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\begin{document}
\numberwithin{equation}{subsection}
\lhead{Mop} 
\rhead{a random uni} 
{\centering\section{Connections}}
\subsection{Connections}

\begin{defn}{General Connections}{} 
A connection $\nabla$ on a smooth manifold $(M, \topo, \atlas)$ is a map that takes a pair consisting of a vector (field) $X$ and a $(p,q)$-tensor field $T$ and sends them to a $(p,q)$-tensor (field) $\nabla_X T$. Satisfying 
\begin{enumerate}[i]
  \item $\nabla_X f = Xf \hspace{1em} \forall f \in C^{\infty}(M)$;
  \item $\nabla_{X}(T+S) = \nabla_X T + \nabla_X S$;
  \item $\nabla_{X}(T \otimes S) = T \otimes \nabla_X S + \nabla_X T \otimes S$;
  \item $\nabla_{fX+Y} T = f  \nabla_X T + \nabla_Y T \hspace{1em} \forall f \in C^{\infty}(M)$.
\end{enumerate}\end{defn}

\begin{rem}{Terminology}{}
A manifold with connection is a quadruple of structures $(M, \topo, \atlas, \nabla)$. $\nabla_X$ is the extension of $X$, and $\nabla$ is a extension of $\dd$.
\end{rem}

\subsection{New Structure on $(M, \topo, \atlas)$ Required to Fix $\nabla$}

\begin{defn}{Connection Coefficient Functions}{}
Let $(U,x)$ be a chart of a smooth manifold with connection $(M, \topo, \atlas, \nabla)$. Then define the $\dim(M)^3$ connection coefficents as smooth maps 
\begin{align*}
    \tensor{\Gamma}{^\alpha_{\beta\gamma}}:\hspace{0.3em}& U \to \tensor{\Gamma}{^\alpha_{\beta\gamma}}(U) \subset \R^d,  \\
    &p \mapsto \tensor{\Gamma}{^\alpha_{\beta\gamma}}(p) \equiv \dd{x^\alpha} (\nabla_\gamma \partial_\beta) (p).
\end{align*}
    
\end{defn}

\begin{thm}{Action of $\nabla_X$ on a Vector Field}{}
Let $(U,x)$ be a chart, and $X,Y$ vector fields. In $(U,x)$, one may explicitly write the components of $\nabla_X Y$ as
\begin{equation} \label{eq:1}
    (\nabla_X Y)^\nu = X^\mu \partial_\mu Y^\nu + X^\mu Y^\lambda \tensor{\Gamma}{^\nu_{\lambda \mu}}.
\end{equation}
\begin{boof}
\proof Proof follows from the application of Definition 1.1. 
\begin{align*}
    \nabla_X Y &= \nabla_{X^\mu \partial_\mu} Y^\nu \partial_\nu \\
    &= X^\mu ((\nabla_\mu Y^\nu)\partial_v + Y^\nu (\nabla_\mu \partial_\nu) ) \\
    &= X^\mu \partial_\mu Y^\nu \partial_\nu + X^\mu Y^\nu \tensor{\Gamma}{^\lambda_{\nu \mu}} \partial_\lambda.
\end{align*}
On a renaming of indices, the above yields \ref{eq:1}. \qed
\end{boof}
\end{thm}

\begin{thm}{Action of $\nabla_X$ on a Covector Field}{}
We may write $\nabla_\mu \dd{x^\lambda} = \tensor{\Sigma}{^\lambda_{\nu \mu}} \dd{x^\nu}$. Then $\tensor{\Sigma}{^\lambda_{\nu \mu}}$ is not independent of $\tensor{\Gamma}{^\lambda_{\nu \mu}}$, such that generally, for a covector $\omega$,
\begin{equation}
    (\nabla_X \omega)_\mu = X^\lambda \omega_\mu \partial_\lambda - \Gamma^\nu_{\mu \lambda} \omega_\nu X^\lambda.
\end{equation}
\begin{boof}
\proof Proof follows from the previous definitions. Let $\dd{x^\lambda}$ act on $\partial_\nu$ so
\begin{align*}
    \nabla_\mu \dd{x^\lambda} \partial_\nu &= \nabla_\mu \tensor{\delta}{^\lambda_\nu} = \partial_\mu \tensor{\delta}{^\lambda_\nu} = 0. 
\end{align*}
Then by (iii) of Definition 1.1, 
\begin{align*}
    (\nabla_\mu \dd{x^\lambda}) \partial_\nu &= -  \dd{x^\lambda}(\nabla_\mu \partial_\nu) = -  \dd{x^\lambda} \tensor{\Gamma}{^\phi_{\nu \mu}} \partial_\phi =- \tensor{\Gamma}{^\lambda_{\nu \mu}}.
\end{align*}
Since $(\nabla_\mu \dd{x^\lambda})_\nu = - \tensor{\Gamma}{^\lambda_{\nu \mu}}$, the result follows. \qed
\end{boof}
\end{thm}

