dpmms.cam.ac.uk/~wtg10/meta.doubledual.html ←Also the first #1 in that does not make sense to me. "Let v(*v) be v*(v)". Hmm?!

Sorry (I wish I could edit typoes). ^^^ "Indeed, given v* in V*, define v(v*) to be v*(v)."

It just looks like an asterisk moved somewhere. Meaning what

(To clarify where I'm coming from, I do get why a covector = a row vector belongs to the dual space, and what that dual space means irrespective of being a matrix transpose.

They should use different letters. You want to give a map $V\to (V^*)^*$. Given an element v\in V$ it has to define a linear functional on $V^*$. That functional $L_v$ is defined by $L_v(\phi) = Phi(v)$ for $\phi\in V^*$.

Back in a bit.

thanks @TedShifrin

I also get the dual basis. You can choose it to be Kronecker to make things easy, or it could be a GL(n) transform of that if you wanted to make life ugly.

(GL(n) excludes det=0, right?)

Hi @TedShifrin.

@TedShifrin: You might like this question I posted earlier. Or you might not, or your feelings might be somewhere in between, and possibly you won't ascribe a value judgement to it.

@MichaelAlbanese Hi. How's things?

Hi @MikeMiller. Things are OK. I'm going on holiday next week so I'm just trying to organise some stuff before I leave. How about you?

I have a talk tomorrow I'm stressing over a couple of the details over, so I'm taking a break for a bit. Should be calmer after tomorrow.

Vacation sounds good. Going home?

No. Going to LA and vegas.

I'm sure your talk will be fine.

Appreciate it. LA is nice right now. You should go to the Santa Monica beach if you have time.

If I get time, I'll swing by. Do you know a graduate student by the name of Casey?

Hi @TedShifrin

Hi @MichaelA !

Hi Karim.

Casey Jao? I don't know him well, but I see him now and then. He's one of the few people usually in the department on weekends.

Not that Casey. I'm thinking of a female, but she may not be at UCLA. I met her at a conference last week and I remember she was from California, but I can't remember exactly where.

@Mike: you've got the attention of arguably the two smartest people on here :)

Details matter, @MichaelA:) Especially if you hope to find her :D

The only other Casey I know is a Bylund. I don't think she's been to conferences lately though.

@TedS: I don't think their suggestions are the right ones here but I need to check the details this weekend. (Admittedly they were the two I was targeting... :))

@TedShifrin I'm not looking to find her really, just thought Mike might know her.

Oh, my bad :)

Oh boy, my copy of "The h-principle for dummies" just came

I'm guessing that's not literally its name, is it?

unfortunately not

Ah, well, maybe you can read it, explain the results to me, and you'll have something that can rightfully called "the h-principle for dummies" :P

@MikeMiller Can I ask you a really basic algebraic topology question?

Sure.

"Non-commutativity of operations on a space may be expressed either using algebras or categories. Thus we may construct an algebra from a category, or vice versa, so let us consider obtaining an algebra from a category. Since composition of arrows in a category is associative we must construct an associative algebra. But an arrow may decompose into the composition of other arrows, usually in more than one way.

Two maps $S^1 \to X$ are freely homotopic if and only if they define conjugate elements in $\pi_1(X, x_0)$. So if $\pi_1(X, x_0)$ is abelian, we can identify it with $[S^1, X]$. Right so far?

Yeah.

What about maps $S^n \to X$ for $n \geq 2$?

The general statement is that $[S^n,X] = \langle S^n, X \rangle$ if $\pi_1$ acts trivially on $\pi_n$. (The action happens when you move the basepoint.) This is obviously true if $X$ is simply connected, but there are cases where you can see that it acts trivially, like if your $X$ is a Lie group or something.

What is the latter notation?

I guess it's probably just something I made up. Basepointed maps up to basepointed homotopy, aka $\pi_n(X)$.

OK. Sorry, but I don't understand what the action of $\pi_1$ on $\pi_n$ is.

@bolbteppa I have a feeling we'd need more context to answer that.

