### BF

While we’re on a cosmological constant kick, I should mention a recent paper by Stephon Alexander. He claims to have found a mechanism by which the cosmological constant can be relaxed to a small value.

Which sounds pretty important. Anyone who’s been in the field for any length of time has tried and failed to find such a mechanism.

It’s well known that four-dimensional gravity has a CP-violating topological term,

$S_\theta = \frac{\theta}{348\pi^2} \int \tr R\wedge R$
which is very analogous to the $\theta$-term in QCD. If you have a Dirac fermion, chiral rotations are anomalous in a curved space background. A massless fermion makes the $\theta$-angle physically unobservable. If the fermion has a *mass*, then the linear combination
$\overline{\theta}= \theta +\arg(m)$
is observable. In QCD, the apparent smallness of $\overline{\theta}$ is a problem, and the Peccei-Quinn mechanism was invented to solve it. $\overline{\theta}$ is replaced by a pseudoscalar field, the axion. Chiral symmetry breaking induces a potential for the axion which causes it to relax to zero.

Stephon points out that a very similar mechanism can be made to work in gravity, relaxing the gravitational $\theta$-angle.

*“Who cares?”* I hear you cry, *“Gravity is so weak, gravitationally-induced CP-violating effects are unmeasurably small. Besides, I thought you were going to tell us about the cosmological constant.”*

Stephon’s idea to connect the gravitational $\theta$-angle and the cosmological constant comes from a formulation of General Relativity due to Freidel and Starodubtsev.

Consider an oriented (and, for Stephon’s purposes, spin) four-manifold, $X$. Let $V\to X$ be a real, oriented rank-5 vector bundle with a *fixed* orthogonal structure. The structure group of $V$ is either $SO(5)$ or $SO(1,4)$, depending on whether we wish to think of $X$ as Euclidean or Minkowskian. As a topological restriction on $V$, we will demand that there exists an embedding of the tangent bundle, $TX\hookrightarrow V$.

The fields of our theory will consist of

- $B\in \Omega^2(\wedge^2 V)$
- $A$, an $SO(5)$ or $SO(1,4)$ connection on $V$. $F$ is its curvature.
- $\Phi\in\Gamma(V)$
- $\mu$ a volume-form on $X$.

The action is

$S= \int B^{A B}\wedge F_{A B} - \frac{\alpha}{4l} \epsilon_{A B C D E}B^{A B}\wedge B^{C D} \Phi^E - \frac{\beta}{2} B^{A B} \wedge B_{A B} +\mu\, (\Phi^A\Phi_A -l^2)$
$\alpha$ and $\beta$ are dimensionless constants and $l$ is a constant with units of length. The field $\mu$ acts as a Lagrange multiplier, forcing a nonzero expectation value for $\Phi$. We can use the $SO(5)$ gauge invariance to rotate $\Phi$ to point in some particular direction. The residual unbroken gauge group is $SO(4)$. Truth be told, Freidel and Starodubtsev *start* with this latter formulation in which $\Phi$ is just a constant $SO(5)$ vector. But, as we shall shortly see, it’s helpful to have a manifestly $SO(5)$-invariant formalism.

Once you’ve imposed the Lagrange multiplier constraint, $e \equiv D \Phi : TX \to V$ plays the role of the vierbein. Since $e^A\Phi_A=0$, we have an orthogonal decomposition (when $e$ is nondegenerate), $V= im(e)\oplus span(\Phi)$, and an orthogonal projector $P_{A B} = \delta_{A B} - \Phi_A \Phi_B /l^2$ Denoting the projected indices with lowercase latin letters, and setting $\omega^{a b} = A^{a b}$, $R^{a b}(\omega) = F^{a b}(A) + \frac{1}{l^2} e^a\wedge e^b$ will be identified as the Riemann curvature. Integrating out the $B$ auxiliary field, (2) becomes

$\array{ \arrayopts{\colalign{right left}} S =& -\frac{1}{2G}\int R^{a b}\wedge e^c \wedge e^d \epsilon_{a b c d} - \frac{\Lambda_0}{6} e^a\wedge e^b\wedge e^c\wedge e^d \epsilon_{a b c d}\\ & - \frac{2}{\gamma} R^{a b} \wedge e_a \wedge e_b - \frac{6(1-\gamma^2)}{\gamma\Lambda_0}(D e^a\wedge D e_a - R^{a b}\wedge e_a \wedge e_b) \\ & - \frac{3}{2\Lambda_0} R^{a b}\wedge R^{c d}\epsilon_{a b c d} + \frac{3\gamma}{\Lambda_0} R^{a b}\wedge R_{a b} }$ where $\Lambda_0 =\frac{3}{l^2},\quad G = \frac{\alpha^2-\beta^2}{\alpha}l^2,\quad \gamma =\frac{\beta}{\alpha}$

The first line, you recognize as the Einstein-Hilbert term (in Palatini form) and a cosmological constant term. The second line contains terms which vanish for the Levi-Cevita connection: the second term is the Nieh-Yan class. The third line contains the Euler density and the aforementioned gravitational $\theta$-term.

Comparing (1) and (3) you find $\frac{\theta}{348\pi^2} =\frac{3\gamma}{G\Lambda_0}$ So, says Stephon, relaxing $\theta$ is tantamount to relaxing the cosmological constant.

