Previously I asked a question Question on dimensions of CFT operators (ref: MAGOO, hep-th/9905111) here and it was (correctly of course) answered by Motl. I realized I didn't understand a part of it even though I thought I did.
It was about dimensionlessness of bulk scalars from the perpective of boundary conformal transformation. If we have a bulk scalar $\phi$ it is dimensionless from the point of boundary transformations as (as answered by Motl)
``it is an actual field in the AdS space (the bulk), unlike $\phi_0$ that is defined on the boundary. They mean that it is dimensionless under the conformal transformations of the boundary - and that's true because the conformal transformations are realized as simple isometries in the bulk, so they can't rescale the AdS field $\phi$ - at most, they move it to another point. ''
But the way we talk about conformal dimension of a boundary operator is through it's transformation under the diffeomorphism $x\to\lambda x$ and then if a boundary field $f(x)\to f^\prime(x)=\lambda^hf(\lambda x)$ (or the relation written later by Motl in that answer), then $h$ is it's conformal dimension. I hope what I said so far is right. But, isn't conformal transformation at the boundary also act as a diffeomorphism in the bulk (at least for boundary part of the coordinates of the bulk field). Then why will it not transform in such a way that it has a dimension?
I believe, e.g. spin-1 gauge field in the bulk has (confomal/scaling/mass) dimension 1. So, how does it have a dimension? Sorry for the novice question and thanks in advance!