# When can one continuously prescribe a unit vector orthogonal to a given orthonormal system?

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.

• 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. – Loop Space Nov 5 '18 at 17:41
• 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. – 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 $$k$$.

• The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$. – R. van Dobben de Bruyn Nov 5 '18 at 15:58
• 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. – Terry Tao Nov 5 '18 at 16:19
• 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. – Thomas Rot Nov 5 '18 at 18:03
• 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. – Michael Nov 6 '18 at 22:25
• @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 $n=3, k=1$. – 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.

• 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. – David E Speyer Nov 6 '18 at 16:02
• This would seem to give a section of $\gamma_2^{\perp} \to G^+(2,7)$. What did I miss? – David E Speyer Nov 6 '18 at 16:02
• @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. – Michael Albanese Nov 6 '18 at 16:32
• @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. – Michael Albanese Nov 6 '18 at 17:37
• @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. – 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 $$1, even without requiring $$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}))$$.