Mathematics - 2024-05-01 (page 2 of 3)

EE18

3:00 PM

I do, especially as it relates to the canonical one (our "real world")

Xander Henderson

I.e. the key example is a vector subspaces of $\mathbb{R}^n$, translated away from the origin.

EE18

I see my earlier claim regarding the zero vector space was nonsense now

We would have the fact that for $Q,P \in E$, there is a $v \in V$ such that $Q = P + v$ but, since $v = 0$, this means $Q = P + 0 = P$

But I suppose for $\dim V > 0$, then there is no limit on the set of points $E$

Thorgott

isn't $E$ supposed to be the ambient space

Xander Henderson

Again, you seem to be working very abstractly. Do you understand the simple examples, i.e. affine spaces seen as subsets of $\mathbb{R}^n$?

EE18

I think I am being confusing so I will just get back to reading rather than asking highly imperfect questions and wasting both of your times

Sort of, Xander, But this is how my book introduced it. Anyway, I won't belabor this right now, sorry again about the confusion.

Xander Henderson

3:06 PM

Okay, but any time you are hit with a new definition, you're very first action should be to cook up a simple example.

You have a definition of an affine space, which is cool and all. Now try to come up with some very concrete examples.

EE18

That's well taken, thanks Xander

user 70432

The more general learning principle is to try to make connections between what you know and what is new :^)

psie

3:30 PM

In Folland's book, there is a simplification made that I do not understand. After showing $\mu_0$ is well-defined, we want to show $\mu_0$ is a premeasure on $\mathcal A$ (the algebra of all finite disjoint union of half-open intervals or 'h-intervals'). In particular, we want to show countable additivity. Folland considers a sequence of disjoint h-intervals $(I_n)_{n\in\mathbb N}$ with $I=\bigcup_1^\infty I_n\in\mathcal A$.

Then it is claimed that since $I$ is a finite union of h-intervals, the sequence can be partitioned into finitely many subsequences such that the union of the intervals in each subsequence is a single h-interval. By using finite additivity of $\mu_0$, we may simply assume $I$ is an h-interval of the form $(a,b]$. Is there any loss of generality by not assuming $I$ to be of the form $(a,\infty)$? After all, this is also an h-interval.

EE18

OK I am back lol...

So in the very next sentence after I got back to reading I am told (as is clear) that I can consider any vector space as an affine space in particular

Is that the concrete example you were alluding to @XanderHenderson?

Xander Henderson

@EE18 Sure, any vector space is an affine space, but that example might be too narrow.

Instead, consider, perhaps, a line in the plane.

(Specifically, a line which does not pass through the origin.)

EE18

Oh ya, I see what you mean. Take the vector (direction) space as a one-dimensional subspace of $V$ and then fix $E$ as some singleton from $V$ in order to get your case

I did have a question about considering $V$ itself as an affine space. The book says "We choose, of course, the zero vector to be the origin." I want to be sure since sometimes language like "we choose" can mean "we must choose" or "we can choose". This is a case of the latter right? i.e. choosing the zero vector as the origin is the obvious choice, but not the only one?

Xander Henderson

3:46 PM

@EE18 I mean, maybe? But, again, that seems too abstract...

For example, in $\mathbb{R}^2$, take $A$ to be the set $\{ (x,y) : y = 2x + 1\}$. Then you can take $\vec{A}$ to be the vector subspaces of $\mathbb{R}^2$ generated by $\langle 1, 2\rangle$.

@EE18 You can choose the zero vector to be the origin, which is how that particular example is chosen. But consider the real number line, translated to the left by one unit. This is an affine space which isn't quite the original vector space.

EE18

Fair, thanks Xander

My book finishes the section with the following (which was why I came back here for follow up):

The interpretation of finite dimensional vector spaces as affine spaces has also an extremely important computational aspect. The introduction of coordinate systems leads to concrete descriptions of geometric objects in terms of equations and inequalities for the coordinates. A coordinate system is determined by the choice of an origin and basis, and it is essential to make these choices so that the calculations are as simple as possible. The right choice of the coordinate system can be decisive for a successful solution of a given problem.

