String Theory - Winding Strings

sneutron

The answers will probably be "it ain't as simple as that" but here's trying, anyway.

1. Is current thinking that strings have to wind "around" something? If so, must the compactified space have at least one hole?

2. With just a single hole, presumably there can be just one topologically distinct string configuration (ignoring winding number and sense)?

3. With two holes, three string configurations - or maybe more with "twists"?

4. Three holes give seven ways of winding - plus twisted ways?

5. How many holes are favoured, and are there enough winding configurations to accommodate SuSy?

Thanks in advance.

Paul

suprised

Sure - you need topologically non-trivial, non-contractible cycles, otherwise th e cycle shrinks to nothing.

yep

Usualy one determines a basis of cycles (generators of the homology group), and
considers all other non-trivial cycles as linear combinations of the basic wrappings.

For your question of two holes, the answer is essentially yes, but to be precise, you must specify more data (like if the space is compact or not). For example, a compact Rieman surface with two holes has four basis cycles - two "around" the holes and one "around" each neck. Lets denote this basis by (b1,b2,a1,a2). Then a general string wraps around a cycle of the form n1 b1+ n2 b2+m1 a1 + m2 a2. One says that the string has wrapping numbers (n1,n2,m1,m2) with respect to that basis; these numbers can also be viewed as certain charges.


There is a priori no preferred number of holes. And whether the theory is SUSY or not, depends on various properties of the compactication space. If you consider 2 dimensional Riemann surfaces, then only the space with one hole, ie the torus, will give rise to a SUSY spectrum.


It primarily depends on whether such a decay is energetically favored. For wrappings on flat tori the mass of a string wrapping two times the _same_ cycle is indeed the same as two strings wrapping each one time. So one cannot really distinguish these, and calls this a "bound state at threshold". The situation is more interesting if there are several cycles. Then in general a string with wrappings (n1,n2,m1,m2) has less mass than two strings A,B whose wrappings add up like n1A+n1B=n1, etc. Thus one considers this string as bound state of those others. When the wrapping numbers are non coprime (ie, have a common factor), then the situation is again degenerate and one has a bound state at threshold.

sneutron

@surprised, that was really helpful - thanks a lot! It's moved me on a fraction and even nudged me gently to go back to my 30-year-old linear algebra, group theory and diff. geometry. They may just make a bit more sense this time around...

inflector

This is interesting.

This question may be just a sign of my extreme ignorance, but is there some reason that one couldn't have strings connected to each other at vertices in some sort of giant branching tree in order to keep the individual strings from shrinking to nothing?

sneutron

@surprised, you've made me think even more about "twists". Are your wrapping numbers allowed to be negative? Even though it may be impossible to specify winding sense around an isolated hole, there must be configurations of several holes where the net result of a cycle does depend on the (relative) individual senses.

Sorry i can't give you a picture, but think of an east-west double bagel - two holes and a neck. Wind the string over the north-west, under the neck and onwards under and into the eastern hole, over the top of the south-eastern limb, down again to go under the south-western limb on the western bagel and back to the start.

Now i'm fairly convinced this cycle has a different handedness from the one where all the overs and unders are reversed. So the quantum numbers need to be distinguishable (even if, as might be the case here, the two configurations have the same energy).

My next problem might be dismissed on the grounds of compactness or some other "nice" topological requirement? Find a bit of the space where two bagel-like pieces come close but don't meet. (#1) Wrap the string over the left hole, over the right, wind back down and under both to complete the cycle. Give this wrapping numbers (1,1). (#2) Wrap over the left, dip under the right, complete a figure-eight path. Give this wrapping numbers (1,-1) and think of it as #1 with a 180 twist. (#3) Make as with #1 but give it a 360 twist so the wrapping numbers are again (1,1). Now #1 and #3 are obviously not equivalent in handedness or energetically, so i guess it's anathema to string theory. If i pumped up the shape enough, the bagels would separate and the twisted bits of string would ping back and wind around *something"?

Conclusion: strings aren't allowed to twist around themselves (to the extent of touching, anyway)?

suprised

Yes, there are string configurations like that, made out of so called (p,q) strings which may be thought of a bound states of p ordinary and q "D-strings". These are intrinsically non-perturbative objects and thus do not have a world-sheet description (let's not go into issues of duality frames, mutual non-locality etc).

There is a number of papers on such "string junctions", one is for example: http://arXiv.org/pdf/hep-th/9804210

suprised

Of course the n_i and m_i can be negative, they correspond to objects with negative charges. Flipping the sign of all of them describes an anti-particle, which has the same mass indeed. What you address, essentially, is whether there is an absolute definition of a particle versus an antiparticle - there is none. The signs are only fixed once you have defined a basis of cycles.

I couldn't quite figure out what you were asking in the first part, but be assured that there is nothing deep behind all of this, it just boils down to write any winding string, no matter how complicated, as a combination of the basic wrappings. There is no twist of strings, and everything is completely specified by the wrapping numbers; so if you have two seeemingly different configurations with the same wrapping numbers, then they are topologically equivalent.

Note that for your example with two holes, there are 4 basic cycles and thus 4 wrapping numbers to be specified, so in order to determine whether two configurations are the equivalent, you must compare all four wrapping numbers.

sneutron

Lovely, ta very much!