12:11 AM
What Serre does is take a finite group $G$ with irreducible characters $\chi_1,\dots,\chi_n$ and assumes $f$ is in the orthogonal complement of the $\chi_1^{\ast},\dots,\chi_n^{\ast}$. For a rep $\rho$ on $V$, he then looks at the endomorphism $\sum_{g\in G}f(g)\rho(g)$ of $V$. This is seen to be equivariant, whence a homothety by Schur's lemma and its ratio is computed as $\frac{|G|}{\dim(v)}\langle f,\chi_{\rho}^{\ast}\rangle$. This vanishes for all irreps by hypothesis, whence for all reps by additivity. You then apply this to the regular rep and obtain $0=\sum_{g\in G}f(g)g$, whence $f\ i agree that that argument doesn't convey the more detailed information. i'd forgotten where it appeared in the book. but 'average [or sum] over G' is hardly an ad hoc trick. if serre were here he would be disappointed in you. it would be very funny if a mathematician of serre's vintage suddenly appeared on math.SE and this chat beefing with people. averaging things is one of the most important things in rep theory, granted, but why average an arbitrary rep weighted by$f$? i've done versions of that so many times i don't even know. it isn't exactly common knowledge and i can't give a one-paragraph explanation for why you'd try it. my answer might be just 'you just do this and see what happens.' which i guess is your point yeah also, for some we have to wait till chapter 7 to be taught the incredibly useful fact that the inner product of two characters counts the number of equivariant homomorphisms between their representations even though this fact is completely elementary what's worse, this fact can be deduced independently from Schur orthogonality and be given as a conceptual justification of Schur orthogonality (it then is immediate from Schur's lemma) yet Serre argues this fact is true because of Schur orthogonality which he proved by some incomprehensible matrix calculations it's so unenlightening @AndrewMicallef just read about the loan. i laughed out loud. refresh your knowledge of your jurisdiction's laws on usury, if they exist. they tend not to anymore. that sounds borderline illegal. so called "payday" lending in the US is the absolute worst. if your bill is due on the 29th and you won't be paid until the 1st. you can end up paying a whole lot for a very short term loan of a small amount of money. i see the economic side of it. it's so much more expensive to have less money than it is to have more money. my credit score was pretty bad until i got a mortgage on a house, now it's stellar, even though i now owe a whole lot more money. figure that out. 1:07 AM A measure of reliability, not of wealth. i should inform the credit agencies that i also own a very difficult cat. that is an indicator of reliability. for a while, she even went to an eye specialist. a cat with an eye specialist should be worth 100 points. Only if you’ve been bludgeoned by claws. once or twice a day. we keep them trimmed to limit the damage. I might ask you trim my kitten’s. if they ever get in a docile mood and just collapse in your lap, that's the time to do it. 1:12 AM Yeah, right. I uses to do my adult cats’ all the time in GA. we needed about two years before olivia got to that point. Right. Anyone happened to have Evan's pde book on them who might be able to answer a question...? i thought i had a PDF of it, but i don't think i do. at least not on this computer. evans is a great lecturer btw, see him if you get a chance. By the way, the name is Evans. 1 hour later… 2:27 AM my chateau lynch moussas 2016 finally arrived... 2:44 AM @leslietownes I was contemplating cancelling it. But I just more or less agreed in the form of paying the first installment. The way it went down just reminded me of the chat we were having the other day. I felt all the social pressure and felt no chance to make a rational assessment. But at the end of the day, my new situation requires a car, so pay through the teeth I will to earn more in the immediate term :) extremely quick question, does anyone know why we require compactness in locally integrable functions? 3:00 AM the very good news is that cars apparently do not depreciate in value. hawk that is an interesting question. i think in the topological context compact sets often play a role analogous to closed/bounded sets in R^n, even if there is no notion of boundedness. i would expect that results about such functions are not 'tuned' to be optimal about the hypotheses on the functions they talk about, but that it just seems like a general enough starting point that people use it. in functional analysis they axiomatized a set of properties that are like having bounded sets, even if you have no metric. bornological spaces. in my view a historical side show and waste of time. Another quick question, if a function is smooth, then it must have a weak derivative right (the ordinary derivative)? In Evans page 251, he makes a distinction when he proves global approximations by smooth functions where he let a sequence of functions$u_m \in C^\infty (U) \cap W(U)_{k,p}$. Why does he have intersection when it should be immediately that we have inclusion of smooth functions in Sobolev spaces? @AndrewMicallef people feel pressure because they are considerate. in retail negotiations i find it helpful to be brutally honest (not rude, just straight up). so, if i feel uncomfortable, i will say so and i will say why. i find that takes a lot of pressure off. the last time i bought a car, i said "i'm paying cash for X price and i'm not going to listen to any of the financing stuff. no offense." was done in about 15 minutes. obviously a different conversation if i didn't have the cash. i think it helps to find outside financing. i was going to add same. get financing elsewhere, separate the concerns i think a lot of it is a volume business and if they don't make much on one sale they tend not to care, they just want to move the car off the lot. 3:13 AM i have bought my cars at year end, with cash. same. they can quickly tell if you know what you want. that is what retail sales folk do best. hmm, after many gyrations trying to source a wine i just found out that the bottle i had months ago was imported by a place in danville. i learned a bit about the realities of buying through a wine merchant. are smooth functions automatically Lp? kermit lynch on san pablo is good. and paul marcus in market hall by rockridge bart. they tend to know what is niche and requires effort vs. what they can get cheap. there was also a spot, vino, across from safeway on college. there was a guy there who knew weird ways of finding things. i just shop at trader joes now and enjoy Chateau D' I-5. @Hawk Is$x\in L^p(\Bbb R)$? 3:26 AM what an ornate example, when constant functions were available. Oh, silly me. Maybe my remark at the top of the starboard is à-propos. it's not a porpoise, it's a turtle. Or a slug. Clever how I got to units of mass. groucho lives. I've always been at least a couple of Marx brothers. 3:30 AM i love their work, even the bad ones. looking forward to introducing my daughter to it. And don't forget to introduce her to Jacques Tati, too. my favorite is 'playtime.' it's a little long for her attention span at the moment. it might work though because it is less verbal. one of my favorite movies of all time is laurel and hardy's short 'tit for tat.' it has a bit that has been coopted a million times, including on sesame street, where at first, two people are fighting and destroying things that the other one regards as valuable, and then people eventually are just grabbing whatever is in sight and breaking it even if it is their own. it's one of the best humor premises of all time. Mon oncle is best, then M Hulot's Holiday. Then Playtime. i'd forgotten about mon oncle. kermit lynch is next to acme. i have some connection there if only my grey matter worked to spec... 3:42 AM one time i was in line at acme and heard they weren't doing credit cards, or only did cash, or something, and i had to walk a few blocks down the street and get a$4.50 ATM surcharge just to get some paper to get me some bread.
and i lost my place in line.

