Why does bosonic string theory require 26 spacetime dimensions? - MathOverflow

Why does bosonic string theory require 26 spacetime dimensions?

2012-06-14T19:59:34Z

I do not think it is possible really believe or experimentally check (now), but all modern physical doctrines suggest that out world is NOT 4-dimensional, but higher. The least sophisticated candidate - bosonic string theory says that out world is 26 dimensional (it is not realistic due to presence of tachion, and so there are super strings with 10 dimensions, M-theory with 11, F-theory with 12).

Let us do not care about physical realities and ask: what mathematics stands behind the fact that 26 is the only dimension where bosonic string theory can live ? Definitely there is some mathematics e.g. 26 in that MO question is surely related.

Let me recall the bosonic string theory background. Our real world is some Riemannian manifold M which is called TS (target space). We consider the space of all maps from the circle to M, actually we need to consider how the circle is moving inside M, so we get maps from $S^1\times [0~ T]$ to M ( here $S^1\times [0~ T]$ is called WS - world sheet); we identify the maps which differs by parametrization (that is how Virasoro comes into game and hence relation with Leonid's question).

That was pretty mathematical, but now ill-defined physics begin - we need integrate over this infinite-dimensional space of maps/parametrizations with measure corresponding to exp( i/h volume_{2d}(image(WS))). This measure is known NOT to exist mathematically, but somehow this does not stop physists they do what they call regularization or renormalization or something like that and 26 appears...

Answer by Chris Gerig for Why does bosonic string theory require 26 spacetime dimensions?

2012-06-14T21:35:21Z

I think this is standard in some String Theory textbooks:
The quantum operators form the Virasoro algebra, where the generators obey $[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}m(m^2-1)\delta_{m+n,0}$. Here "c" is the central charge, which is the space-time dimension we are working over. We need this algebra to interact appropriately with the physical states of the system (i.e. $L_m|\phi\rangle$ information), and only when $c=26$ do we guarantee that there are no negative-norm states in the complete physical system.

[Addendum] In the method I described, $c=26$ arises correctly as the critical dimension so that no absurdities occur. What I believe David Roberts is thinking about (in his comment below) is another way to get the same answer: You consider light-cone coordinates and write down the mass-shell condition (summing over the worldsheet dimension $D−2$), and you end up with the requirement $\frac{D-2}{24}=1$. In other words, $c=D=26$.

Answer by Tom Dickens for Why does bosonic string theory require 26 spacetime dimensions?

2012-06-15T01:32:25Z

The value of 26 ultimately comes from the need to rid the theory of negative-norm states, as previously noted. This involves regularizing the sum $ \sum_{n=0}^\infty n, $ which of course diverges. One obtains a finite part equaling $-\frac{1}{12},$ which leads to the 24, and from there to 26 for the number of dimensions (25 space, 1 time). (Analytic continuation of the zeta function gives $\zeta(-1) = -\frac{1}{12} $).

Rather than me writing the details in here, I suggest this nice introductory reference, where the number of dimensions required for consistency of the bosonic string is derived in Ch. 2:

http://www.damtp.cam.ac.uk/user/tong/string/string.pdf

Answer by Pavel Safronov for Why does bosonic string theory require 26 spacetime dimensions?

2012-06-15T01:44:23Z

In addition to Chris Gerig's operator-language approach, let me also show how this magical number appears in the path integral approach.

Let $\Sigma$ be a compact surface (worldsheet) and $M$ a Riemannian manifold (spacetime). The string partition function looks like $$Z_{string}=\int_{g\in Met(\Sigma)}dg\int_{\sigma\in Map(\Sigma,M)}d\sigma\exp(iS(g,\sigma)).$$ Here $Met(\Sigma)$ is the space of Riemannian metrics on $\Sigma$ and $S(g,\sigma)$ is the standard $\sigma$-model action $S(g,\sigma)=\int_{\Sigma} dvol_\Sigma \langle d\sigma,d\sigma\rangle$. In particular, $S$ is quadratic in $\sigma$, so the second integral $Z_{matter}$ does not pose any difficulty and one can write it in terms of the determinant of the Laplace operator on $\Sigma$. Note that the determinant of the Laplace operator is a section of the determinant line bundle $L_{det}\rightarrow Met(\Sigma)$. The measure $dg$ is a 'section' of the bundle of top forms $L_g\rightarrow Met(\Sigma)$. Both line bundles carry natural connections.