I don't think there is any more context, it should be clear from that but I don't see the convolution idea being 100% clear

@MichaelAlbanese: Think of elements of $\pi_n(X,x)$ as (classes of) maps from $(D^n,\partial D^n) \to (X,x)$. Pick $\alpha: [0,1] \to X$, $\alpha(0)=\alpha(1)=x$ as a representative of your element of $\pi_1$, $\beta: D^n \to X$ as your $\pi_n$ representative, then define $\alpha \cdot \beta(r,\theta) = \alpha(2-2\theta)$ for $1/2 \leq r \leq 1$, and $\beta(r,\theta/2)$ for $0 \leq r \leq 1/2$. The picture is that you do $\alpha$ on the outer "shell" and then $\beta$ on the inner one.

This is somewhere in Hatcher ch 4. An equivalent and I find more conceptually valuable picture is as follows.

$\pi_n(X) = \pi_n(\tilde X)$. The action of $\pi_1(X)$ on $\tilde X$ defines the action of $\pi_1(X)$ on $\pi_n(\tilde X)$.

(Push spheres to where they go, but follow the loop $\alpha$ back to the basepoint you should be at in the first "half" of the sphere's "time".)

This is probably incoherent. Sorry.

No, it's OK. I appreciate the effort. It will take me some time to digest it.

Seeing the picture makes that coherent, but I can't draw right now.

closer to fun mathematical game

@MikeMiller: I can kind of see what's going on here, but it doesn't appear very natural. Is it just a matter of making sense of moving a sphere around a loop?

Yes, essentially. You just collapsed the outer half-radius into a string. So it's a ball on a string. You moved the ball to the second position and the string came back.

Oh, I see what you mean. Nevermind.

It's just to account for basepoints, I think, sort of. The $\pi_1$ action measures how much the basepoint matters. But there are plenty of things that happen in homotopy theory that you need trivial $\pi_1$ action to do.

Spectral sequences (though I don't know why, since I haven't looked at the proof that the serre spectral sequence exists yet), obstruction theory.

@anon That's good to know, I always imagine it differently. I don't know why they bum me out so much, there was a similar question a few months ago.

OK. The reason I ask is that people will refer to $[S^n, X]$ as a homotopy group, but that isn't true.

Not in full generality, no. For instance $[S^2,\Bbb{RP}^2] = \Bbb N$, each map is determined by the absolute value of the degree of its lift to the universal cover.

So, contrary to what Drew Carey used to say, the points do matter.

At least on this gameshow.

Yay, you got the reference.

Thanks for your help yet again. The specific instance that I was thinking about was $[S^n, Gr_k(\mathbb{R}^{\infty})]$, but this is just $\pi_n(Gr_k(\mathbb{R}^{\infty}))$ because $Gr_k(\mathbb{R}^{\infty})$ is simply connected.

I see, gotcha.

@pjs36 The question is, what's the simplest way to define them? The geometric ones aren't formal, the infinite series ones aren't intuitive, the differential equations ones use heavy machinery.

Here's something interesting about Grassmannians. $Gr_k$ is always connected, but if you instead consider the oriented Grassmannian $\tilde{Gr}_k$ is connected precisely when $k=1$ and is otherwise disconnected.

@pjs36 But it turns out you only need the power of equalities and inequalities.

@pjs36 (...which can be shown, quite easily, to be equivalent to the geometric definitions, by using the usual proofs...)

Ehh, if you say so. I still vastly prefer the geometric definitions as a starting point.

Recalling that the Grassmannian is the classifying space for oriented vector bundles, this corresponds to the fact that the every oriented line bundle is isomorphic to itself with the negative orientation, but there is always an oriented $n$-plane bundle when $n>1$ that is not isomorphic to itself with the negative orientation.

That's today's cool Grassmannian fact.

If the geometric way isn't formal, how can you deduce it from the uninspired definitions?

Is there a way to construct this vector bundle?

Hm, I think the above is wrong. What I want to say is how many oriented bundles there are on the point. And that should be 1 if $k$ is odd, and 2 if $k$ is even, I think.

My cool Grassmannian fact wasn't much of a fact, I guess.