weeps

this might be the worst thing that has ever happened to anybody.

4:00 AM
2

Let $G$ be a group of order $15$ then prove that it is cyclic. By Lagrange's theorem, its non identity elements can only have order $3$ or $5$. Not all non identity elements are of order $\displaystyle 5$ because $\displaystyle \phi ( 5) =4$ does not divide $14$. If all non identity elements are ...

Hello. :)

I am trying to understand the answer here :-)
If I allow using class equations, then clearly existence of elements of order 5 and 3 is established.
assuming G to be non-abelian
Going further, I'll still have to prove $<a>$ and <b> are normal in G. From there, I am able to prove that $ab=ba$ (using Fermat's theorem) whence, showing that $ab$ has order 15 thus proving the group cyclic.
My question is: can class equation do more here? That is, using this can cyclic nature of the group be deduced without proving <a> and <b> normal in G?

4:18 AM
is it true that $\partial_x f(x-y) = -\partial_y f(x-y)$?

Use the chain rule?

When i take \partial_y(x-y) = -1, i just get =\partial_f*-1, how do I relate the original derivatives, \partial_xf and \partial_yf?

Huh? $f$ is a function of a single variable. Write the chain rule correctly.
Evans PDE book is awfully advanced if you can’t do this right.

y isn't dependent on x.

Of course not.
What is the correct formula for $\partial_x f(x-y)$?

4:30 AM
we are talking about the one-variable one right? F'[x] = f(g(x))g'(x), if we let g = x -y, there is still a problem.

f'(x-y)*\partial_x(x-y)

Not quite …

yo
as usual

OK. Good. Now the $y$ partial?

