> > This is about Witten's paper on "Mirror Manifolds and Topological
String
> > Theory" , 9112056.

> > What is the best way to visualise powers, both positive and
negative, of
> > the canonical line bundle? What are the meanings of "K^1/2" and
"K^-1"?

>    Remember that for line bundles, dual and inverse are the same
> thing.
>    For one dimensional complex curves, K = T^*X.  So K^{-1} = TX.
> And K^{-1/2} is a square root of the tangent bundle, one of the
> conventional spinor bundles, so K^{1/2} is the dual of

> > What is the relation between twisting the fermions and rotational
symmetry
> > on the world sheet?

>    Twisting the fermions corresponds to declaring a subgroup of SO(2)
> x U(1)_L x U(1)_R other than SO(2) to be the rotational symmetry
group.
>    The twistings correspond to using, instead of J,

>    J - 1/2 (J_L - J_R)  (A-model)
>    J - 1/2 (J_L + J_R)  (B-model)

> > Why are the twisting the only ones we could do? If we took the
fermons to
> > be sections of other powers of the canonical line bundle on sigma
(of
> > course tensored with the pullback of the tangent bundle on X) what
would
> > be the effect on the world sheet? Would the sigma-model lagrangian
remain
> > invariant?

>    The point of doing this is that we want to treat Q as a BRST
> operator, on any curve Sigma.  Q has to act globally, so we need the
> fermionic parameters of the Lie algebra to be well-defined sections
of
> some line bundles.  This is easy locally, in some small neighborhood.
> But it's difficult globally: many line bundles don't have _any_
global
> sections.  So we twist and put our fermionic parameters in the
functions,
> where there is always at least the constant sections.

>    It's a lovely coincidence that this forces us to choose the A or B
> twistings.  Any twisting would be a symmetry fo the classical
Lagrangian.
> But only the A and B twistings don't become anomalous when we move to
the
> quantum theory.

> > Does twisting the fermions change the spins of the G+- in the
> > superconformal algebra? What exactly is the relation between
changing
> > the relevant bundles of the fermions and the effect on the
> > superconformal algebra- for example the shift in the
energy-momentum
> > tensor?

>    Yes, twisting changes the spins.  The spins are encoded in the
> eigenvalues h and \bar{h} of the grading operators L_0 and \bar{L}_0.
> spin = h - \bar{h}.  So twisting alters the line bundles by changing
how
> their fibers transform under rotations; this means that the spins of
> fields which arise as sections of these line bundles are altered.
> Changing the spins really means changing how we measure them, so we
must
> alter the components L and \Lbar of the energy momentum tensor.

>    Hope that helped.

> --A.J.