[proofplan] The strategy proceeds in three linked moves. First, we package the Ricci tensor at a point $m$ as a self-adjoint endomorphism $\theta_m$ of $T_mM$, using the symmetry of Ricci (a consequence of the first Bianchi identity). Second, we use the fact that the stabiliser $H$ acts by isometries to see that $\theta_m$ is $H$-equivariant, so its eigenspaces are $H$-invariant subspaces of $T_mM$. Irreducibility of the $H$-action then forces $\theta_m$ to be a scalar multiple $\lambda(m)\,\mathrm{Id}$, yielding $\mathrm{Ric}_m = \lambda(m)\, g_m$. Third, we use transitivity of $G$ by isometries to transport this pointwise identity across $M$ and show that $\lambda$ is a constant function on $M$, giving the Einstein condition $\mathrm{Ric} = \lambda\, g$ globally. [/proofplan]

[step:Package the Ricci tensor at $m$ as a self-adjoint endomorphism $\theta_m$ of $T_mM$]The Ricci tensor $\mathrm{Ric}_m: T_mM \times T_mM \to \mathbb{R}$ is a symmetric bilinear form (a consequence of the first Bianchi identity applied to the Riemann curvature tensor). Using the non-degenerate inner product $g_m$ on $T_mM$, we define the associated self-adjoint endomorphism \begin{align*} \theta_m: T_mM &\to T_mM \\ v &\mapsto \theta_m(v), \end{align*} characterised by the identity \begin{align*} g_m(\theta_m(v), w) = \mathrm{Ric}_m(v, w) \qquad \text{for all } v, w \in T_mM. \end{align*} Existence and uniqueness of $\theta_m$ follow from non-degeneracy of $g_m$ (the Riesz-type identification of bilinear forms with endomorphisms in finite dimensions). Self-adjointness with respect to $g_m$ is the symmetry of $\mathrm{Ric}_m$: for all $v, w \in T_mM$, \begin{align*} g_m(\theta_m(v), w) = \mathrm{Ric}_m(v, w) = \mathrm{Ric}_m(w, v) = g_m(\theta_m(w), v) = g_m(v, \theta_m(w)). \end{align*}[/step]

[guided]Why do we convert $\mathrm{Ric}_m$ from a bilinear form into an endomorphism? Because the irreducibility hypothesis is phrased in terms of $H$-invariant subspaces of $T_mM$, and the cleanest way to extract subspaces from a bilinear form is to convert it into a linear operator and look at its eigenspaces. The conversion uses the metric $g_m$ in exactly the same way that the Riesz representation theorem identifies a bilinear form on a Hilbert space with a linear operator. Concretely, for each fixed $v \in T_mM$, the map $w \mapsto \mathrm{Ric}_m(v, w)$ is a linear functional on $T_mM$. By non-degeneracy of $g_m$, there is a unique vector, call it $\theta_m(v)$, such that $g_m(\theta_m(v), w) = \mathrm{Ric}_m(v, w)$ for all $w$. The map $v \mapsto \theta_m(v)$ is linear by bilinearity of $\mathrm{Ric}_m$, so $\theta_m$ is a well-defined endomorphism of $T_mM$. Why is $\theta_m$ self-adjoint with respect to $g_m$? Because $\mathrm{Ric}_m$ is symmetric: $\mathrm{Ric}_m(v, w) = \mathrm{Ric}_m(w, v)$. This symmetry is a standard consequence of the first Bianchi identity combined with the symmetries of the Riemann curvature tensor $R$ (see the [symmetries of the Riemann curvature tensor](/theorems/???)). The computation \begin{align*} g_m(\theta_m(v), w) = \mathrm{Ric}_m(v, w) = \mathrm{Ric}_m(w, v) = g_m(\theta_m(w), v) = g_m(v, \theta_m(w)) \end{align*} is precisely the self-adjointness condition $g_m(\theta_m(v), w) = g_m(v, \theta_m(w))$. Self-adjoint endomorphisms on a real inner product space are diagonalisable with real eigenvalues (the [spectral theorem](/theorems/???)), so $T_mM$ decomposes as an orthogonal direct sum of eigenspaces of $\theta_m$.[/guided]