@pjs36 Once you have the geometric definitions, you get to derive important identities such as the sum formulae. Which is basically what's being offered in the answers to that question. Or are you asking about the reverse?

Ok, now I agree with what I just said.

And the $\displaystyle\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n$ one isn't uninspired, if you think about things the right way...

A few months ago, after thinking about complex numbers for a while, the reasoning behind that expression got a whole lot clearer to me.

I still just don't see the point of it all, that's been my (perhaps unexpressed) point. I must not do enough analysis, I guess.

@MichaelA: Ok, I'm confused again. I think what I said up there is probably not worth thinking about. Sorry.

@pjs36 This is math — what's the point of anything we do?

@MikeMiller No worries.

(That ended up sounding existential.)

...$\exists$

Yeah, I know, it's silly. Like I said, the questions have always just bugged me, because I get the impression that it's implied that the geometric definitions aren't good enough, and that's really the heart of the matter.

Why on earth is my post here attracting so many downvotes? Is something wrong with my math? math.stackexchange.com/questions/1303417/…

I suppose some people are incensed by incomplete answers. I did what I could! :)

@KajHansen the OP has two downvotes and a close vote - they are -1ing you as well for rewarding low-effort/context-like questions

@anon, I suppose that's the most plausible explanation. It didn't have a downvote/close vote when I answered, but I admit that it's low-effort. Thanks!

although reading your answer I also don't understand your last sentence (unless in addition to thinking about degrees you also want the reader to observe that they aren't the same extension because, say, one is real while the other isn't)

and looking at Gregory's answer, his doesn't have a downvote, so I rescind my explanation

@pjs36 Don't know or care if what I'm saying applies to sin and cos, but you'd want to do that type of thing to be able to abstract the notion.

oh never mind it's x^3-3 not x^2-3

walks away

@anon, because if we consider those two fields over the rationals, we have $\mathbb{Q} \subset \mathbb{Q}[\alpha] \subset \mathbb{Q}[\sqrt{-3}]$. Then the last one is a degree $2$ extension, and the middle is a degree $3$.

haha

Hello everyone!

@anon, I think there might just be a mean-spirited person out tonight. This completely legitimate question got immediately downvoted: math.stackexchange.com/questions/1303436/…

That's a valid point, @KarlKronenfeld. I'm comfortable with that sort abstraction in a more general setting, I should probably just pretend it never happened, and admit I'm a philistine :)

hi chat

Hello. :)

What is everyone working on? I'm quite curious as to what kinda math you all are tinkering with.

well, i've got a definite integral i'm trying to grok for math-physics purposes

Neat! :0 Care to show it?

$\displaystyle \int_0^1 \frac{\ln(1+t)}{t(1+t^r)}\,dt$ for $r>0$

@PerplexedGuest I am studying the rational $\sigma$-algebras whose elements are 3-connected hypergraphs.

I do not know what those are!

I suppose I picked my name aptly after all.

I'm introducing myself to the $p$-adic integers.

@MikeMiller I'm slightly amused by the fact that you're talking about Grassmannians tonight, seeing as I was just doing some reading on them yesterday for research reasons

@PerplexedGuest randomly generated mathematical term

$p$-adic numbers look quite interesting, Kaj.

Unfortunately your research is not so easily Googled, Karl.

@PerplexedGuest I'm afraid I am the only person studying these things.

That would explain it!

I myself am studying a certain aspect of self-avoiding walks. P:

I suggest the book generatingfunctionology because it has a section on stuff like that.

(Though it might take a lot of reading to get to that point.)

Thanks, I shall take a look. P:

Yeah, pretty sure the relevant part is pretty far in, so you might not know what's going on... but I still recommend it anyway, because it's a good book.

I believe it's available online for free.

I am getting it from UPenn right now. :)

@PerplexedGuest Prerequisite: You need to know how to differentiate.

I do in fact know how to differentiate. xD

OK, good, just making sure.

(Couldn't tell — "On the internet, no one knows you're a dog" and all that.)

It's fine. xD

$xD$ being shorthand for $x\frac d{dx}$, I assume.