\begin{defn}{Divergence of a Vector Field}{}
Let $X$ be a vector field on a smooth manifold with connection $(M, \topo, \atlas, \nabla)$. Then the divergence of $X$ is the function
\begin{equation}
    \text{div}(X) = (\nabla_\lambda X)^\lambda.
\end{equation}
\end{defn}

\begin{thm}{Transformation of the Connection Coefficients under a Chart}{}
Let $(U,x)$ and $(V, \Bar{x})$ be charts of $(M, \topo, \atlas, \nabla)$, with $\bigcup \{U, V\} \neq \varnothing$. Then generally, the connection coefficients are not tensorial.
\begin{boof}
\proof Let $\tensor{\Bar{\Gamma}}{^\alpha_{\beta\gamma}}$ denote $\tensor{\Gamma}{^\alpha_{\beta\gamma}}$ in $(V, \Bar{x})$. Then, by virtue of the above definitions,
\begin{align*}
    \tensor{\Bar{\Gamma}}{^\alpha_{\beta\gamma}} &= \dd{\Bar{x}^\alpha}\left(\nabla_{\frac{\partial}{\partial \Bar{x}^\gamma}} \frac{\partial}{\partial \Bar{x}^\beta}\right) \\
    &= \dd{x^\lambda} \frac{\partial \Bar{x}^\alpha}{\partial x^\lambda} \left(\nabla_{\frac{\partial x^\mu}{\partial \Bar{x}^\gamma} \frac{\partial}{\partial x^\mu}} \frac{\partial x^\nu}{\partial \Bar{x}^\beta} \frac{\partial}{\partial x^\nu} \right) \\
    &= \dd{x^\lambda} \frac{\partial \Bar{x}^\alpha}{\partial x^\lambda} \left(\frac{\partial x^\mu}{\partial \Bar{x}^\gamma}\left(\nabla_{\frac{\partial}{\partial x^\mu}} \frac{\partial x^\nu}{\partial \Bar{x}^\beta}  \right) \frac{\partial}{\partial x^\nu} + \frac{\partial x^\mu}{\partial \Bar{x}^\gamma} \frac{\partial x^\nu}{\partial \Bar{x}^\beta} \left(\nabla_{\frac{\partial}{\partial x^\mu}} \frac{\partial}{\partial x^\nu} \right)  \right) \\
    &= \dd{x^\lambda} \frac{\partial \Bar{x}^\alpha}{\partial x^\lambda} \frac{\partial x^\mu}{\partial \Bar{x}^\gamma} \frac{\partial x^\nu}{\partial \Bar{x}^\beta} \frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^v} + \dd{x^\lambda} \frac{\partial \Bar{x}^\alpha}{\partial x^\lambda} \frac{\partial x^\mu}{\partial \Bar{x}^\gamma} \frac{\partial x^\nu}{\partial \Bar{x}^\beta} \tensor{\Gamma}{^\phi_{\nu \mu}} \frac{\partial}{\partial x^\phi} \\
    & = \frac{\partial \Bar{x}^\alpha}{\partial x^\lambda} \frac{\partial^2 x^\lambda}{\partial \Bar{x}^\beta \partial \Bar{x}^\gamma} + \frac{\partial \Bar{x}^\alpha}{\partial x^\lambda} \frac{\partial x^\mu}{\partial \Bar{x}^\gamma} \frac{\partial x^\nu}{\partial \Bar{x}^\beta} \tensor{\Gamma}{^\lambda_{\nu \mu}}.
\end{align*}
The connection coefficients are only tensorial when the symmetric term vanishes. \qed
\end{boof}
\end{thm}

\begin{rem}{'Span' of the Connection Coefficients}{}
On a chart domain $U$, choice of the $\dim(M)^3$ functions $\tensor{\Gamma}{^\alpha_{\beta\gamma}}$ suffices to fix the action of $\nabla$ on a vector field. Fortunately, the same functions fix the action of $\nabla$ of any tensor field.

