Mathematics - 2021-09-17 (page 2 of 2)

6:02 AM

@copper.hat Night Mr. Joe!

Suppose V is a vector space such that dim V>1 and suppose W is infinite dimesnional then vector space of all linear transformations from V to W is infinite dimensional

Proof: Let dim V=m and let $v_1,v_2,...,v_m$ be a basis for V. Since every linear transformation T: V$\to W$ is completely determined by its values on basis of V, each of $T_1(v_1), T_1(v_2), ..., T_1(v_m)$ has infinite choices. Any choice for $T_1(v_1), T_1(v_2), ..., T_1(v_m)$ will completely determine $T_1$

SmokenSieEinBitteChebaHitBits

Suppose $V \to W$ is finite dim

leslie townes

koro: is V assumed finite dimensional? one way to do this is to exhibit arbitrarily large collections of linearly independent elements of L(V,W)

Koro

yes, V is finite dimensional, also dim V>0 (I wrongly wrote earlier that dim V>1)

leslie townes

it's true if V is infinite dimensional too, but you'll probably need some form of the axiom of choice and sometimes books try to avoid that

SmokenSieEinBitteChebaHitBits

You want to find a way such that $W \subset (V \to W)$ I think

leslie townes

6:13 AM

koro, note that you have "infinite choices" for a linear map from R to R because you have infinitely many k for which T(x) = kx defines a linear map. and dim L(R,R) = 1

Koro

@leslietownes hmm, showing collection of linearly independent vectors seems to be the key to solve the problem

leslie townes

see if you can construct a linearly independent collection of maps from V to W of rank 1

SmokenSieEinBitteChebaHitBits

If you take $f : W \to \text{Hom}(V,W)$ given by sending every element $x \in W$ to the map $\phi : V \to W, \phi(v) = x$, i.e. $f(x)(v) = x$. Then you get that the functions $\phi_x, x \in W$ are linearly independent as well. And so you you have an infinite linearly independent subset.

Now to prove the part about linear independence

@Koro

should be easy since it's the precise isomorphic image of $W$

To prove linear independence, all you need is to show that there is a map from $g:f(W) \xrightarrow{\sim} W$

That is the the two-sided inverse of $f$

Koro

@leslietownes Let w be any fixed element in W. Then I define $T(v_i)=w$ for all $1\le i \le m$, where $v_i$'s are basis for V

SmokenSieEinBitteChebaHitBits

Clearly there is, you simply take the value of that map for any of its arguments

*There is such a g

Koro

6:23 AM

@SmokenSieEinBitteChebaHitBits Hi, I'm afraid I don't yet understand Hom (V,W) :(

SmokenSieEinBitteChebaHitBits

That's just the set of all linear maps from $V \to W$

Koro

Ok. So it's same as our L(V,W). :)

SmokenSieEinBitteChebaHitBits

What you said "fix $w \in W$. Then define $T(v_i) = w$ for all $i$ is the same thing

You've defined $T = f(w)$

$f(w)$ is my notation and $T$ is yours

So we're essentially on the same wavelength

Koro

and hence on same frequency $n=v/\lambda$ :)

SmokenSieEinBitteChebaHitBits

So now, I say there's an easy inverse from $\{ T = f(w) : w \in W\} \subset L(V,W)$

The image of your map made by fixing $w$ and so on...

Which is $g(y) = y(v_0)$ for any $v_0 \neq 0$ which must exist since $\dim V \gt 0$.

I.e. simply take the value of $y$, then you've inverted $f$, both sides

$fg = \text{id}_{\{T\}}$ and $gf = \text{id}_W$

Koro

6:28 AM

@SmokenSieEinBitteChebaHitBits I see, that's very nice actually. So we have $f(x)=\phi_x$, where $\phi_x:V\to W$ is being defined as $\phi_x(v)=x$

SmokenSieEinBitteChebaHitBits

So we can say that the image is also linearly independent and so the mother space $L(V,W)$ is as well

So there's no proving of linear independence as once thought!

required

the problem inherits linear indep from $W$

Koro

do you know what a category is?

