Bosonic action in M Theory -> IIA Theory

[AlphaNumeric]

In various textbooks like Polchinski and in papers like Witten hep-th/9503124 they go through the process of starting with the bosonic action (in p-form notation) of M Theory and turning it into the action for IIA string theory.

I understand the process (ie dimensional reduction) which converts one action into another, but I don't understand why looking at just the bosonic part of the action is sufficent, given IIA includes fermions. During the process of compactifying and dimensional reduction the M Theory action splits into various parts, the NS-NS part, the R-R part and the CS part (ie a mixture of NS and R terms). Is it the case that M theory is a purely bosonic theory which then turns into a fermionic one when you reduce to 10 dimensions?

Thanks for any help :)

 

[josh1]

Yes, IIA and M-theory are related, but where have you read that they're equivalent? One cannot fully obtain one from the other.

 

[AlphaNumeric]

Not so much directly equivalent, but that IIA is partly derived from taking an 11 dimensional theory and compactifying it on a circle in one dimension.

For instance, in the http://arxiv.org/PS_cache/hep-th/pdf/9503/9503124.pdf he considers the bosonic action for 11 dimensional super gravity, compactifies it and then turns it into the IIA bosonic action. This then shows that they are related through the various transformations (Paragraph starting 'So we need an eleven-dimensional...' on Page 10).

I'm confused why it's okay only to talk about the bosonic part. In the various bits of literature I've read they always start with the bosonic action for the 11 dimensional theory and don't mention fermions at all.

/edit

Actually, have I just got this all muddled up. It's not that M Theory reduces to IIA, but that you can turn 11d super gravity into IIA via dimensional reduction, and that super gravity is the low energy, long distance limit of M Theory, which is an as yet incomplete theory since short distance physics at strong coupling is not yet understood.

 

[josh1]

Not all of the objects in IIA can be obtained by dimensional reduction of ordinary eleven dimensional supergravity. In particular, this is true of D0-branes. Now, it is an R-R gauge boson that couples to D0-branes, so D0-branes are in the bosonic sector of IIA. In the limit of large string coupling g>>1, the spectrum of D0-branes becomes a continuum of light states. This is characteristic of a system acquiring an additional large spacetime dimension. Thus there is an eleventh dimension that appears in IIA for g>>1 that doesn’t appear in the perturbative sector of IIA - the sector in which g<1 when D0-branes don’t appear. Thus M-theory (by which we mean the stringy eleven dimensional supergravity) may be discovered in the strong coupling limit of the bosonic sector of IIA, so - and this is the point - although IIA has a fermionic sector, it's irrelevant for this.

 

[AlphaNumeric]

Thanks Josh. You've pretty much confirmed (with a tweak or two) what I had in my head for how they all sort of mesh together.