\end{rem}

\begin{defn}{Torsion}{}
The torsion tensor is a $(1,2)$-tensor field, which transforms by
\begin{equation}
    \tensor{\Bar{T}}{^\alpha_{\beta\gamma}} = \frac{\partial \Bar{x}^\alpha}{\partial x^\lambda} \frac{\partial x^\mu}{\partial \Bar{x}^\gamma} \frac{\partial x^\nu}{\partial \Bar{x}^\beta} \tensor{\Gamma}{^\lambda_{\nu \mu}}.
\end{equation}
A connection $\nabla$ is called torsion-free if for each chart, $\tensor{{T}}{^\alpha_{\beta\gamma}}=0$.
\end{defn}

\begin{thm}{Normal Coordinates}{}
    Let $(M, \mathcal{O}, \mathcal{A}, \nabla)$ be a smooth manifold with connection $\nabla$, and $p \in M$. Then one can construct a chart $(U,x)$ with $p \in U$ such that
    \begin{equation}
        \tensor{\Gamma}{^\lambda_{(\mu\nu)}}(p) = 0.
    \end{equation}
    Such a $(U,x)$ is called a normal coordinate chart.

    \begin{boof}
    \proof Let $(V,\Bar{x})$  with $p \in V$. In general, 1.2.5 does not hold. Then consider $(U, x)$, to which one transits by virtue of
    \begin{equation*}
        (x \circ \Bar{x}^{-1})^\lambda (\alpha^1, \hdots, \alpha^d) \equiv \alpha^\lambda - \frac12 \tensor{\bar{\Gamma}}{^\lambda_{(\mu\nu)}}(p)  \alpha^\mu \alpha^\nu.
    \end{equation*}
    The commutativity of $\mathbb{R}$-multiplication automatically symmetrises the lower indices. Then,
    \begin{align*}
            \left( \frac{\partial x^\lambda}{\partial \Bar{x}^\phi} \right)_p &= \tensor{\delta}{^\lambda_\phi} - \frac12 \tensor{\Bar{\Gamma}}{^\lambda_{(\mu\nu)}}(p) (\tensor{\delta}{^\mu_\phi} \alpha^\nu + \alpha^\mu \tensor{\delta}{^\nu_\phi}) \\
            &= \tensor{\delta}{^\lambda_\phi} - \frac12 (\tensor{\Bar{\Gamma}}{^\lambda_{(\phi \nu)}}(p) \alpha^\nu + \tensor{\Bar{\Gamma}}{^i_{(\mu \phi)}}(p) \alpha^\mu) \\
            &= \tensor{\delta}{^\lambda_\phi} - \frac12 (\tensor{\Bar{\Gamma}}{^i_{(\phi\nu)}}(p) \alpha^\nu + \tensor{\Bar{\Gamma}}{^\lambda_{(\phi \mu)}}(p) \alpha^\mu) \\
            &= \tensor{\delta}{^\lambda_\phi} - \tensor{\Bar{\Gamma}}{^\lambda_{(\phi \mu)}}(p) \alpha^\mu.
    \end{align*}
    
    In a similar manner,
    \begin{equation*}
       \left( \frac{\partial^2 x^\lambda}{\partial \Bar{x}^\phi \partial \Bar{x}^\eta} \right)_p = - \tensor{\Bar{\Gamma}}{^\lambda_{(\phi\mu)}}(p)  \tensor{\delta}{^\mu_\eta} = - \tensor{\Bar{\Gamma}}{^\lambda_{(\phi \eta)}}(p).
    \end{equation*}


    Without loss of generality, fix $x(p) = (0, \hdots, 0)$. Then by the $\Gamma$ transformation rule,
    \begin{align*}
        \tensor{\Gamma}{^\lambda_{\mu\nu}}(p) &= \tensor{\delta}{^{\beta\gamma\lambda}_{\mu\nu\alpha}}\tensor{\Bar{\Gamma}}{^\alpha_{\beta\gamma}}(p) -  \tensor{\delta}{^\eta_\mu} \tensor{\Bar{\Gamma}}{^\lambda_{(\mu \eta)}}(p) \\
        &= \tensor{{\Bar{\Gamma}}}{^\lambda_{\mu\nu}}(p) - \tensor{{\Bar{\Gamma}}}{^\lambda_{(\mu\nu)}}(p)\\
        &= \tensor{\Bar{T}}{^\lambda_{\mu\nu}}.
    \end{align*} 
    The result follows. \qed
    \end{boof}
\end{thm}

\end{document}