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Boldface notation $\boldsymbol R$ is reserved for objects that are best-viewed as elements of $\mathbb{R}^{n \times n}$, or functions on a manifold valued in $\mathbb{R}^{n \times n}$.
> [!definition] Definition. ([[Riemannian curvature]]) — the (1,3) tensor
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> Let $(M,g)$ be a [[Riemannian manifold|Riemannian]] [[smooth manifold|manifold]] of dimension $n$. The **(full) (1,3) Riemannian curvature ([[tensor of type (p,q)|tensor]][^9]) of $g$** is the minus[^2] [[covariant derivative on a vector bundle|of]] [[higher covariant derivative|the]] [[Levi-Civita connection|Levi-Civita]] [[connection on a manifold|connection]] [[curvature form|curvature]]: $\begin{align}
> R &\in \Omega^{2}_{M}(\text{End }TM) \\
> R &:= -D \circ D.
> \end{align}$
> Locally ($U \subset M$), $R$ is a [[differential form with values in a vector bundle|matrix of]] [[differential form|two-forms]], variously organized as $\begin{align}
> R |_{U}&=\boldsymbol R_{\ k \ell} \ dx^{k} \wedge dx^{\ell} \\
> &= [R^{i}_{j, k \ell}]_{j\in [n]}^{i \in [n]}\ dx^{k} \wedge dx^{\ell} \\ \\
> &= [R^{i}_{j, k \ell} \ dx^{k} \wedge dx^{\ell}]_{j\in [n]}^{i \in [n]}\ \\
> &= [R^{i}_{j}]_{j \in [n]}^{i \in [n]}
> \end{align}$where $\boldsymbol R_{ \ k \ell} \in C^{\infty}(U; \mathbb{R}^{n \times n})$, $\underbrace{ R^{i}_{j, k \ell} }_{( R_{\ k\ell})^{i}_{j} } \in C^{\infty}(U)$, $R^{i}_{j} \in \Omega^{2}(M)$.
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> Depending on whether $k,\ell$ are being indexed as $k,\ell \in [n]$ or $1 \leq k < \ell \leq n$, $R^{i}_{j, k \ell}$ might be scaled by $\frac{1}{2}$.[^1]
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> The **curvature endomorphism of $g$** is the[^8] $C^{\infty}$-[[bilinear map|bilinear map]] on [[vector field|vector fields]] $\begin{align}
> \Gamma(TM) \times \Gamma(TM) &\to \Gamma(\text{End }TM) \\
> X, Y & \mapsto R(X,Y) ,\\
> \text{where } R(X,Y)(p) &:= R_{p}(X_{p},Y_{p}).
> \end{align}$
> Here, [[differential form with values in a vector bundle|we have identified]] $\Omega^{2}_{M}(\text{End }TM)_{p} \cong A(T_{p}M, T_{p}M; \text{End }T_{p}M)$. Locally:
>
> $\begin{align}
> R\left( \frac{ \partial }{ \partial x^{k} }, \frac{ \partial }{ \partial x^{\ell} } \right) &= \boldsymbol R_{\ k \ell} = [R^{i}_{j, k \ell}]_{i=1,\dots,n}^{j=1,\dots,n}
> \\
> R\left( X^{k} \frac{ \partial }{ \partial x^{k} } , Y^{\ell} \frac{ \partial }{ \partial x^{\ell} } \right) &= X^{k}Y^{\ell} \boldsymbol R_{ \ k \ell}.
> \end{align}$
>
$R(X,Y)$ evaluates against a [[vector field]] $Z \in \Gamma(TM)$ as $R(X,Y)Z= \big( \ \ p \mapsto R(X,Y)(p) (Z_{p}) \ \ \big) \in \Gamma(TM).$
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> So far this is all just notations and conventions for applicable to any element of $\Omega^{2}_{M}(\text{End }TM)$. At last invoking the definition of the [[curvature form]] with $A=D$ the [[Levi-Civita connection]] (see below), one finds $R(X,Y)=D_{[X,Y]}-[D_{X}, D_{Y}],$
where $D_{X}, D_{Y}: \Gamma(TM) \to \Gamma(TM)$ denote the [[partial covariant derivative|partial covariant differential operators]] in the directions of $X$, $Y$ respectively and $[-,-]$ the [[Lie algebra|Lie bracket]] (of vector fields or of differential operators thereof).
[^8]: Exercise to check $C^{\infty}$-bilinearity.
