Path integrals and the sum of surfaces, is this general?

[Mike2]

Main Question or Discussion Point

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

Thanks.

 

[lethe]

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?

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

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?

 

[Mike2]

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 R² can be describes by a function. Any given function can be described as the infinite weighted sum of exponential, the exponential expansion of a function. Each of these basis functions have different curvature and zero cross nodes, etc. I was wondering if there isn't a similar principle for surfaces?


QUOTE]
more general than what? [/QUOTE]
More general than its use in quantum physics.

 

[lethe]

Originally 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?

yes

I was wondering if there isn't a similar principle for surfaces?

i don t think that such a thing exists.

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?

 

[Mike2]

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.

 

[lethe]

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 functions in higher dimensions.

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

 

[Mike2]

Originally posted by lethe
there certainly is Taylor series for functions in higher dimensions.

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

 

[lethe]

Originally posted by Mike2
Are the series expansions of a surface only for open surfaces as functions of (x,y)?

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.

If not, then are the expansion of surfaces into the sum of basis surfaces only for genus 0 surfaces?

basis surfaces?? i cannot imagine what that means, since surfaces are not algebraic objects, and i cannot add them

 

[Mike2]

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?

 

[Hurkyl]

Aren't surfaces described by functions?

Technically, no; they're described by equations.

If you meant that a surface (in $$\mathbf{R}^3$$) can be represented in the form $$z=f(x, y)$$, 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 surface¹...

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:

¹: An example of a suitable definition of a function prescribing a surface is:

Definition: The function $$f(\vec{x})$$ is said to prescribe the surface $$S$$ if and only if the points of $$S$$ coincide with the solution set of $$f(\vec{x})=0$$.

 

[lethe]

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.

 

[Mike2]

I also have to wonder whether it is possible to describe a "hole" of genus 1 and higher surfaces. If I were tempted to describe a hole with f(x,y)=0 there, that is still a level plain, not a hole. How do you describe a state of something not existing there with a function everywhere? I'm just going be intuition here. I perfer to get into detail as necessary.

 

[selfAdjoint]

Trying to get an equation for a worldsheet is the wrong way to go. String world sheets are Riemann surfaces, and their topology is determined, Riemann fashion, by the singularities of the meromorphic functions they describe. And the singularities are determined from the vertices, places where strings enter or leave the interaction. A beautiful theorem lets physicists calculate their operator expansions from mathematical functions defined around these vertices. It's these mathematical things that are quantized by the path integrals.

 

[Mike2]

Originally posted by selfAdjoint
Trying to get an equation for a worldsheet is the wrong way to go. String world sheets are Riemann surfaces, and their topology is determined, Riemann fashion, by the singularities of the meromorphic functions they describe. And the singularities are determined from the vertices, places where strings enter or leave the interaction. A beautiful theorem lets physicists calculate their operator expansions from mathematical functions defined around these vertices. It's these mathematical things that are quantized by the path integrals.

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.

 

[selfAdjoint]

The string is osciliating and moving. Also it is splitting off strings and joining with other strings in interactions. The interactions are what are interesting to the physicists. The motion of the string in spacetime is represented by potentials on the worldsheet. These are universally designated the X_i, and the number of them is one less than the number of spacelike dimensions of the background. One less, because one of the spacelike dimensions is taken up on the worldsheet by the length of the string.

Each of the X_i is a real valued potential on the worldsheet - the coordinate in that dimension which that point of the string occupies at that moment. As the string moves the coordinates, and hence the potentials on the worldsheet, change. Oh the worldsheet is a busy, complicated place. It takes Polchinski several chapters to introduce all the math and physics.

 

[Mike2]

Was it that capital X_i are potentials and lowercase x_i were space-time coordinates? Maybe that's what confuses me.

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.

 

[lethe]

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 physics prerequisites. if you haven't studied quantum field theory for quite a while, you may find polchinski's book illegible.

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...

 

[selfAdjoint]

Someone who hasn't had complex analysis might be snowed by all the Conformal stuff, though I suppose Laurent series could be taken as introduced. You are very right about the physics, though. I think the requirement is to be mature in your understanding of field theory. Meaning not only do you know the facts, but you are right there with their interelationships. Down in the middle of Polchinski is no time to be having mysteries about how that integral transformed or where that bracket came from in the operator product expansion.

 

[Mike2]

Originally posted by selfAdjoint
Someone who hasn't had complex analysis might be snowed by all the Conformal stuff, though I suppose Laurent series could be taken as introduced. You are very right about the physics, though. I think the requirement is to be mature in your understanding of field theory. Meaning not only do you know the facts, but you are right there with their interelationships. Down in the middle of Polchinski is no time to be having mysteries about how that integral transformed or where that bracket came from in the operator product expansion.

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

 

[lethe]

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.