Let's take off from where we were last time...
$E$ is a manifold. Therefore, there exists a local chart with $p\in\mathcal{U}_\alpha$ such that the transition functions $\varphi_\beta\circ\varphi_\alpha^{-1}\circ\varphi^{-1}_\alpha$ is a differentiable function on $\varphi_\alpha(\mathcal{U}_\alpha)$. Consider $E$. For this section, we will disregard the proof shown in the last post, and assume that $E\subseteq\mathcal{E}$. (Note that taking the equality part of the $\subseteq$ gives back the result in the last post.) Let $V$ be a compact subset of $\mathcal{E}$ such that $E\subseteq V$. Then there exists a $\varrho\in C^\infty(\mathcal{E})$ which is identically $1$ on $E$, identically $0$ outside $E$. $\varrho$ is a metric on $E$ since $\forall x\in E,y\in\mathcal{E}\backslash E,\varrho(x,y)\geq0,\varrho(x,y)=\varrho(y,x),\mbox{ if and only if }x=y,\varrho(x,y)=0$. The triangle equality can be shown to be saturated. $\varrho$ can be the Ricci flow, therefore define the local Ricci curvature of $\mathcal{E}$. In fact, the Ricci flow states that $\mathcal{E}$ is locally Ricci flat.
Consider a derivation (in the general case written as $\mathcal{D}$) $X$ on $C^\infty(E)$. Let $X$ be such that $X\varrho=0$. $X$ is therefore the Killing vector field defined locally on $E$. On $E$, let us define an evolution parameter $s$ (physically interpreted as time). Then, upto a constant, $X=\partial/\partial s$. Let $\{X_i\}$ be the set of all Killing vector fields on $E$. Let us write $X_p:\varrho\to(X\varrho)(p)$. Then, $\{X_i\}\subset\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}$. Consider a mapping $\iota:E\to E$. Write $\mathcal{G}_\iota=\mathcal{G}\circ\iota^{-1}\forall\mathcal{G}\in C^\infty(E)$. Then, $\forall X$, $(\mathcal{G}X)_\iota=\mathcal{G}_\iota X_\iota, (X\mathcal{G})_\iota=0=X_\iota\mathcal{G}_\iota$. Let $\mathcal{G}=\varrho$. Then, $(\varrho X)_\iota=\varrho(\iota^{-1}(X(\iota^{-1})))$. $\iota$ is a homeomorphism, therefore, $\iota^{-1}\in C^\infty(E)$. Therefore, $X(\iota^{-1})=0$. Since $\iota$ is the inclusion map, $\iota^{-1}(0)=0$, and therefore, $(\varrho X)_\iota=\varrho(0)$. This shows us that using just the Killing vector field $X$, the metric at $s=0$ can be found. By means of differential equations and functional analysis, it is possible to determine $\varrho(s)$.
Let us define an affine connection on $E$ as $\Delta:\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}\to\bigcup_p\{X_p|X\in\{\mathcal{D}\}\},\Delta_{\mathcal{F}X+\mathcal{G}Y}=\mathcal{F}\Delta_X+\mathcal{G}
\Delta_Y,\Delta_{X}(\mathcal{F}Y)=\mathcal{F}\Delta_X(Y)$
for $X,Y\in \bigcup_p\{X_p|X\in\{\mathcal{D}\}\},\mathcal{F,G}\in C^\infty(E)$. Another affine connection can defined on $\mathcal{E}$ as $(\Delta^{-1}_X(Y))^\iota=\Delta_{X\circ\iota^{-1}}(Y\circ\iota^{-1})$. $\iota^{-1}$ is continuous and thus, $Y\circ\iota^{-1}=0$. Therefore, at a point $p\in E\exists x\in\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}$ such that $X(p)=0$. There also exists an open neighborhood $N$ of $0$ (defined on a coordinate system on $E$) in $E$ and another open neighborhood $M$ of $p\in E$ such that for some maximal geodesic $\gamma$ along $X$, $X\to\gamma_X$ is a diffeomorphism of $N$ onto $M$ such that $\gamma(0)=p,\dot{\gamma}(0)=Y\neq0$. Thus, $\exists$ neighborhood $\mathfrak{N}$ of $X\in\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}$ and interval $\mathfrak{I}$ about $0\in\mathbb{R}$ such that $(\mathfrak{n},t)\mapsto\gamma_\mathfrak{n}(t)$ is smooth on $\mathfrak{N}\times\mathfrak{I}$. Therefore, $M=\exp(N)$. If the evolutionary parameter $s$ is continuous, then we should be able to take $M$ arbitrarily close to $N$, so $M=N\cup\partial N$. Also, there exists $\mathcal{M}=M\cup\partial M,\mathcal{M}=\exp(\exp(N))$. Substituting this equation into $\mathcal{M}=M\cup\partial M$ gives $\exp(\exp(N))=\exp(N)\cup\partial\exp(N)$. Using the methods of boundary operators provides $N=\emptyset$. To resolve this, the evolutionary parameter $s$ must not be continuous. Therefore, if one considers a local tangent vector $\omega$ (Ricci flatness implies same direction) on $\mathcal{E}$, then since $s=||\omega||$, $\omega$ provides a direction of $s$ and is therefore quantized. This can therefore be generalized to conclude that \emph{all elements of $\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}$ are quantized.} This provides some method of quantizing the metric $g$ defined on $E$ in the following manner. $X$ is quantized. Therefore, $\exists$ a smooth vector field $Y$ such that $X=nY$. Since $X\varrho=0$, $nY\varrho=Y(n\varrho)=0$. Therefore, the Killing vector field can be made smooth for the price of quantizing the metric ($\varrho$ is a semi-classical metric and therefore satisfies the flat-space Einstein equations. Solving provides one with the solution: $\forall s,\varrho(s)=\varrho(0)$. This means that the ``graviton'' is constant through general relativistic ``time''.).
