Mathematics - 2015-12-02 (page 2 of 3)

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3:00 PM

@Huy: I'm a bit frustrated because I can't do this :) It should be easy, but I'm not able to produce a proof.

I thinkt he idea is to pick $f_n \to f$, pick compact sets $C_n$ outside of which $f_n$ vanishes, to get some control of $f$ on the boundary of $C_n$ with something like the integration by parts formula; then once you have that, use the fact that the volume of $\partial C_n$ is bounded (because you're in a bounded open set) or something to get that $f$ must vanish on the boundary?

yeah I recall something similar

definitely integration by parts

I don't have to prove that right now, maybe at a later point

but those completions as I stated them should be that way, right?

$H^1_0$ should be the completion of $C^\infty_c$ and $H^1$ should be the completion of $C^\infty_{H^1}$ (that is, of the smooth functions with bounded $H^1$ norm)

first completion wrt to ?

Same norm.

$H^1_0$ is just a special subspace of $H^1$

ah

3:05 PM

Now if you're on a complete manifold I think $H^1$ and $H^1_0$ coincide, because I think you should be able to $H^1$ approximate a bounded-$H^1$ smooth function by compactly supported ones as you go off to $\infty$

My excuse is that I usually think about this latter framework :)

ok I think I got it again

one more thing

Rosenberg generalizes the gradient by composing $d: C^\infty(M) \to \Gamma(T^*M)$ with $\alpha^{-1}_g: \Gamma(T^*M) \to TM$. the only $\alpha$ he defines before is $\alpha: TM \to T^*M$ (I assume the canonical map). how does he get from $\Gamma(T^*M)$ to $T^*M$? did I miss something here or is a different $\alpha$ meant?

@Huy: There's no canonical map $TM \to T^*M$. He's using a metric to get an isomorphism $\alpha_g$, and then is taking its inverse to go backwards.

can you slightly (but not completely) elaborate?

I've been reading this part several times in the book and don't see which map that is

Explicitly, $\alpha(v)$ is defined by $\alpha(v)=g(v,-)$.

(Because $\alpha(v)$ is a 1-form on a certain vector space $T^*_pM$.)

ah, ok

ok, so that's $\alpha$

3:12 PM

Then $\alpha_g^{-1}$ is just the one in the other direction.

$g(\alpha_g^{-1}(\beta),w)=\beta(w)$.

but $d$ sends me to a section of $T^*M$, so I'd need to evaluate at some point to get to $T^*M$, no?

@r9m nice avatar

I don't understand what that means. This is a bundle map, not just a smooth map, so it sends sections of $T^*M$ to sections of $TM$.

What he's defining here is precisely $\nabla f$.

I mean that by definition, $\alpha: TM \to T^*M$, right?

wait

no I still don't get it

Yes, but I picked the inverse map.

3:17 PM

yeah but why is that one defined for any section of $T^*M$?

Because $\alpha$ is an isomorphism.

(Which is ultimately because $g$ is a nondegenerate bilinear form.)

I think I'm missing something

I'll take a 5min break and I'll think it over again

(cause it still doesn't make sense to me. brB)

You need to prove that $\alpha_g$ is an isomorphism. Once you do, you can define its inverse. That's what he's doing.

Hello!! Does someone of you have an idea about my question:

0

Q: Can we express this condition using only these operations?

Suppose we have the differential equation $y'=y$. It's solution is $y=ce^x$. So that we get the solution $y=e^x$, we need the condition $y(0)=1$. Can we express this condition using only the operations $\{+, \cdot , \frac{d}{dx}\}$ ? For example if we have the differential equation $xy'=k...

differential-equations

?

@MikeMiller: About the isomorphism thing, isn't that just Riesz?

3:27 PM

Yes.

but my problem is that the inverse would be $\alpha^{-1}: T^*M \to TM$

and not $\Gamma(T^*M) \to TM$

He's not writing $\Gamma(TM)$? That's a typo.

he just wrote $\alpha: TM \to T^*M$

and then $\alpha_g^{-1}: \Lambda^1 T^*M \to TM$

You said he has a map $\Gamma(T^*M) \to TM$. That's a typo. He gets from the inverse a map $\Gamma(T^*M) \to \Gamma(TM)$.