[^9]: Identifying $\text{End}(TM) \cong T^{*}M \otimes TM$ and thus $\Omega^{2}_{B}(\operatorname{End}TM) \cong \Lambda^{2} T^{*}M \otimes T^{*}M \otimes TM \subset (T^{*}M)^{\otimes 3} \otimes TM,$we see that $R \in \Omega^{2}_{B}(\operatorname{End}TM)$ is indeed a $(1,3)$-[[tensor of type (p,q)|tensor]].
one finds in coordinates
$\boldsymbol R_{ \ k \ell}= \frac{ \partial \boldsymbol A_{\ell} }{ \partial x^{k} } - \frac{ \partial \boldsymbol A_{k} }{ \partial x^{\ell} } + \boldsymbol A_{k} \boldsymbol A_{\ell} - \boldsymbol A _{\ell} \boldsymbol A_{k} $
and
$R^{i}_{j, k \ell}= \frac{ \partial \Gamma ^i_{j \ell} }{ \partial x^{k} } - \frac{ \partial \Gamma^{i}_{jk} }{ \partial x^{\ell} } + \Gamma^{i}_{pk} \Gamma^{p}_{j\ell} - \Gamma^{i}_{p \ell} \Gamma^{p}_{jk} $
and the coordinate-free expression
> [!definition] Definition. ([[Riemannian curvature]]) — the (0,4) tensor
> Define a map on 4-tuples of [[vector field|vector fields]] thus: $(X,Y,Z,T) \mapsto g\big( R(X,Y)Z , T \big)$
> Locally, have $\begin{align}
> g\big( R(X^{k} \frac{ \partial }{ \partial x^{k} },& Y^{\ell} \frac{ \partial }{ \partial x^{\ell} } ) Z^{j}\frac{ \partial }{ \partial x^{j} } , T^{i} \frac{ \partial }{ \partial x^{i} } \big)\\&= X^{k} Y^{\ell} Z^{j} T^{i}\underbrace{ g(\boldsymbol R_{\ k \ell} \frac{ \partial }{ \partial x^{j} }, \frac{ \partial }{ \partial x^{i} } ) }_{ := R_{ij,k\ell} }.
> \end{align}$
> Observe that, in terms of the (1,3) tensor above, we have[^3]
> $R_{ij, k\ell}=g_{iq}R^{q}_{j, k \ell};$
> This is called the **(0,4) curvature tensor**.
>
By invoking the [[symmetries of the Riemann curvature tensor]], we see that the Riemann curvature tensor $(R_{ij,k \ell})_{p}$ defines, at any point $p \in M$, a [[symmetric multilinear map|symmetric]] [[bilinear map|bilinear form]] on $\Lambda^{2}T_{p}^{*}M$ as $(\omega_{ij} dx^{i} \wedge dx^{j}, \eta_{k \ell}dx^{k} \wedge dx^{\ell}) \mapsto R_{ij, k \ell}\omega_{ij} \eta_{k \ell}$.
> [!basicproperties]
> - [[symmetries of the Riemann curvature tensor]]
^properties
[^1]: Don't worry too much about this.
[^2]: This is a convention, and unfortunately our course appears to go back and forth on using it...
[^3]: Indeed, $\boldsymbol R_{\ k \ell} \frac{ \partial }{ \partial x^{j} }$ is the $j$th column $\boldsymbol R^{:}_{j, k \ell}$ of $\boldsymbol R_{k \ell}$; in the standard coordinate basis it is $R^{q}_{j, k \ell} \frac{ \partial }{ \partial x^{q} }$. Now, $\begin{align}
g\big(\overbrace{ \boldsymbol R_{\ k\ell} \frac{ \partial }{ \partial x^{j} } }^{ j\text{th col.} } , \frac{ \partial }{ \partial x^{i} } \big) &= g(R^{q}_{j, k \ell} \frac{ \partial }{ \partial x^{q} } , \frac{ \partial }{ \partial x^{i} } ) \\
&= R_{j, k \ell}^{q} g_{qi} \\
&= g_{iq} R^{q}_{j, k \ell}.
\end{align}$
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####
##### Scratchwork
~~operator can be looked at as a field of matrices of bilinear forms... so compute in terms of $R_{k \ell} \in \Gamma(\text{End }TM)$ 'basis curvature endomorphisms': $R_{k \ell}= R\left( \frac{ \partial }{ \partial x^{k} } , \frac{ \partial }{ \partial x^{\ell} }\right):U \subset M \to \text{End } TM$. ~~
~~$^{i}_{j}$th component of $R_{k \ell}$ corresponds to a single field of bilinear forms, ~~
~~$R\left( \frac{ \partial }{ \partial x^{k} }, \frac{ \partial }{ \partial x^{\ell} } \right)^{i}_{j}=$ ~~
Have a bilinear map $\Gamma(TM) \times \Gamma(TM) \to \Gamma(\text{End }TM)$ " **curvature endomorphism** "
$R(X,Y)(p) \in \text{End }T_{p}M$ is given by $R(X,Y)(p)=R(p)(X_{p}, Y_{p})$
If $Z \in \Gamma(TM)$ a vector field, $R(X,Y)Z=(p \mapsto R(X,Y)(p)(Z_{p}))$
or something (should be obvious)
can look at it like $R=[R^{i}_{j}]_{j \in [n]}^{i \in [n]} , \text{ where } R^{i}_{j}= \frac{1}{2} R^{i}_{j, k \ell} \ dx^{k} \wedge dx^{\ell}, \ \ R^{i}_{j} \in \Omega^{2}(M), \tag{$*$}$
for $R^{i}_{j, k \ell} \in C^{\infty}(U \subset M)$, or can "pull the $dx^{k} \wedge dx^{\ell}