However, the space $Met(\Sigma)$ is enormous: for example, it has a free action by the group of rescalings $Weyl(\Sigma)$ ($g\mapsto \phi g$ for $\phi\in Weyl(\Sigma)$ a positive function). It also carries an action of the diffeomorphism group. The quotient $\mathcal{M}$ of $Met(\Sigma)$ by the action of both groups is finite-dimensional, it is the moduli space of conformal (or complex) structures, so you would like to rewrite $Z_{string}$ as an integral over $\mathcal{M}$.

Everything in sight is diffeomorphism-invariant, so the only question is how does the integrand change under $Weyl(\Sigma)$. To descend the integral from $Met(\Sigma)$ to $Met(\Sigma)/Weyl(\Sigma)$ you need to trivialize the bundle $L_{det}\otimes L_g$ along the orbits of $Weyl(\Sigma)$. This is where the critical dimension comes in: the curvature of the natural connection on $L_{det}\otimes L_g$ (local anomaly) vanishes precisely when $d=26$. After that one also needs to check that the connection is actually flat along the orbits, so that you can indeed trivialize it.

Two references for this approach are D'Hoker's lectures on string theory in "Quantum Fields and Strings" and Freed's "Determinants, Torsion, and Strings".

Answer by userN for Why does bosonic string theory require 26 spacetime dimensions?

2012-06-15T12:28:05Z

Just to correct a small misconception: Physicists aren't claiming that an integral over the space of smooth maps from $WS$ to $X$ exists. The path integral measure for the bosonic string is defined on something more like the "space of distributions on $WS$ valued in $X$". It's not a problem that no integral exists on the space of smooth maps, because this space shows up only as a convenient shorthand for discussing the renormalization procedure which is used to define the path integral measure.

The quantum mechanics of the simple harmonic oscillator is subject to a similar abuse of language. You discuss the theory in terms of functions on the timeline $[0,t]$, but if you're careful, the path integral measure (aka, the Ornstein-Uhlenbeck measure) is actually defined on the space of distributions on $[0,t]$. (What makes life easy in this case is that the Wiener measure is supported on distributions which are almost everywhere continuous functions.)

The situation is somewhat more complicated for the 2d nonlinear sigma model, because there isn't really anything you'd want to call the distributions valued in $X$. Instead you try to define the measure as a linear functional on observables which are well approximated by functions of the form $\phi \mapsto ev_{\sigma} \phi^*f$, where $f$ is a function on $X$ and $ev_\sigma$ evaluation at a point $\sigma \in WS$. The correlation function of observables $\hat{\mathcal{O}}_1$, $\hat{\mathcal{O}}_2$ should be approximated by integrals of the form

$\int_{Map(L,X)} \mathcal{O}_1(\phi) \mathcal{O}_1(\phi) e^{i S_L(\phi)} d\phi$

for some finite set of points $L \subset WS$, and some approximation $S_L$ of the classical action $S$ defined using only finite differences among the points in $L$. When you refine the set of points $L$ to fill in $WS$, you get an expectational value functional on the set of observables. This expectation value functional should have the same properties (like OPE) that you see in QFTs where the classical fields take values in linear spaces.

Answer by Kevin Wray for Why does bosonic string theory require 26 spacetime dimensions?

2012-09-01T03:59:31Z

I'm not an expert on this so bear with me, but I don't think you must require $\dim(M) = 26$, you must only require that the worldsheet is conformally invariant - i.e., the Weyl anomaly vanishes. You can do this by adding 26 bosons (which represent the coordinates of $M$) - which is called critical string theory - or you can turn on the dilaton expectation value - which is then called non-critical string theory. There's a lot of interesting research involving these non-critical string theories, for e.g. check out $c=1$ matrix models and type 0 string theories.