I frequently use emoticons to express how I feel over text. :P

Also, that is objectively terrible notation.

it's fine

Am a dog :D

I mean Bark

@PerplexedGuest I was kidding. Also, it's used in the book, I think.

I know you were kidding. xD Oh, interesting. I'm quite liking the book so far - I finally know what a generating function is.

(There's a chance I'm confusing it with another thing that used the notation.)

...a lot.

It's quite difficult, and I believe still open.

But most of all it's interesting!

There's a chance that's an open problem, but I might be confusing it with something else.

EDIT: I am a slow typer, apparently.

I am $\geq$ 50% sure it's open.

I need to go to bed

With respect to what topology?

Badum-tss.

(I'm really tired, clearly...)

bye

Adios. ~

hey

I don't understand the following

The structure of the group G is reflected in the structure of the quotient groups and subgroups of G. For example we shall see that the lattice of subgroups for a quotient of G is reflected at the "top"

what is the "top"

if you write down the lattice of subgroups of G on paper, you should have G at the top and go down from there.

yes

then the lattice of subgroups of (say) G/N has an isomorphic copy within the lattice of subgroups of G

you just copy and paste it at the top of the lattice, with G/N (the top of the lattice of subgroups of G/N) going to G (the top of the lattice of subgroups of G) and N/N (the bottom of the lattice of subgroups of G/N) going to N

oh I see

given any ordered set X and two elements a and b, one can speak of the interval [a,b] as the set of all x between a and b (including a and b themselves). the lattice correspondence theorem says that the lattice of subgroups of G/N is isomorphic to the interval between G and N in the lattice of subgroups of G

I se

I see makes sense

yeah because your breaking G Into chunks of N

good thank you @anon

Hello@Karim

hi @Rememberme

Hi@anon

hi

Hi, I'm reluctant to re-ask a question that's been posted so many times, but I'm having trouble understanding the double dual V** of a vector space V.

Thinking of V* ≝ { V → k } would make V** ≝ { {V→k} → k }. Which seems wrong.

what seems wrong?

Well, I can't picture that.

it sends functionals to scalars

Like if V→k were just a scalar field, ...

So sending each of the basis functions to a scalar?

as particular cases, yes. it sends all functionals to scalars.

in a linear way

which, incidentally, is exactly what "evaluation at [this vector]" does

What do you mean "as a particular case"?

@isomorphismes I explained what I meant in the very next sentence

ok, thanks anon, I keep getting confused whenever people bring up the evaluation functional.

it doesn't just send basis elements of {V->k} to scalars (there isn't even a canonical basis), it sends all functionals to scalars

if you call evaluation a functional you might confuse yourself

What is getting evaluated?

have you ever evaluated a function at an input before? that's evaluation.

Over what space are we talking? V?

there are many spaces being talked about here

Where is evaluation happening over

I am using vector to refer to an element of V, and functional to refer to an element of {V->k}.

Given a functional f:V->k and a vector v, you can evaluate f at v

that's evaluation

ok great

So why is it bad to think of that as a functional?

it is not an element of {V->k}

and we are already using the word functional to refer to elements of {V->k}

so let's not overuse the word

hmm ok

it's like if you have functions, and functions of functions, and functions of functions of functions, and so on, you don't want to simply refer to all of these things just as "functions" because that makes it hard to distinguish which of the classes its in (i.e. a function, or a function of functions, or...?)

ok that seems reasonable

/ like a good idea

@anon hmm, ok, I think this is making sense now ...

Why does it fail for ∞-dimensional V?

(or, not always hold)

since the act of evaluating various functionals f:V->k at a (fixed, chosen) vector v is a linear map {V->k}->k (which is technically a functional, just as the elements of {V->k} are technically vectors, but let's keep the terms organized), each individual vector can be "reimagined" as sending functionals to scalars. this amounts to sending elements of V to elements of {{V->k}->k}.

(((((For reference, I'm coming to this question after reading github.com/JuliaLang/julia/issues/4774 where they debate canonically isomorphic vs isomorphic and data-type definition of vectors, covectors, etc for a programming language)))))

@isomorphismes I recommend thinking in terms of coordinate vector spaces to see why. Do you understand the difference between direct product and direct sum (of an infinite number of vector spaces)?