The set of vector spaces forms a category

$\text{Hom}_{C}(V,W)$ is how you'd write the analogous space of morphisms from object $V$ to object $W$ of the category

*The class of vector spaces

not necc. a set

You can also restrict to $k$-vector spaces for a field $k$

and that would be denoted $\text{Hom}_k(V, W)$ understood to mean $k$-linear maps

Koro

@SmokenSieEinBitteChebaHitBits not yet, I'm afraid. :(

SmokenSieEinBitteChebaHitBits

Usually though we drop any subscript in $\text{Hom}$

I'm currently designing software that hopefully speeds up learning of advanced category theory stuff

So the classes of abelian groups, groups, rings, fields, vector spaces, modules, magmas, etc. Any algebraic object forms a category

But so too form categories the cat of topological spaces

Etc.

For each area of math, there are categories behind them - an abstract framework to keep track of results essentially and to come to related conclusions in once seemingly unrelated areas of math

This is because in addition to morphisms of the objects of study in the categories. Recursively, but up one level, we have morphisms of categories themselves, called functors

$\text{Hom}(\cdot, W)$ is a contravariant functor (because of the way it acts on linear maps)

$\text{Hom}(V, \cdot)$ is a covariant functor.

Endo functor in the case of $\textbf{Vect}$

That's like an endomorphism or a linear map $V \to V$ but instead it's $C \to C$ between categories

What gets preserved since morphimsms are supposed to preserve something, right? Like linear maps preserve addition and scalar mult. A solution is that we preserve composition of maps within the category i.e. $F(f)F(g) = F(gf)$ for any composable maps $g,f$ and functor $F$.

You can also write composition as $\circ$ but $fg$ saves space

Koro

6:44 AM

@SmokenSieEinBitteChebaHitBits Another way: I suppose on the contrary that dimension of L(V,W) is m, then since W is infinite dimensional, there exist m+1 linearly independent vectors in W. Let's call these $x_1,x_2,...,x_m,x_{m+1}$. Let $v_1,v_2,...,v_n$ be basis of V then $T_i:V\to W$ defined as $T_i (v_j)=x_i$ for all $1\le j\le n$ defines an LT. Now, {$T_i: 1\le i\le m+1$} is to be proven to be LI. Let there exist scalers $c_i$ such that $\sum c_i T_i(x)=0$ for all x in V. It can be show that all $c_i$'s

are $0$

SmokenSieEinBitteChebaHitBits

Too much confusion in it!

Koro

for example?

SmokenSieEinBitteChebaHitBits

You have to recall several things that might require you to look them up. Mine was easy and fundamental : isomorphism means isomorphic bases

the second isomorphism is just in the category sets or usually called a bijection there

Koro

I'm currently at chapter 3.A of LADR by Professor Sheldon Axler

leslie townes

you could clean the argument up a tiny bit by defining T_i by sending v_1 to w_i and the rest of the v's to zero. then from sum c_i T_i = 0, apply to v_1, deduce sum c_i w_i = 0, from linear independence of the w_i, deduce c_i = 0 for all i

SmokenSieEinBitteChebaHitBits

6:47 AM

What's it about?

Koro

and by that chapter isomorphism has not yet beet covered.

SmokenSieEinBitteChebaHitBits

It should be covered in the first chapter, I would switch books lol, sry!

leslie townes

you've got the right idea, koro. i wouldn't bother 'optimizing' this proof. it's not onerous to work in terms of the definitions although there are more general perspectives you could fit this argument into

SmokenSieEinBitteChebaHitBits

Or you can simply skip ahead yourself

Koro

@SmokenSieEinBitteChebaHitBits definition of linear maps has been defined and L(V,W) has been defined. Also, the fact that T(0)=0 has been proved.

SmokenSieEinBitteChebaHitBits

6:48 AM

some mathebooks are not meant to be read linearly

Koro

But I know isomorphism from group theory point of view.

