That is why I write v[f] = Yeah, that is nice. I borrowed that in my report :) --> NullPtr (~NullPtr@cs6668166-88.austin.rr.com) has joined #mathematics --- NOR-Gate gives channel operator status to NullPtr --- Cauchy gives channel operator status to NullPtr It's kind of nice, you like a copy of the fresh report. It is much nicer now :) report. = report? OK dcc it. I don't know why the .pdf output is so crappy, but there it is :) use .ps dcc me a .ps :) Ok :) your connection is slow I particular like equation (5) and I use it alot :) --- Ping reply from DrFurious : 2.15 second(s) No kidding :) Anyone want my short blurb on dynamics, while were dccing? sure i'll take it -Simplex-- DCC Send dynamics.ps (208.180.17.244) haha, simplex's just overcame drf's oh, and it's done :) I gotta take a shower. BBL. I'll look at your thing, DrF. What can I say, I'm on a modem, I can't compete with cable :) kewl. Thanks :) cable modems are cool. :) I still haven't address the chain rule yet though :| Yeah, I had one for a while but I cancelled it because the techs were so horrible The box didn't have a power button. They said, "If it stops working, just unplug it for a few seconds and then plug it back in." :B Everytime it stopped working and I had to unplug it I'd get this huge shower of sparks I said, forget that! :) nerdy You might like Eq (6) :) hehe Kind of nifty :) wanna learn more sheaves? ok, i'll take that as a yes :) I still haven't learned those 3 pages yet! :) ok, we'll go over that to start with :) Can I take a rain check? :) I'd like to go over it, but I really have to finish this report to pacify my advisor given a topological space X, a sheaf O of abelian groups is an assignment, to each open set U of X, an abelian group O(U), and for each inclusion U subset V, a restriction map r_{U,V} : O(V) -> O(U) such that (a) r_{U,U} = id (b) r_{W,V} o r_{V,U} = r_{W,U} (c) if U = union U_i is an open cover of U, and s in O(U) and r_{U_i, U}(s) = 0, then s = 0 (d) if U = union U_i is an open cover, given s_i in U_i with r_{U_i intersect U_j, U_i}( s_i) = r_{U_i intersect U_j, U_j}(s_j) then there is an s in O(U) with s_i = r_{U_i, U}(s) (by map i mean homomorphism) basically you should think 'regular' functions on some space What is the difference between a sheaf and a bundle? if M is a manifold, the assignment C(U) = { C^infty functions U -> R } makes a sheaf of rings drf, well they are a little different, basically because the bundle has trivializations Would C(U) be a sheaf of algebras? drf, R-algebras, yes Oh R is a ring, I was thinking R is reals R is the reals Oh :) Ok an algebra has to be over something :) aight if V is a complex analytic space, then H(U) = { holomorphic functions U -> C } makes a sheaf of rings now also, if x in X, then O_x = lim O(U) over all U containing x = { : f in O(U), U contains x } modulo the equivalence relation that ~ if there is an open W subset U intersect V still containing x where r_{W,U}(f) = r_{W,V}(g) (also it is common to use the normal restriction notation for restriction, i.e. f|_W, even if it isn't normal restriction that is the homomorphism) Is O_x the germs? :) yes exactly Ok (aside) a local ring is one with a unique maximal ideal we call (X,O) where X is a topological space, and O is a sheaf of rings, a ringed space, if all the O_x are local rings, we call it a locally ringed space or a local ringed space so in terms of manifolds, germs are a local construction, so we can focus on the euclidean case, create a function O_x -> R, -> f(x) this is obviously onto (constant germs), and so O_x / kernel = R, and the kernel is maximal the kernel is exactly the germs in O_x which vanish at x we call it M_x in fact, i claim it is the unique maximal ideal --> Hafen (~dolex@HubD-mcr-24-95-117-8.midsouth.rr.com) has joined #mathematics I'll take your word for it :) --> Galois (scythe@DOMINIA.MIT.EDU) has joined #mathematics suppose we take a germ outside of M_x, then it doesn't vanish at x, so in some open set around x, we can invert it <-- Hafen (~dolex@HubD-mcr-24-95-117-8.midsouth.rr.com) has left #mathematics so everything not in M_x is invertible, which means it can't be in any proper ideal --- Evariste gives channel operator status to Galois --- Galois has changed the topic to: Facieaux so every proper ideal must be a subset of M_x, and it shows that M_x is the unique maximal ideal and also we have shown that O_x / M_x = R now let's consider a first possible definition of tangent space on a manifold, in this more abstract setting... R-derivations of O_x that is derivations O_x -> R --- deddisney is now known as dedekind even