[step:Show that $\theta_m$ commutes with the isotropy action, so its eigenspaces are $H$-invariant]For each $h \in H$ the isometry $h: M \to M$ fixes $m$, so its differential $dh_m: T_mM \to T_mM$ is a linear isometry of $(T_mM, g_m)$. Since $h$ is an isometry of $(M, g)$, it preserves the Levi-Civita connection and hence the Riemann curvature tensor $R$; by contraction it preserves the Ricci tensor: \begin{align*} \mathrm{Ric}_m(dh_m(v), dh_m(w)) = \mathrm{Ric}_{h(m)}(dh_m(v), dh_m(w)) = \mathrm{Ric}_m(v, w) \end{align*} for all $v, w \in T_mM$, where the second equality uses $h(m) = m$ (so $\mathrm{Ric}_{h(m)} = \mathrm{Ric}_m$) and the isometry invariance of $\mathrm{Ric}$ under $h$. Equivalently, for all $v, w \in T_mM$, \begin{align*} g_m(\theta_m(dh_m(v)), dh_m(w)) = \mathrm{Ric}_m(dh_m(v), dh_m(w)) = \mathrm{Ric}_m(v, w) = g_m(\theta_m(v), w). \end{align*} Using that $dh_m$ is a $g_m$-isometry, the left-hand side equals $g_m((dh_m)^{-1} \theta_m\, dh_m(v), w)$. Non-degeneracy of $g_m$ gives $(dh_m)^{-1} \theta_m\, dh_m = \theta_m$, i.e., \begin{align*} \theta_m \circ dh_m = dh_m \circ \theta_m \qquad \text{for all } h \in H. \end{align*} Now let $E_\lambda \subseteq T_mM$ be an eigenspace of $\theta_m$ with eigenvalue $\lambda \in \mathbb{R}$. For any $v \in E_\lambda$ and any $h \in H$, \begin{align*} \theta_m(dh_m(v)) = dh_m(\theta_m(v)) = dh_m(\lambda v) = \lambda\, dh_m(v), \end{align*} so $dh_m(v) \in E_\lambda$. Hence every eigenspace $E_\lambda$ is $H$-invariant.[/step]

[guided]The core question: how does the hypothesis that $H$ acts by isometries get used? It enters through the fact that isometries of $(M, g)$ preserve the Riemann curvature tensor, and hence preserve every tensor derived from $R$ by algebraic operations — in particular, Ricci. Let $h \in H$. Because $h$ is an isometry, $h^* g = g$, and an isometry preserves the Levi-Civita connection (the Levi-Civita connection is uniquely determined by the metric, so an isometry intertwines the Levi-Civita connections of source and target; when the source and target are the same manifold with the same metric, this intertwining becomes equivariance). From connection-preservation we get Riemann-preservation: $h^* R = R$. Taking a trace, $h^* \mathrm{Ric} = \mathrm{Ric}$. Evaluated at $m$, using $h(m) = m$, this reads \begin{align*} \mathrm{Ric}_m(dh_m(v), dh_m(w)) = \mathrm{Ric}_m(v, w), \qquad v, w \in T_mM. \end{align*} Rewriting via $\theta_m$ and using that $dh_m$ preserves $g_m$: \begin{align*} g_m(\theta_m(dh_m(v)), dh_m(w)) = g_m(\theta_m(v), w) = g_m(dh_m(\theta_m(v)), dh_m(w)), \end{align*} where the last equality is $g_m$-isometry of $dh_m$. Non-degeneracy of $g_m$ on $T_mM$ gives $\theta_m \circ dh_m = dh_m \circ \theta_m$. Why does this equivariance make eigenspaces invariant? Because if $\theta_m(v) = \lambda v$, then \begin{align*} \theta_m(dh_m(v)) = dh_m(\theta_m(v)) = dh_m(\lambda v) = \lambda\, dh_m(v), \end{align*} so $dh_m(v)$ is again an eigenvector with the same eigenvalue. This is the standard principle: a commuting operator preserves each eigenspace of any other operator.[/guided]

[step:Conclude from irreducibility that $\theta_m = \lambda(m)\,\mathrm{Id}$]By self-adjointness, the [spectral theorem](/theorems/???) for self-adjoint operators on a real inner product space gives an orthogonal decomposition \begin{align*} T_mM = \bigoplus_{\lambda \in \mathrm{spec}(\theta_m)} E_\lambda, \end{align*} where each eigenspace $E_\lambda$ is a non-zero linear subspace of $T_mM$. By the previous step, each $E_\lambda$ is $H$-invariant. By the irreducibility hypothesis, the only $H$-invariant subspaces of $T_mM$ are $\{0\}$ and $T_mM$. Since $E_\lambda \neq \{0\}$, we must have $E_\lambda = T_mM$, and hence $\theta_m$ has a single eigenvalue. Denote it $\lambda(m) \in \mathbb{R}$, so that \begin{align*} \theta_m = \lambda(m)\,\mathrm{Id}_{T_mM}. \end{align*} Equivalently, $\mathrm{Ric}_m(v, w) = g_m(\theta_m(v), w) = \lambda(m)\, g_m(v, w)$ for all $v, w \in T_mM$, i.e., \begin{align*} \mathrm{Ric}_m = \lambda(m)\, g_m. \end{align*}[/step]

