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A Finite Orbit with Two Gauge-Condition Intersections

Physical scope

A finite-dimensional regulated gauge-orbit example only; it is not a non-Abelian Yang–Mills field theory or a continuum functional measure.

Example

All coordinates $x, y$ , radii $r$ , and angles $θ$ in this finite-dimensional toy are dimensionless; the stated coordinate densities and $Δ_{F P}$ therefore carry no physical units.

Let $X = R^{2} ∖ {(0, 0)}$ have positive density $∣ d x d y ∣$ , and let $K = S^{1}$ have angle coordinate $θ$ and right-invariant density $∣ d θ ∣$ . Let $K$ act on the right by $(x, y) \cdot e^{i θ} = (x \cos θ - y \sin θ, x \sin θ + y \cos θ) .$ For $r > 0$ , set $s_{r} = (r, 0)$ , choose an open neighborhood $U_{r} \subset {x > 0}$ of $s_{r}$ , let $S_{r} = U_{r} \cap {y = 0}$ have density $∣ d x ∣$ , and set $χ (x, y) = y$ on $U_{r}$ . The unit angular generator $1 \in k = T_{e} K ≅ R$ has infinitesimal orbit vector $R_{s_{r}} (1) : = {\frac{d}{d θ} ∣}_{0} (s_{r} \cdot e^{i θ}) = (0, r) .$ Write $O_{r} : = {s_{r} \cdot k : k \in K} \subset X$ for the right orbit of $s_{r}$ , and write $S_{g l o b a l} : = {(x, y) \in X : y = 0}$ for the global zero set of the coordinate function $y$ . The Faddeev–Popov operator is the linear map $M_{s_{r}} : = d χ_{s_{r}} \circ R_{s_{r}} : k \to R$ , and its density Jacobian relative to $∣ d θ ∣$ and $∣ d y ∣$ is $Δ_{F P} (s_{r}) = ∣ \det M_{s_{r}} ∣ = r > 0$ . The full orbit $O_{r}$ nevertheless meets $S_{g l o b a l}$ at both $(r, 0)$ and $(- r, 0)$ . For each $s^{'} = (x, 0) \in S_{r}$ , the same calculation gives $M_{s^{'}} (1) = x > 0$ , so the local slice is transverse throughout $S_{r}$ . Thus this local transversality does not make $S_{g l o b a l}$ intersect the full orbit at only one point. The regulator group $K$ in this finite-dimensional toy is not identified with the model's non-Abelian structure group $G$ .

Example references

Finite-dimensional regulated gauge-slice and Jacobian definition

Facts & Assumptions

Given: The finite-dimensional right action, local gauge condition, and regulated densities in the example statement.

[F1]

A finite-dimensional regulated gauge-slice datum has infinitesimal orbit map $R_{s} = d α_{s} ∣_{e}$ , Faddeev–Popov operator $M_{s} = d χ_{s} \circ R_{s}$ , and Jacobian $Δ_{F P} = ∣ \det M_{s} ∣$ when domain and codomain have the specified positive densities. Finite-dimensional gauge slices and Faddeev–Popov operators

Verification

Given: [F1], $r > 0$ , and the rotation action on $X$ .

1.1givenF1algebra

Differentiating $(r, 0) \cdot e^{i θ} = (r \cos θ, r \sin θ)$ at $θ = 0$ gives $R_{s_{r}} (1) = (0, r)$ . Since $χ (x, y) = y$ , its differential sends $(u, v)$ to $v$ , so $M_{s_{r}} (1) = r$ and $Δ_{F P} (s_{r}) = ∣ r ∣ > 0$ . More generally, for $s^{'} = (x, 0) \in S_{r}$ , differentiating $(x, 0) \cdot e^{i θ} = (x \cos θ, x \sin θ)$ gives $R_{s^{'}} (1) = (0, x)$ and $M_{s^{'}} (1) = x > 0$ because $U_{r} \subset {x > 0}$ . Thus every point of $S_{r}$ is transverse.

2.1step 1.1algebra∎

For $k = e^{i θ}$ , the second coordinate of $s_{r} \cdot k$ is $r \sin θ$ . It vanishes exactly at $θ = 0$ and $θ = π$ modulo $2 π$ , which give the distinct points $(r, 0)$ and $(- r, 0)$ . Hence $O_{r} \cap S_{g l o b a l} = {(r, 0), (- r, 0)}$ , even though every point of the local slice $S_{r}$ is transverse by step 1.1.

Remarks

The sketch rescales the radius to $1.1$ drawing units; it represents the orbit circle $x^{2} + y^{2} = r^{2}$ and the horizontal axis $S_{g l o b a l}$ . The dashed short segment schematically highlights $S_{r} = U_{r} \cap {y = 0}$ near $s_{r}$ ; it is not intended to fix the unspecified size or shape of $U_{r}$ . The marked positive crossing is the local slice point; the opposite crossing is the other point of $O_{r} \cap S_{g l o b a l}$ . The proof uses the displayed action and derivative, not the drawing.

Depends on

Finite-dimensional gauge slices and Faddeev–Popov operators