SmokenSieEinBitteChebaHitBits

Yes, it's the same thing here

actually

it's an abelian group isomorphism

with one additional axiom / constraint if you will

That is preserves also scalar mul or $f(\alpha x) = \alpha f(x)$

Then you have a vector space isomorphism defined via the abelian group mentioning method

Since $+$ on $V$ or $W$ each form abelian groups

Koro

@leslietownes Thanks a lot Leslie and @SmokenSieEinBitteChebaHitBits. I think I understand now.

SmokenSieEinBitteChebaHitBits

There is a $k$-linear map $G$ taking $W \to L(V,W)$, given by $g(w)(v) = w, \forall w \neq 0$. This map is clearly invertible and thus there exists an $\infty$-dimensional subspace of $L(V,W)$. $\blacksquare$

@Koro you're welcome

If on the otherhand $\dim V = 0$ we'd have a non-injective map since $\ker g = 0 \iff g$ is injective for any vector space morphism $g$.

In that case you don't neccesarily have an embedding of $W \to L(V,W)$.

If $\dim V = 0$ you'd simply have that $L(V, W) = 0$

Since $V = \{0\}$ and as you said $T(0) = 0$

So $L(V,W) \approx 0$ is an isomorphism i.e. the single zero-map on the left and the zero module on the right

Koro

7:15 AM

Given two subspaces $W$ and $U$ of a vector space $V$, does $W+U$ have a geometric meaning?

In particular, I know that in $\mathbb R^2$, if $W=\{c(0,1): c\in \mathbb R\}$ and $U=\{c(1,0): c\in \mathbb R\}$ then $W+U$ is $\mathbb R^2$

shintuku

8:10 AM

@Koro yeah, two subspaces that are lines become a plane if they're not parallel, two subspaces that are planes become a 3d space if they're not parallel

since a vector space is just a set tuples with some added algebra, you can be given a subspace that is the set of vectors that form a line, or the set of vectors that forms a plane ,etc.

Koro

8:44 AM

@shintuku I was thinking on almost the similar lines. Thank you :)

shintuku

np

user178758

9:33 AM

Thanu Padmanabhan is dead :(

Typo

11:47 AM

f maps (a,b) to (c,d) where

How can I write down algebraically if the relation is well-defined?

(I am literally asking for a friend, any questions directed to me, I'll just ask him and give his response here. He can't be bothered to make a SE for some reason..)

Typo

12:11 PM

His working is that it can't be well-defined as the input (2,-3) gives an output which is undefined. So not all of the input domain gets mapped to an output.

Thorgott

2:21 PM

that question does not make sense

UserX

https://math.stackexchange.com/questions/3613704/what-is-the-matrix-derivative-of-a-symmetric-bilinear-form-mathbfat-x-math

I have a question regarding this. Let's say we're differentiating a symmetric billinear form $\mathbb{u}^{\top}\mathbb{X}\mathbb{u}$ wrt to the symmetric matrix $X$. If we expand this to its sum and try to differentiate wrt to the matrix we end up at some point with this $\frac{\partial}{\partial X_{i j}} \sum_{p, q} u_{p} X_{p q} u_{q}=\sum_{p, q} u_{p}
\frac{\partial X_{p q}}{\partial X_{i j}} u_{q}$. Regarding the $\frac{\partial X_{p q}}{\partial X_{i j}}$ term, i…

Alan P.

I posted this question yesterday, pitifully it did not get many views
https://math.stackexchange.com/questions/4252548/recursive-definitions-and-proposition-2-1-16-in-taos-analysis-i

UserX

What am I missing? Where do these 2s go?

geocalc33

2:39 PM

does the def. of foliation require a group structure

it says that you need to be able to decompose the real coordinate space into cosets $x+\mathbf{R}^p$

Foliation

In mathematics (differential geometry), a foliation is an equivalence relation on an n-manifold, the equivalence classes being connected, injectively immersed submanifolds, all of the same dimension p, modeled on the decomposition of the real coordinate space Rn into the cosets x + Rp of the standardly embedded subspace Rp. The equivalence classes are called the leaves of the foliation. If the manifold and/or the submanifolds are required to have a piecewise-linear, differentiable (of class Cr), or analytic structure then one defines piecewise-linear, differentiable, or analytic foliations,...

can you do something of a foliation of a component of a manifold, where a structure available is a commutative semigroup?

also (though I don't know the details) I understand that one can do a Fourier transform of a Riemannian or pseudo riemannian manifold. What's the benefit of this?