Obtained by applying the bundle map $\alpha_g^{-1}$ to a section of $T^*M$.

Why is $\mathbb{Q}(\alpha, \sqrt{-23})$ Galois, where $\alpha^3 = \alpha + 1$?

3:36 PM

@MikeMiller: so I get a map $\Gamma(T^*M) \to \Gamma(TM)$ by "applying $\alpha^{-1}$ at every point in $M$" or how do you mean that?

Yes, @Huy.

ok, that makes more sense

I'm awful at combinatorics and was hoping someone could shed some light on a problem I have. I had a chess board (8x8) of 64 squares and seven white pieces and seven black pieces. I can place any number of blacks and/or whites (i.e. I could place all fourteen or none or three of one and two of the other) anywhere on the board with no rules other than that they cannot share a space.

I don't need to know the exact number just was hoping someone could verify this isn't a problem I could probably just generate every solution for computationally (i.e. more than, say, 100,000,000 combinations)

and then I have a section of $TM$

ah

the gradient should be a map $C^\infty(M) \to \Gamma(TM)$, right?

or to $TM$?

$\Gamma(TM)$.

3:38 PM

he writes $TM$ and that's what I copied

Horrible typo.

that's a lot of things that confused me

thanks for the clarification

ok now let's see how to show that this is closable and we can thus take the adjoint

what's a "formal adjoint" btw ?

@r9m Have you heard anything since this?

We have an inner product on all the various bundles we're working with, incl $TM$ and $T^*M$. (You can either think of the one on $T^*M$ as coming from the Hodge star or as coming from pulling back the inner product on $TM$.) If $D: \Gamma(E) \to \Gamma(F)$ is some map (say, differential operator) then its formal adjoint is a map in the other direction such that $\langle Df,g\rangle = \langle f,D^*g\rangle$.

I think it's only called a formal adjoint because the spaces of sections are not complete, and that you can push it to the level of whatever your favorite completions are.

@MikeMiller hmm, dunno.

3:45 PM

@anon: I later decided you probably could by some sort of standard averaging out $h$ business but didn't carry it out

@AshleyDavies are the pieces distinguishable (other than by color)?

now if $H_0^1$ is the closure of $C_c^\infty$ wrt. to the $H^1$ norm and I restrict $D$ (the gradient map I just obtained) to $C_c^\infty$, its closure wrt. the $H^1$ norm will be defined on $H_0^1$ and not on all of $L^2$, right?

Yes.

and I guess it's not possible in general to close it wrt to the $L^2$ norm?

No, it's not continuous w/r/t the $L^2$ norm. It has a derivative in its definition, so you'll need control on convergence of the derivative.

3:49 PM

@r9m: Ack! I just saw this.

I guess maybe you could pull it off with some negative index Sobolev spaces, but I haven't thought about whether it works

why the anti-social comment, I wonder?

@robjohn once bitten, twice shy? :P

ah, I see

@Huy: Write $C_j$ for $C^\infty_c$ with the $H^j$-norm. Then I am now confident that $D: C_j \to C_{j-1}$ is always continuous, even if $j$ is not an integer. So you can extend it to the Sobolev completions.

3:55 PM

@user685252 Did something happen that I haven't heard of? In any case, why no math community anywhere?

Could someone help me understand what exactly is a "matrix associated to an endomorphism $f: V \to V$ *with respect to a reference $R$*"?
I know what is a matrix (of course), I know what is a reference but I have no idea what it means "associating a matrix to something *with respect to a reference*.

@Overflowh by reference do you mean ordered basis?

@robjohn something about somebody ruining/using her original work...

@anon Yes exactly

okay, you know what a matrix is - do you know why they are, what we do with them?

they represent linear transformations

3:57 PM

Yes I know, and an endomorphism is a linear application where the domain and the co-domain are the same.

