being used in the book I'm reading because it's developing everything chronologically

Gotcha. Well, it's a good experience for you :)

Yeah definitely :) I used a few chapers later on in the book for my disseration because it describes lots of what is going on in Kummer's $\Bbb Z[\zeta_p]$ work

rather than just using post-Dedekind ring theoretic language I guess

Didn't Kummer actually prove Fermat for cubes?

Euler proved it for cubes, Kummer proved it for a huge swathe of primes

Oh, Euler. I knew it was someone important :P

heya @Eric

Right lol, Kummer's idea was reaaaally really nice, I remember having a huge smile reading it for the first time hahaha

Oh, both skulls are here.

Missing having @Mathein around to chat to about number theory!

Yes, whatever has become of @Mathein?

This chatroom may wither up and die.

Hopefully not, it was difficult during university finding someone enthusiastic enough about mathematics to actually want to talk about it

I'll just create more accounts to keep it alive :-)

That's crazy, @ÍgjøgnumMeg. Math majors at UGA had plenty of chats with one another. And my students usually talked a lot of math together.

@Ted unfortunately very very few people were in it for the mathematics

I think I had a proper chat with 2 or 3 students in the whole 3 years, and that was in the final year

That's very disquieting.

I expect Heidelberg will be much different!

Or maybe you're antisocial and unapproachable.

That doesn't apply to me thankfully

LOL

lol

Maybe it applies to me :D

Probably not..

It definitely applies to me.

Sad

Be approachable!

nah

I'm not a people person.

Yeah, @ÍgjøgnumMeg, probably not in my case :)

@skull: if that's so, why are you here so much?