Gauge fields in Polyakov's treatment of renormalization for nonlinear sigma model

Question

I am deriving the results of renormalization for nonlinear sigma model using Polyakov approach. I am closely following chapter 2 of Polyakov's book--- ``Gauge fields and strings''.

The action for the nonlinear sigma model (NLSM) is \begin{equation} S= \frac{1}{2e_0^2}\int d^2x (\partial_\mu \mathbf{n})^2. \end{equation}

Polyakov breaks the $N$-dimensional unit vector $\mathbf{n}$ up into slow ($e_a,\mathbf{n}_0$) and fast ($\varphi_a$) variables as \begin{equation} \mathbf{n}(x) = \sqrt{1-\varphi^2}\mathbf{n}_0(x)+\sum_{a=1}^{N-1}\varphi_ae_a(x) \end{equation} where $\varphi^2 = \sum_{a=1}^{N-1}(\varphi_a)^2$. The vectors $e_a$ and $\mathbf{n}_0$ are orthogonal unit vectors.

He then introduces the gauge fields $A_\mu^{ab}$ and $B_\mu^a$ by \begin{eqnarray} \partial_\mu \mathbf{n}_0 &=& \sum_{a}^{}B_\mu^a e_a\\ \partial_\mu e_a &=& \sum_{b}^{} A_\mu^{ab} e_b - B_\mu^a\mathbf{n}_0 \end{eqnarray} where $a,b=1,2,\dots,N-1$ denote the transverse directions; $\mathbf{n}_0 \cdot e_a=0$ and $e_a \cdot e_b=0$.

Using this parametrization, the action of NLSM becomes \begin{equation} S= \frac{1}{2e_0^2}\int \left\{ \left( \partial_\mu \sqrt{1-{\varphi}^2} -B_\mu^a\varphi^a\right)^2 + \left( \partial_\mu \varphi^a- A_\mu^{ab}\varphi^b +B_\mu^a\sqrt{1-{\varphi}^2}\right)^2\right\}d^2x \end{equation}

The second order correction is given as \begin{equation} S^{(II)}= \frac{1}{2e_0^2}\int \left\{ \left( \partial_\mu \varphi^a -A_\mu^{ab}\varphi^b\right)^2 + B_\mu^a B_\mu^b\left(\varphi^a \varphi^b-{\varphi}^2 \delta^{ab}\right)\right\}d^2x+\frac{1}{2e_0^2}\int \left(B_\mu^a \right)^2 d^2x \end{equation} At this level, he clearly ignores terms like \begin{equation} B_\mu^a \partial_\mu \varphi^a\quad \text{and}\quad B_\mu^a A_\mu^{ab}\varphi^b. \end{equation} On which basis, these terms can be ignored? This is my first question. Secondly, how to perform integration over ${\varphi}$?

Moreover...

• How the terms in action $S^{(II)}$ change under the continuous rotation of transverse coordinate system?
• How can we prove that the Lagrangian can only depend on derivatives of gauge fields?
• What is importance of gauge fields?

Hints of these questions are given Assa Auerbach book---``Interacting Electrons and Quantum Magnetism '' [Chapter 13; section 13.3 Poor Man's renormalization]. But it is not clear to me. I would appreciate very much if some one help me in understanding the Mathematics and Physics related to my questions.

Answer 1

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 B_μ and A_μ 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 B_μ s, but only (N―1)(N―2)/2 A_μ 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 A_μ 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 B_μ s are not quite gauge fields in minimal coupling. What to do? I’ll be schematic. Think of the slow B_μ 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 B_μ 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 B_μ 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.)

Answer 2

One thing that I found is that for the background field $\mathbf{n}_0$ to be stationary this equation has to be satisfied: $(\delta^{ij}\partial_{\mu}+A^{ij}_{\mu})B^i_{\mu}=0$. You can show this by demanding the action be stationary at first order in the fields $\varphi$. Then you can integrate the action functional by parts to see that those two types of terms cancel each other as they appear with the same coefficient.