Gravity and Strings

11.5 Generalized dimensional reduction 341
2. The converse is not true: the invariance of the action under constant shifts of ˆϕ does
not mean that the field lives on a circle and GDR makes sense. Formally, the GDR
procedure can be performed, but the result could be meaningless since no vacuum
solution associated with the GDR Ansatz is guaranteed to exist. We are going to see
an example of this fact in Section 11.5.1.
3. Under U(1) gauge transformations
δz =−(x), δ Aµ = ∂µ(x), δϕ = mN
(x), (11.222)
ℓ
i.e. the lower-dimensional scalar field transforms by shifts of the gauge parameter!
This kind of gauge transformation is called a massive gauge transformation and allows
us to eliminate ϕ completely by fixing the gauge. ϕ plays the role of Stückelberg
field for Aµ [871]. KK gauge invariance is broken after this gauge fixing and this is
reflected, as we will see, in a new mass term for the vector field. It is usually said
that the vector has “eaten” the scalar, becoming massive. This is a sort of Higgs phenomenon,
the difference being that there is no scalar potential. Observe that ϕ can be
removed consistently by a gauge transformation if both and ϕ live in circles, as we
have assumed.
In the next sections we are going to see some examples of GDR that illustrate these
ideas. In the first example we perform the complete GDR of the real scalar field that we
have discussed above and give an alternative interpretation.
11.5.1 Example 1: a real scalar
Let us consider the simple model
ˆS = d ˆd
ˆx |ˆg| ˆR + 1
(∂ ˆϕ)2,
(11.223)
where ˆϕ is a real scalar field. This action is invariant under constant shifts of the scalar and
therefore it is possible to use the standard recipe for GDR: we perform now a z-dependent
shift of the usual z-independent Ansatz ˆϕ(x, z) = ϕ(x) + mNz/ℓ, which will lead us to a
d-dimensional theory with no dependence on z.
However, as we have stressed repeatedly, this recipe makes real sense only if the scalar
field lives in a circle and is identified periodically, ˆϕ ∼ˆϕ + 2πm. Although it looks as if we
can simply decree that identification, the above action does not contain enough structure
to enforce it and we will see that, in particular, there is no vacuum solution with ˆϕ(x, z) =
mNz/ℓ. This example is therefore just an academic exercise.
Using the standard Ansatz for the Vielbein Eq. (11.33) but adding a subscript (1) to the
KK scalar field, we find
S =
d d x |g| k
R − 1
4k2 F 2 1
(2) + 2 (Dφ)2 − 1
2
2
mN
ℓ
2
k −2
, (11.224)