Typo

3:46 PM

@Thorgott Mind elaborating?

Thorgott

what relation are you talking about

Typo

Like, a function of (a,b) to (c,d)

But reluctant to call it a function since that already assumes it is well-defined, so I wrote 'relation'

Thorgott

4:02 PM

what's the domain and the codomain supposed to be

Typo

Both real numbers I suppose

shintuku

4:19 PM

a function of (a,b) to (c,d) is unclear: is it a function from any ordered pair to any ordered pair of real numbers? in that case it is a function from $\mathbb{R}^2 \to \mathbb{R}^2$

the place where it does not seem to be defined is whenever the denominator of either element of the output is 0, otherwise it seems fine

Avra

Hello if we have an input of size $n_1$, if we accelerate reading by 10 times we could finish it in $10n_1$, why we finish $(n_2)^2$ in $\sqrt{10}(n_2)^2$ please?

Typo

I just asked him, he said R^2 -> R^2

So yeah, ordered pair to ordered pair

@shintuku Yes, that is what he guessed as well. What he wanted to know was how he can write the answer in set notation/algebra.

shintuku

all you need to do is figure out what is the domain more precisely. you'll need to write $f:\mathbb{R}^2 \setminus X \to \mathbb{R}^2$, where $X$ is wherever you obtain a $0$ in the denominator

Typo

4:35 PM

Right, thank you. I will tell him about that. Another thing, how can he formulate/prove in math that it is well-defined for all other ordered pairs except (2,-3)? Like, we disproved that a specific pair didn't work but what about proving that the rest will work? Does it just go without saying?

shintuku

it's well defined if you remove all places in the domain where it outputs a $0$ in the denominator

Typo

That's true but how do we know that?

shintuku

(2,-3) is a single pair where it outputs $0$ in the denominator, there are infinitely many others

Typo

Ah, I see.

shintuku

so if you remove all those infinitely many places where it outputs $0$ in the denominator, you're fine

Typo

4:39 PM

I guess it could also be seen that c and d have patterns (either increasing or decreasing) after it crosses that singularity.

shintuku

you can sort of see that a certain type of combination will output 0 in the denominator

why don't you try plotting it? it will be easier to see how it behaves

Typo

Alright I'll tell the friend. Multivariable stuff is way out of my league (I'm just in a levels). I think he has enough info to figure out what to do next.

Thank you for the time shintuku and Thorgott

shintuku

np

Koro

4:54 PM

how many subgroups of Q (set of rationals) are there?

countable or uncountable?

Q has uncountably many subsets as power set of Q is of cardinality c

leslie townes

there are uncountably many.

Avra

do you have any though about my question please @leslietownes?

What happens to the allowable size of the problem if we increase the speed of reading **tenfold**?

- In linear case, then for algorithm A with time complexity $n$ and
size $s_1$, the allowable size becomes $10s_1$.
- In quadratic case, then for algorithm B with time complexity $n^2$ and size $s_2$, the allowable size becomes $\sqrt{10}s_2 = 3.16s_2$.
- etc.

**Problem**: can you please show how that is the case that size of algorithm B increases to $\sqrt{10}s_2 = 3.16s_2$ if we increase reading speed by 10 fold?

leslie townes

5:52 PM

this isn't really my field but i can make some guesses.

Avra

@leslietownes. That's fine. Thank you anyway.