4:37 AM
its just
f'(x-y)*\partial_y(x-y)
wait, we are relating f'(x-y)

Huh?

i meant that \patial_y f(x-y) = f'(x-y)*\partial_y(x-y), but \partial_x f = f'(x-y)*\partial_x(x-y). so we are relating the two partials with f'(x-y) is what i meant.

No. $\partial_x = f’(x-y) = -\partial y$.

yeah that's what i meant, we use f'(x-y) to relate the two partials
which was my original identityr

OK. Just do it correctly and there is no doubt.

4:44 AM
this is a long shot, you showed me eariler that smooth functions are no necessarily Lp, so this actually means smooth functions do not necessarily have weak derivtives right (unless we add further conditions)?

Distributions pair with compactly supported functions. So smooth functions certainly have weak derivatives.

only if we add the compact support condition right?

That's the definition.

of?

Distributions.

4:49 AM
what about non-test functions, i m jut talking about infinite diff functions f: U \to R, C^\infty(U) with no assumption on its suppf
oh wait, are we not able to define weak derivatives without compact support assumption since that integral might not vanish on the boundary

You’d better learn basics. What does the weak derivative mean?

it has to satisfy that integral \int u d\phi = \int \phi du for all test functions with compact support in U.

So what is the issue with true integration by parts?

compact support ensures the vanishing on the boudnary

Right. So no problems.
Not sure what you mean by boundary, but we reduce to a compact set.

5:15 AM
when it comes to the support set of a function f, are there any properties regarding its algebraic operations? for example, can we say anything about supp(f + g), supp(fg) or supp(f/g)? assuming none of them are constants? i m guessing not since there is nothing on wikipedia that says there is

