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.

  • $\begingroup$ It's been a loooong time since I've thought about such things, but I think that the obstruction you want is the Euler class, on the Wikipedia page it says "Note that "Normalization" is a distinguishing feature of the Euler class, so that it detects the existence of a non-vanishing section". So you'd need to look at the pull-back of the Euler class under the map $Gr_{n,k} \to Gr_{n,n-k}$ which maps $E$ to its orthogonal complement. The issue is a little more complicated if you're starting with a basis as then you need to pull back further to the frame bundle. $\endgroup$ – Loop Space Nov 5 '18 at 17:41
  • 9
    $\begingroup$ Just a comment on your intuition (now that you have a full answer). Having lots of choice doesn't make it any easier to find a continuous choice because the act of making a choice is not continuous. So after making lots of local choices, you are confronted with the problem of patching them together and that is where the obstructions lie. $\endgroup$ – Loop Space Nov 6 '18 at 7:21

$\def\RR{\mathbb{R}}$ 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 $k$.

  • 9
    $\begingroup$ The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$. $\endgroup$ – R. van Dobben de Bruyn Nov 5 '18 at 15:58
  • 18
    $\begingroup$ Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently. $\endgroup$ – Terry Tao Nov 5 '18 at 16:19
  • 9
    $\begingroup$ Write $n=2^{c+4d} a$, with $a$ odd and $0\leq c \leq 3$. Then there are $2^c+8 d -1$ linear independent sections of the tangent bundle of the sphere, which you can make orthogonal I think. This is a famous theorem of Adams. $\endgroup$ – Thomas Rot Nov 5 '18 at 18:03
  • 1
    $\begingroup$ This is a terrific answer for $\mathbb{R}^3-\{0\}$, but I'm not sure that it applies to $\mathbb{R}^3$, for there's no non-degenerate vector field in $\mathbb{R}^3$ that would be normal to the sphere. $\endgroup$ – Michael Nov 6 '18 at 22:25
  • 3
    $\begingroup$ @Michael The question is not asking for complementation of a vector field in $\mathbb{R}^n$. It is asking for complementation when "the tuple ranges over all possible orthonormal systems". This is precisely the sphere $S^2$ when $n=3, k=1$. $\endgroup$ – Aloizio Macedo Nov 7 '18 at 19:09

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 in general. However, if $n - k \leq k$, then $w_{n-k}(\gamma_{n-k}) \neq 0$, so for values of $k$ and $n$ with $2k \leq n$, there are no $SO(k)$-invariant such maps.

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 T\operatorname{Gr}^+(n-1, n)\\ &\cong \operatorname{Hom}(\gamma_{n-1}, \gamma_{n-1}^{\perp})\\ &\cong \gamma_{n-1}^*\otimes\gamma_{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.

  • $\begingroup$ It seems to me that something is wrong with this argument, because of the existence of the 7 dimensional cross product: en.wikipedia.org/wiki/Seven-dimensional_cross_product . This is an antisymmetric bilinear map $R^7 \times R^7 \to R^7$ such that $u \times v$ is always perpendicular to $u$ and $v$ and, if $u$ and $v$ are orthogonal, then $|u \times v| = |u| |v|$. Using bilinearity and antisymmetry, I get $(\cos \theta u + \sin \theta v) \times (-\sin \theta u + \cos \theta v) = (\cos^2 \theta + \sin^2 \theta) (u \times v) = u \times v$, meaning this map is $SO(2)$ invariant. $\endgroup$ – David E Speyer Nov 6 '18 at 16:02
  • 1
    $\begingroup$ This would seem to give a section of $\gamma_2^{\perp} \to G^+(2,7)$. What did I miss? $\endgroup$ – David E Speyer Nov 6 '18 at 16:02
  • $\begingroup$ @DavidESpeyer: I can't see anything wrong with what you've written, so there must be something wrong with my answer. Maybe the cohomology ring of $\operatorname{Gr}^+(n-k, n)$ is not correct. I don't know of a reference, but I was under the impression that this was correct. $\endgroup$ – Michael Albanese Nov 6 '18 at 16:32
  • 2
    $\begingroup$ @DavidESpeyer: The cohomology ring is wrong as can be seen in the case $k = 2$, $n = 4$ where $\operatorname{Gr}^+(2, 4) = S^2\times S^2$. I will try to think about this and see what can be salvaged. $\endgroup$ – Michael Albanese Nov 6 '18 at 17:37
  • $\begingroup$ @DavidESpeyer: It seems that the values of $k$ and $n$ for which $w_{n-k}(\gamma_{n-k}) \neq 0$ is not known; see this question. $\endgroup$ – Michael Albanese Nov 8 '18 at 23:10

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 $H^{i+1}(V(k,n);\pi_i(S^{n-k-1}))$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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