1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.