Let the weight of a spanning tree be the weight of the maximum weight edge
in the tree (now the weight is not the sum of all weights of all edges in the tree). Give an algorithm to
compute a minimum weight spanning tree under this denition of weight of a spanning tree. Analyze
its complexity.

leslie townes

say i give you a fixed time T. to say the "maximum size" at one speed is an input of length s_1 is maybe to say that K s_1^2 = T (approximate equality and K some constant). i'm assuming "size" is something like a count of operations.

so maybe now i'm going 10x the speed and it now takes me only T/10 to process the same number of operations. being able to do an input of length s_1 in the shorter time T/10 now tells me that K s_1^2 = T/10 or K 10s_1^2 = T or K (sqrt(10) s_1)^2 = T. this tells me that given the same unit time T, i can now handle an input of length s_2 = sqrt(10) s_1.

i never formally studied this and don't know what 'size' means in this context, pure guesswork on my part

similar formalism would give you a size increase of a factor of 10^(1/k) for an O(n^k) algorithm. and a piddly plus log_k(10) size increase for an O(k^n) algorithm, which feels right.

Avra

6:09 PM

@leslietownes. Thanks

Ted Shifrin

6:45 PM

@Xander (or whatever mod may know): Who is out of the blue activating years-old questions today? Why? This, this have a common theme. But what makes them active now?

Xander Henderson

@TedShifrin They are getting bumped by edits. The editor is can be seen in the edit history or timeline of the question or answer being edited.

Ted Shifrin

Michael Hardy changes "any" to "every" because this is earth-shattering after so many years. Great.

A really substantive edit.

I'm an idiot for not paying attention to that instead of bothering you, though. I apologize.

Xander Henderson

Oh, indeed. Utterly necessary.

No, thanks for pointing it out.

Neither edit is, in my opinion, of any value.

Ted Shifrin

I know that some low-rep people go through editing to get rep, but for him???!!!

I wouldn't care, but we have enough low quality distractions without adding these.

Oh, and he "modified" a third similar post, too.

leslie townes

haha, i kinda like the varepsilon edit. i realize that comedy is not the point but when a bit works, it works.

Semiclassical

7:00 PM

o/

Ted Shifrin

Revolting.

I realize that you need bad comedy because of munchkin.

Xander Henderson

Ugh... I was going to write a couple of exams today, and I just can't bring myself to care...

I even brought my computer home from the office. :/

leslie townes

i'm gonna turn all of those curvy arrows you use into \mapsto. i think that would be a good use of my time.

2

Ted Shifrin

I don't think you'll find very many. I use function notation, not mapsto.

@Xander Are these courses you've taught a bunch of times?

I seem to have embarrassed an OP into removing his post (which showed that he hadn't the vaguest idea of what a geodesic is).

The bad homework posts are getting worser and worser.

leslie townes

@Ted that makes a lot of sense. mapsto is one of those weird notations, where it's very helpful, if one is used to it and can trust the audience to be used to it. otherwise, it opens up an abyss of confusion.

Ted Shifrin

7:09 PM

To wit: this and this. At least the second shows a modicum of knowledge.

@leslie The fact that some people write $f\colon X\mapsto Y$ distresses me. So, yes, the notation is not understood.

user193319

7:26 PM

0

Q: Convergence of sequence of functions $\{f_n\}$ to $f$ in $L^p$ if and only if $\|f_n\| \to \|f\|$

Let $\{f_n\}$ be a sequence of functions in $L^p( [0,1])$, $1 \leq p < \infty$, which converges almost everywhere to a function $f$ in $L^p$. Show that $\{f_n\}$ converges to $f$ in $L^p$ norm if and only if $\|f_n\| \to \|f\|$. What does $\|f_n\| \to \|f\|$ mean and which convergence theorem i...

Would someone mind explaining Kavi's answer? See my comments on it.

$\lim \inf$ is superadditive, so I don't see how to split things up the way he. presumably, wants to.

zingergi

Hey! I hear a lot of people talking about studying math for "pleasure". Is there such a thing, and if so, how does one experience it?

I mean, how does one get to the level where studying maths becomes such a pleasure that you want to do it all day long?

user193319

7:42 PM

Nvw, I figured it out...Use the fact that $\lim \inf \le \lim \sup$

Xander Henderson

@TedShifrin In some sense, yes. But I have only been at this institution for a year, so I have not taught them a lot here.

I also tend not to ever teach exactly the same course twice, and I experiment a lot with the kinds of questions I ask.

For example, I recently came across this, and am toying with the idea of trying it out in a class next spring.