@robjohn yes but what about the limits

5:32 AM
What does 'cobound' mean?
like in a statement '...they cobound a smoothly embedded annulus in S^3\times[0,1]$Two things together bound the annulus… co = together, not math dual stuff 5:47 AM ah right. @satan29 I do the heavy lifting and the finishing work, too? @robjohn that wasnt heavy lifting :p. I managed to do that too. The only issue was with the limits @satan29 You integrate and plug in the limits. sir but the function is defined piece-wise. 6:07 AM So integrate on the pieces. think about what you're doing $$\int_0^t\sin(t-x)\overbrace{\sin(x)[\pi\le x\le2\pi]}^{f(x)}\,\mathrm{d}x$$ You have the intersection of two intervals:$[0,t]$and$[\pi,2\pi]$and possibly$[3\pi,4\pi]$if$t$is big enough, etc sir will you check out my approach 14 hours ago, by satan 29 i think we can proceed by writing t=$2\pi k + b$, where k is an integer and b is in$[0,\pi)$sure the integral can then be written as k integrals of over an interval width of 2pi and an extra integral, but that is zero . but you're missing the intervals where$f$is non-zero how? f is non zero in pi,2pi right? 6:17 AM yes, but your$b$is in$[0,\pi]$hmm now that I think about it, b should be in$[0, 2\pi)$did you mean$b\in[0,2\pi)$yes wat so doesnt that mean simply k+1 integrals instead of k yes. but$k$of the integrals will be equal so you end up doing two integrals why not k+1? in light of the new limits, it appears that the last integral==the k integrals 6:26 AM The integrals over each of the$[(2j-1)\pi,2j\pi]$for$1\le j\le k$will be equal. then there is the integral over$[(2k+1)\pi,t]$sir 2kpi,t it is$0$on$[(2n-2)\pi,(2n-1)\pi]$, so you only integrate on$[(2n-1)\pi,t]$if$t\in[(2n-1)\pi,2n\pi]$hang on. we are integrating on [0, 2pik +b] so we can split them into THINK 5 [0, 2pi k],[2pik , 2pik +b] which is equivalent to k[0,2\pi] , [0,b] is that not right? OHH i get you now. The use of n confused me at first (as opposed to k) sorry the upper limmit of the interval, itself lies in an interval 6:35 AM I have fixed the limits since my numbering was different: 8 mins ago, by robjohn The integrals over each of the$[(2j-1)\pi,2j\pi]$for$1\le j\le k$will be equal. 8 mins ago, by robjohn then there is the integral over$[(2k+1)\pi,t]$thanks a lot sir. 6:54 AM if$t\in[2k\pi,(2k+1)\pi]$, the last integral is$0$i am getting$-\pi*kcos(t) $+ the last integral That looks right, wait, there is a$\frac12$factor missing, I think the interval is width$\pi$and there is a factor of$\frac12$on the$\cos(t)$so$-\frac{k\pi}2\cos(t)+\dots$3 hours later… 10:12 AM Can anyone help me with some Lagrangian coordinates how does the author go from (8) to (9) arxiv.org/pdf/2003.03803.pdf 10:24 AM @TedShifrin FYI, I quoted a short passage from your Multivariable Mathematics in this answer. I assumed you wouldn't mind, but wanted to let you know just in case (and apologies if it's not OK!) I made my answer a CW since they're your words, not mine. 11:20 AM I am trying to compute the intersections of Schubert cells in G(2,5). What is$\sigma_{1,1}^2$?.$\sigma_{1,1}$denotes the cell$\begin{pmatrix} \ast & \ast & 1 & 0 & 0 \\ \ast & \ast & 0 & 1 & 0 \end{pmatrix}$This is complex Grassmannian. So we have Poincaré duality. And$\sigma_{1,1}$is an 8-cell. So the intersection product$\sigma_{1,1}^2$is a$4$-cell 12:11 PM Well of course I have Giambellis' formula,$\sigma_{1,1} = \sigma_{1}^2 - \sigma_{2}$. So$\sigma_{1,1}^2 = (\sigma_{1}^2 - \sigma_{2})^2$, but... there has to be a simpler expression right? 2 hours later… 1:50 PM Hi Everyone :) on wikipedia the following is stated: By the portmanteau lemma (part C), if Xn converges in distribution to c, then the limsup of the latter probability must be less than or equal to Pr(c ∈ Bε(c)c), which is obviously equal to zero. https://en.wikipedia.org/wiki/Proofs_of_convergence_of_random_variables#Convergence_in_distribution_to_a_constant_implies_convergence_in_probability Could someone please give me some collour where does it follow from? I just can't figure out why X_n -w-> 0 => lim P(|Y_n| >= \epsilon) Every portmonteau theorem definition on the internet is different I mean, they're probably the same, just formulated very diffrently :D Given a curve $$curve(v)=(c_1,c_2,c_3) \text{ point on the curve}$$ $$(x,y,z)=([c_1\cos(u)-c_2\sin(u)] \vec e_1 + [c_1\sin(u)+c_2\cos(u)]\vec e_2 + \vec e_3\cdot \vec e_3 + O$$ How do I find the farthest distance of the curve from its axis? 2:10 PM The function$f(x)=|px-q|+r|x|$,x€ (-infinity, infinity) where p>0,q>0,r>0 assume its minimum value only on one point if (a)$p\not=q$(b)$r\not=q$(c)$r\not=p$(d)p=q=r it looks like you've mistook this chatroom for your google search bar 3 2:28 PM How do the authors know this? they dont give a definition of what$\delta \mathcal{E}/\delta \rho$is... ah its the functional derivative what is the difference between functional derivaitve and first variation? @Thorgott It's like the main site, except you don't have to provide context! and those who answer don't get any reputation. I think that's what they call a win-win situation 2:47 PM @Rover Have you tried plotting this in each of the cases (a), (b), (c), and (d)? @Monty without the book, or even without knowing who the authors are, it is pretty hard to know what they meant. @robjohn yes that was a terribly formulated questions sorry! Privyet @Ted @Eminem Can you find the axis? What is$O$? Hey @feynhat @Ted @SayanChattopadhyay Hi. How are things at Mohali? Is the semester over? 2:57 PM @robjohn not exactly, but yes virtually and now I got it , it says ," minimum value only at one point " , when r=p , there will be only one minimum, but at two points, and in other cases there comes one minimum. @feynhat Yeah the semester is over. Things went a little crazy with an on-campus death, but they recently held vaccination camps so stuff is better now. @robjohn thanks :-) @Rover so that gives your answer. If they want a proof, there is some more work. What about you? Did you get to move? @feynhat However, I think it should be$|r|\ne|p|$2:59 PM Yes Ah, they are both positive, so that is irrelevant. Yes, How can we prove it , exactly ? I would start by looking at the triangle inequality. Damn... @robjohn Okay , because it involves mods 3:04 PM Things never got that grim in my campus. That's probably because as soon as the second wave started they sent a mail saying "We strongly urge you to reconsider your choice of staying back in the campus as we don't have the facilities to handle too many critical cases". Many people (including me) chickened out and left. @feynhat "chickened out"? sounds like "were sensible" Well people who did stay back are now telling us, "See? Nothing happened, chickens!" Nothing happened because those who left left Also, the campus is in a state which is not very badly affected as opposed to the one where I have moved to now (which is one of the worst affected). People were telling me atleast I was safer at campus than at home. Second guessing in situations like this is pointless. People did what seemed best at the time and if they had done differently, who knows what might have happened. 