I don't at all see why interpreting finite-dim vec spaces as affine spaces matters at all here. The canonical isomorphism between a finite-dim space and $K^m$ has nothing to do with affine spaces?

Xander Henderson

@EE18 I don't have the physics background to really have the intuition for this, but the basic idea is that there is no "real" origin, so you can choose the origin to be whatever you like.

Suppose that you are trying to model the motions of two bodies (say, two boats on an ocean). You could choose some origin independent of the two ships, but you might choose the origin to actually be one of the ships, because then you only have to track the relative motion of the other ship. Or vice versa.

Thorgott

@psie not sure I follow your suggestion, there are intervals that do not look like $(a,\infty)$

EE18

Are they saying (or, rather, is it possible that they are saying) that sometimes it is fruitful to interpret the vector space as an affine space, use that the choice of an origin induces an isomorphism between the affine space and the vector space, and then use the canonical isomorphism between that vec space and $K^m$?

leslie townes

imvho that comment also seems overblown, unless by "computational aspect" they just mean "the handwritten formulas look nicer"

Xander Henderson

3:55 PM

@leslietownes That, too.

Thorgott

@EE18 what is $E$

Xander Henderson

@Thorgott The fifth letter of the English alphabet!

leslie townes

i don't think there's a "real world" problem where a team full of people in front of NASA computers is high fiving like, oh, thank god we found the right origin

EE18

Oops, sorry Thorgott. I realize now I forgot to include this at the beginning of the discussion:

psie

@Thorgott yes, but there are h-intervals that do not look like $(a,b]$ too, or?

EE18

3:56 PM

$E$ is the "set of points", part of the structure that defines the affine space

Xander Henderson

@leslietownes No, but I don't think that the author is quite claiming that, either.

leslie townes

note that "the right choice of the coordinate system can be decisive for a successful solution of a given problem" sounds like something in the spirit of, say, a schaums outline in college physics

and not a statement about any kind of non hand calculational difficulty

EE18

@leslietownes OOF, a brutal takedown ;)

leslie townes

i'm not trying to be brutal! i'm trying to put it in context

EE18

No that's fair

leslie townes

3:59 PM

changing the choice of origin is not going to affect whether or not you can successfully solve a problem, unless a "successful" solution is just one that looks nicer and maybe took less paper to find by hand

choice of origin is important, but it's just a choice of origin

psie

@Thorgott what I'm trying to say is that Folland assumes $I$ to be an h-interval. He writes $I=(a,b]$, but h-intervals can also be of the form $(a,\infty)$.

leslie townes

to be brutal, you might google the background of the author and see if they've ever worked on applied problems in their own "real life," or if they're the kind of author who repeats the stuff that they've read in their textbooks about "here is where you say that this is sometimes really important for applications"

EE18

That's fair. I guess what I'm hoping to clarify is that in the paragraph I copy-pasted above, there seem to be two things at play: the vector space, and the vector space as an affine space. Am I right to say that given a vector space, you would never go back and consider it an affine space. But, instead, you would take an affine space, fix an origin and so be able to reegard it via isomorphism as a vector space, and then take coordinates etc.

Thorgott

@psie read till the end, he addresses this

EE18

@leslietownes This is the chap: scholar.google.com/… I don't know if these sorts of problems are considered applied but what I am sure of is that they're way above my pay grade!

Thorgott

4:02 PM

@EE18 oh, that's how they define it

@EE18 so this does not make sense

leslie townes

EE18 looks like a pure mathematician to me (PDE/ODE people are not uncommonly accustomed to getting a pass, where everyone in their department thinks they are "applied" because, hey, who knows) :)

EE18

@Thorgott Why not? $E = \{a\}$ for some $a$ in $V$, and $E = W$ for some 1D subspace $W$ of $V$?

leslie townes

remember kids, rationalize the denominator because it can be decisive for the solution of your problem

psie

@Thorgott well, he addresses the case when $b$ is infinite in $(a,b]$. Is this the same as addressing $(a,\infty)$?

EE18

Leslie on a ROLL this morning

Thorgott

4:07 PM

@psie that's the same interval, yes

@EE18 now you're saying $E$ is two different things?