Given a basis $\{e_i\}$, we know that $f(e_i)=\sum_j a_{ij}e_j$ for some array $a_{ij}$ of coefficients.

that's where matrices come from

(linear application or transformation, I don't know what's the standard way to call them in English)

so if you represent a vector $v=\sum v_ie_i$ as a column vector and apply the matrix $[a_{ij}]$, you will get the same thing as if you wrote $f(v)$ as a column vector

@user685252 If she published it first, even here, that is copyright. Other than that, I don't think knowledge should be patented, only copyrighted.

but of course the actual matrix associated to $f$ depends on what basis you used.

4:00 PM

@robjohn I don't believe she is aware of that.

@anon What do you mean by "apply the matrix $[a_{ij}]$"? I assume you mean to apply it to the column vector, but what does the "apply" action actually do?

you've never multiplied a column vector by a matrix?

Yes, I did

so you know what I mean

Oh

4:04 PM

an ordered basis induces an isomorphism $V\to F^n$ (called "coordinates") where $F$ is the scalar field. A matrix defines a linear map $F^n\to F^m$ where we treat elements of $F^n$, $F^m$ as column vectors. If we put $V\xrightarrow{f}W$ up top and $F^n\to F^m$ down below, and connect $V$ and $W$ above to the $F^n$s down below using coordinates, then the map $F^n\to F^n$ is the matrix defining $V$.

yesterday, by OFFSHARING

@r9m hehe, at least for a good while I wanna share nothing about my work, and here I'm tempted to do it.

@anon I'm sorry but I missed the last one almost completely.

If I would write $f \in X \to Y$ as shorthand for $f$ is a function with domain $X$ and codomain $Y$, would this be confusing in some cases?

So where some would write the set as $Y^X$, I would write it as $X \to Y$

That is not in any way shorthand, given that $f: X \to Y$ means precisely that.

Certainly, but this makes it more clear that $f$ is a member on the set

So the notation puts emphasis on the set $X \to Y$ instead of the function $f$

4:15 PM

I would have no idea what you're saying. I would also not expect to see $X \to Y$ named a set.

I'm just wondering if there are any objections to using it as a name

@MikeMiller: Can I just integrate by parts with my defined $D$ to show its closable? I'm hesitating because I only defined it as a composition of two maps.

I object because there's essentially no value to it.

@Krijn Seems reasonable to me, but you do need to explicitly define it. As in "I'm using $X \to Y$ to denote the set of functions from $X$ to $Y$." You might also want to mention that it's conventionally denoted $Y^X$, but that you don't like that notation because <reasons>.

also I'm not taking the $L^p$ norm as usually when we showed differential operators are closable in functional analysis

4:18 PM

Wikipedia says that it is already used, so Im fine with that

@IlmariKaronen Oh definitely!

Cool. I just won't read your questions if their notations annoy me, and there may be others who also find nonstandard notation obnoxious. :)

Oh I wasnt planning on using it on Math.SE, I had a discussion with a roommate because I thought this was more intuitive

And he said that I don't understand set theory and that I should feel bad :(

But Wikipedia agrees with me

That's a strange discussion.

We often have strange discussions

@MikeMiller Just a little note: keep in mind though that there could come someone who does not know the standard notation simply because they have been taught nonostandard ones (like me calling "reference" an "ordered basis" of a vector space few lines above).

4:20 PM

I don't understand why he likes proof theory and such, and he doesn't understand my enthusiasm about algebra and topology

@Overflowh: I wasn't talking to you there.

@MikeMiller I know I know, I was just reading reading the conversation.

I solved my math research.

Time to research more haha.

Actually, I kind of agree. Whenever I see $Y^X$ used for general functions, I have to stop and think to remember if it means $\{f:X \to Y\}$ or $\{f:Y \to X\}$. (I can remember it for, say, $\{0,1\}^X$ and generalize from there, but I don't see the general $Y^X$ notation often enough to have in fully ingrained.) Your notation makes it obvious which one is the domain and which is the codomain.