tho the R kinda comes out of nowhere in here now, we can restrict any derivation to a derivation M_x -> R (and we can go backwards, quotient rule !!) any derivation on M_x will vanish on M_x^2, because if f,g are in M_x, d(f*g) = f(x) d(g) + g(x) d(f) (liebniz rule) = 0 because both f,g vanish at x so we can even factor that out, and look at derivations M_x/M_x^2 -> R f*g in M_x^2? but wait, we found out that O_x/M_x = R, and it's also apparent that M_x/M_x^2 is an O_x/M_x vector space, so let's replace the R now drfurious, yes drfurious, the square of an ideal is the ideal generated by products of elements in the original ideal products of two elements Ok I had never heard of that so we have arrived at the proper definition, derivations M_x/M_x^2 -> O_x/M_x with the advantage that this is completely general, i can apply it to any locally ringed space (schemes, complex analytic spaces, complex manifolds, real manifolds, etc) now, i claim that the word 'derivation' can be omitted, that any O_x/M_x-linear map will do in fact, because on M_x the liebniz rule specifies how it should act on M_x^2 and specifies nothing for M_x Dropping product rule? so since we mod'd out M_x^2 we don't have any products, so we don't have to impose the product rule ! isn't that neat so an even shorter definition, O_x/M_x linear maps M_x/M_x^2 -> O_x/M_x, or (M_x/M_x^2)^* oh shit later :) huh I dun giddit So the tangent space is the dual of M_x/M_x^2? That might be neat if I understood what the hell M_x/M_x^2 is :) germs which vanish at x with the proviso that the product of any two is 0 Wouldn't the dual of a germ be a germ? it makes no sense to speak of the dual of one vector Wouldn't the dual of the space of germs be a space of germs? :) how in the world is that? a germ is a collection of functions an element of the dual is a function mapping the entire collective of germs to R, in a linear way Ok That sounds a lot like a tangent vector :B it almost is a tangent vector is something in (M_x/M_x^2)^* the thing you were speaking of is M_x^* oops I'm not good with quotients saying "fg = 0" is no worse than imposing a leibniz condition A germ is an equivalence class of functions at agree on an open neighborhood of some point, right? at = that yes kewl I've got some weird notation problem again hmm. is there a way to analytically solve stuff like x^2 + x + 1 = m^3 in integers Let id:R->R be the identity function Does it make sense to define d(id)/dt |_t0 ? yes. This is high school calculus. =0 =) Ok, then if id:R^n -> R^n then it makes sense to define @(id)@x^i |_x^i_0 right? anyone? :P Unless I am doing something lame, @(id)/@x^1 = (1,0,0,...) that particular equation is an elliptic curve, and techniques for finding integer points on them are known ah. where does one learn about elliptic curves eric, @f/@x at a point is always a scalar silverman/tate, Rational Points on Elliptic Curves analytic number theory? or something for a general polynomial equation, it has been proved that no algorithm exists which will take an equation as input and output whether it has integer solutions or not --> frackois (~god@server10.suhfields1.com) has joined #mathematics sounds like a tough proof Galois That makes sense if f:R->R, but what if f:R^n->R^n? Galois: how do you prove something like that drf, it makes no sense to talk of @f/@x unless f: R^n -> R para-dox, carefully would it be fruitful to repost a question here that went unanswered in #math? has it been proven that x^2 + x + 1 takes on infinitely many prime numbers for integersx ? very carefully for f: R^n -> R^m, your choices are, either take a component of f, or take the 1st partials of all components at once Ok, how does the latter work? It results in an element of R^m, right? it does, but I consider this element a rather useless element what you really want is the m by n matrix of all partials of all components of f that matrix represents a linear transformation on tangent spaces, which is df (or maybe f_* in your notation) Here's what I was thinking. I hope it is correct: id: R^n -> R^n @(id)/@x^1 |_{x^1_0} := lim_{delx^1->0} [id(x^1+delx^1,x^2,x^3)-id(x^1,x^2,x^3)]/delx^1 = (1,0,0) (for n = 3) that mess may be correct but it is utterly pointless Is there a problem with that? you're losing the whole point of the df construction It's not! :) <-- NullPtr has quit (Idle time limit exceeded) the point is to get a matrix map between tangent spaces all else is mypoic myopic I have a method to my madness :) well, out with it fast, because I really do have to do things that do not involve IRC Ok c:R->M a cruve M, phi:M->R^3 a coordinate map