@zingergi How does anyone get the the point of doing anything all day long?

Ted Shifrin

@Xander We try to change things up a lot (for one thing to stop cheating), but sometimes it can be difficult to keep the level right. I made up a novel max/min problem (a can problem, but minimizing the length of the seams) and more than half the students didn’t even bother to work the right problem. They just did the one they’d memorized.

Xander Henderson

Yikes!

Ted Shifrin

The problem was way easier than the standard one. I was proud of it. :)

Xander Henderson

Heh. Nice.

zingergi

7:52 PM

@XanderHenderson Haha, I guess I meant like doing it a lot, definitely not all day long tho.

Xander Henderson

@zingergi I'll repeat the question: how does anyone get to the point where they enjoy doing anything?

zingergi

Hmmm

Xander Henderson

Like, I don't enjoy writing poetry or running or interviewing folk for ethnographic research. But other people really seem to enjoy that stuff. How? Why?

(The answer is that they are not me, I am not the center of the universe, and other people have other tastes.) :P

zingergi

Uh, well, I'm not sure I can say with conviction whether I like or dislike maths because I haven't really tasted it yet.

Maybe I have to invest more time into it, and then decide?

Alan P.

Hey, I was starting to read a textbook, I was wondering if someone could give a look to my question?
Thanks
https://math.stackexchange.com/questions/4252548/recursive-definitions-and-proposition-2-1-16-in-taos-analysis-i

Xander Henderson

7:56 PM

I mean, I think that is true of anything. There are a lot of authors I am familiar with who didn't start writing until their 30s or 40s.

zingergi

Like all the math I've taken is from high school and a couple of college classes

leslie townes

i wouldn't force yourself to keep with it unless you had some reason to, whether curiosity or interest in some other subject that might require it. much of hs and early college is not representative of the kind of math people do for fun.

at least in most schools in the US.

zingergi

What kind of math do people do for fun?

leslie townes

number theory is a popular subject because the barriers to entry are very low and there is a lot to explore.

plane and solid geometry is another one. high school geometry is not very representative of that, unless you had a very good instructor

zingergi

Cool.

Semiclassical

8:01 PM

@leslietownes low barriers to understanding the questions, at any rate

The barriers to actually proving things can be far higher

leslie townes

yeah, when i talk about barriers to entry, i mean, these subjects don't have a lot of prerequisites where you always have to understand a long list of X, Y, and Z before you can even understand the statement of every problem. i'm not saying these subjects are easy.

Semiclassical

Yeah

leslie townes

but you can find elementary number theory books that do get you quite a bit into the subject without assuming basically anything in the way of preliminaries. PDE, not so much.

Semiclassical

The simplest contrast I can think of is infinitude of primes vs infinitude of twin primes

Xander Henderson

@Semiclassical I firmly believe that number theory (particularly algebraic number theory) is where young careers go to die.

Semiclassical

8:04 PM

First was known even in antiquityt, second is still conjecture

leslie townes

it's tough to work in a field where there are a large number of outstanding problems that many people have worked on for decades and a lot of good results are basically highly incremental tweaks and minor improvements, or stepping stones to conjectured approaches that might not pan out. because it's easy for your peers/superiors to identify that.

Semiclassical

Yeah. Vs fertile ground in some newer field

leslie townes

you want to work in a new field, where any research plan sounds promising because nobody knows what's true yet. and if you're lucky you can rediscover results from 1921, in new language, and get them published in 2021 because everyone in the field is a blank slate and none of the reviewers bother to read old math.

Xander Henderson

@Semiclassical *cough* category theory *cough*

Semiclassical

Problem is that it gets harder and harder to find new ground

leslie townes

8:07 PM

i use quantum computing as an example because that's where i ran into it.