EE18

OMG sorry that was me typing too fast. The second $E$ should be $V$ (the vector space in the affine space definition, not the underlying $V$ for this specific example)

Thorgott

yeah, this still doesn't work

note if $E$ is an affine space over $V$, then $v\mapsto P+v$ defines a bijection $V\rightarrow E$ for any $P\in E$

EE18

Yup, the authors mention as much (this is "taking an origin")

But I don't see why my example fails?

Or how my examples is related to this comment (there is no mention of an origin in my example, merely a construction of a particular--or so i thought--affine space)

Thorgott

your $E$ has a single element and your $V$ is a 1-dimensional vector space

there is no bijection between those things

so $E$ cannot be an affine space over $V$

EE18

Hmm, let me think on this but at first glance I see I've erred somewhere

Ah OK, my error was that what I really mean was $E =\{a + w \mid w \in W \}$

That's where you and Xander were leading me

Thorgott

4:15 PM

that works

to be honest, I'm not a big fan of this definition, but it is what it is

EE18

Wikipedia gives some more abstract definition in terms of a free and transitive group action which maybe you prefer?

If possible can someone confirm if this (chat.stackexchange.com/transcript/message/65587999#65587999) suspicion was correct?

Thorgott

nah, that's the same thing

EE18

Oh, true (at least to the experts like you!)

Thorgott

what they're considering is what's called a torsor of the underlying abelian group of a vector space

I don't really find this notion geometrically useful

but don't mind my preferences

Bml

4:41 PM

Hi everyone. I would like to know if two bases of different sizes could be added together (in linear algebra). If you have a 5×1 matrix (5 rows and 1 column) of the type A = (-1, 2, -t, 0, -(t+1)), and a 4×1 matrix (4 rows and 1 column) of the type B = (-t, 1, 0, 0), can A and B be summed if, for example, t= -1?

Soumik Mukherjee

4:53 PM

@Bml no

Bml

@SoumikMukherjee Why?

Soumik Mukherjee

That's how matrix addition works, you can add an m×n matrix to a p×r matrix iff m=p and n=r

leslie townes

5:14 PM

you can certainly 'make it happen,' but in more than one way, and only at the cost of making choices of how to get those things in a form where you can add them together. the data itself won't just make that choice for you

in situations where you want to do this, the reason that you want to do this is usually some signal about how to approach

peek-a-boo

5:32 PM

lolol referring to Amann’s words as something sounding like “Schaums outline in college physics”

EE18

@peek-a-boo Leslie was "in their bag" this morning (Leslie not sure if that reference hits or not)

Have you read Amann Escher peekaboo?

maybe ive asked you before tbh i can't remember

peek-a-boo

@EE18 we do this all the time in math though. Sometimes, we have extra structure at our disposal, and we frequently disregard extra structure if it’s not “relevant”, or sometimes

yea I know about their books

and I read vol III in a fair amount of detail… but not the first two, because I had already learnt that material elsewhere

EE18

Ah gotcha gotcha

@peek-a-boo As for this, I'm not sure I follow so I'll try to clarify what I mean in the hope it helps (me). I see them as referring to two (related but distinct) things in this paragraph. (1) they say consider a vector space $V$ as an affine space (not sure why we'd do this given the former is "simpler" than the latter, in the sense I'll describe in (2)) and then what? (2) given an affine space, use a choice of origin to thereby identify $E$ with $V$, and now you can use isomorphism to $K^m$.

peek-a-boo

also to leslie’s point above about NASA not celebrating at finding the right origin… ok sure if taken literally, but if we put on our narcissistic hats on and pretend the Earth and the Sun are the only things in the solar system (i.e a 2-body problem), then working in a coordinate system where the center of mass is at the origin is absolutely a win (granted here, the origin ‘moves’ relative to a ‘fixed’ frame but still…)

EE18

I understand (2) but not what they mean with (1)

That is, I don't understand why you would do (1). I understand that you can do it, but not why you would do it

peek-a-boo

5:43 PM

you might want to forget the origin because perhaps the origin you chose was ‘wrong’, or perhaps it was an extraneous choice. For example in physics, we really only care about time differences, but yet we call this the year 2024. Obviously in the grand scheme of things, this is not really ‘significant’

Similarly, sometimes if we have an inner product space, we might want to forget the inner product and just focus on the vector space aspect. Similarly, on $\Bbb{R}^n$, which has the structure of a smooth manifold, we might want to ignore the smooth structure, and just focus on the topology.

