----- Warning re: sign conventions is probably wise. > [!proposition] Proposition. ([[symmetries of the Riemann curvature tensor]]) > Let $(M,g)$ be a [[Riemannian manifold]] of dimension $n$. Consider the [[Riemannian curvature|Riemann curvature tensor]] $R$. It has the following symmetries in its indices $i,j,k,\ell$: > > 1. **(Intra-comma skew-swaps)** $R_{ij,k\ell}=-R_{ji, k \ell}=-R_{ij, \ell k}$ ; > 2. **(1st Bianchi identity)** $R^{i}_{j, k \ell}+R^{i}_{\ell, jk}+R^{i}_{k, \ell j}=0$ ; > 1. Equivalently: $R_{ij, k\ell}+R_{ik, \ell j}+R_{i \ell, j k}=0$; > 3. **(Inter-comma swaps)** $R_{ij, k \ell}=R_{k \ell, ij}$. > > > [!proposition] In coordinate-free language. > > In coordinate-free language, these are > > 1. **(Intra)** $R(X,Y,Z,T)=-R(Y,X,Z,T)=R(Y,X,T,Z)$ > 2. **(1st Bianchi identity)** $R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$ > 1. Applying $g(\cdot, W)$ gives $R(X,Y,Z, W)+R(Y,Z,X,W)+R(Z,X,Y,W)=0$ > 3. **(Inter)** $R(X,Y,Z,W)=R(Z,W,X,Y)$ > > Also note the obvious but useful symmetry $R(X,Y)=-R(Y,X)$.[^1] > [!proof]- Proof. ([[symmetries of the Riemann curvature tensor]]) > ~ *Bianchi.* Write out $\begin{align} \textcolor{Skyblue}{\nabla_{X} \nabla_{Y} Z} - \textcolor{Apricot}{\nabla_{Y} \nabla _{X} Z } - \nabla_{[X,Y]}Z &= R(X,Y)Z \\ \textcolor{Apricot}{\nabla_{Y} \nabla_{Z} X }-\textcolor{LimeGreen}{ \nabla_{Z}\nabla_{Y} X } - \nabla_{[Y,Z] } X &= R(Y,Z) X \\ \textcolor{LimeGreen}{\nabla_{Z} \nabla_{X} Y }- \textcolor{Skyblue}{\nabla_{X} \nabla_{Z} Y }- \nabla_{[Z,X] }Y &= R(Z,X)Y \end{align}$ now summing and using torsion-freeness of the [[Levi-Civita connection]] get that the sum of curvature terms equals $\begin{align} &\textcolor{Skyblue}{\nabla_{X}(\overbrace{ \nabla_{Y}Z- \nabla_{Z}Y }^{ [Y,Z] })} + \textcolor{Apricot}{\nabla_{Y}(\overbrace{ \nabla_{Z}X - \nabla_{X}Z }^{ [Z,X] })} + \textcolor{LimeGreen}{\nabla_{Z}(\overbrace{ \nabla_{X}Y - \nabla_{Y}X }^{ [X,Y] })} \\ & - \textcolor{Skyblue}{\nabla_{[Y,Z] }X} - \textcolor{Apricot}{\nabla_{[Z,X]}Y} - \textcolor{LimeGreen}{\nabla_{[X,Y] }Z} \end{align}$ which applying torsion-freeness *again* becomes $[X, [Y,Z]]+[Y,[Z,X]]+[Z, [X,Y]]$ and this equals $0$ by the [[Lie algebra|Jacobi identity]]. *Intra.* Enough to show $R(X,Y,Z,Z)=0$, i.e. that $g\big( R(X,Y)Z, Z \big)=g(\nabla_{X}\nabla_{Y} Z, Z)-g(\nabla_{Y}\nabla_{X}Z, Z) - g(\nabla_{[X,Y]}Z, Z)=0.$ The main trick is to iterate metric-compatibility: $g(\nabla_{X}\nabla_{Y}Z, Z)=Xg(\nabla_{Y}Z, Z)-g(\nabla_{Y}Z, \nabla_{X}Z)=\frac{1}{2}X Y g(Z,Z) - g(\nabla_{Y}Z, \nabla_{X}Z)$ and $g(\nabla_{Y}\nabla_{X}Z, Z)=\frac{1}{2} YX g(Z,Z)- g(\nabla_{Y}Z, \nabla_{X}Z).$ So $\begin{align} g(\nabla_{X}\nabla_{Y}Z, Z) - g(\nabla_{Y}\nabla_{X}Z, Z)= \frac{1}{2} [X,Y] g(Z,Z). \end{align}$ And then again by metric compatibility $g(\nabla_{[X,Y]}Z, Z)=\frac{1}{2}[X,Y]g(Z,Z)$. Thus $\begin{align} g\big( R(X,Y)Z, Z \big) &= \frac{1}{2}[X,Y]g(Z,Z) - g\big( \nabla_{[X,Y]}Z, Z \big) \\ &= 0. \end{align}$ For *(inter)* we cycle the four arguments within the *(1st Bianchi identity)* to obtain $\begin{align} \textcolor{Apricot}{\cancel{ R(X,Y,Z,W) }} + \textcolor{LimeGreen}{\cancel{ R(Y,Z,X,W) }} + \textcolor{Skyblue}{R(Z,X,Y,W)}& \\ +\textcolor{LimeGreen}{\cancel{ R(Y,Z,W,X) }} + \textcolor{blue}{\cancel{ R(Z,W,Y,X) }} + \textcolor{Thistle}{R(W,Y,Z,X)} & \\ +\textcolor{blue}{\cancel{ R(Z,W,X,Y) }} + \textcolor{green}{\cancel{ R(W,X,Z,Y) }} +\textcolor{Skyblue}{R(X,Z,W,Y)} & \\+\textcolor{green}{\cancel{ R(W,X,Y,Z) }} + \textcolor{Apricot}{\cancel{ R(X,Y,W,Z) }} + \textcolor{Thistle}{R(Y,W,X,Z)} &=0, \end{align}$ where the color-coded cancellations and aggregations come from repeated applications of *(intra)*. This yields $2R(Z,X,Y,W)+2R(Y,W,X,Z)=0,$ and therefore $R(Z,X,Y,W)=-R(Y,W,X,Z).$ Applying *(intra)* once more to the latter two arguments, this becomes $R(Z,X,Y,W)=R(Y,W,Z,X)$ as desired. ----- #### [^1]: Indeed, have $\begin{align} R(X,Y)Z&= \nabla_{[X, Y]}Z -\nabla_{X}\nabla_{Y}Z + \nabla_{Y}\nabla_{X}Z \\ &= -\nabla_{[Y,X]} Z + \nabla_{Y}\nabla_{X}Z - \nabla_{X}\nabla_{Y} Z \\ &= - (\nabla_{[Y,X]}Z - \nabla_{Y}\nabla_{X}Z +\nabla_{X}\nabla_{Y}Z) \\ &= -R(Y,X)Z \end{align}.$ ---- #### 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 > ```