@Huy: I think you would have to carefully write down the proof.

4:23 PM

@IlmariKaronen Ha, yes, that was my argument

That $Y^X$ is not intuitive enough for me

I'm actually kinda upset about solving my research problem. It was way too easy and I can't even write about it. :(

@Overflowh: I see your point. There are hundreds upon hundreds of MSE questions per day, and whether or not nonstandard notation is the OP's fault, I could just look at any number of questions with notation I'm used to instead of spending time figuring out an unclear question.

@MikeMiller Of course, I never meant the opposite.

@Krijn: If you want to avoid this perceived ambiguity whilst not using nonstandard notation you could write $f \in \text{Map}(X,Y)$.

@MikeMiller: is there a more direct way maybe? the way my prof talked about it, it sounded like it should be very obvious the map is closable.

4:26 PM

@Huy: Local coordinates.

he never even talked about being closable, just that "we'll take the adjoint of the closure"

but if we do it locally, will it globally be a closure?

You're thinking too hard about the notion of closure. Closable with respect to certain norms just means that $D$ is continuous with respect to those.

@MikeMiller That's possibly even better for my purposes

and?

why does that make it easier?

Just take a partition of unity and work in local coordinates. In local coordinates this is the literal gradient which clearly is continuous as a map $H^1 \to L^2$.

4:31 PM

just take a partition of unity. I wish I was as comfortable with partitions of unity as you are :/

Write it down

so should I express $D$ in a chart now?

Yeah

iwriteonbananas

Is $\Bbb Z_m / 2 \Bbb Z_m \cong \Bbb Z_2$ for all $m$?

What is $2\mathbb{Z}_3$?

iwriteonbananas

4:41 PM

whoops.

@Huy: I'm sorry, I've been leading you astray. Remind me how you define the $H^1$ norm on $C^\infty_c$.

iwriteonbananas

Don't mind me I'm talking nonsense again.

@MikeMiller: The sum of the norm of the function and the norm of the derivative

What's the derivative mean? We're on a manifold, right?

yes

well up to now my prof always told me I could just use the complex derivative because we'll be applying it to the upper half plane

but now that I defined this map $D$ I'm confused as to what to take

because it would make sense to take $D$ but that's something different, no?

4:49 PM

@Huy: Nope! One way of defining $H^1$ is as $\|f\|_{L^2}+\|Df\|_{L^2}$. Note that $Df$ is a section of the tangent bundle which has a fiberwise norm so I can make sense of its L^2 norm.

isn't there a relationship between closable or closed and the graph norm?

urm

Now, with this norm, it should be obvious that $C^\infty_c \to \Gamma(TM)_{L^2}$ is continuous.

should

let me think about it

ok I'm understanding your second last statement

(at least I think so)

@anon No, with regards to the problem I featured earlier; just by color. Each cell can be empty, black or white; there's no other distinguishing features.

I tried generating them and had to stop it when it got to 100GB of files, so I think I've came to the conclusion that it's not viable :P

@MikeMiller: because wrt. this norm it's bounded?

by definition

4:53 PM

@AshleyDavies you need to sum the multinomial coefficient (64 choose a,b,64-a-b) for 0<=a,b<=8

I think

@Huy: Yup.

It's gonna be huge though, right?

Talking maybe 100,000,000+ boards

oh, you have 7 of each piece, not 8

sure, but a computer can do it in seconds

I typed Sum[Sum[Multinomial[a, b, 64 - a - b], {a, 0, 7}], {b, 0, 7}] into Mathematica

it spit out 218894030249902545

@Huy: Sorry I've been slow to respond, I've been getting ready to face the cupcakes.

Ah I'm actually needing to keep these around if I do this approach

Basically trying to generate every single perfect solution to a game by generating every board and working back from the winning points to every other board position

5:00 PM

There are about a hundred different equivalent Sobolev norms. Some use local coordinates, some use the gradient.

@robjohn I didn't know that either!!

