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A smooth nondegenerate component of the fixed-point set of an isometry is a totally geodesic submanifold. If a geodesic starts tangent to the fixed component, its image under the isometry has identical position and tangent. Uniqueness of the geodesic equation makes the two geodesics coincide throughout their common domain, so the original geodesic stays in the fixed set.

An anti-symplectic involution is an anti-symplectic map whose square is the identity. Its nonempty fixed-point set is a Lagrangian submanifold; the linear version has complementary $+1$ and $-1$ Lagrangian subspaces.

A nonempty fixed-point set of an anti-symplectic involution is a Lagrangian submanifold. Local linearization identifies its tangent space with the $+1$ eigenspace of the derivative.

Consider the discrete mapIt is an isometry: $\cos \theta$ and $d\varphi$ both change sign, leaving $\cos \theta d\varphi$ unchanged, while $d{\theta }^2$ and ${\sin }^2\theta d{\varphi }^2$ are also unchanged. A local component of its fixed-point set is $\theta =\pi /2$, $\varphi ={\varphi }_0$.

If a geodesic starts tangent to that component, the isometry fixes both its initial position and its initial tangent. Applying the isometry therefore gives a geodesic with the same initial data. Uniqueness of the geodesic equation makes it the same curve, so it remains in the fixed component. This proves that the selected submanifold is a totally geodesic submanifold; it has dimension three, not two. The proof continues through a horizon when expressed in regular coordinates, so coordinate singularities of the original chart do not invalidate it.

The geodesic conserved quantities from Killing vectors associated with ${\partial }_{\psi }$ and ${\partial }_{\varphi }$ areDots denote derivatives with respect to an affine parameter. On the chosen submanifold $\dot{\varphi }=0$ and $\cos \theta =0$, so $L_{\varphi }=0$. Requiring also $L_{\psi }=0$ givesMore generally, away from the polar-coordinate axes, both displayed angular charges being zero imply $\dot{\varphi }=0$ and the same relation. In particular, zero conserved angular momentum does not imply constant $\psi$: the motion follows the local dragging of the angular coordinate by the rotating geometry.

For $p\in \mathrm{Fix}(\sigma )$, differentiation gives $(d{\sigma }_p{)}^2=I$ and $(d{\sigma }_p{)}^{\ast }{\omega }_p=-{\omega }_p$. Thus $d{\sigma }_p$ is an anti-symplectic involution of the symplectic vector space $T_{p}M$. By part (a), its $+1$ eigenspace is a Lagrangian subspace.

In the local manifold chart supplied in the question, $\sigma$ is linear, so its fixed-point set is the intersection of the chart with this $+1$ eigenspace. Therefore the fixed-point set is an embedded submanifold andEvery such tangent space has half the ambient dimension and the symplectic form vanishes on it. Hence every nonempty fixed-point set is a Lagrangian submanifold. The fixed-point set itself need not be connected.

Let $\sigma (z)=\overline{z}$, componentwise complex conjugation. Interpret the printed $|z^2|$ in the potential as $|z{|}^2={\sum }_j|z_j{|}^2$. With $D=1+|z{|}^2$, the local Fubini-Study form isIts pullback of a differential form under $\sigma$ iswhere the last equality swaps $j,k$ and uses antisymmetry of the wedge product of differential forms. Since ${\sigma }^2=I$, it is an anti-symplectic involution, with fixed-point set ${\mathbb{R}}^n$. Part (c) gives

As usual for Complex projective space and Real projective space, homogeneous coordinates are represented by nonzero vectors. Componentwise complex conjugation descends to the diffeomorphismIt is well defined because $\overline{\lambda z}=\overline{\lambda }\overline{z}$, and ${\tau }^2=I$. Every standard manifold chart is invariant under $\tau$, with ${\phi }_i\circ \tau =\sigma \circ {\phi }_i$. Part (d) and the gluing of the Fubini-Study form therefore imply ${\tau }^{\ast }{\omega }_{FS}=-{\omega }_{FS}$.

A point $[z]$ in the fixed-point set satisfies $\overline{z}=\lambda z$. Applying complex conjugation again gives $|\lambda |=1$. Choose $a\ne 0$ with $a/\overline{a}=\lambda$; then $\overline{az}=az$, so $[z]$ has a real representative. Conversely every real representative determines a point in the fixed-point set. Hence $\mathrm{Fix}(\tau )={\mathbb{RP}}^n$, and part (c) gives

Use the paper's Fubini-Study form normalization ${\tilde{\omega }}_{FS}=\frac{i}{2}\partial \overline{\partial }\log (1+|z{|}^2)$, the Hamiltonian vector field convention ${\iota }_{X_H}\omega =dH$, and the Lie algebra generator whose circle action is $e^{i\theta }$. Write $D=|z_0{|}^2+|z_1{|}^2+|z_2{|}^2$. The normalized moment map for the torus action isThe moment map image is the filled triangleIndeed, the three quantities $|z_j{|}^2/D$ are nonnegative and add to one, and any such triple is realized by choosing their square roots as homogeneous coordinates.

For the diagonal circle action, a point in the fixed-point set satisfies $[z_0,t{z}_1,t{z}_2]=[z_0,z_1,z_2]$ for every $t\in S^1$. If $z_0\ne 0$, the projective scaling must be one, forcing $z_1=z_2=0$. If $z_0=0$, all coordinates are scaled together. ThereforeBy part (b), the diagonal moment map isThe factor $1/2$ follows from the specified Fubini-Study form, not the integral normalization in the linked general article. For example, in one affine complex coordinate $w=r{e}^{i\theta }$ the symplectic form is $r(1+r^2{)}^{-2}dr\wedge d\theta$, whose interior product of a differential form with ${\partial }_{\theta }$ is $-r(1+r^2{)}^{-2}dr=d[-r^2/(2(1+r^2))]$. Choosing the generator $e^{2\pi it}$ rescales all moment maps by $2\pi$; reversing the defining sign reverses their signs. The negative level in parts (d) and (e) uses the convention displayed here.

On $N={\mu }^{-1}(-1/4)$, the diagonal fundamental vector field $X$ is nonzero, since the fixed-point set lies at moment map values $0$ and $-1/2$. The nondegenerate bilinear form ${\omega }_p$ and $d\mu ={\iota }_X\omega$ imply $d\mu \ne 0$, so $-1/4$ is a regular value. By the regular level set theorem, $N$ is a three-dimensional embedded submanifold of the four-dimensional Complex projective plane.

At $p\in N$, set $W=T_{p}N=\ker d{\mu }_p$. Then $\omega (X,v)=0$ for every $v\in W$, so $X\in W^{\omega }$, the symplectic orthogonal complement. This complement is one-dimensional and $d\mu (X)=\omega (X,X)=0$, givingThus the level set is a coisotropic submanifold, neither an isotropic submanifold nor a Lagrangian submanifold. Indeed an isotropic subspace of a symplectic vector space in dimension four has dimension at most two, whereas $\dim N=3$. Its characteristic line field is generated by the diagonal circle action.

The equivariant supertrace is independent of $\beta$ because positive-energy bosonic and fermionic states pair under the supercharge. Take the short-time limit $\beta \to 0$. A finite-action path becomes constant, but the twisted boundary condition then requires $x=f(x)$; its constant saddles are exactly the fixed-point set $M^f$. Supersymmetry cancels the nonzero bosonic and fermionic fluctuations away from their zero modes. Consequently the path integral localizes to a tubular neighborhood of $M^f$, with the remaining Gaussian determinants giving the local fixed-point contribution to the equivariant index theorem.