# Orthonormal frame bundle orthogonal to a curve

Let $M$ be a $n$-dimensional smooth riemannian manifold and $\varphi\colon(-\varepsilon,\varepsilon)\rightarrow M$ an embedding. $\varphi$ will denote the image of $\varphi$, too.

Consider the bundle of orthonormal $(n-k)$-frames in $M$ over $\varphi$, where $k>2$, which are orthogonal to $\varphi$. The fiber of this bundle is the orthonormal stiefel manifold $V_{n-k}(\mathbb{R}^{n-1})$, which is connected, since $n-k<n-1$.

My first idea to show, that this bundle is smoothly trivial was, to transport one fixed fiber via parallel transport along $\varphi$ of the Levi-Civita-connection of $M$. This works, because parallel transport along $\varphi$ gives me isometries, which leave vectors tangent to $\varphi$ fixed. (Just to be sure, this works, right?)

Now take a global section $\eta$ of this bundle and consider the bundle of orthonormal $(k-1)$-frames in $M$ over $\varphi$, which are orthogonal to $\varphi$ and to $\eta$. This gives me a bundle with fiber $V_{k-1}(\mathbb{R}^{n-1-(n-k)})=V_{k-1}(\mathbb{R}^{k+1})$.

I am searching for an argument to show, that this second bundle is also smoothly trivial. The proof considered before doesn't work, because parallel transport along $\varphi$ doesn't preserve beeing orthogonal to $\eta$ in general. Can someone help me out by giving me a hint how to prove that?

I don't want to use any classification results or any sophisticated bundle-theory. I think there must be an more-or-less elementary argument like the one of my first idea.

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