@MikeMiller: no problem. it made sense just a few seconds ago but now I'm confused again. so we wanted to show that $D: C_c^\infty(M)_{H^1} \to \Gamma(TM)_{L^2}$ is closable wrt. $H^1$-norm, and that just means that it is continuous wrt. $H^1$ norm. it is continuous wrt. $H^1$-norm if it is bounded wrt. $H^1$ norm. it is bounded, if $$\| Df \|_{L^2} \leq M \|f\|_{H^1} = M (\|f\|_{L^2} + \|Df\|_{L^2})$$ which obviously is true. is that correct? just to be sure I'm using the correct norms etc.

Ayup.

@MikeMiller: is really being closable wrt a norm the same as being continuous, or is it just continuous $\implies$ closable?

(not that it matters here)

Clovable doesn't really make sense unless you're already continuous I think because I think closable means "you can extend it continuously to the completions". But If you're continuous and $f_n \to f$ then define $Df = \lim Df_n$. Continuity is precisely what you need for this to be possible and well defined.

5:16 PM

we defined closable as "the closure of the corresponding graph is a linear graph"

I think that's the same as what I said, maybe.

my brain is totally fried.

I vote for $\text{Map}(X, Y)$, @Krijn.

@BalarkaSen It is probably the best out of the three, yes

Are you working on cohomology, @Krijn?

Not yet

The project should be finished somewhere in January, and I'm quite busy up until Chritsmas

5:28 PM

Ah, ok. Let me know when you need help/clarification.

Thanks for the offer! I will

user image

I wish I can mathematically check my guess
If my guess is correct, then slicing the figure 8 klein bottle this way shoudl yield two moebius stripes that can be deformed slightly to form the Sudanese moebius band parametrisation

Still years away from fully learning topology...

Thanks for the help @anon! :) I think I'm just gonna have to write an AI for it instead :'(

@Krijn Have you learnt the degree theoretic proof of Brouwer fixed pt theorem yet?

@BalarkaSen I have seen a short sketch of the proof I believe

I could be mistaking though

5:39 PM

Which one?

Nope nevermind, I have not seen it

I was mistaking

Yeah, so, let $f : D^n \to D^n$ be a continuous map. You can obtain a map $S^n \to S^n$ which sends both the hemispheres to the northern hemisphere by $f$.

This is not surjective, so degree = 0. However, if it has no fixed point then degree would be $(-1)^{n+1} \neq 0$.

Ah, and I did indeed prove the last statement last week

Hence there is a fixed point. And since the (open) southern hemisphere to(open) northern hemisphere bit cannot have any fixed point, any fixed point appears on the northern hemisphere. So, $f$ has a fixed point.

@Krijn Which statement? No fixed point implies degree is $(-1)^{n+1}$?

Yes

5:46 PM

You homotope it to the antipodal map, that's essentially it.

Just a small question, how did you obtain the map $S^n \to S^n$?

Nevermind

Visualised it

Send northern hemisphere to northern hemisphere by $f$. Send southern hemisphere to southern hemisphere by $f$.

Gluing lemma says continuous.

Even after so many years, the notation $D^n$ and $S^n$ still confuses me

Indeed?

@BalarkaSen ?

5:50 PM

I was expressing mild surprise about the confusion between $D^n$ and $S^n$.

Not which one is which

But which one is in $\mathbb{R}^n$ and which one is not

No scrap that

Neither is. $D^n$ is closed here.

Which one has dimension $n$

Both has dimension $n$?

Yes... My brain is fried

5:52 PM

Fair enough :)

Mine is fried too.

I did algebraic number theory all day and it got way too difficult

But I'm finished with mathematics for today

algebraic number theory is tough stuff.

Yeah, it is

It started with extensions $K$ of $\mathbb{Q}$

But now we're looking at extensions $L$ over $K$ where $K$ is an extension over $\mathbb{Q}$ and there are just so many details to keep track of

Yikes.

Anyway, I'm off

Have a nice day!