Semiclassical

Yeah, I believe it

And novelty depends on the audience

It might not be novel to the original math community, but it can still be novel to the audience that finds a use for it

Xander Henderson

Or you can just kind of invent your own new ground. A former collaborator of mine (who I should really send an email to...) has a small cottage industry working in a very small part of fractal geometry, based on a couple of papers written in the 1920s, and a PhD thesis from 1979. No one cared 10 years ago. Now I see all kinds of people citing his work.

leslie townes

i was asked to referee a paper. i had no business doing it but didn't say no. it took a while to translate the result to linear algebra language, at which point i realized it was something 100+ years old that appears, in linear algebra language, in many books on matrix theory.

i flagged this for the editor and suggested including citation to earlier exposition. the editor accepted the paper but it was published without any citation.

i guess results don't count if they appear in publications that don't have 'quantum' in the title.

at least it was just a technical result and not something with the guy's name associated to it - yet.

i like reading new proofs of old results and think journals should publish more of them. novelty is overrated. but no need to publish old proofs of old results. and maybe cite the old results.

Ted Shifrin

8:26 PM

Or you could be like a referee of one of my NSF grant proposals, who dismissively said that what we were trying to do was in old-style Italian algebraic geometry papers of the 20s or so. He was silly enough to give a particular reference. We read the paper (in Italian). It had nothing regarding what we were interested in pursuing. At this point, I've even forgotten what that particular problem was.

These sorts of problems were good enough for A'rnold and other average mathematicians, but not good enough for NSF reviewers.

leslie townes

haha

geocalc33

9:21 PM

Hi I have

geocalc33

9:32 PM

complex conjugate

that's just $z$ and $\bar z$ right?

leslie townes

people do often write it with a bar. superscript asterisk is also sometimes done

if you do use the bar, in latex i find \overline{z} a little better looking than \bar{z}; the \bar strikes me as too narrow

geocalc33

where did the word "conjugate" come from?

$\overline {z}$

copper.hat

some sex thing

leslie townes

i don't know the origin of the word, but the concept generalizes. like people will talk about a + b sqrt(2) and a - b sqrt(2) as conjugates. in this setting in the background there is an order-2 element of a group of permutations of the roots of a polynomial that leaves the polynomial invariant.

copper.hat

all jokes aside, it means a pair with things in common but some particular features being opposite.

leslie townes

9:39 PM

i don't know if that's reconstructing an idealized history. i'd very much guess people were calling z-bar something like a conjugate well before field theory / galois theory.

geocalc33

so could you do a conjugate with same dimensional manifolds with a diffeomorphism?

maybe $N$ has a linear structure and $M$ has a linear structure. and $f$ is a diffeo. could you package a new object $(P,g,\mathfrak{B})$ which inherits the linear structure of $M$ but inherits the metric of $N$?

leslie townes

the word wouldn't come with a pre-made meaning. you could certainly adapt the word for some use. i haven't seen it used in that context.

geocalc33

$g$ is the metric on $N$

$\mathfrak{B}$ is the linear structure of choice on $M.$ When it's embedded in $N$ (assuming that's possible), then you should be able to tell whether the linear structure is linear with respect it's embedding in $N,$ right?

geocalc33

9:59 PM

I guess you could say "permuting." like you have $A=(N, g_1, \mathfrak{A})$ and $B=(M,g_2, \mathfrak{B}),$ which are diffeomorphic. And you form the object $(P, g_1, \mathfrak{B}).$ And then I guess $P$ would have to be properly embeddable in $N$ in some cases

SmokenSieEinBitteChebaHitBits

Hey @geocalc33

geocalc33

Hey @smoke

SmokenSieEinBitteChebaHitBits

Any hits on your question about manifolds?

geocalc33

this one?

SmokenSieEinBitteChebaHitBits

Yep

geocalc33

10:11 PM

I didn't ask this on the main site

SmokenSieEinBitteChebaHitBits

What's the motivation for your question?

@geocalc33 I think you problem is about topological vector spaces, not manifolds

@geocalc33 do you know what that is?

topological vector space

Because on the one hand you have linearity

And on the other hand you have a metric (which means there's a topology induced by the metric)

Thus you have a vector space and a topology. It would be worthwhile to ask whether $+$ and scalar $\cdot$ are continuous operators

If so, then that gives you a topological vector space (by its definition)

geocalc33

@SmokenSieEinBitteChebaHitBits okay

SmokenSieEinBitteChebaHitBits

To prove continuity of $+ : P\times P \to P$ you only have to prove that $c + x = f(x)$ is continuous for each fixed $c$

But first you should show that what I just said is valid

so you'll need to know what a product topology is

*is defined as

geocalc33

10:30 PM

@SmokenSieEinBitteChebaHitBits I see, but a TVS is not the same structure as a manifold

SmokenSieEinBitteChebaHitBits

Yes, but its tangent space is always a topological vector space

What is linear about your manifolds?