[guided]This is the pivotal use of the irreducibility hypothesis. The spectral theorem produces eigenspaces $E_\lambda$; the previous step showed they are $H$-invariant; irreducibility has only two $H$-invariant subspaces available, namely $\{0\}$ and the whole $T_mM$; since eigenspaces are non-zero by definition, each one must be the whole $T_mM$; so there is only one eigenvalue. A sanity check: if $H$ acted reducibly, decomposing $T_mM = V_1 \oplus V_2$ into proper $H$-invariant pieces, then $\theta_m$ could in principle act with different eigenvalues $\lambda_1, \lambda_2$ on $V_1, V_2$, and the argument would fail. The irreducibility hypothesis is precisely what rules this out. The conclusion $\mathrm{Ric}_m = \lambda(m)\, g_m$ is the Einstein condition at a single point. The task of the final step is to upgrade "at a single point" to "on all of $M$" and check that $\lambda$ is constant.[/guided]

[step:Propagate the pointwise Einstein identity to all of $M$ using transitivity of $G$]Let $m' \in M$. Since $G$ acts transitively on $M$, there exists $\varphi \in G$ with $\varphi(m) = m'$. Because $\varphi$ is an isometry of $(M, g)$, it preserves both the metric and the Ricci tensor: $\varphi^* g = g$ and $\varphi^* \mathrm{Ric} = \mathrm{Ric}$. Concretely, for $v', w' \in T_{m'}M$, setting $v := (d\varphi_m)^{-1}(v') \in T_mM$ and $w := (d\varphi_m)^{-1}(w') \in T_mM$, \begin{align*} \mathrm{Ric}_{m'}(v', w') &= \mathrm{Ric}_{m'}(d\varphi_m(v), d\varphi_m(w)) \\ &= \mathrm{Ric}_m(v, w) \\ &= \lambda(m)\, g_m(v, w) \\ &= \lambda(m)\, g_{m'}(d\varphi_m(v), d\varphi_m(w)) \\ &= \lambda(m)\, g_{m'}(v', w'), \end{align*} where line 2 uses $\varphi^* \mathrm{Ric} = \mathrm{Ric}$ at $m$, line 3 is the pointwise Einstein identity at $m$ from the previous step, and line 4 uses $\varphi^* g = g$ at $m$. Hence \begin{align*} \mathrm{Ric}_{m'} = \lambda(m)\, g_{m'} \qquad \text{for every } m' \in M. \end{align*} In particular, the same scalar $\lambda := \lambda(m)$ works at every point. We have proved the identity of $(0,2)$-tensor fields \begin{align*} \mathrm{Ric} = \lambda\, g \qquad \text{on } M, \end{align*} with $\lambda \in \mathbb{R}$ constant. This is precisely the Einstein condition, completing the proof.[/step]

[guided]Two things needed verification at this stage: (a) that the Einstein condition holds at every point of $M$, not just at $m$; and (b) that the scalar $\lambda$ is the same at every point (i.e., a constant function of $M$, not just a smooth function). Both are handled by transitivity of the $G$-action by isometries. Transitivity supplies, for any $m' \in M$, an isometry $\varphi \in G$ sending $m$ to $m'$. An isometry pulls back Ricci to Ricci and the metric to the metric, and these identities are equivalent at points via their differentials. So the Einstein identity at $m$ transports to the Einstein identity at $m'$, with the same scalar $\lambda(m)$. What would happen if $G$ acted only locally transitively, or merely smoothly? Then $\lambda$ would a priori be a smooth function $\lambda: M \to \mathbb{R}$ rather than a constant. The Einstein condition with a non-constant function is called a quasi-Einstein condition; it requires extra work (such as differentiating $\mathrm{Ric} = \lambda\, g$ and using the contracted second Bianchi identity) to upgrade to true Einstein. Here, transitive isometric action does the job for free. Combining the estimates: Step 1 packaged $\mathrm{Ric}_m$ as a self-adjoint operator $\theta_m$; Step 2 used isometry to show $H$-equivariance of $\theta_m$; Step 3 used irreducibility to force $\theta_m$ to be a scalar; Step 4 used transitivity to propagate the scalar identity globally with a constant scalar. This gives $\mathrm{Ric} = \lambda\, g$ with $\lambda \in \mathbb{R}$, i.e., $(M, g)$ is Einstein.[/guided]