5:58 PM

Bye.

hi guys! I am searching for nice mathematic websites to keep on track with the current development of scientific research! I am mostly interested in applied mathematics! Do you have some links for me?

6:30 PM

0

Q: Convergence of a sequence in complete metric space when distance between subsequent terms is $< \alpha^{n}$, $0<\alpha<1$.

I am attempting to solve the following: Let $0<\alpha<1$. Suppose that $\{x_{n}\}$ is a sequence in a complete metric space $(X,\rho)$, and for each $n$, $\rho (x_{n},x_{n+1}) \leq \alpha^{n}$. Show that $\{ x_{n}\}$ converges. Does $\{x_{n}\}$ necessarily converge if we only require that f...

real-analysis convergence metric-spaces cauchy-sequences

@the.polo arxiv.org

Please help your dear, darling Jessy Cat!

7:05 PM

@PVAL cool thanks :)

@MikeMiller: so now that I have a densely defined closable operator $(D_T, T): H \to H'$, I think there's a proposition that says that there exists a densely defined closed adjoint. how do I figure out its domain?

@FrankScience: every linear operator has a linear graph. conversely, every linear graph induces a linear operator. sometimes, the closure of the linear graph of an operator is still linear. then, you'd call that operator closable and the induced operator its closure.

I have no idea on unbounded operators.

Great, stumped by $e^{x}$ yet again

Stupid integrals.

$\int{e^{x}sin(e^{x})dx}$, trying IBP..

7:22 PM

try something else

Tobias Kildetoft

@Owatch Just do substitution

Maybe I over thought it.

Is there any geometric meaning for Artin-Rees lemma?

$\int{usin(u)*\frac{du}{e^{x}}dx}$...

Okay.

Let me try something else..

There's some geometric meaning of Krull intersection theorem.

But for its proof, I cannot see any geometry, along with Artin-Rees.

Tobias Kildetoft

7:25 PM

@Owatch You really ought to see immediately why substitution is the right thing here

the integral has precisely the form specified in the theorem

..

$\int{f(g(x))g`(x)}dx = \int{f(u)du}$

standby

@Huy: I don't really like to talk about closable operators since I've never heard of them before. Let's talk precisely in the level of generality of vector bundles and differential operators. If I have an order $k$ linear differential operator $D: \Gamma(E) \to \Gamma(E')$, then (because I'm taking $d$ derivatives) it extends continuously to a map of Sobolev spaces $H^s(E) \to H^{s-k}(E')$.

$= -cos(e^{x})$

The adjoint is usually just a $k$ order differential operator in the other direction. Model case: $E = \Bbb R$, $E' = T^*M$, $D = d$. The adjoint is $\pm *d*$, sign depending on various dimensions. This still takes a derivative, so you should expect it to lose $k$ sobolev indices.

What you get is a differential operator $D^*: H^s(E') \to H^{s-k}(E)$.

Think back to our model case: all I've told you is that the Laplacian $d^*d$ on functions loses two derivatives. (That is, in the model case $M$, or even more of a model case $\Bbb R^n$, the Laplacian $\Delta$ can be extended to map $H^s \to H^{s-2}$.)

7:43 PM

Hi guys

Hello

Does anyone know of an elegant construction of $f_C$ where $f\,:\,\mathbb{R}\in[-\infty,\,\infty]\to\mathbb{Z}\in[0,\,C]$ ?

I'm afraid I have no idea what that means.

??

I don't know how to imply that I want to not lose information, despite the requirement to lose information

I am writting my abstract for my Ulam Borsuk theorem what people normally begin with ?

when writing an abstract?

7:48 PM

"We present...", "We develop...", "We introduce..."

"With <interesting condition>, we obtain <amazing result>"

I've seen some though with a totally different style, where it's a little humorous.

Here are some sample abstracts of talks people have given here, say.

cool thank you

Has anyone seen Tee-dog lately?

no

> Last seen Jul 2 at 23:37

:<

One day I will find him

7:55 PM

@MikeMiller: you've never heard of them before? but surely you've worked with densely defined operators, right?

is Brown representability about that theorem which says any generalized cohomology theory is representable by spectra?