@geocalc33 what if you generalize definition of manifold from looking locally like a Euclidean vector space to a $V^n$ for some vector space $V$ not necc. related to $\Bbb{R}$ and then generalize further from vector space to module. So a generalized manifold is a space that looks locally like some free module (i.e. with a basis) $M$.

What is that called in literature if they've already done it

?

Xander Henderson

@SmokenSieEinBitteChebaHitBits That is the question I screwed up on my topology qual. :\

SmokenSieEinBitteChebaHitBits

@XanderHenderson seriously? Then what is the answer?

Xander Henderson

@SmokenSieEinBitteChebaHitBits Oh, the question was to prove it. Show that the tangent bundle had an appropriate topological structure, etc etc.

But I learned differential topology from a homotopy type theorist, so I think that maybe she didn't understand it any better than I did.

SmokenSieEinBitteChebaHitBits

Well, since nothing is formally checked by a comptuer you could just pretend you proved it by skipping over the parts that you don't know how to prove with "clearly, obviously, or we know that".

Xander Henderson

10:41 PM

BUT... I managed to get a masters pass, which turned out to be good enough, so I was able to get on with my life.

@SmokenSieEinBitteChebaHitBits That isn't how qualifying exams work.

SmokenSieEinBitteChebaHitBits

@XanderHenderson what about my question. What is manifold when you generalize from vector space to module

Xander Henderson

@SmokenSieEinBitteChebaHitBits No idea. Like I said, I f'd up the section on differential topology. Manifolds suck.

SmokenSieEinBitteChebaHitBits

Manfolds are a worm hole

Ted Shifrin

smacks Xander with great intensity

Xander Henderson

I am sure that there is a notion of a topological module (I mean, such a thing would fit reasonably "between" a Lie group and a topological vector space. I just don't know if anyone cares...

SmokenSieEinBitteChebaHitBits

10:44 PM

@XanderHenderson @TedShifrin en.wikipedia.org/wiki/Ringed_space

Ted Shifrin

topological group would be want you're thinking of there?

leslie townes

time to categorify everything, xander. to be extra sure that nobody will care.

SmokenSieEinBitteChebaHitBits

It's a locallly ringed space, found out on math overflow

Ted Shifrin

locally ringed spaces is how algebraic geometers think, not smooth people

SmokenSieEinBitteChebaHitBits

Yes, but they're related categorically - both locally ringed space studiers

Ted Shifrin

10:47 PM

Yes, when we think sheafically, sure. Which I am prone to do from time to time because I was a complex geometer.

Ted Shifrin

11:25 PM

This was actually an interesting question. Too bad no response …

leslie townes

yeah, that's a good question. not phrased very well. good example of where initial segments of $\mathbb{N}$ are maybe not the best way of indexing a finite set because you basically have to name the last element. although arbitrary index sets in beginning linear algebra maybe also not a good idea.

Ted Shifrin

11:43 PM

I thought it was too ambiguous whether fixed or all $n$.

But I’d never seen this sort of thing priorly.

leslie townes

i've seen it somewhere, but i don't remember where. not any of the usual books.

Ted Shifrin

As I said, it seems sophisticated for a test.

leslie townes

yeah. better as a homework problem. and maybe not the week after someone learns what a vector space is. you never know where this is happening within someone's syllabus.

maybe something in a C* algebras book involving the geometry of the unit ball or state space. that is where i used to see affine combinations.

it's been asked before. math.stackexchange.com/questions/1383957/…

there, n seems to fixed at the dimension of the space, which seems to be missing your insight.

Ted Shifrin

I would say that solution missed the points I made.

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