3:11 PM But thankfully (and surprisingly) the vaccination is going on at a much faster rate here, and I was able to get the first dose soon after I arrived here. Many people who stayed back still haven't been able to get an appointment. UCLA shut down most operations for the past 15 months. They will soon open up. I don't know if the second summer session will be in person or by Zoom, but I think the Fall quarter will be in person. Unless something bad happens 3:29 PM Let$L$be a holomorphic bundle on a compact Riemann surface$X$and$p\in X$. Can anybody help me figure out what "the exact sequence$0\rightarrow\Gamma(-,L(-2p))\rightarrow\Gamma(-,L)\rightarrow\mathcal{O}_{X,p}/\mathfrak{m}_{X,p}^2\rightarrow0$" is supposed to be?$L(-2p)$is the line bundle twisted by$\mathcal{O}_X(-2p)$(holomorphic functions vanishing at least by order two at$p$),$\Gamma$denotes sheaf of sections as usual and the last thing should be interpreted as a skyscraper at$p$Looking at this SES supposedly demonstrates that if$H^1(X,L(-2p))=0$, the differential of the associated map into projective space is injective, but I'm really seeing what sequence accomplishes this Long horse is the Saviour good boi 4:13 PM Meh, I spend the entire time trying to think of something intrinsic, but I think you just have to fix a bundle chart around$p$, identify a section with a holomorphic map locally and take the germ at$p$. This obviously does the job of making the sequence exact, though I'm not quite sure yet how it implies the desired result. 5:05 PM Long exact sequence on cohomology? Show you get surjection … Although usually we use$H^1$vanishing to do that. You need some hypotheses on$L$here, for sure. I don't think any further hypothesis on$L$is necessary, I just don't see how the surjectivity implied by$H^1$vanishing implies injectivity of the differential I think I can see it on the vector space level after fixing a chart, but not why it remains injective when passing to projective space I probably know more than you do. That$H^1$does not always vanish. What if$L$has no nontrivial sections at all? my cat just, out of the blue, arose from slumber in a sunny spot, ran down the stairs, tackled her scratcher, and did quite a lot of scratching on it. being a cat must be an interesting mental experience. Your cat doesn't need to give my kitten any advice. is screech still at the stage where you get one or two hours of the day like that? we're down to one hour, because we have a mature, 4.5 year old cat. 5:19 PM oh,$H^1$vanishing is my assumption One or two? That sounds more like it, @Thor. Care to reboot? @leslie I'm just fortunate if those hours don't occur in the middle of the night with me the scratching post. olivia still does that with my wife. decides that 3am is the time to begin pawing at her head like it's one of her toys. she's so weird about that. a lot of her most annoying behavior is directed at my wife, but she also spends almost all day with her, and usually sleeps by her side and not mine. I can relate. we let our cat sit on the dinner table while we eat because historically she was very well behaved in just sitting there and watching and not interfering with mealtime. even if we were having fresh fish. now she begins roving the table in search of scraps when she senses someone is done eating. we probably never should have allowed her onto the table. I could have told you that. 5:26 PM it doesn't help that our daughter offers her food. we didn't have a daughter when all of this started. she'll eat absolutely anything that my daughter offers her, even if it's not something an obligate carnivore would ever touch in real life. she's fed the cat celery sticks. anyway, long story short, i'm an idiot. I hate it when I am compelled to agree with you. Hello chat on the math side, my daughter can now count to around twenty before she starts improvising with numbers like eleventeen and sixtyten. Salut M Astyx! The entire statement is that if$L$is a holomorphic line bundle on a compact Riemann surface$X$(I think this might be true on any complex manifold) and$H^1(X,L(-p-q))=0$for any$p,q\in X$, then$L$is very ample, meaning the associated map into projective space is an embedding. To be clear, the associated map (I prefer the coordinate-free version) is$\alpha\colon X\rightarrow\mathbb{P}(H^0(X,L)^{\ast})$given by mapping$x\in X$to$[s\mapsto\varphi(s(x))]$, where$\varphi\colon L\vert_U\rightarrow U\times\mathbb{C}$is some bundle chart about$x$. 5:31 PM Yes, this is the proof of Kodaira embedding. Higher dimensions as well. (I didn't read past the first few lines.) Re your last point: Do a separate exercise of understanding when a map$f$to$\Bbb P^N$is an immersion in terms of any local holomorphic lifting$\tilde f$to$\Bbb C^{N+1}$. It actually helps to think a little geometrically here. @Bungo Absolutely fine, of course. This is always a point that confuses people. I think the OP is still confused. I was under the impression the harder part for Kodaira embedding is to actually exhibit line bundles satisfying these hypotheses (I only know this stuff for Riemann surfaces, where you need Riemann-Roch to argue that any divisor of high enough degree does the job). Re the second point: It's an immersion iff the lifting is an immersion, the image of whose differential at any point$v\in\mathbb{C}^n$does not meet the subspace$\langle v\rangle$(identifying$T_v\mathbb{C}^n\cong\mathbb{C}^n$as always, of course). I'm not sure what more to say. For Riemann surfaces all that matters is degree. In higher dimensions you need the notion of a positive line bundle and then the Kodaira vanishing theorem is the relevant tool. aha, I hope we'll do that towards the end of this complex geometry lecture Does not meet isn't correct. Here's the right statement. You want$d\tilde f_p \not\equiv 0\pmod{\tilde f(p)}$. So you should unscramble your earlier stuff to see why you in fact know this. @TedShifrin That would be the condition for a non-vanishing differential, not an injective one, no? I mean, we agree that the tangential component killed by projecting to projective space is the radial component, and remaining injective after quotienting that out is the same as asking that the map does not meet the thing were quotienting out by non-trivially? 5:49 PM (Oh, I guess I was doing the R.S. case. You're right in higher dimensions.) Oh, the section$s$in my last paragraph automatically satisfies$s(p)=0$(since it maps to$\mathfrak{m}_{X,p}/\mathfrak{m}_{X,p}^2$) and, which was the point,$d(\varphi\circ s)\vert_p(X)\neq0$. Then, in particular, if$\tilde{\alpha}\colon U\rightarrow H^0(X,L)^{\ast}$is the considered lift,$d\tilde{\alpha}\vert_p(X)$is identified with the map$t\mapsto d(\varphi\circ t)\vert_p(X)$, whereas$\alpha(p)$is the map$t\mapsto\varphi(t(p))$. For$s$, one of these vanishes and the other does not by the observations just made, so they cannot be proportional, whence$d\tilde{\alpha}\vert_p(X
so I had to get my hands slightly dirty, but it does work out nicely