@MichaelAlbanese easier way : $\pi_1$ naturally acts on $\widetilde{X}$ by deck transformations. and $\pi_n(\widetilde{X}) \cong \pi_n(X)$. this doesn't make the action explicit, however.

@anon Maybe

@isomorphismes I'll take that as a no.

do you have latex in chat working?

@anon Haha

@anon No

see "latex in chat" on the starboard -->

and get it working first

@anon oh nice! thanks!

aight. let $\Bbb Q^{\oplus\Bbb N}$ be the space of infinite sequences $(x_0,x_1,\cdots)$ of rationals all but finitely many of which are zero. do you see how this is a vector space? (this is a direct sum of countably infinitely many copies of $\Bbb Q$)

@anon Is there sometimes a good basis, and just not in general? I'm thinking in case of $V \overset{let}{=} \mathbb{E}^2$, then let $f_1(x)=3x$ and $f_2(y)=1y$ and $V^*$ would then seem "naturally" to have a basis aligned with the original axes.

@isomorphismes yes, there can be good choices of basis in particular cases

@anon What is another functional that would not be like the example I gave?

huh?

@anon You said "it sends all functionals to scalars". I'm still having trouble imagining a $V^*$ that doesn't decompose into a sum of basis functionals.

a few moments ago you just gave me two functionals, not one, but the comment of mine that you're currently pinging is where I say evaluation-at-a-vector sends every functional and not just basis functionals to scalars (and back there you did not give any examples of functionals)

@isomorphismes given any basis, every vector is a sum of basis elements. I don't understand what you're talking about.

@anon Sorry, I meant to imply that $\mathbb{E}^2 = (x,y)$. And $V^*$ decomposes as $3x+1y$ along the original definition of $x$ and $y$.

I think you mean $f_1(x,y)=3x$ and $f_2(x,y)=y$ and you're saying that every element of $V^*$ can be written as $g(x,y)=af_1(x,y)+bf_2(x,y)$ for some scalars $a,b$. Which is correct. What about it?

brave question, i guess

@anon well in the earlier message you were clarifying something about evaluation.

evaluation-at-a-vector doesn't just send the basis functionals to scalars (assuming you've chosen a basis for V^* in the first place), it sends all functionals to scalars

of course, it is determined by where it sends basis functionals

@anon ooook. Gotcha

@anon thanks

hello, @iwriteonbananas

@anon ok yes, cool

now let $\Bbb Q^{\Bbb N}$ (with no $\oplus$ symbol) be the space of all sequences of rationals, no restrictions

@anon 👍

you should think of $\Bbb Q^{\Bbb N}$ as the dual space of $\Bbb Q^{\oplus\Bbb N}$, because given any sequence $(x_i)\in\Bbb Q^{\Bbb N}$ and $(y_i)\in\Bbb Q^{\oplus\Bbb N}$ we get the scalar $\sum_i x_iy_i\in\Bbb Q$. this sum is always finite, since all but finitely many $y_i$s are zero.

@anon wow interesting

(I will not explain why $\Bbb Q^{\Bbb N}$ is the whole dual space, but just think of it as a space of at least some of the dual vectors i.e. functionals)

now $\Bbb Q^{\Bbb N}$ is bigger than $\Bbb Q^{\oplus\Bbb N}$ is. in fact, the first is uncountable (it contains all infinite binary sequences, for instance) while the second is countable (it is a countable union of the countable subsets $\Bbb Q^n$ for $n\in\Bbb N$).

Who is starring everything ??

this illustrates a general fact: taking the dual of an infinite-dimensional vector space "expands" it to an objectively bigger space. so taking the double dual will be even bigger still than V is, which is why V->V^** will not be surjective then.

@anon that is weird.

@Hippalectryon Sorry, that's me, I thought it just worked for me (like questions on the site)....