Brown representability means a number of things. That's one.

@Huy: I tend not to think about that, no. I just extend my operators to Sobolev completions and be merry.

ok, just a bit surprising because I've looked at 3 lecture notes on FA from my uni and they all work with densely defined operators and their closure

maybe it's just a popular way here

"I will introduce Ulam Borsuk theorem in dimension 2 and prove it.
I will also give some applications that uses this theorem such as under some assumptions at any given moment on earth surface there always exists two opposite antipodal points with same pressure and barometric pressure!.
"

what do you guys think ?

*Borsuk-Ulam

and put a -, they are not the same person.

8:01 PM

alright

also, I think any abstract of a survey of [some random theorem] should state, if possible, what [that random theorem] is.

should I state it first ?

i.e before the introduction above ?

if you wish. I was just expressing my opinion.

I think that might be how people with more of a functional analysis or geometric background might think of things that way, @Huy. Like maybe it's helpful to think in terms of unbounded operators.

ic

8:03 PM

Alas, I am a humble topologist.

Alas, I am a humble studyist.

hello I'm back

I am having a look at inequalities

as you all know cauchy-swartz is a real famous one...

$$\sum a_ib_i\le \sqrt{\sum a_i^2}\sqrt{\sum b_i^2}$$

I wish to prove it, can i square both sides, since... -3<1 but 9>1

phew, that first one scared me, I thought I was reeeeaaaly out of touch :P

hehe

my bad

@ADG The point is that, without generality, you can assume that $a_k,b_k\ge0$

8:09 PM

If we can square both sides, then I get for n=2 $(a_1b_2-a_2b_1)^2\ge0$

and hence $$\frac{a_i}{b_i}=\lambda\forall i$$

@MikeMiller what was the Brown representability you talked about?

Tobias Kildetoft

@FrankScience Heh, I think you mean without loss of generality.

that spectra version is the only one I'm aware of.

how tobias?

Tobias Kildetoft

Though I will need to assume something without generality at some point.

8:11 PM

@TobiasKildetoft Thanks for the comment. Yes, WLOG.

(though, admittedly, I don't understand it)

@FrankScience I don't get it. Why WLOG?

@ADG Do you know that $\sum_k\alpha_k\le\sum_k\lvert\alpha_k\rvert$?

yup, I think what you're telling . if i can prove $\sum |a_kb_k|\le\sqrt{\sum a_k^2}\sqrt{\sum b_k^2}$ then I can say : $\sum a_kb_k\le\sum |a_kb_k|\le\sqrt{\sum a_k^2}\sqrt{\sum b_k^2}$

which i can prove by $\sum^2a_kb_k\le \sum a_k^2\sum b_k^2$

Now given that $a_k,b_k\ge0$. Ex1. Show that $\sum_ka_k^2+\sum_kb_k^2\ge2\sum_ka_kb_k$.

8:20 PM

AM-GM

actually I did it like:

$$\sum a_ib_i\le \sqrt{\sum a_i^2}\sqrt{\sum b_i^2}\iff \sum a_i^2b_i^2+2\sum a_ib_ia_jb_j\le \sum a_i^2bj^2$$

Ex2. Deduce that $\lambda^2\sum_ka_k^2+\lambda^{-2}\sum_kb_k^2\ge2\sum_ka_kb_k$.

