Towards Holography in the BV-BFV Setting

Abstract

We show how the BV-BFV formalism provides natural solutions to descent equations and discuss how it relates to the emergence of holographic counterparts of given gauge theories. Furthermore, by means of an AKSZ-type construction we reproduce the Chern–Simons to Wess–Zumino–Witten correspondence from infinitesimal local data and show an analogous correspondence for BF theory. We discuss how holographic correspondences relate to choices of polarisation relevant for quantisation, proposing a semi-classical interpretation of the quantum holographic principle.

Appendix A. Proofs of Section 2.3

Proof of Lemma 53

The first statement follows from a standard computation, of which we report only a few steps. Considering first the classical (i.e. degree-0) part, we have

$$\begin{aligned} S[A^g]&= \int _M \frac{1}{2}\langle A,dA \rangle + \frac{1}{6}\langle A, [A,A]\rangle -\frac{1}{12}\langle g^{-1}dg, [g^{-1}dg,g^{-1}dg] \rangle \\&\quad - \frac{1}{2}\langle g^{-1}Ag, dg^{-1}dg \rangle + \frac{1}{2}\langle g^{-1}dg, d(g^{-1}Ag) \rangle \\&\quad + \frac{1}{2}\langle g^{-1}Ag, dg^{-1}Ag - g^{-1}Adg\rangle + \frac{1}{2}\langle g^{-1}dg,[g^{-1}Ag,g^{-1}Ag]\rangle . \end{aligned}$$

In the first line, we find the classical CS action and the WZ functional. The terms in the second line combine into a total derivative and yield a boundary term

$$\begin{aligned} \frac{1}{2}\int _{\partial M} \langle A,dg g^{-1} \rangle . \end{aligned}$$

The last line vanishes due to the invariance of the inner product. Finally, turning to the extended BV action we recall that the covariant derivative of a graded field $\omega$ satisfies $d_A^g\omega ^g = (d_A\omega )^g$. It follows immediately from invariance of the inner product that the remaining terms in the extended action (56) are gauge invariant. The claim follows.

In the case of the polarised action, we first compute the effect of a gauge transformation on the polarising functional²⁸$f_\mathrm{min}^{1,0}$:

$$\begin{aligned} \int \limits _{\partial M} f_\mathrm{min}^{1,0}[{\mathcal {A}}^g] - f_\mathrm{min}^{1,0}[{\mathcal {A}}]&= \frac{1}{2} \int _{\partial M} \Big \{ \langle g^{-1}A^{1,0}g, g^{-1}A^{0,1}g\rangle + \langle g^{-1}\partial g, g^{-1}A^{0,1}g\rangle \\&\quad + \langle g^{-1}A^{1,0}g, g^{-1}{\bar{\partial }} g \rangle \\&\quad + \langle g^{-1} \partial g, g^{-1} {\bar{\partial }} g\rangle - \langle A^{1,0},A^{0,1}\rangle \Big \}\\&= \frac{1}{2}\int _{\partial M} \langle g^{-1}\partial g, g^{-1}A^{0,1}g\rangle + \langle g^{-1}A^{1,0}g, g^{-1}{\bar{\partial }} g \rangle \\&\quad + \langle g^{-1} \partial g, g^{-1}{\bar{\partial }} g\rangle . \end{aligned}$$

Then,

$$\begin{aligned} S^{1,0}[{\mathcal {A}}^g]-S^{1,0}[{\mathcal {A}}]= & {} S[A^g]- S[A] + \int \limits _{\partial M} f_\mathrm{min}^{1,0}[A^g] - f_\mathrm{min}^{1,0}[A] \\= & {} \int _{\partial M} \frac{1}{2}\langle g^{-1}Ag, g^{-1}dg \rangle - \int _M \frac{1}{12} \langle g^{-1}dg,[g^{-1}dg,g^{-1}dg]\rangle \\&+ \frac{1}{2}\int _{\partial M} \langle g^{-1}\partial g, g^{-1}A^{0,1}g\rangle + \langle g^{-1}A^{1,0}g, g^{-1}{\bar{\partial }} g \rangle \\&+ \langle g^{-1} \partial g, g^{-1}{\bar{\partial }} g\rangle \\= & {} \int _{\partial M} \langle g^{-1}A^{1,0}g, g^{-1}{\bar{\partial }} g \rangle + \frac{1}{2} \langle g^{-1} \partial g, g^{-1}{\bar{\partial }} g\rangle \\&-\int _M \frac{1}{12}\langle g^{-1}dg,[g^{-1}dg,g^{-1}dg]\rangle . \end{aligned}$$

$\square$

Proof of Lemma 54

This follows immediately from

$$\begin{aligned} S_\mathrm{gWZ}(h^{-1}g,A^h)= & {} S_\mathrm{CS}(A^g) - S_\mathrm{CS}(A^h) =\nonumber \\= & {} \Big (S_\mathrm{CS}(^g A)-S_\mathrm{CS}(A)\Big ) - \Big (S_\mathrm{CS}(^h A) +S_\mathrm{CS}(A)\Big ) \nonumber \\= & {} S_\mathrm{gWZ}(g,A) - S_\mathrm{gWZ}(h,A). \end{aligned}$$

(133)

