[proofplan] Work in [geodesic normal coordinates](/theorems/???) at an arbitrary point $p$ so that $g_{ij}(p) = \delta_{ij}$ and all Christoffel symbols vanish at $p$. Apply the second Bianchi identity for the Riemann tensor, contract two indices to produce a divergence-type identity for the Ricci tensor, and substitute the Einstein condition $R_{ij} = \lambda \delta_{ij}$. The result is $(2 - n)\,\partial_m \lambda = 0$ at $p$. For $n \ne 2$ this forces $\partial_m \lambda(p) = 0$ for every $m$; since $p$ is arbitrary, $d\lambda \equiv 0$ on $M$, and connectedness gives $\lambda$ globally constant. [/proofplan]

[step:Fix geodesic normal coordinates and record their properties] Fix $p \in M$ and let $n := \dim_{\mathbb R} M$. By the [existence of geodesic normal coordinates](/theorems/???), there is a chart $(U, \varphi)$ around $p$ with coordinates $(x_1, \dots, x_n)$ such that \begin{align*} g_{ij}(p) = \delta_{ij}, \qquad \Gamma^k_{ij}(p) = 0, \qquad \partial_k g_{ij}(p) = 0. \end{align*} In these coordinates, covariant derivatives at $p$ reduce to partial derivatives, and raising/lowering indices with $g$ at $p$ is trivial. We use the comma notation $T_{\ldots, m} := \nabla_m T_{\ldots}$ for covariant derivatives throughout, and work at the single point $p$. [/step]

[step:Apply the second Bianchi identity and contract]The [second Bianchi identity](/theorems/???) for the [Riemann curvature tensor](/page/Riemann%20Curvature%20Tensor) reads \begin{align*} R^i{}_{jkl;m} + R^i{}_{jmk;l} + R^i{}_{jlm;k} = 0. \end{align*} Contract with $k$ and $j$ equal (i.e.\ set $k = j$ and sum over $j$): \begin{align*} \sum_{j=1}^n \Bigl(R^i{}_{jjl;m} + R^i{}_{jmj;l} + R^i{}_{jlm;j}\Bigr) = 0. \end{align*} At $p$, since $g_{ij}(p) = \delta_{ij}$, we identify $R^i{}_{jkl}(p) = R_{ijkl}(p)$. The [Ricci tensor](/page/Ricci%20Tensor) is defined as the contraction \begin{align*} R_{il} := \sum_{j=1}^n R^j{}_{ijl} = \sum_{j=1}^n g^{jk} R_{kijl}, \end{align*} which at $p$ (where $g^{jk} = \delta^{jk}$) simplifies to $R_{il}(p) = \sum_j R_{jijl}(p)$. Using the Riemann tensor symmetries $R_{ijkl} = -R_{jikl} = -R_{ijlk} = R_{klij}$: - $\sum_j R^i{}_{jjl;m}(p) = \sum_j R_{ijjl;m}(p) = -\sum_j R_{jijl;m}(p) = -R_{il;m}(p)$. - $\sum_j R^i{}_{jmj;l}(p) = \sum_j R_{ijmj;l}(p) = \sum_j R_{jmij;l}(p) = R_{im;l}(p)$ (using $R_{ijmj} = -R_{ijjm} = R_{jijm} = R_{jmji} \cdot \text{sign} \dots$; equivalently $\sum_j R_{ijmj} = \sum_j R^j{}_{jim}$... let us simplify directly). By the symmetry $R_{ijmj} = R_{mjij}$ (pair swap) and re-indexing, $\sum_j R_{ijmj} = \sum_j R_{mjij} = R_{mi}$, and by symmetry of Ricci $R_{mi} = R_{im}$. So $\sum_j R^i{}_{jmj;l}(p) = R_{im;l}(p)$. - $\sum_j R^i{}_{jlm;j}(p) = \sum_j R_{ijlm;j}(p)$, which we denote $\sum_j R_{ijlm;j}(p)$. Substituting into the contracted Bianchi identity, \begin{align*} -R_{il;m}(p) + R_{im;l}(p) + \sum_{j=1}^n R_{ijlm;j}(p) = 0. \tag{$\ast$} \end{align*}[/step]