Lukas Heger

because you can apply theorems or operations that are defined for affine spaces, but not necessarily just vector spaces. Let me give an example. Consider the vector space $\Bbb R^3$ and let $V$ be the xy-plane inside $\Bbb R^2$. Now consider a plane $P$ inside $\Bbb R^3$ that does not include the origin.
We would like to say that $P \cap V$ is again an affine space. It's easy to prove that an intersection of two affine subspaces of an affine space is again an affine space. To apply this here, though you need to treat $\Bbb R^3$ and $V$ as affine spaces. But you might say, they are not affin…

peek-a-boo

etc. forgetting extra structure is a very common thing to do in math

Lukas Heger

this is a really trivial example, there are more sophisticated things you can do with affine spaces. Their join operation is a bit more subtle than just sum of vector spaces iirc

EE18

That's well taken Lukas. I guess what was confusing was that in their paragraph (I'll include it again below) they seem to be alluding to the use of considering $V$ an affine space as arising from the ability to take coordinates, whereas this has nothing to do with $V$ being an affine space

"The interpretation of finite dimensional vector spaces as affine spaces has also an extremely important computational aspect. The introduction of coordinate systems leads to concrete descriptions of geometric objects in terms of equations and inequalities for the coordinates.

A coordinate system is determined by the choice of an origin and basis, and it is essential to make these choices so that the calculations are as simple as possible. The right choice of the coordinate system can be decisive for a successful solution of a given problem."

peek-a-boo

oh yes, the thing I was implicitly getting at (and which thankfully Lukas explicitly said) is that by forgetting extra structure, you can apply all the theorems available to the previous structure

Lukas Heger

5:48 PM

this is getting into physicsy terminology right now, but I guess you might say that an affine space is something that can take linear coordinates (whatever that means). Technically introducing an isomorphism with a vector space is not the same as choosing (linear) coordinates, choosing linear coordinates would be choosing an isomorphism with a vector space with a fixed basis

so the bottom line is that I agree that that quote is confusing

EE18

@LukasHeger That's what I guess I'm saying/confused about with what Amann Escher are saying. Choosing coordinates is just choosing a basis (i.e. the particular isomorphism with $K^m$), and we can only do it in the case that the affine space is itself a vector space AFAIK (or, if not, if we've chosen an origin to the affine space and so defined an isomorphism from $E$ to $V$)

oh ok, thank you!

good to know my confusion is somewhat warranted

Lukas Heger

dictionary:
affine space <-> something that can take linear coordinates
vector space <-> something that can take linear coordinates and has a chosen origin
vector space + basis <-> actual concrete linear coordinates

EE18

@peek-a-boo This is a good point which I will keep in mind going forward. Definitely a lot of AE proofs of larger structures do this trick of considering only substructure and then using theorems already developed therefor (thinking vector space proof using that vec addition makes $V$ an abelian group for e.g.)

@LukasHeger Possible to say more about what you mean by "can take linear coordinates"? Does that just mean that by the procedure I outlined above (first choose origin, then a basis of the vec space which we obtain), it is possible to induce an isomorphism to $K^m$?

Lukas Heger

@EE18 basically yes. I thought that people more trained in physics than I am would have a more intuitive understanding of "linear coordinates" as well

Thorgott

there is a notion of affine basis

EE18

5:53 PM

I am not sure I am more trained in physics than you ;)

But thank you very much for all the help Lukas and peekaboo!

Bml

@leslietownes Are you referring to me?

leslie townes

@Bml yes

Lukas Heger

@EE18 one thing I wanted to add is that if you noticed it or not, you probably have practiced forgetting structure before: like when you treat a vector space as a set or a linear transformation as a mapping (something which can be injective or surjective for example)

Bml

@leslietownes But the discussion below refers to something else entirely, right?

EE18

@LukasHeger Yup, that's well taken too. Even though it somewhat triggers the set theory demons, I do do this :)

peek-a-boo

5:58 PM

and for more demons: en.wikipedia.org/wiki/Forgetful_functor

EE18

I'm going to be sick^

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