[proofplan] At a fixed positive time, the assumed parallel-transport invariance of the curvature-null spaces gives fiberwise invariance under parallel transport in $\Lambda^2T^*M$. Transport around loops then makes the null space at each basepoint invariant under the corresponding holonomy representation. The holonomy hypothesis forces that null space to be zero, and nonnegativity of the self-adjoint curvature operator upgrades zero kernel to positive definiteness. Since the positive time was arbitrary, the curvature operator is positive at every point for every positive time. [/proofplan]

[step:Define the curvature-null distribution at a fixed positive time] Fix $t_0 \in (0,T]$. For each $x \in M$, define the finite-dimensional real inner-product space \begin{align*} E_{x,t_0}:=\Lambda^2T_x^*M \end{align*} with the [inner product](/page/Inner%20Product) induced by $g(t_0)$. Let \begin{align*} \mathcal{R}_{x,t_0}: E_{x,t_0} \to E_{x,t_0} \end{align*} denote the curvature operator of $g(t_0)$ at $x$, acting on two-forms. Define the curvature-null space \begin{align*} \mathcal{N}_{x,t_0}:=\ker \mathcal{R}_{x,t_0}\subseteq E_{x,t_0}. \end{align*} Since $\mathcal{R}_{x,t_0}$ is self-adjoint and nonnegative, $\mathcal{R}_{x,t_0}$ is positive definite if and only if $\mathcal{N}_{x,t_0}=\{0\}$. Therefore it suffices to prove that $\mathcal{N}_{x,t_0}=\{0\}$ for every $x \in M$. [/step]

[step:Use the assumed parallel transport invariance to obtain fiberwise null-space invariance]By the positive-time parallel-transport invariance assumed in the statement, the family \begin{align*} \mathcal{N}_{t_0}:=\bigsqcup_{x \in M}\mathcal{N}_{x,t_0} \end{align*} is invariant fiber by fiber under parallel transport with respect to the Levi-Civita connection of $g(t_0)$. Equivalently, if $\gamma:[0,1]\to M$ is a smooth path and \begin{align*} P_{\gamma,0,1}: \Lambda^2T_{\gamma(0)}^*M \to \Lambda^2T_{\gamma(1)}^*M \end{align*} denotes parallel transport induced by $g(t_0)$ from time parameter $0$ to time parameter $1$ along $\gamma$, then \begin{align*} P_{\gamma,0,1}\bigl(\mathcal{N}_{\gamma(0),t_0}\bigr)=\mathcal{N}_{\gamma(1),t_0}. \end{align*}[/step]

[guided]The purpose of this step is to turn the positive-time hypothesis into fiberwise transport invariance. For the fixed time $t_0$, define the family of null spaces \begin{align*} \mathcal{N}_{t_0}:=\bigsqcup_{x \in M}\ker \mathcal{R}_{x,t_0}. \end{align*} The statement assumes precisely that these curvature-null spaces are invariant under parallel transport. Thus, for every smooth path $\gamma:[0,1]\to M$, the parallel transport map \begin{align*} P_{\gamma,0,1}: \Lambda^2T_{\gamma(0)}^*M \to \Lambda^2T_{\gamma(1)}^*M \end{align*} induced by the Levi-Civita connection of $g(t_0)$ from parameter $0$ to parameter $1$ maps curvature-null directions exactly to curvature-null directions: \begin{align*} P_{\gamma,0,1}\bigl(\mathcal{N}_{\gamma(0),t_0}\bigr)=\mathcal{N}_{\gamma(1),t_0}. \end{align*} Thus $\mathcal{N}_{t_0}$ is not merely a pointwise collection of kernels for this argument; its fibers are preserved exactly by parallel transport at time $t_0$, which is the only structure needed for the holonomy argument below.[/guided]

[step:Convert parallel invariance into holonomy invariance] Fix a point $x \in M$. Let $O(\Lambda^2T_x^*M)$ denote the group of linear isometries of the finite-dimensional inner-product space $\Lambda^2T_x^*M$ with the inner product induced by $g(t_0)$. Let \begin{align*} \operatorname{Hol}_{x}(g(t_0)) \subseteq O(\Lambda^2T_x^*M) \end{align*} denote the holonomy group acting on $\Lambda^2T_x^*M$ by parallel transport around loops based at $x$. If $A \in \operatorname{Hol}_{x}(g(t_0))$, then there is a loop $\gamma:[0,1]\to M$ with $\gamma(0)=\gamma(1)=x$ such that $A=P_{\gamma,0,1}$. Since the fibers of $\mathcal{N}_{t_0}$ are preserved by parallel transport, \begin{align*} A(\mathcal{N}_{x,t_0}) = P_{\gamma,0,1}(\mathcal{N}_{x,t_0}) = \mathcal{N}_{x,t_0}. \end{align*} Hence $\mathcal{N}_{x,t_0}$ is an invariant subspace for the holonomy representation on $\Lambda^2T_x^*M$. [/step]

