---- > [!definition] Definition. ([[Levi-Civita connection]]) > > > Let $(M,g)$ be a [[Riemannian manifold]]. There is a unique [[connection on a manifold|connection]] $D$ on $M$ satisfying > 1. $D$ is [[orthogonal connection on a vector bundle|orthogonal]], in the sense that $Zg(X,Y)=g(D_{Z}X, Y)+g(X, D_{Z}Y)$ for all [[vector field|vector fields]] $X,Y,Z \in \mathscr{V}(M)$; > 2. $D$ is [[connection on a manifold|symmetric]]. > > > $D$ is called the **Levi-Civita connection of $g$ on $M$**. Its [[connection on a manifold|Christoffel symbols]] are given by $\Gamma^{i}_{jk}=\frac{1}{2}g^{iq}\left( \frac{ \partial g_{qj} }{ \partial x^{k} } + \frac{ \partial g_{kq} }{ \partial x^{j} } - \frac{ \partial g_{jk} }{ \partial x^{q} } \right),$ > where the coefficients $g^{iq}$ are those of the inverse to the [[nondegenerate bilinear form]] $g(-,-)$. Alternatively, the Levi-Civita connection may be characterized by the **Koszul formula**[^1] $\begin{align} > 2g(D_{X}Y, Z) &= Xg(Y,Z) + Yg(Z,X) -Zg(X,Y) \\ > &- \big( g(Y, [X,Z]) + g(Z, [Y,X]) - g(X, [Z,Y]) \big). > \end{align}$ - [ ] viewing orthogonality as being 'covaraint constant' after discussing how to extend LC connection to general tensor fields [^1]: A way to remember this formula is by listing out the cyclic permutations $(X,Y,Z), (Y,Z,X), (Z,X,Y)$. The first term applies the first vector field to the pairing of the latter two with signs $+ +-$. Then we subtract off the second term, which pairs the middle vector field with the Lie bracket of the outer two, again with signs $++-$. > [!justification] > **Uniqueness.** > > Summary: > - Apply orthogonality condition with $X=\frac{ \partial }{ \partial x^{i} }$, $Y=\frac{ \partial }{ \partial x^{j} }$, $Z=\frac{ \partial }{ \partial x^{k} }$. A tip is to first compute what $D \partial_{i}$ looks like, hence what $D_{k}\partial_{i}$ looks like, hence what $g(D_{k} \partial_{i}, \partial_{j})$ looks like. Then carefully transfer the computation to get also $g(\partial_{i}, D_{k} \partial_{j})$ > - Cycle $i,j,k$ to get three formulae. > - Add the first two and subtract the third formula. The symmetry condition enforces a bunch of cancellations happen. Then manipulate with $g ^{-1}$ to get the desired expression > > > > > > Suppose $D$ satisfies properties (1), (2). We show that $D$ must then take the form given above. Working locally, the section $\frac{ \partial }{ \partial x^{i} }$ of $TM$ identifies with constant vector $\boldsymbol e_{i} \in \mathbb{R}^{m}$. Hence $D \frac{ \partial }{ \partial x^{i} } = \cancel{ d \frac{ \partial }{ \partial x^{i} } }^{=0, \text{const.}} + (\Gamma^{p}_{jk} \delta_{j}^{i} \ dx^{k})_{p=1}^{m}=(\Gamma^{p}_{ik} \ dx^{k})_{p=1}^{m}.$ > > Now apply condition $(1)$ with $X=\frac{ \partial }{ \partial x^{i} }$, $Y=\frac{ \partial }{ \partial x^{j} }$, $Z=\frac{ \partial }{ \partial x^{k} }$. Note e.g. that $D_{Z}X= \big( ( \Gamma^{p}_{ik} dx^{k} )_{p=1}^{m}\big)\frac{ \partial }{ \partial x^{k} }=(\Gamma^{p}_{ik})_{p=1}^{m}$, so e.g. $g(D_{Z}X, Y)=g\left( \Gamma^{p}_{ik}\ \frac{ \partial }{ \partial x^{p} }, \frac{ \partial }{ \partial x^{j} } \right)=\Gamma^{p}_{ik}g_{pj}$. > > We get that $Zg(X,Y)=\begin{align} > \frac{ \partial }{ \partial x^{k} } g\left( \frac{ \partial }{ \partial x^{i} } ,\frac{ \partial }{ \partial x^{j} } \right) = \frac{ \partial }{ \partial x^{k} } g_{ij} > \end{align}$ > equals $g(D_{Z}X, Y)+g(X, D_{Z}Y)=\Gamma^{p}_{ik}g_{pj}+ \Gamma^{p}_{jk}g_{ip},$ > that is, condition (1), and then cycling $i,j,k$, gives three formulae $\frac{ \partial }{ \partial x^{k} } g_{ij}=\Gamma^{p}_{ik} g_{pj}+\Gamma^{p}_{jk}g_{ip}$$\begin{align} > \frac{ \partial }{ \partial x^{j} } g_{ki}&= \Gamma^{p}_{kj} g_{pi} + \Gamma^{p}_{ij} g_{kp} \\ > \frac{ \partial }{ \partial x^{i} } g_{jk} &= \Gamma^{p}_{ji} g_{pk} + \Gamma^{p}_{ki} g_{jp}. > \end{align}$ > Let $(g^{iq})$ denote the [[inverse matrix]] to $(g_{iq})$, so $\Gamma^{p}_{jk}g_{qp}g^{iq}= \Gamma^{p}_{jk} \delta^{i}_{p}=\Gamma^{i}_{jk}$. Now adding the first two equations and subtracting the third, obtain: $\begin{align} > \frac{ \partial }{ \partial x^{k} } &g_{ij} + \frac{ \partial }{ \partial x^{j} } g_{ki} - \frac{ \partial }{ \partial x^{i} }g_{jk}\\&=\cancel{ \Gamma^{p}_{ik} g_{pj} }+ \Gamma^{p}_{jk}g_{ip} + \Gamma^{p}_{kj} g_{pi} +\cancel{ \Gamma^{p}_{ij} g_{kp} - \Gamma^{p}_{ji} g_{pk} }- \cancel{ \Gamma^{p}_{ki} g_{jp} } \\ > &= 2 \Gamma^{p}_{jk} g_{ip} > \end{align}$ > where the cancellations and doubling are due to symmetry of the connection $D$ and symmetry of the [[bilinear map|bilinear form]] $g(-,-)$. Now halving and multiplying through by $(g^{iq})$ gives the claimed expression for $\Gamma^{i}_{jk}$. > > > > **Existence.** One way to show existence is to demonstrate the local formula above is invariant under a change of coordinates. This is annoying. Instead we will note that one can formulate the derivation of $\Gamma^{i}_{jk}$ above to arrive at the **Koszul formula** $\begin{align} > 2g(D_{X}Y, Z) &= Xg(Y,Z) + Yg(Z,X) -Zg(X,Y) \\ > &- \big( g(Y, [X,Z]) + g(Z, [Y,X]) - g(X, [Z,Y]) \big). > \end{align}$ > From there, one can directly check that the Koszul formula entails a [[covariant derivative on a vector bundle|covariant derivative]] that is [[connection on a manifold|symmetric]] + [[orthogonal connection on a vector bundle|orthogonal]]. (TODO, I guess) > > - [ ] also todo is the way which avoids coordinates altogether and rather just immediately deduced koszul formula from uniqueness condition ---- #### [[connection on a vector bundle]] [[covariant derivative on a vector bundle]] ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```