6:16 PM
Right. If you pick a section vanishing at $p$, then mod $\tilde f(p)$ is irrelevant and it's just what you expect.

6:28 PM
@feynhat Yeah we got such an email as well. But I stayed considering I had recovered from the virus once before. Travelling seemed too much of a risk.

@TedShifrin ah, that's a nice way of putting it, thanks for the clarifications

Oh you guys are proving Kodaira, nice. Another way to state would be to say that the first chern class is positive (I think). One way is should be easy I think. If it is very ample then there is s_0,..,s_n which generate the global sections. This would imply that curvature form is non-negative. Then I have to argue that the chern class is positive

I'm only doing the baby preliminaries, not the hard stuff
what does it even mean for a form to be positive?

6:44 PM
Positive definite (1,1) form @Thorgott
Here that is, in general you call (p,p) forms positive
This comes from the natural splitting of complexified cotangent bundle caused by the endomorphism J on each tangent space, with J^2 = -1

that only depends on the cohomology class?

Well, you still have to choose a "correct" representative of the class.

For instance if you have a hermitian vector bundle with the connection compatible with the complex structure (which you can always do) ( D = D' + $\tilde{D}$ and $\tilde{D} = \partial^{ \overline}$) you can show that the curvature form is always (1,1)
Yeah I gotta get a new keyboard. It's hard managing without a functioning backslash

A new keyboard with no backslash?