$\square$

Proof of Lemma 55

Using the defining property of the path-ordered exponential, $\frac{d}{dt}\mathrm {Pexp}(\int _0^t \gamma _s ds)= \mathrm {Pexp}(\int _0^t \gamma _s ds) \gamma _t$, we have that $g^{-1}_t\dot{g}_t = \gamma _t$. Hence,

$$\begin{aligned} \frac{d}{dt} A^{g_t}= & {} \frac{d}{dt} (g^{-1}_tA\,g_t+ g^{-1}_tdg_t) = [g^{-1}_tA\,g_t, \gamma _t] - \gamma _t g^{-1}_tdg_t + g^{-1}_td\dot{g}_t \\= & {} [g^{-1}_tA\,g_t, \gamma _t] + [g^{-1}_tdg_t,\gamma _t] + d\gamma _t = d_{A^{g_t}}\gamma _t. \end{aligned}$$

The second claim follows from a simple direct calculation: denoting $\phi _t\equiv g_t^{-1}dg_t$

$$\begin{aligned} \dot{\phi }= & {} \frac{d}{dt}g_t^{-1} dg_t + g^{-1}_t d (\dot{g}_t) = -g_t^{-1}\dot{g}_t g_t^{-1} dg_t + g_t^{-1}d (\dot{g}_t) \\= & {} - \gamma _t g_t^{-1} dg_t + g_t^{-1} dg_t \gamma _t + d\gamma _t = d\gamma _t + [g_t^{-1} dg_t,\gamma _t]= d_{\phi _t}\gamma _t. \end{aligned}$$

$\square$

Proof of Lemma 56

The Wess–Zumino functional in Eq. (57) does not depend on the extension $\widetilde{g}$: choosing a different extension changes $S_{WZ}$ by a constant. In particular, this is irrelevant when taking a time derivative. Hence, let us choose an extension $\widetilde{g}_t:=\mathrm {Pexp}(\int _0^t \widetilde{\gamma }_s ds)$, with $\widetilde{\gamma }_t:M \rightarrow \mathfrak {g}$ an extension of $\gamma _t$, i.e. $\widetilde{\gamma }_t\vert _{\partial M}=\gamma _t$. For simplicity of notation, we drop the tildes in what follows. Let us denote again $\phi _t\equiv g_t^{-1}dg_t$. Because $\phi _t$ is the (pullback of the) Maurer–Cartan form on G, in addition to Lemma 55 we have that

$$\begin{aligned} d\phi _t = -\frac{1}{2}[\phi _t,\phi _t]. \end{aligned}$$

Then, we can directly compute

$$\begin{aligned} \frac{d}{dt}S_{WZ}[g_t]= & {} -\frac{d}{dt} \int _M \frac{1}{12}\langle \phi _t,[\phi _t,\phi _t]\rangle = \frac{d}{dt} \int _M \frac{1}{6}\langle \phi _t,d\phi _t\rangle \\= & {} \frac{1}{6}\int _M \langle \dot{\phi }_t,d\phi _t\rangle + \langle {\phi }_t,d\dot{\phi }_t\rangle \\= & {} \frac{1}{6}\int \limits _M \langle d\gamma _t, d\phi _t\rangle + \langle [\phi _t,\gamma _t],d\phi _t\rangle + \langle \phi _t,d(d\gamma _t + [\phi _t,\gamma _t])\rangle \\= & {} \frac{1}{6}\int \limits _M \langle d\gamma _t, d\phi _t\rangle - \langle [\phi _t,\phi _t],d\gamma _t\rangle = \frac{1}{2}\int \limits _M \langle d\gamma _t, d\phi _t\rangle \\= & {} -\frac{1}{2}\int \limits _M d \left[ \langle d\gamma _t, \phi _t\rangle \right] = \frac{1}{2}\int \limits _{\partial M} \langle \phi _t, d\gamma _t\rangle . \end{aligned}$$

$\square$

Proof of Proposition 57

Using Lemma 55, Lemma 56 and denoting again $\phi _t\equiv g_t^{-1}dg_t$, we compute

$$\begin{aligned} \frac{d}{dt} S_\mathrm{gWZ}= & {} \frac{1}{2}\int _{\partial M} \left\langle \frac{d}{dt}\left( g_t^{-1}A\,g_t\right) , \phi _t\right\rangle + \langle g_t^{-1}A\,g_t, \dot{\phi }_t\rangle - \frac{d}{dt}\int _M \frac{1}{12}\langle \phi _t,[\phi _t,\phi _t]\rangle \\= & {} \frac{1}{2} \int _{\partial M}\langle -\gamma _t\,g_t^{-1}A\,g_t, \phi _t\rangle + \langle g_t^{-1}A\,g_t\,\gamma _t, \phi _t\rangle + \langle g_t^{-1}A\,g_t,d\gamma _t\rangle \\&+ \langle g_t^{-1}A\,g_t,[\phi _t,\gamma _t]\rangle + \langle \phi _t, d\gamma _t \rangle \\= & {} \frac{1}{2} \int _{\partial M}\langle [g_t^{-1}A\,g_t,\gamma _t], \phi _t\rangle + \langle g_t^{-1}A\,g_t,[\phi _t,\gamma _t] \rangle \\&+ \langle \left( g_t^{-1}A\,g_t + \phi _t\right) , d\gamma _t\rangle = \frac{1}{2} \int _{\partial M} \langle A^{g_t}, d\gamma _t\rangle \end{aligned}$$

where we used $\langle g_t^{-1}A\,g_t,[\phi _t,\gamma _t]\rangle = - \langle [g_t^{-1}A\,g_t,\gamma _t],\phi _t\rangle$.

The details of the calculation for $S_\mathrm{gWZW}^{1,0}$ are identical, upon replacing $g^{-1}dg$ with $g^{-1}\bar{\partial }g$, the connection A with $A^{1,0}$, and expanding $d=\partial + \bar{\partial }$ in the right-hand side of formula (64). $\square$