nLab
examples for Lagrangian BV

The following lists some detailed examples for Lagrangian BV formalism.

BV integration on cylinder modulo rotation

(… warning, this is unpolished and unfinished at the moment, I am running out of time…)

To illustrate the BV integration method, choose an example which is trivial as an ordinary integration problem and artificially add a bit of redundancy to see how the BV machinery in turn gets rid of that.

So consider the real line by regarding it as the quotient of the action

$$a:(ℝ\times S^1)\times U(1)\to (ℝ\times S^1)$$a : (\mathbb{R}\times S^1) \times U(1) \to (\mathbb{R}\times S^1)

of $U(1)$ on the cylinder $X:=ℝ\times S^1$ by rotation around the main axis.

The corresponding action Lie groupoid is

$$(ℝ\times S^1)//U(1):=(X\times U(1)\stackrel{\stackrel{s:=p_1}{\to }}{\overrightarrow{t:=a}}).$$(\mathbb{R}\times S^1)//U(1) := (X \times U(1) \stackrel{\stackrel{s := p_1}{\to}}{\stackrel{\to}{t:= a}} ) \,.

Its Lie algebroid is given by the anchor map

$$𝔤:=\begin{array}{ccc}X\times ℝ& \hookrightarrow & Tx\\ & \searrow \swarrow \\ & X\end{array}$$\mathfrak{g} := \array{ X \times \mathbb{R} &\hookrightarrow& T x \\ & \searrow \swarrow \\ & X }

which includes the bundle of vectors that circle around the cylinder into the full tangent bundle of the cylinder.

The Lie algebra on the sections restricts over each point of $ℝ$ to the Lie algebra of vector fields on $S^1$.

The Chevalley-Eilenberg algebra of functions on this Lie algebroid is generated over $C^{\mathrm{\infty }}(X)$ from a single generator $\theta$ in degree 1, which we can think of as a multiple of the the canonical 1-form on $S^1$ pulled back along the canonical projection $p_2:ℝ\times S^1\to S^1$:

$$\mathrm{CE}(𝔤)=({\wedge }_{C^{\mathrm{\infty }}(X)⟨\theta ⟩}^{•},d_{𝔤})=(C^{\mathrm{\infty }}(X)\oplus C^{\mathrm{\infty }}(X)\otimes \theta ,d_{𝔤}).$$\mathrm{CE}(\mathfrak{g}) = ( \wedge^\bullet_{C^\infty(X) \langle \theta\rangle}, d_{\mathfrak{g}}) = ( C^\infty(X) \oplus C^\infty(X)\otimes \theta, d_{\mathfrak{g}}) \,.

If we write $v\in \Gamma (TX)$ for the canonical vector field running around the cylinder, i.e. the push-forward of the canonical vector field on $S^1$ along the action $a$, then the differential here is given by

$$d_{𝔤}f=v(f)\cdot \theta$$d_{\mathfrak{g}} f = v(f) \cdot \theta

for all $f\in C^{\mathrm{\infty }}(X)$, and

$$d_{𝔤}\theta =0.$$d_{\mathfrak{g}} \theta = 0 \,.

We say

• $f\in C^{\mathrm{\infty }}(X)$ are the fields

• $\theta$ is the ghost

• $d_{𝔤}$ is the BRST operator

• $\mathrm{CE}(𝔤)$ is the BRST complex

Next there is the corresponding Weil algebra, equivalently the algebra of differential forms on our Lie algebroid, equivalently the algebra of functions on the shifted tangent bundle on our Lie algebroid:

$$\mathrm{W}(𝔤):={\Omega }^{•}(𝔤):=C^{\mathrm{\infty }}(T[1]𝔤):=({\Omega }^{•}(X){\otimes }_{C^{\mathrm{\infty }}(X)}(⟨\theta ⟩\oplus ⟨d\theta ⟩),d_{\mathrm{W}(𝔤)}),$$\mathrm{W}(\mathfrak{g}) := \Omega^\bullet(\mathfrak{g}) := C^\infty(T[1]\mathfrak{g}) := ( \Omega^\bullet(X)\otimes_{C^\infty(X)} (\langle \theta\rangle \oplus \langle d\theta\rangle) , d_{\mathrm{W}(\mathfrak{g})} ) \,,