Let the affine connection be the Levi-Civita connection $\Delta$. Obviously, $X=\Delta$, therefore $\Delta\varrho=0$. By the definition of the Levi-Civita connection, $\Delta:\mathrm{Riem}(E)\to\mathrm{Conn}(E)$, where $\mathrm{Conn}(E)$ is the space of all torsionless connections on $E$. All $\varrho$ are in $C^\infty(E)$, so $X\varrho=\Delta\varrho=0$. Thus, $\mathrm{Conn}(E)=\{0\}$, so the only torsionless connection on $E$ is $\Delta=X=0$. The Ricci curvature $\mathrm{Ric}$ is therefore $0$. Let $A$ be a torsionless tensor field so that $\Delta^{'}=\Delta+A=0$. Therefore, $\mathrm{d}\mathrm{Ric}=0$. Let $\hat{\varrho}$ be the scalar product induced by $\varrho$ on $T^{*}E$. Then, $\int_E\hat{\varrho}\mathrm{d}\mathrm{Ric}g=0$. Let $\mathcal{S}=-\int_E\hat{\varrho}\mathrm{Ric}g$. Then, evaluating and using Hamilton's least action principle gives $\hat{\varrho}\circ\hat{\varrho}=-2$. Writing as an integral gives $\int_E\mathrm{d}\hat{\varrho}\circ\varrho=-1$. If $\hat{\varrho}$ was a Grassmann variable, then this integral becomes $\int_E\varrho\circ\mathrm{d}\hat{\varrho}=1$, showing that $\hat{\varrho}$ is indeed a Grassmann variable. This is plausible since $\varrho$ is a $C^\infty$-multilinear function on $E$.
$\varrho$ is the creation operator of a classical fermionic field, and its complex conjugate is the annihilation operator of the same classical fermionic field. Since $\varrho$ is a creation operator, it must have only positive eigenvalues, so $\varrho$ is the $++++$ metric. I.e., $E$ is a homogenous Riemannian manifold. Therefore, $E$ can be embedded in $\mathbb{R}^n$. Hence, it is locally homeomorphic to $\mathbb{C}^{n/2}$ via charts $\varphi_i:U_i \to V_i,U_i \subset E, V_i \subset \mathbb{C}^{n/2}$ open and such that $\varphi_j \circ \varphi_i^{-1}: V_i \cap V_j \to V_i \cap V_j$ is holomorphic. $E$ is therefore
homeomorphic via charts on $\mathbb{C}^{n/2}$ to $\underbrace{\mathcal{R}\times\mathcal{R}\times\mathcal{R}\times\ldots\times\mathcal{R}}_{n/2\mbox{-times}}$. This space is unique in the sense explained above. Therefore, the Tychonoff topology induced on $E$ via $\mathcal{R}$ is unique upto a homeomorphism. We ask - what is the genus of $\mathcal{R}$? The only Riemann surface that is unique for low genus is the torus (genus $1$). Therefore, $E=\mathrm{T}^n$. This is compact (as we wanted) and is also an Abelian Lie group. Therefore, there exists a finite extension $1\to E\to\pi_0(E)\to 1$. If $n=4$, as in conventional general relativity, then $E$ is hyperK\"{a}hler \footnote{An algebraic geometric intervention: A Hilbert scheme of $E$ has a natural morphism to the infinite symmetric product of $E$.} (and thus symplectic) and has its holonomy group contained in $SU(2)$. Since $E$ is hyperK\"{a}hler, it is the Pontryagin dual of a discrete torsion-free Abelian group. An $SU(2)$ holonomy would produce $\mathcal{N}=2$ supersymmetry on $E$. We will not look into supersymmetric spacetime in this post, but we have shown that supersymmetry must be an intrinsic property of spacetime.
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