Uh, not really, it stars it for everyone to see

@Hippalectryon Sorry

Well you can always unstar them :-)

@Hippalectryon """It is too late to undo this operation"""

Ah, ok

@isomorphismes if you're interested in using it for documentation purposes, you can just star one of the comments, then at any point in the future you can click the "show all [number]" link above the starboard, and once you're in the log of all starred messages you can select "those starred by me" and see the original star you made. then you will be able to click the timestamp to be linked back to the original chat we had in the chatroom at that time.

@anon oh nice! That's exactly what I needed

I've cancelled out all your stars but the shortest one so as to reduce clutter.

@anon ok chat.stackexchange.com/rooms/info/36/mathematics/… I think I follow your chat instructions.

Hint : show that $(p, f)$ is a maximal ideal of $\Bbb Z[x]$, and use the fact that maximal ideals are prime.

key facts to know: I is a prime/maximal ideal iff R/I is a domain/field (respectively).

no.

e.g. (2,x) is not principal

no zero divisors

@AlexClark exercise you might want to think about : prove that if $R[x]$ is a pid, then $R$ must be a field.

yes.

@anon This is not $g(a,b)$ but $g(x,y)$?

it's always a prime ideal, isn't it?

?

@isomorphismes Huh?

Write $g=af_1+bf_2$ without writing the arguments of the functions if you want to.

the idea is that if $g$ is a linear functional then $g$ is a linear combination of $f_1$ and $f_2$

what the hell is that.

it's a fact that $(p, f)$ for prime $p$ and irred $f$ is always a prime.

so i dunno what you mean

@BalarkaSen hey balarka

yeah, (2,x^2-2x-4) is not prime in Z[x]

@anon I dunno why we need the $a$ and $b$ since $f_1$ and $f_2$ already contained some constants.

ok, right

$f$ must be irred modulo $p$ (look at the quotient to see why it's the case, @AlexClark)

sorry.

@isomorphismes okay, consider the elements $v_1=(3,0)$ and $v_2=(0,1)$ of $\Bbb R^2$. I claim that every element of $\Bbb R^2$ can be written as $av_1+bv_2$ for some scalars $a,b\in\Bbb R$. do you disagree with me? it's important you understand what a basis is: it's just a handful of things that everything else is a linear combination of.

@anon Ah ok, sure.

@iwriteonbananas doing anything interesting?

your $f_1$ and $f_2$ are just two functions out of infinitely many. all the rest are linear combinations of $f_1$ and $f_2$.

right now im integrating on manifolds

:'(

im gonna study degree theory when i solve this damn integral

:)

@anon Your example is making a basis for $V$, not for $V^*$, though, correct?

@isomorphismes correct.

to help you understand what a basis is using something more familiar to you

it's pretty cool to note that $S^1 \vee S^1 \vee S^2$ has the same homology as $S^1\times S^1$

nah, it's easy to construct such spaces.

yeah, i guess so

homology of a wedge of spaces is always direct sum of each space right?

right

the whole point is that homology is abelian, which makes it easier to construct spaces with same homology in all dimensions

all homology groups are abelian?

@anon ok, thanks. But let's now transition to $V^*$. So let $f_1(x,y) \overset{def}{=} 3x + 0y$ and $f_2(x,y) \overset{def}{=} 0x + 1y$ as before. Now I'm imagining two "abstract indices" $a,b$ over which $V^*$ splits. (abstract so that I'm not specifying a basis)

i only know that $H_1$ is the abelianization of $\pi_1$

by definition, yes, @iwriteonbananas

@isomorphismes abstract indices? what?

@anon Like in tensor notation

fair enough

$C_\bullet$s are all free abelian groups

@isomorphismes tensor notation is not conducive for this discussion (you're misunderstanding how indices work anyway)

and image of free abelian groups under homs are abelian

oh yeah, silly me

so are the kernels

after quotienting, you get back an abelian groups

i've only been thinking in general homology theories lately, forgetting the whole cycle stuff

the whole point of singular homology is that it's abelian, @iwriteonbananas

:P

you didn't read the introduction of chapter 2 in hatcher, didja?

hmm, is that the whole point?

yeah, i did read it...very often in fact

it's excellent

@iwriteonbananas well, almost the whole point. for example, consider you have two surfaces, one of genus 2 and one of genus 3

$\pi_1$ of them are crazy : you can never say if they are isomorphic by direct hands-on computation

but $H_1$ are abelian, and clearly nonisomorphic. thus, $\Sigma_2 \not \cong \Sigma_3$

@anon ok well, pick some basis over $V^*$ and can we call it $(a,b)$?

right

@isomorphismes what does (a,b) mean? bases are sets of things, and that's not even a set.