[guided]The Bianchi identity expresses a deep integrability property of the Riemann tensor: the cyclic sum of covariant derivatives of $R$ over the three "trailing" indices vanishes. Contracting with the metric produces contracted Bianchi identities for the Ricci tensor and the scalar curvature — in particular the identity known as the *twice-contracted Bianchi identity*, which is the source of the divergence-free property of the Einstein tensor. Here we contract once (on the pair $(k, j)$) to get a relation involving the Ricci tensor and a single remaining Riemann-tensor term $\sum_j R_{ijlm;j}$. The algebra is bookkeeping with the Riemann symmetries: whenever we contract an index pair against a covariantly-constant $\delta^{ij}$ (which at $p$ is what $g^{ij}$ is), we can rename indices and use the $(12)$-antisymmetry $R_{ijkl} = -R_{jikl}$ and the pair-swap $R_{ijkl} = R_{klij}$. The payoff is the identity $(\ast)$: the skew-symmetric "Ricci gradient" combination $-R_{il;m} + R_{im;l}$ equals the divergence-like sum $-\sum_j R_{ijlm;j}$. This pattern — divergence of $R$ controls antisymmetric derivatives of $\mathrm{Ric}$ — is the engine of Schur-type rigidity results.[/guided]

[step:Substitute the Einstein condition and contract again]Since $(M, g)$ is Einstein, $\mathrm{Ric} = \lambda g$, so at $p$, \begin{align*} R_{ij}(p) = \lambda(p)\, \delta_{ij}, \qquad R_{ij;m}(p) = \partial_m \lambda(p)\, \delta_{ij}, \end{align*} where the second identity uses that $g_{ij;m} = 0$ (metric compatibility) and that at $p$ the covariant derivative of a tensor equals the partial derivative of its components plus Christoffel correction terms, which vanish at $p$. Now set $i = l$ in $(\ast)$ and sum over $i$: \begin{align*} \sum_{i=1}^n\Bigl(-R_{ii;m}(p) + R_{im;i}(p) + \sum_{j=1}^n R_{ijim;j}(p)\Bigr) = 0. \end{align*} Term by term, using $R_{ij}(p) = \lambda(p) \delta_{ij}$: \begin{align*} \sum_i R_{ii;m}(p) &= \sum_i \partial_m \lambda(p)\, \delta_{ii} = n\, \partial_m \lambda(p), \\ \sum_i R_{im;i}(p) &= \sum_i \partial_i \lambda(p)\, \delta_{im} = \partial_m \lambda(p), \\ \sum_{i,j} R_{ijim;j}(p) &= \sum_j \Bigl(\sum_i R_{ijim}(p)\Bigr)_{;j} = \sum_j R_{jm;j}(p), \end{align*} where the last equality uses the definition of the Ricci tensor $R_{jm} = \sum_i R_{ijim}$ (equivalent to the earlier contraction after applying the Riemann symmetries; at $p$, $g^{ii} = 1$ eliminates index raising). Substituting $R_{jm}(p) = \lambda(p)\delta_{jm}$: \begin{align*} \sum_j R_{jm;j}(p) = \sum_j \partial_j \lambda(p)\, \delta_{jm} = \partial_m \lambda(p). \end{align*} Assembling, \begin{align*} -n\,\partial_m\lambda(p) + \partial_m\lambda(p) + \partial_m\lambda(p) = (2 - n)\,\partial_m\lambda(p) = 0. \end{align*}[/step]

[guided]The Einstein condition $\mathrm{Ric} = \lambda g$ is a scalar curvature condition on the Ricci tensor. Taking a covariant derivative and using $\nabla g = 0$ (metric compatibility of the Levi-Civita connection), \begin{align*} \nabla_m R_{ij} = (\partial_m \lambda)\, g_{ij}, \end{align*} so the "Ricci gradient" is controlled entirely by $d\lambda$. In normal coordinates at $p$, $g_{ij}(p) = \delta_{ij}$, giving $R_{ij;m}(p) = \partial_m\lambda(p)\, \delta_{ij}$. Setting $i = l$ and summing in $(\ast)$ traces the identity over the index $i$, converting the two Ricci gradient terms into $-n\partial_m\lambda + \partial_m\lambda$ (the first from $\sum_i \delta_{ii} = n$, the second from $\sum_i \delta_{im} = 1$). The Riemann-tensor divergence term collapses (after renaming indices) to one more copy of $\partial_m\lambda$, for a net coefficient of $-n + 1 + 1 = 2 - n$. The factor $(2 - n)$ is the dimensional obstruction: in dimension $n = 2$, every metric is automatically Einstein and the scalar $\lambda$ carries no constraint. In dimensions $n \geq 3$, the obstruction is non-zero, forcing $\partial_m\lambda(p) = 0$.[/guided]

[step:Conclude $\lambda$ is constant using $n \ne 2$ and connectedness] Since $\dim_{\mathbb R} M = n \ne 2$ by hypothesis, the coefficient $2 - n \ne 0$, so $\partial_m \lambda(p) = 0$ for every $m = 1, \dots, n$. Equivalently $d\lambda(p) = 0$. The point $p \in M$ was arbitrary, so $d\lambda \equiv 0$ on $M$. The function $\lambda \in C^\infty(M)$ has vanishing differential on the connected manifold $M$, hence is constant by the [Constancy Criterion on Connected Manifolds](/theorems/???). This is the cosmological constant. $\square$ [/step]