# A curve in the bundle of two forms

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.

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 $$d\pi(\tilde\gamma'(0))=\gamma'(0).$$