[step:Apply the holonomy hypothesis to force the null space to vanish]The holonomy hypothesis says that no nonzero holonomy-invariant subspace is contained in $\mathcal{N}_{x,t_0}$. Since the previous step proves that $\mathcal{N}_{x,t_0}$ itself is holonomy invariant, the only possible conclusion is \begin{align*} \mathcal{N}_{x,t_0}=\{0\}. \end{align*} Because $x \in M$ was arbitrary, $\mathcal{N}_{x,t_0}=\{0\}$ for every $x \in M$.[/step]

[guided]Fix a point $x \in M$. The previous step proved that $\mathcal{N}_{x,t_0}$ is invariant under the holonomy representation of $\operatorname{Hol}_{x}(g(t_0))$ on $\Lambda^2T_x^*M$. The holonomy hypothesis is formulated precisely to rule out such equality directions: it says that there is no nonzero holonomy-invariant subspace contained in the curvature-null distribution. We now apply that hypothesis to the specific subspace \begin{align*} \mathcal{N}_{x,t_0}\subseteq \Lambda^2T_x^*M. \end{align*} It is contained in itself, and it is holonomy invariant by the loop-parallel-transport argument. Therefore, if $\mathcal{N}_{x,t_0}$ were nonzero, it would be a nonzero holonomy-invariant subspace contained in the null distribution, contradicting the hypothesis. Hence \begin{align*} \mathcal{N}_{x,t_0}=\{0\}. \end{align*} Because the point $x \in M$ was arbitrary, this proves \begin{align*} \mathcal{N}_{y,t_0}=\{0\} \end{align*} for every $y \in M$.[/guided]

[step:Conclude positivity of the curvature operator at every point and positive time]We have shown that, for the arbitrary fixed time $t_0 \in (0,T]$ and every $x \in M$, \begin{align*} \ker \mathcal{R}_{x,t_0}=\{0\}. \end{align*} Since each $\mathcal{R}_{x,t_0}$ is self-adjoint and nonnegative, zero kernel is equivalent to positive definiteness: \begin{align*} \langle \mathcal{R}_{x,t_0}\omega,\omega\rangle_{g(t_0)} > 0 \end{align*} for every nonzero $\omega \in \Lambda^2T_x^*M$. As $t_0 \in (0,T]$ was arbitrary, $\mathcal{R}_{x,t}$ is positive definite for every $x \in M$ and every $t \in (0,T]$.[/step]

[guided]We have proved that, at the fixed but arbitrary time $t_0 \in (0,T]$, the curvature-null space vanishes at every point: \begin{align*} \ker \mathcal{R}_{x,t_0}=\{0\} \end{align*} for every $x \in M$. Now take a point $x \in M$ and a nonzero two-form $\omega \in \Lambda^2T_x^*M$. Because $\mathcal{R}_{x,t_0}$ is nonnegative, we have \begin{align*} \langle \mathcal{R}_{x,t_0}\omega,\omega\rangle_{g(t_0)} \geq 0. \end{align*} If equality held for this nonzero $\omega$, then the finite-dimensional spectral theorem for a self-adjoint nonnegative operator would imply $\omega \in \ker \mathcal{R}_{x,t_0}$. Equivalently, using an orthonormal eigenbasis, all nonnegative eigenvalue contributions to $\langle \mathcal{R}_{x,t_0}\omega,\omega\rangle_{g(t_0)}$ must vanish, so $\mathcal{R}_{x,t_0}\omega=0$. This contradicts \begin{align*} \ker \mathcal{R}_{x,t_0}=\{0\}. \end{align*} Therefore \begin{align*} \langle \mathcal{R}_{x,t_0}\omega,\omega\rangle_{g(t_0)} > 0 \end{align*} for every nonzero $\omega \in \Lambda^2T_x^*M$, so $\mathcal{R}_{x,t_0}$ is positive definite. Since $t_0 \in (0,T]$ was arbitrary, the same argument applies at every positive time $t \in (0,T]$ and every point $x \in M$.[/guided]