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Mike2

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Thanks.

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Mike2

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Thanks.

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lethe

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weighted sum?? how do you "add" surfaces?Originally posted by Mike2

I'm wondering if it is true that any surface can be equated to a weighted sum of a basis of surfaces differring only by genus?

as far as i know, there are no sensible algebraic operations defined on surfaces

I think this is asking whether the path integral formulation for strings is more general.

more general than what?

- #3

Mike2

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Originally posted by lethe

weighted sum?? how do you "add" surfaces?

as far as i know, there are no sensible algebraic operations defined on surfaces

So what you are saying is that a surface can only be conformally equivalent to a surface of one genus type, either it has a hole, or it does not, right?

A line in R

QUOTE]

more general than what?

More general than its use in quantum physics.

- #4

lethe

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yesOriginally posted by Mike2

So what you are saying is that a surface can only be conformally equivalent to a surface of one genus type, either it has a hole, or it does not, right?

i don t think that such a thing exists.I was wondering if there isn't a similar principle for surfaces?

More general than its use in quantum physics.

where are higher genus surfaces used in physics other than in string theory? how can i compare their use in string theory?

- #5

Mike2

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Originally posted by lethe

yes

i don t think that such a thing exists.

So what you are saying is that there is no higher dimensional application of the Taylor series expansion, where, for example a surface in 3D being describe by f(x,y)=z is not equivalent to a sum of other functions of (x,y)? I don't remember seeing any examples of this either, but I've not seen anything that rules it out. It seems to me one could easily expand the f(x,yc) with y a constant and then expand f(xc,y) with x constant separately, and then combine these separate expansions into various surfaces. The original surface would then be the result of a sum of surfaces.

- #6

lethe

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Originally posted by Mike2

So what you are saying is that there is no higher dimensional application of the Taylor series expansion, where, for example a surface in 3D being describe by f(x,y)=z is not equivalent to a sum of other functions of (x,y)? I don't remember seeing any examples of this either, but I've not seen anything that rules it out. It seems to me one could easily expand the f(x,yc) with y a constant and then expand f(xc,y) with x constant separately, and then combine these separate expansions into various surfaces. The original surface would then be the result of a sum of surfaces.

there certainly is Taylor series for

i do not see what that has to do with your idea about adding

- #7

Mike2

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Originally posted by lethe

there certainly is Taylor series forfunctionsin higher dimensions.

i do not see what that has to do with your idea about addingtopological spaces

Are the series expansions of a surface only for open surfaces as functions of (x,y)? If not, then are the expansion of surfaces into the sum of basis surfaces only for genus 0 surfaces?

- #8

lethe

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i cannot answer this, because there is no such thing as a series expansion of a surface.Originally posted by Mike2

Are the series expansions of a surface only for open surfaces as functions of (x,y)?

taylor expansion is only defined for real valued functions, not for surfaces.

basis surfaces?? i cannot imagine what that means, since surfaces are not algebraic objects, and i cannot add themIf not, then are the expansion of surfaces into the sum of basis surfaces only for genus 0 surfaces?

- #9

Mike2

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Originally posted by lethe

i cannot answer this, because there is no such thing as a series expansion of a surface.

taylor expansion is only defined for real valued functions, not for surfaces.

Aren't surfaces described by functions? Don't functions have expansions? Can't functions, these expansion functions, perscribe surfaces? What?

- #10

Hurkyl

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Aren't surfaces described by functions?

Technically, no; they're described by equations.

If you meant that a surface (in [tex]\mathbf{R}^3[/tex]) can be represented in the form [tex]z=f(x, y)[/tex], then the answer is, in general, definitely no.

Don't functions have expansions?

Generally, no. And even when they do, they generally only on a small piece of the function.

Can't functions, these expansion functions, perscribe surfaces?

Using a suitable interpretation of a function prescribing a surface

The expansion (if it exists) will prescribe a surface, a piece of which is the same as a piece of the surface prescribed by the original function; it is possible that they may only have a single point in common!

Each partial sum of the expansion will prescribe a surface. The surfaces prescribed by the partial sums may or may not eventually look like the surface prescribed by the expansion.

Each term of the expansion will prescribe a surface. However, because we cannot add surfaces, this does not prescribe a perturbative expansion of the surface.

What?

The point is that perturbation is done on functions, not surfaces.

Footnotes:

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lethe

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Hurkyl has explained this very nicely, and captured the essence of my point.

let me say it one more way: a function z=f(x,y) is a one-to-one valued map, which means that it can never "double back" on itself. yet, i can conceive of many surfaces which do "double back", like the circle, or the sphere. these manifolds can never be represented in their entirety by a function, and i can ascribe no meaningful content to the notion of a perturbation expansion of these surfaces.

let me say it one more way: a function z=f(x,y) is a one-to-one valued map, which means that it can never "double back" on itself. yet, i can conceive of many surfaces which do "double back", like the circle, or the sphere. these manifolds can never be represented in their entirety by a function, and i can ascribe no meaningful content to the notion of a perturbation expansion of these surfaces.

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Mike2

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selfAdjoint

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Mike2

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Originally posted by selfAdjoint

I'm still not quite sure what's vibrating. Are the various points along the string oscillating in space with time, or is it some function on the string that is changing at various points on the string?

Thanks.

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selfAdjoint

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Each of the X

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Mike2

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Also, do you consider Polaski's book complete? Or is there some graduate math needed as a prerequisite to this book? Also, how many chapters must you read through before you get to Superstring theory and not just practicing on a faulty theory?

Thanks.

- #17

lethe

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Originally posted by Mike2

Also, do you consider Polaski's book complete? Or is there some graduate math needed as a prerequisite to this book? Also, how many chapters must you read through before you get to Superstring theory and not just practicing on a faulty theory?

Polchinski's book is rather complete. no graduate math is needed as a prerequisite, as this book is intended for physicists. maths you would need to know would include tensor index notations, complex analysis, and it is beneficial to know as much differential geometry and topology as possible, although not a prerequisite, a little group theory, and some Lie theory. i dunno... basically, it assumes you know a bit more math than your average physics grad student, but not too much more...

on the other hand, it does assume a lot of graduate level

as for how many chapters you spend on a toy model before you get to the superstring, well the book is in two volumes, and all of volume I is on the nonsupersymmetric string. so volume I is 9 chapters, although you don t necessarily have to finish all of those before moving to volume II.

but umm... i don t know... what do you have against spending some time with a toy model? it can be quite useful...

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selfAdjoint

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- #19

Mike2

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Originally posted by selfAdjoint

Do you think Hatfield book, "Quantum Field Theory of Point Particles and Strings" would be sufficient to read Polaski?

- #20

lethe

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Originally posted by Mike2

Do you think Hatfield book, "Quantum Field Theory of Point Particles and Strings" would be sufficient to read Polaski?

yes. if you can read Hatfield, then you will be able to read Polchinski.

and please note: his name is Polchinski, not Polaski.

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