whose differential acts as

$$d_{\mathrm{W}(𝔤)}f=v(f)\theta +d_{\mathrm{dR}}f$$d_{\mathrm{W}(\mathfrak{g})} f = v(f)\theta + d_{\mathrm{dR}}f

$$d_{\mathrm{W}(𝔤)}df=0$$d_{\mathrm{W}(\mathfrak{g})} d f = 0

for all $f\in C^{\mathrm{\infty }}(X)$ and

$$d_{\mathrm{W}(𝔤)}\theta =d\theta$$d_{\mathrm{W}(\mathfrak{g})} \theta = d\theta

$$d_{\mathrm{W}(𝔤)}d\theta =0$$d_{\mathrm{W}(\mathfrak{g})} d\theta = 0

The ordinary canonical form on the cylinder, $\omega \in {\Omega }^2(X)$ corresponds here to the element

$$\omega =dx\wedge \theta$$\omega = d x\wedge \theta

in $\mathrm{W}(𝔤)$ (with $x$ the canonical coordinate function pulled back from $ℝ$). This, however, is not a closed form in $\mathrm{W}(𝔤)$, as we have

$$d_{𝔤}(\mathrm{dx}\wedge \theta )=-\mathrm{dx}\wedge d\theta \ne 0.$$d_{\mathfrak{g}} (dx \wedge \theta) = - dx \wedge d\theta \neq 0 \,.

To do BV integration we need to choose some closed form $\mu \in \mathrm{W}(𝔤)$. A given choice will allow us to integrate $\mu$ or any form obatained from it by contraction with a multivector field, over suitable sub-supermanifolds.

We want to recover from all this machinery the ordinary integral of a function

$$\exp (S_0)\in C^{\mathrm{\infty }}(ℝ)$$\exp(S_0) \in C^\infty(\mathbb{R})

on the quotient $ℝ$, but regarded now as a function on the cylinder, but independent – gauge invariant– of the canonical coordinate running around the cylinder.

This can be done by choosing as a reference 2-form the closed form

$$\mu :=\mathrm{dx}\wedge d\theta .$$\mu := dx \wedge d\theta \,.

We regard the exponentiated action, which is really just a function on the line, now as a function on the cylinder

$$\exp (S_0)\in C^{\mathrm{\infty }}(X)\subset \mathrm{CE}(𝔤).$$\exp(S_0) \in C^\infty(X) \subset \mathrm{CE}(\mathfrak{g}) \,.

The fact that this is really just a function on the line is remembered by its gauge invariance, namely the exponentiated action is invariant under rotation of the cylinder.

More precisely, let ${\theta }^{\#}:=\frac{\partial }{\partial \theta }$ be the multivector field coming from differentiation by the canonical coordinaty running along the circle. Then the gauge invaariance of the exponentiated action is expressed by the fact that

$${\theta }^{\#}\exp (S_0)=0$$\theta^\# \exp(S_0) = 0

(… to be continued …)

This is great! This is the kind of down-to-earth example which ‘joe the plumber’ mathematicians need to get on board. – Bruce

Urs Schreiber say: Good, that’s what I thought, that it might be useful. I ran out of time last Friday, and then over the weekend became a bit sick. Maybe you can try to continue this exercise, and I’ll check later how you are doing. Next task is to add the terms to the action that makes the Poisson bracket with it encode the BRST differential, and then add furthermore a “gauge fixing fermion”. Finally to find a Lagrangian submanifold and show that integration of the the master action over that yields the plain integral of $\exp (S_0)$ over the line…

I wondered why things were so quiet here! Hope you're better now, or if not then soon. —Toby