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This is a whole other question that you maybe expect from me, but it is bugging me since I heard of a long time ago. Bosonic string theory (to take the simplest one of string theories) is consistent in only 26 dimensions. This critical dimension is needed to rule out things like negative propabilities, etc.

The way most physisists are calculating the 26 always boggles down to the analytic continuation for the $\zeta(s)$ function that means that $\zeta(-1)=-\tfrac{1}{12}$ and the "definition" $\zeta(-1)=\sum_{k=0}^{\infty} n$. If you don't want to use this, most physisist are relying on a technique called zeta-renormalization, which pretty much looks like the time I set down for my analysis lecture and saw $"\ldots = \tfrac{\infty}{\infty}=1$ on the blackboard (a leftover from a quantum field lecture that wasdn't erased).

Question: Are there mathematically sound proofs for the critical dimensions (for instance, based on some consistency rules of the inproduct on Hilbert spaces) ?

My sincere appologies for bashing theoretical physisists this way, but the leftover was really what I saw (I did even once see $\infty - \infty = 0$ in very much the same context, for that matter).

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Since there is no axiomatic approach through "constructive quantum field theory" (a la Jaffe and Glimm, etc.) to strings or string field theory, I'd say it's pretty unlikely that you'll find such a mathematically rigorous calculation. –  user02138 Dec 2 '11 at 12:25
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I assume you are familiar with Polyakov's "Quantum geometry of bosonic strings" article? The Polyakov formula has been fairly well justified mathematically (and the zeta functional determinant is of great use in studying the spectrum of Riemannian manifolds), so I think your characterisation of zeta-renormalisation is unfair, at least in this particular instance. –  Willie Wong Dec 2 '11 at 14:37
    
I can't put my hand on the bibliographical details right now, but Floyd Williams of U Mass Amherst has published some rigorous stuff on this question, you might look for that. –  user20911 Dec 9 '11 at 5:15
    
And as usual, Absence of proof is not a proof of absence. Whatever the context is, it might not be rigorous just yet. As far as I know from my colleagues, most of their time is spent on making sense out of calculations that lead to $\infty$. –  user13838 Dec 9 '11 at 5:39
    
What I understand back then (1990 or so, when I was studying maths), most physics calculations with infinites in them, which must be right (because they can be shown in labs), are much later rigously proved by mathematicians, not seldom by discovering new areas of maths. –  Willem Noorduin Dec 11 '11 at 20:50
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