Start from appreciating the simple geometry of the unit N-vectors n, which comprise a hypersphere $S^{N-1}$ embedded in N-space. (Think of a beach-ball $S^2$ embedded in our 3-space.)
The symmetry of the hypersphere is $O(N-1)$, that is the $O(N)$ symmetry of the n s has been broken down to it, and so N―1 of the scalar degrees of freedom involved have to be massless Goldstone bosons, by Goldstone’s theorem.
You note that for a unit vector u , you always have $\partial ( {\mathbf u}\cdot {\mathbf u})=0 = {\mathbf u}\cdot \partial {\mathbf u} $, so the gradients of your relabeled unit vectors must live in the space orthogonal to each: so must be linear combinations of all unit vectors but themselves.
The vector (potential) functions Bμ and Aμ are then the vector components in these orthogonal subspaces, as you may check, and nothing else. They are, of course, slowly varying—properties of the slowly varying fields, so will serve as some sort of decoupled semiclassical background field when the fast modes φ are integrated over, only to have their coupling coefficients be modified by this integration.
There are N―1 Bμ s, but only (N―1)(N―2)/2 Aμ s, as they have to be antisymmetric in their two indices, as you may check by dotting the ∂e expression by another e and utilizing orthogonality. So they look like rotations on the N―1-dim subspace, the surface of the hypersphere, so far.
Further check that
$$
{\mathbf n}\cdot \partial {\mathbf n}=0
$$
is automatically satisfied by this reparameterization, as it should, by virtue of
$$
\partial {\mathbf n}= {\mathbf n}_0 \left (\partial \sqrt{1-\varphi^2} - B^a\varphi_a\right ) +e_a \left (\sqrt{1-\varphi^2}~B^a + \partial \varphi_a -A^{ab}\varphi_b\right ),
$$
in summation convention.
Finally observe that here, as in S, the (N―1)(N―2)/2 coefficients Aμ enter in full covariant derivative combinations on the φ s, so, indeed, they are $O(N-1)$ gauge rotations on the hypersphere. The action S is gauge invariant under rotations of the hypersphere. Any operations on it will also net a gauge invariant result. As Polyakov says, the effective action, after integrating out the φ s, has to also be gauge-invariant, so it can only involve the gauge invariant Y-M field strengths traced (what else could it be?), thus dismissible from dimensional grounds――he is looking for only logarithmic divergences!
But the Bμ s are not quite gauge fields in minimal coupling. What to do? I’ll be schematic. Think of the slow Bμ s as background fields, fixed vectors, and the φ s as the fluctuating variables whose expectation values, propagators, etc… you are evaluating. The terms linear in Bμ are fixed vectors for rapidly rotating vectors around them, so by symmetry average out to zero, like $\langle r\cdot R\rangle$. Recall $\langle \varphi_a\rangle=0$ and $\langle {\mathbf n} \rangle= {\mathbf n}_0$, metastable against slow rotations on the hypersphere.
Polyakov shows you how to average the bilinears in the rapidly fluctuating φ s, in S(II), $ B_\mu^a B_\mu^b\langle \varphi^a \varphi^b-{\varphi}^2 \delta^{ab}\rangle$; of course, the answer must be an $O(N-1)$-invariant tensor, hence $\propto B_\mu^aB_\mu^a$.
If you rewrite the last term in S(II) as an effective action of the N-1 slowly varying Bμ s, the massless Goldstone modes $B^a_\mu= e_a\cdot \partial_\mu {\mathbf n}_0$, you see that the leading terms averaged effectively modify the respective original coupling $e_0$: it has increased—poor man’s infrared slavery.
(Alert: he is a little sloppy. For constrained unit N-vector n s, the sphere is $S^{N-1}$, not $S^{N}$, so for 3 space, N = 3, so a beach-ball, there are 2 Goldstons not 3.)
Best Answer
In a sum on polarizations, like $\sum_\lambda~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)$, there is a fundamental difference if you are considering all the polarizations, or only the physical polarizations .
If you take all polarizations, the sum is equals to $g_{\mu\nu}$, and it is in fact a normalizations of the $e_{\mu}^{\lambda}(k)$. This sum is non-physical.
$$\sum_{all ~polarizations~~ \lambda}~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)=-g_{\mu\nu} \tag{1}$$
If you consider only the physical polarizations, this sum is physical, and you will get the pole of the propagator, which is a physical quantity too :
$$\sum_{physical ~polarizations~~ \lambda}~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)=-(g_{\mu\nu} - \frac{k^\mu k^\nu}{m^2})\tag{2}$$
The propagator here is :
$$D_{\mu\nu}(k) = \frac {-(g_{\mu\nu} - \frac{k^\mu k^\nu}{m^2})}{k^2-m^2} \tag{3}$$