6:59 PM
for a while i had a, s, and d go out. i made up for it with alt codes. until the repetitive stress injury forced me to break down and buy a new keyboard.

@TedShifrin That would be ideal. I will live my life in the mountains

Sounds pacific!

7:25 PM
Btw @Ted, I was trying to show that for a compact Kahler metric, the vanishing of the chern class implies that there is a kahler metric with vanishing ricci curvature. I have that the ricci form is 0, so maybe I have to somehow invoke stokes theorem?

7:42 PM
Hey @TedShifrin I was just wrapping up the question that David K provided me an explanation for. Using his advice and letting $z$ be a constant temporarily I was able to show that $x$ and $y$ were individually proportional to $z$, but I didn't get the same scalar for each of them.
What I ended up getting was:

$x_{+} = \frac{z(ac+b)}{(a^{2}+b^{2})}$ or $x_{-} = \frac{z(ac-b)}{(a^{2}+b^{2})}$

and

$y_{+} = \frac{z(ac+b)}{(a^{2}+b^{2})}$ or $y_{-} = \frac{z(ac+b)}{(a^{2}+b^{2})}$

where $x_{+}$ and $x_{-}$ correspond to taking the $+$ or $-$ from my application of the quadratic formula.
so the scalars don't match up.....so I wouldn't be able to express it as $t(a,b,c)$ form. Or am I expecting too much?
On another note...for some reason people have been downvoting David's answer...

@SayanChattopadhyay isn't this the Calabi conjecture, proved by Yau in the late 70s?

yes, I think so too

@dc3rd I don’t remember any details. Of course it needs to be the ruling if the cone, so your algebra is not sufficient/correct.

lol what

In mathematics, the Calabi conjecture was a conjecture about the existence of certain "nice" Riemannian metrics on certain complex manifolds, made by Eugenio Calabi (1954, 1957) and proved by Shing-Tung Yau (1977, 1978). Yau received the Fields Medal in 1982 in part for this proof. The Calabi conjecture states that a compact Kähler manifold has a unique Kähler metric in the same class whose Ricci form is any given 2-form representing the first Chern class. In particular if the first Chern class vanishes there is a unique Kähler metric in the same class with vanishing Ricci curvature; these are...
seems like this is actually hard functional analysis, yikes

7:55 PM
Jeez yeah, I thought this would be fun algebraic geometry

Lots of third order PDE estimates.
I sat through a bunch of Yau's lectures in grad school.

Just for clarity it is showing tangent plane to $f(x,y,z) = x^2+y^2-z^2 = 0$ intersects the cone in a line.

I was trying to do this because I wanted a Kahler metric on x^4 + y^4 + z^4 + w^4 = 0 in P^3, it has chern class 0 and I thought some theorem would do it for me

I'll rework the algebra, but it doesn't seem wrong. Maybe I'm missing the last step

Yes, I know, @dc3rd. No quadratic formula. When you eliminate $z$, you have a perfect square like $(bx-ay)^2=0$.

7:59 PM
I think this is the Fermat Quartic or something

This? @Sayan
You asked a totally general question.

1 min ago, by Sayan Chattopadhyay
I was trying to do this because I wanted a Kahler metric on x^4 + y^4 + z^4 + w^4 = 0 in P^3, it has chern class 0 and I thought some theorem would do it for me
That hypersurface

Oh, I missed that.
The canonical bundle is trivial. Doesn’t the induced Kähler metric do it?

Oh right, the induced two form will be closed because the bundle is trivial right?

Closed?
The Kähler form is always closed.

8:05 PM
The hermitian metric, who's associated two form has to be closed for it to be Kahler right

Just restrict the Kähler form in the first place!

Okay yeah that was stupid

8:50 PM
Is it possible for a function, that not all points of the domain are mapped by the function?

no, by definition: a function has a domain and each element of the domain is mapped by the function.

Thank you @Quin !

anytime :)
if youre curious, you can dig back to the definition involving relations on sets and this will, in some sense, show that every input must be mapped somewhere (thought im not sure this is necessary for understanding)
anyway, im off again for a few more weeks! (in the middle of a roadtrip up the west coast!)

Bon voyage @Quin