@anon Sorry {a,b} but ordered

and what are your a and b?

@iwriteonbananas yeah, hatcher provides excellent geometric intuition.

@anon Well since $(x,y) \in V$ were fixed when we defined $f_1,f_2$, then to get the value at a particular point in $V^*$ I thought we would need two numbers which come from $V^*$, not from $V$.

@anon $V^* \ni g := a \cdot f_1 + b \cdot f_2$

neither elements of V nor elements of V^* are numbers

and we defined $f_1$ and $f_2$ as functions, so $(x,y)$ was not fixed when we were doing that

@anon $(x,y) \in V$, each of $x$ and $y$ is a scalr

@isomorphismes yes

eh, i need a few more upvotes to get the close votes privilege, but i can't find any good questions to answer. i guess i'll dig up old questions from the algebraic topology tag and answer them later on.

Given your two functions $f_1,f_2\in V^*$, every element of $V^*$ looks like $af_1+bf_2$ for some scalars $a,b$. You needn't have ever heard the word "evaluation" in your life to understand this.

@anon Well, why did you need $a,b$ in your formula up there?

@isomorphismes let's go back to vectors again. Consider $v_1=(3,0)$ and $v_2=(0,1)$ in $\Bbb R^2$. Do you agree that every element of $\Bbb R^2$ looks like $av_1+bv_2$ for some $a,b\in \Bbb R$

@anon $a(3x+0y) + b(0x+1y)$

@isomorphismes Right. For instance, consider $g(x,y)=x+y$. We can write this as $g=\frac{1}{3}f_1+f_2$.

@anon ok, sure.

@iwriteonbananas make sure you lecture me about forms after you learn those stuff. i badly want to know what those are.

@anon But I thought I was already "covering" $V^*$ with just the $f_1$ and $f_2$, without needing to span(f₁,f₂) ?

@BalarkaSen my course in weird in that respect. we prove/compute a bunch of things for general homology theories, but we don't yet know that such an object exists

@isomorphismes what do you mean by covering? do you think every element of R^2 is either (3,0) or (0,1)? do you think R^2 has only two elements in it?

same idea for $V^*$. we have two functions $f_1$ and $f_2$. every other function is a linear combination of $f_1$ and $f_2$

@anon Well, $f_1(x,y)$ and $f_2(x,y)$ both run over $V=(x,y)$.

hmm, that's not good @iwriteonbananas. most people won't get the motivation to think about general homology theories if they don't know about singular homology.

@BalarkaSen will do, i assume im gonna learn about them soon. im suprised they havent been introduced in my course yet

@isomorphismes both $f_1$ and $f_2$ can be applied to any element $(x,y)\in V$. but $\{f_1,f_2\}$ are just two of all of the infinitely many functions...

yeah, im not fond of how my alg top course is structured

Hello @Balarka

@anon OK, so we are talking about all possibilities i.e. the whole space $V^*$ and I was confusing that with an element of $V^*$. Thanks

hi

@iwriteonbananas to hell with it, learn in your own way. whatever the class is doing, you know that those exist. :)

@isomorphismes I said repeatedly "every function is a linear combination of $f_1$ and $f_2$" and "every function $g$ looks like $g=af_1+bf_2$ for some $a,b$"

@anon I see what you mean now. My choice of $f_1$ and $f_2$ was just one of the possibilities for basis functions, just as span{ (3,0), (0,1) } would be over $\mathbb{E}^2$. Thank you

@anon You know what confusion is, right?

@BalarkaSen im more or less just studying hatcher, and occasionally having a look at tom dieck

tom dieck is a good book.

i'm gonna study some bordism homology theory from it after i get to know a bit of differential topology

cool