That is why I write v[f] = 
 Yeah, that is nice. I borrowed that in my report :)
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 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 :)
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 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
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 I'll take your word for it :)
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 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
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 so everything not in M_x is invertible,  which means it can't be in any proper ideal
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 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
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 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
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 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! :)
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 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
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 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.
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 and the quicker you get that into your head, the better off you are to handle the rest that comes down the line
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 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
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 Maybe, but I'm atheist. Don't expect me to bow :)
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 in fact, he's a little too applied for my taste :)
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 howdy Simplex-
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 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
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 kinda ironic eh
 Why ironic?
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--- 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

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