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Let $(M,g)$ be a closed Riemannian manifold. Fix a point $m\in M$ and a $2$-form $\omega$ at $m.$ Take a curve $\gamma$ in $M$ such that $\gamma(0)=m.$ Now we can get a $2$-form along $\gamma$ by taking a parallel transform of $\omega$ along $\gamma$. Notice the parallel transform $\omega(t)$ at $\gamma(t)$ is defined as \begin{equation*} \omega(t)(v,w):= \omega\big(P_t^{-1}(v),P_t^{-1}(w)\big),\hspace{2 ex}\text{where $P_t$ is parallel transform along $\gamma: T_mM\rightarrow T_{\gamma(t)}M$ } \end{equation*} One can check $\nabla_{\gamma'(t)}\omega(t)=0$ at all points of $\gamma(t)$ following the definition: $(\nabla_v\theta)(a,b)$ at a point $m\in M$ can be defined as $\frac{d}{dt}|_0\theta|_{\xi(t)}\big(P_t(a),P_t(b)\big),$ where $\xi$ is a curve in $M$ with $\xi'(0)=(m,v).$ Notice this parallel transform also preserves norm. So if $|\omega|_m=1,$ then $|\omega(t)|=1$ for all $t$. So it makes sense to define a curve in the sphere bundle of two-forms: $S(\wedge^2 M):$ \begin{equation*} \tilde\gamma(t)=(\gamma(t),\omega(t)\big) \end{equation*}

I want to understand $\tilde\gamma'(0)$. Here we want to describe the tangent space at a point in the sphere as the space of forms perpendicular to the form denoting the point in the sphere.

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Using the connection, the tangent bundle $T S(\wedge^2M)$ splits as a direct sum into the vertical part $$\mathcal VS(\wedge^2M):=\ker d\pi,$$ where $\pi\colon S(\wedge^2M)\to M$ is the projection, and into the horizontal part $$\mathcal H S(\wedge^2M).$$ We have $$\pi^* TM\cong\mathcal H S(\wedge^2M)$$ via $d\pi.$ Then, $\tilde\gamma'(0)\in \mathcal H S(\wedge^2M)$ such that $d\pi(\tilde\gamma'(0))=\gamma'(0).$

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