*“Bollocks!”* you say, *“There’s no symmetry relating those terms in the action. Why should their coefficients be related in this way?”*

And, indeed, you’re right. There are 6 terms in (3), whereas there were only 3 independent coupling constants in (2). Of course we got relations. But that’s just because Freidel and Starodubtsev failed to write down the most general action consistent with the symmetries.

Consider adding to the action,

$\array{ \arrayopts{\colalign{right left}} S' =&\int \frac{c_1}{2l^3} \epsilon_{A B C D E} \Phi^A D\Phi^B \wedge D\Phi^C \wedge F^{D E}\\ & + \frac{c_2}{2 l^5} \epsilon_{A B C D E} \Phi^A D\Phi^B\wedge D\Phi^C\wedge D\Phi^D\wedge D\Phi^E\\ & + \frac{c_3}{l^2} D\Phi^A\wedge D\Phi^B \wedge F_{A B} }$

These contribute terms of exactly the same form as existing first three terms in (3). With this addition, all six terms in (3) have independently variable coefficients. In particular, the first two terms in (4) contribute a shift in the cosmological constant, $\Lambda=\Lambda_0+\Lambda_1$. Relaxing the $\theta$ angle has, as far as I can tell, nothing to do with relaxing the cosmological constant.

It was a very clever paper. Everything Stephon did was OK, except for starting with the action of Freidel and Starodubtsev.

#### Update (3/31/2005):

In response to Urs, let me elaborate a bit on*why*Freidel and Starodubtsev recast GR in this fashion. In the $\alpha\to 0$ limit, the action $S$ becomes topological

^{1}. It is invariant under a much larger set of gauge transformations, namely $A\to A+ \beta\Psi,\quad B\to B +(D\Psi + \Psi\wedge\Psi)$ for arbitrary $\Psi\in \Omega^1(\wedge^2 V)$, which (for nonzero $\beta$) allows you to gauge away all the local degrees of freedom. They then hope to introduce nonzero $\alpha$ as a small perturbation. That’s essentially treating $1/G\Lambda$ as a small parameter, and doing perturbation theory in it. Never mind that, in the real world, this “small parameter” is $10^{120}$.

#### Update (4/2/2005):

Since many people may be bored and fail to read through all the comments below, I think I should bump this up “above the fold.” In response to Lee Smolin, I explained why I find Freidel and Starodubtsev’s work interesting.Despite the fact that F&S failed to write down the full list of terms in the action compatible with the SO(5) symmetry, the full list is nonetheless finite. All of the fields (after gauge-fixing SO(5) to SO(4)) have positive form-degree, and there are only a finite number of SO(4)-invariant 4-forms that one can write down using them.

Thus,

ifa sensible perturbation theory can be constructed for this theory (e.g., preserving gauge invariance of the quantum effective action), then this theory is guaranteed to be renormalizable, as there can only be a finite number of primitive divergences to cancel andonce you’ve remembered to write down all the terms consistent with the symmetriesthere are counterterms available to cancel all of them.That’s very, very interesting.

#### Update (4/3/2005):

I couldn’t help myself. If we’re gonna have a discussion about this, it’s best to have the relevant formulæ at hand. So (with a slight change in notation in (4)) here’s the effect of adding it to the action.- The coefficients of the topological terms in (3) are
*exactly*the same functions of $\alpha,\beta,l$ as they were before. The coefficient of the Nieh-Yan class is still $\frac{1}{\beta l^2}$, the coefficient of the Euler class is still $\frac{32\pi^2\alpha}{\alpha^2-\beta^2}$ and, most importantly for Stephon, $\frac{\theta}{348\pi^2}=\frac{\beta}{\alpha^2-\beta^2}$ as before. - The coefficients of the non-topological terms are, however, shifted.

$\array{ \arrayopts{\colalign{right left}} G =& l^2 \frac{\alpha^2-\beta^2}{\alpha} \frac{1}{1-c_1(\alpha^2-\beta^2)/\alpha}\\ \Lambda =& \frac{3}{l^2} \frac{1+2(c_2-c_1)(\alpha^2-\beta^2)/\alpha}{1-c_1(\alpha^2-\beta^2)/\alpha}\\ \gamma =& \frac{\beta}{\alpha} \frac{1-c_1(\alpha^2-\beta^2)/\alpha}{1+c_3\frac{\beta}{\alpha}(\alpha^2-\beta^2)/\alpha} }$ In particular, the previous relation between $\theta$ and $\gamma/G\Lambda$ no longer holds.

These words were for the Minkowskian ($SO(1,4)$) theory. In the Euclidean ($SO(5)$) theory, we should take $\beta\to i\beta$ and $c_3\to i c_3$ in the above formulæ (as I explain below).

^{1} Their *actual* calculations mostly take place at $\beta=0$, where, instead, $G\Lambda\to 0$ as $\alpha\to 0$. That theory has the same lousy UV behaviour as plain-old GR. They’d really like to work at *nonzero* $\beta$, which is what I’ve described.

## Re: BF

I had looked into Freidel & Starodubtsev’s paper a while ago after having seen people getting excited about the rumor which said that

Apparently Cumrum Vafa explictly said so.

But really what happened is that BF theory shows up in topological M-theory and that BF theory has always been a preferred toy example for some constructions in LQG.

In particular Smolin, Freidel and others are expressing hopes that by reformulating GR as BF plus additional terms it might be possible to quantize GR by ‘perturbing’ about the topological BF theory.

Of course that doesn’t make BF theory the ‘topologivcal sector of LQG’.