Ex3. Conclude.

is there a way to search for newest, unanswered questions?

yes, I don't remember off the top of my head, but stack exchange even lets you query the database directly, I'm sure it's possible

how would I do that?

$$\sum a_i^2b_i^2+2\sum a_ib_ia_jb_j\le \sum a_i^2bj^2\iff \sum_{i\ne j} a_i^2b_j^2-2\sum a_ib_ia_jb_j\ge \sum a_i^2bj^2\iff \sum (a_ib_j-a_jb_i)^2+\underbrace{2\sum a_ia_k(b_jb_l-b_ib_k)}_{0}\iff \sum (a_ib_j-a_jb_i)^2\ge0$$ Equality when $$a_ib_j=a_jb_i$$

8:25 PM

@BalarkaSen math.stackexchange.com/unanswered/tagged/?tab=newest

I mean I want to find the newest unanswered questions on a certain tag

Tobias Kildetoft

@BalarkaSen There is a keyword along the lines of hasanswer:no

and you can search by tag and sort by newest

answers:0

math.stackexchange.com/…

alternatively: hasaccepted:no (I think you can infer their respective functions)

Ex4. Apply the same idea to deduce Hölder's inequality: $(\sum_ka_k^p)^{1/p}(\sum_kb_k^q)^{1/q}\ge\sum_ka_kb_k$ for $a_k,b_k\ge0,p,q>0,1/p+1/q=1$.

@BalarkaSen and the notation [tag-name] can also be appended to the query

8:28 PM

thanks, @Tobias @Mick

that works.

yeah, I figured

Glad I could finally help someone with something here! And you of all people :)

thanks so much @FrankScience

Damn, I am one integral slug...

Hi @PaulPlummer.

Hello @BalarkaSen

8:35 PM

for any real inner product vector space, is ${\bf <v,w>^2\ge<v,v><w,w>}$ true? It seems like cauchy

@PaulPlummer What're you upto?

@BalarkaSen Nothing at the moment, I guess, after I stop procrastinating (and some coffee), I will start reading stuff on the "Poincaré-Hopf index theorem". Yourself?

I am usually pretty busy with my (boring) schoolwork except when I sleep. I'll try to get boring stuff out of the way in multivariable calc tomorrow and finally study the proof of implicit + inverse function theorem. I went to a topology & physics conference about 2 weeks ago, where I did some differential topology from Guillemin-Pollack chapter 1 (had to assume the inverse function theorem, of course).

That's about it. What's the Poincare-Hopf index theorem?

Oh, and heard a lot of praise about PL Morse theory from prof. Said it's a wonderful tool. Also, he says the pipe-dream is to construct an analogue of PL-Morse theory for complex algebraic varieties (not the DIFF Morse theory, which already has an analogue in complex manifolds apparently).

8:50 PM

@BalarkaSen Actually I don't really know, everything I know is from wikipedia. But basically it is looking at vector field on a manifold, where the zeros are isolated and then looking at the degree little closed balls around the zeros, and relating it to Euler characteristic (summing the degrees gives the Euler characteristic apparently).

@Balarka: Instead of considering all $H^n$ at once, say that a functor $h$ from connected spaces to Set is a cohomology functor if it satisfies the usual properties. Then it's represented by a space.

It's just the one on Wikipedia.

Isn't that the $[X, K(G, n)] \cong H^n(X; G)$ one?

@PaulPlummer What do you mean by the zeroes?

Uh, I mean, that's a special case, yes.

@BalarkaSen intags:mine answers:0

Oh by cohomology, you mean a generalized cohomology theory?

I have this bookmarked.

No, I mean a functor satisfying the usual properties one demands something called cohomology have, which if you want explicitly stated, you can find in the aforementioned Wikipedia article.

But note I'm valued in sets instead of some algebraic category.

So I'm talking about some rather general functor. If you have a functor to abelian groups and you want to respect that extra structure, that's also possible with minimal effort.

8:56 PM

Where the vector field is zero @BalarkaSen

@MikeMiller Thanks, that works great.

Ah, @PaulPlummer.

OK, what does "degree little closed balls" mean?

degree of

Tobias Kildetoft

@MikeMiller Interesting. I am not sure I have seen cohomology with "image" in something non-abelian before (apart from a paper by David Steward on non-abelian cohomology)

and there the non-abelianness was in the coefficients (and resulted in the "image" not being a group any longer)

@TobiasKildetoft: It's not really that interesting, I don't think. It just captures some extra nice examples, like say the set of $G$-bundles.

Which yours also captures.

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