dc/dt |_t0 a tangent vector at c(t0) in M phi_* dc/dt |_t0 = d(phi o c)/dt |_t0 Ok... and for any map f:M->N, then I think I can show: d(f o c)/dt |_t0 = @(f o phi^{-1})/@(x^i o c) |_{x^i o c(t0)} d(x^i o c)/dt |_t0 If I let f = phi be my coordinate map, I get: d(phi o c)/dt |_t0 = @(id)/@(x^i o c) |_{x^i o c(t0)} d(x^i o c)/dt |_t0 are you trying to do the chain rule? And if my calculation above is correct, then this works out perfectly Let ~x^i = x^i o c d(phi o c)/dt = (d~x^1/dt,d~x^2/dt,d~x^3/dt) Just like you would want it to but that's a matrix, not a vector Galois yeah, pretty much the chain rule I know what you want. You want gradient vectors that thing you wrote down is a gradient vector --> NullPtr (~NullPtr@cs6668166-88.austin.rr.com) has joined #mathematics --- Cauchy gives channel operator status to NullPtr --- NOR-Gate gives channel operator status to NullPtr take the following from me, who got it from Prof. Bott the master of differential geometry the minute I wrote down a gradient vector in one of my sessions with him, I got castigated for about fifteen minutes these are not gradients. They are matrices. they're not objects. They're maps. --> {Monica31 (~IsabellaB@ppp211134.fx.ro) has joined #mathematics <-- {Monica31 (~IsabellaB@ppp211134.fx.ro) has left #Mathematics and the quicker you get that into your head, the better off you are to handle the rest that comes down the line --> jrt (~e@h-209-202-44-028.fast.escape.ca) has joined #mathematics gradient vector is a matrix? yes it is a 1 by n matrix in some ways your insistence on working with each column or coordinate individually reminds me of cardano's c.1600 derivation of the cubic formula his derivation required 20 separate cases that we would now lump into one, because at the time there were no negative numbers so x^3 + px = q and x^3 = px + q were viewed as separate equations eacch requiring a separate solution method What do you mean by gradient vector? Like grad(f)? d(phi o c)/dt = (d~x^1/dt,d~x^2/dt,d~x^3/dt) the thing is, I don't want to write out any coordinates unless dealing with a specific example Unfortunately, I am an engineer and I'm doing applied problems where the geometry of the physical problem determines the coordinates, so I am forced into coordinates pretty early well, guess what. I am in the pure math division. not applied math so of course that is the answer you get from me :) I am way over an hour online now Get outta here :) Thatnks for the time. I know you are busy Thanks even 'k, I'm out Ciao drfurious, you are forced into coordinates when doing yoour problems drfurious, and you are introducing them in the learning and intuition stage, which is screwing you over :) It's even worse than coordinates back, had to go pick up my sister :) It's a mesh Which is basically a bunch of coordinates patched together nerdy I am making progress. Slow but sure :) Did you look at my report? yes not much, but i glanced at it and, Bott is the god of diff geo <-- NullPtr has quit (ircd.east.gblx.net ircd.lightning.net) --> NOR-Gate_ (norgate@ns2.garen.net) has joined #mathematics <-- Simplex- has quit (ircd.east.gblx.net ircd.lightning.net) <-- Mostow has quit (ircd.east.gblx.net ircd.lightning.net) <-- obliv|on has quit (ircd.east.gblx.net ircd.lightning.net) <-- Kummer has quit (ircd.east.gblx.net ircd.lightning.net) <-- NOR-Gate has quit (ircd.east.gblx.net ircd.lightning.net) Maybe, but I'm atheist. Don't expect me to bow :) --- Cauchy gives channel operator status to NOR-Gate_ --- NOR-Gate_ is now known as NOR-Gate in fact, he's a little too applied for my taste :) --> Simplex- (tjh@cdm-17-244-bcs.cox-internet.com) has joined #mathematics howdy Simplex- --> Mostow (jreno@dentin.eaze.net) has joined #mathematics i read an article by him in honor of weyl and he brought up circuits and electrical stuff and applied topology to it or something --- NOR-Gate gives channel operator status to Mostow --- Tolaria gives channel operator status to Mostow --> q[binary] (binary@m448-mp1-cvx1b.edi.ntl.com) has joined #mathematics <-- q[binary] has quit (irc.exodus.net hub.il) kinda ironic eh Why ironic? --> czth (dbrobins@cr595985-a.ktchnr1.on.wave.home.com) has joined #mathematics --> q[binary] (binary@m448-mp1-cvx1b.edi.ntl.com) has joined #mathematics --- Disconnected (Remote host closed socket). --> nerdy2 (garrett@c1463354-a.cirving1.tx.home.com) has joined #mathematics --- Topic for #mathematics is Facieaux --- Topic for #mathematics set by Galois at Fri May 18 16:15:07 ironic because it was his inspiration Galois was using to get you off of your self-destructive attempt at learning