# Insights Mathematical Quantum Field Theory - Reduced Phase Space - Comments

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1. Dec 18, 2017

### Urs Schreiber

2. Dec 20, 2017

### Greg Bernhardt

8 more to go, what a series!

3. Jan 29, 2018

### Duong

1. In defining the infinitesimal cotangent Lie algebroid, the underlying graded algebra is defined as
$$(T^\ast_{inf} \mathfrak{a})^\ast_\bullet \;:=\; \mathfrak{a}^\ast_\bullet \oplus Der(CE(\mathfrak{a}))_\bullet$$
I understand this is a degree-by-degree direct sum. What is in principle the grading on $Der(CE(\mathfrak{a}))_\bullet$, such that in the next example for the action Lie algebroid, $\frac{\partial}{\partial c^\alpha}$ and $\frac{\partial}{\partial \phi^a}$ have degree $-1$ and $0$ respectively?

2. I am not sure I got why $C^\infty(X)/ (\frac{\partial S}{\partial \phi^a}) \simeq C^\infty(X_{dS=0})$ when $X$ is a superpoint.
My understanding of the given explanation is the following. As $X$ is a superpoint, each generator $f$ in $C^\infty(X) := \mathbb{R} \oplus V$ where $V$ is a finite dimensional vector space that is a nilpotent ideal, and we could write $f = f_0 + f_1a_1 + ... + f_na_n$ where $f_i \in \mathbb{R}$ and $a_i$ are the basis vectors of $V$. Thus as long as $f$ vanishes where $\frac{\partial S}{\partial \psi^a}$ vanishes, we could generate $f$ from $\frac{\partial S}{\partial \psi^a}$ by scaling term by term. Such $f$ is zero in the quotient, and so the quotient is exactly the algebra of functions on $X_{dS=0}$. But
(a) Is $X$ being a superpoint a necessary condition? If $X$ is an usual manifold, we could also do the scaling point by point.
(b) We are not actually allowed to talk about "points" in a superpoint. Are we, again, secretly using a 1-1 correspondence between $\{ X\rightarrow \mathbb{R}^1 \}$ and $\{X_{even} \rightarrow \mathbb{R}^1 \}$ with $C^\infty(X_{even})$ generated from generators of $C^\infty(X)$ regarded in even degree?

4. Jan 29, 2018

### Urs Schreiber

The degree of a derivation $\partial$ on a graded algebra is the amount by which it raises (or lowers) the degree of the algebra elements that it acts on, hence the number $deg(\partial)$ such that for $a_1, a_2$ two elements of homogeous degree in the graded algebra, we have

$$\partial (a_1 a_2) = (\partial a_1) a_2 + (-1)^{deg(\partial) deg(a_1)}\, a_1 (\partial a_2)$$

So $\partial_\phi$ has degree 0 because is "removes" elements of degree zero (field coordinates), while $\partial_c$ has degree -1 because it "removes" elements of degree +1 (namely ghost field coordinates).

The issue is that we want to invoke a theorem about Koszul resolutions (this one) which is proven in algebraic geometry. This does not generally carry over to differential geometry, where one needs to impose and then check subtle regularity conditions for the theorem still to apply. But the synthetic differential super-geometry that we are using gives a partial unification of the worlds of algebraic and of differential geometry: Infinitesimal neighbourhoods in synthetic differential geometry follow the rules of algebraic geometry directly. So here to give a genuine proof (traditionally glossed over in the physics literature) that, after gauge fixing, the local BV-differential really does yield a resolution of the shell inside the jet bundle, we make use of the fact that we need this just for perturbative quantum field theory anyway, where the field histories really are just in the infinitesimal neighbourhood of a fixed on-shell field history (the one we are perturbing around). This allows us to invoke that theorem .

Last edited: Jan 29, 2018
5. Jan 29, 2018

### Duong

Wow I really should have seen this. Thanks!