MTd2 said:
The coupling constant runs so the Planck scale changes. What is it at the AS point?
Newton's G --> 0 at the AS point.
This could be seen as a weakness of the AS approach. The Planck scale fails to exist at the fixed point.
Weinberg alluded to this in his AS talk at Strings 2010. He has been trying to model the early universe using AS---and to get a grip on inflation, for example. He has had difficulty.
That is one reason why I suspect AS might not succeed in producing a model of the early universe.
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You may recall that what converges at the fixed point is the dimensionless version of G, namely if k --> infinity is the momentum/wavenumber scale then the dimensionless coupling number is Gk
2. Therefore G must go to zero as 1/k
2.
The Planck length is sqrt(hG) so unless Planck's constant h also runs and goes to infinity we must face the prospect that as G goes to zero the Planck length also goes to zero.
In other words, the AS theory may be predictive in the sense that once a finite number of constants are determined (like as few as 3 parameters) you get predictions out to arbitrarily high energy. But still the AS predictions may be wrong.

This is just my guess, nothing special to back it up.
I very much admire Weinberg's effort to understand early universe cosmology using AS. This effort must be made, and it still could succeed!
I wish more people would help him get the job done. And I am a Weinberg fan. Cosmology is the testing ground for QG. But Weinberg's effort to get a good cosmology from AS may very well fail and that would mean AS is ultimately no good. This kind of thing happens in theoretical physics. *shrug*