Agfdhyk

Let $1 leq k < n$ be natural numbers. Given orthonormal vectors $u_1,dots,u_k$ in ${bf R}^n$, one can always find an additional unit vector $v in {bf R}^n$ that is orthogonal to the preceding $k$. My question is: under what conditions on $k,n$ is it possible to make $v$ depend continuously on $u_1,dots,u_k$, as the tuple $(u_1,dots,u_k)$ ranges over all possible orthonormal systems? (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.)

When $k=n-1$ then one can just pick the unique unit normal to the span of the $u_1,dots,u_k$ that is consistent with a chosen orientation on ${bf R}^n$ (i.e., take wedge product and then Hodge dual, or just cross product in the $(k,n)=(2,3)$ case). But I don't know what is going on in lower dimension. Intuitively it seems to me that if $n$ is much larger than $k$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise.

It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $Gr(k,n)$, though I don't know how to calculate the space of such sections.

$defRR{mathbb{R}}$ This problem was solved by

Whitehead, G. W., Note on cross-sections in Stiefel manifolds, Comment. Math. Helv. 37, 239-240 (1963). ZBL0118.18702.

Such sections exist only in the cases $(k,n) = (1,2m)$, $(n-1, n)$, $(2,7)$ and $(3,8)$.

All sections can be given by antisymmetric multilinear maps (and thus, in particular, can be taken to be smooth). The $(2,7)$ product is the seven dimensional cross product, which is octonion multiplication restricted to the octonions of trace $0$.

The $(3,8)$ product was computed by

Zvengrowski, P., A 3-fold vector product in $R^8$, Comment. Math. Helv. 40, 149-152 (1966). ZBL0134.38401

to be given by the formula
$$X(a,b,c) = -a (overline{b} c) + a (b cdot c) - b (c cdot a) + c (a cdot b)$$
where $cdot$ is dot product while multiplication with no symbol and $overline{b }$ have their standard octonion meanings. Note that, if $(a,b,c)$ are orthogonal, the last $3$ terms are all $0$, so the expression simplifies to $- a (overline{b} c)$; writing in the formula in the given manner has the advantage that $X(a,b,c)$ is antisymmetric in its arguments and perpendicular to the span of $a$, $b$ and $c$ for all $(a,b,c)$.

Unless I'm missing something, I think that the hairy ball theorem states precisely that you cannot do this when $n = 3$ and $k = 1$. I'm not sure what happens for other values of $n$ and $k$.

The space of orthonormal $k$-frames in $mathbb{R}^n$ is the Stiefel manifold $V(k, n) = SO(n)/SO(n - k)$. There is a natural $SO(k)$ action on $V(k, n)$ and the quotient is the oriented grassmannian $operatorname{Gr}^+(k, n) = SO(n)/(SO(k)times SO(n-k))$. Let $gamma_k to operatorname{Gr}^+(k, n)$ denote the tautological bundle and let $gamma_k^{perp} to operatorname{Gr}^+(k, n)$ denote its orthogonal complement. As you indicated, a nowhere-zero section of $gamma_k^{perp} to operatorname{Gr}^+(k, n)$ would give rise to a map that you desire. In fact, such a map arises this way if and only if it is $SO(k)$-invariant.

The inner product on $mathbb{R}^n$ allows us to define the map $P mapsto P^{perp}$ which induces a diffeomorphism $f : operatorname{Gr}^+(k, n) to operatorname{Gr}^+(n-k, n)$. Under this diffeomorphism we have $f^*gamma_{n-k} cong gamma_k^{perp}$, so $gamma_k^{perp} to operatorname{Gr}^+(k, n)$ admits a nowhere-zero section if and only if $gamma_{n-k} to operatorname{Gr}^+(n-k, n)$ does. Therefore, we would like to know the answer to the following question:

For which values of $k$ and $n$ does $gamma_{n-k} to operatorname{Gr}^+(n-k, n)$ admit a nowhere-zero section?

One necessary condition is that $w_{n-k}(gamma_{n-k}) = 0$. Said another way, if $w_{n-k}(gamma_{n-k}) neq 0$, then there is no $SO(k)$-invariant map.

In a previous version of this answer, I stated what I thought was the $mathbb{Z}_2$ cohomology ring of $operatorname{Gr}^+(k, n)$ - I was incorrect. From this mistake, it followed that for $1 < k < n - 1$, $w_{n-k}(gamma_{n-k}) neq 0$ and hence there were no $SO(k)$-invariant maps for these values of $k$. This conclusion is false; there is a counterexample when $k = 2$ and $n = 7$ as David E Speyer pointed out in the comments below.

Somewhat surprisingly, the $mathbb{Z}_2$ cohomology ring of $operatorname{Gr}^+(k, n)$ is not known in general, see this question. The values of $k$ and $n$ for which $w_{n-k}(gamma_{n-k}) neq 0$ also seems to be unknown.

When $k = n - 1$, you described such a map which is in fact $SO(n-1)$-invariant. By the above correspondence, such maps exist because $gamma_1 to operatorname{Gr}^+(1, n) = S^{n-1}$ is trivial as it is an orientable line bundle (alternatively, $gamma_1$ is trivialised by the Euler vector field).

When $k = 1$, first note that $gamma_{n-1} to operatorname{Gr}^+(n - 1, n) = S^{n-1}$ is isomorphic to the tangent bundle of $S^{n-1}$:

begin{align*}
TS^{n-1} &cong Toperatorname{Gr}^+(n-1, n)\
&cong operatorname{Hom}(gamma_{n-1}, gamma_{n-1}^{perp})\
&cong gamma_{n-1}^*otimesgamma_{n-1}^{perp}\
&cong gamma_{n-1}otimes f^*gamma_1\
&cong gamma_{n-1}
end{align*}

where the last isomorphism uses the fact that $gamma_1$, and hence $f^*gamma_1$, is trivial. By Poincaré-Hopf, $TS^{n-1}$ admits a section if and only if $n$ is even. In this case, the map can be written down explicitly: $(v_1, v_2, dots, v_{n-1}, v_n) mapsto (-v_2, v_1, dots, -v_n, v_{n-1})$. Identifying $mathbb{R}^n$ and $mathbb{C}^{n/2}$ via $(v_1, v_2, dots, v_{n-1}, v_n) mapsto (v_1 + iv_2, dots, v_{n-1} + iv_n)$, the aforementioned map is nothing but multiplication by $i$.

Note, requiring $SO(k)$-invariance for $k = 1$ is not a restriction as $SO(1)$ is the trivial group.

Denoting the Stiefel manifold of orthonormal $k$-frames in $mathbb{R}^n$ by $V(k,n)$ as in Michael Albanese's answer, what you are asking for is a section of the sphere bundle
$$
S^{n-k-1}to V(k+1,n)to V(k,n),
$$
where the projection takes a $(k+1)$-frame to its first $k$ vectors.

According to the paper

Čadek, Martin; Mimura, Mamoru; Vanžura, Jiří, The cohomology rings of real Stiefel manifolds with integer coefficients, J. Math. Kyoto Univ. 43, No. 2, 411-428 (2003). ZBL1061.55015,

the Euler class of this bundle is zero when $n-k$ is odd, and nonzero when $n-k$ is even.

This extends Michael Albanese's answer slightly, showing that the maps cannot exist when $n-k$ is even with $1<k<n$, even without requiring $SO(k)$-invariance. I don't know if the bundle admits a section when $n-k$ is odd. The primary obstruction vanishes in these cases, but note that there may be higher obstructions in the groups $H^{i+1}(V(k,n);pi_i(S^{n-k-1}))$.