# Trivialised sub-bundle of a trivialised bundle and its orthogonal bundle

I'm reading Kirby's The Topology of $4$ Manifolds, and I encountered the following claim:

Recall that a trivialised sub $k$-plane bundle of a trivialised $(n+k)$-plane bundle determines a trivialisation of the orthogonal $n$-plane bundle over the $(n-1)$-skeleton of the base space

How can I prove it?

What I did: My first thought was to use the characterization of Stiefel - Whitney classes as obstruction classes, but when we are dealing with $n$-frames in an $n$-plane bundle, the first class only detects a trivialisation on the $1$-skeleton, (an orientation), and the higher classes detect the presence of $n'$-frames ($n'<n$) on higher skeleta.

The results (that I'm aware of) about extending the $n$-frame on the $1$-skeleton to higher skeleta fail to apply since the fibre ($V_n(\mathbb{R}^n)\cong GL_n(\mathbb{R})$) is not $k$-simple for any $k$. The other approach with obstruction theory was to try proving that the classifying map $f\colon B\to BSO(n)$ is trivial when restricted to the $n-1$ skeleton (that should prove my claim). Even though BSO(n) is simply connected, I don't know higher homotopy groups of it and therefore can t apply obstruction theory.

I tried applying even the prolongation theorem from Husemoller, page $21$, but the fibre is not $m-1$-connected for any $m$, since it is $GL_n(\mathbb{R}^n)$.

From the hypothesis it's easy to see that the Stiefel-Whitney classes of the orthogonal bundle all vanish, but by the remark before, I can't use this information.

What I found strange (about my approaches) is that I'm not using that our bundle is the orthogonal complement of a trivial subbbundle and the "total" bundle where they live is trivial. That's why I fear I'm missing something

This is a special case of the "Real Cancellation Theorem: suppose $X$ is $d$-dimensional CW complex and $E\to X$ an $n$-dimensional real vector bundle with $n>d$. Then, if $F$ is another vector bundle and $E\oplus \mathbb{R}^k\simeq F\oplus \mathbb{R}^k$, then $E\simeq F$."
By choosing $E=\mathbb{R^n}$ you get the thesis. Another way of stating the problem is the following:
"Given a stably trivial vector bundle $E$ on a CW complex $B$ with $\dim E>\dim B$, then $E$ is already trivial".