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Let $\Omega\subseteq\mathbb R^l$ be open and $A:\Omega\to \mathbb R^{n\times n}$ be a matrix valued $C^k$-function, $k\in\mathbb N_0\cup\{\infty\} $ such that for all $p\in\Omega$ $A_p$ is not injective or equivalently $\det (A_p)=0$. Then for all $p$ we can find $x_p\in \mathbb R^n\setminus\{0\}$ with $A_p(x_p)=0$.

The question now is:

Can the $x_p$ be choosen such that $p\mapsto x_p$ is also $C^k$?

$\textbf{Edit:}$

As Aloizio Macedo's answer shows this is not true in general. I am still interested in the case $\Omega=\mathbb R$ though since this is what I originally had in mind.

Aloizio Macedo
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Claire
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1 Answers1

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No. Take $\Omega$ to be a small neighborhood of the sphere $S^{2}$ and $(A(x))(v)=\langle v, x\rangle x$ seen as a matrix in $\mathbb{R}^{3 \times 3}$. The kernel of every $A(x)$ is the tangent space to the sphere centered at the origin and passing through $x$. If we could find such smooth (or even $C^0$) $p \mapsto x_p$, we would be able to find a continuous non-vanishing vector field on $S^2$, which contradicts the Hairy Ball theorem.

EDIT: Taking into account the requests in the comments, let's now prove that if $A: \Omega \to \mathbb{R}^{n \times n}$ is a $C^r$ map, then $\mathcal{E}=(x,\ker A(x))$ is a $C^r$ fiber bundle over $\Omega$ in case $A(x)$ has kernels of constant dimension.

To see this, given a point $p \in \Omega$, consider a maximal set of linearly independent row vectors $l_1(p),\cdots,l_i(p)$ and a neighborhood $U$ of $p$ such that these lines are still independent throughout $U$. (This is possible since linear independency is an open condition.) The kernel is the intersection of the orthogonal spaces relative to each of those. Pick a basis $e_1,\cdots, e_{n-i}$ of the kernel of $A(p)$. Now, consider $$\mathrm{GS}((l_1(x),l_2(x),\cdots,l_i(x),e_1,\cdots,e_{n-i}))=(l_1'(x),l_2'(x),\cdots,l_i'(x), v_1(x),\cdots,v_{n-1}(x)), $$ where $\mathrm{GS}$ is the application of the Gram-Schmidt process. By construction, each $v_j(x)$ is still in the intersection of the orthogonal spaces to the vectors $l_1(x)$, so they are still in the kernel. Since we are assuming that the dimension of the kernel is constant, they span the kernel entirely. This gives a framing of the kernel, and thus we have that $\mathcal{E}$ is indeed a fiber bundle.

Now the fact that a continuous choice of elements in the kernel can be done in the case of $\Omega$ a contractible open set (still under the assumption that all $A$ have kernels of constant dimension) follows due to the fact that every fiber bundle over a contractible set is trivializable. (c.f. https://math.stackexchange.com/a/1085448 .)

Note that the assumption that they have kernels of constant dimension is important. For instance, as in the initial example of this answer and for the same reasons, the map $A: \mathbb{R}^3 \to \mathbb{R}^{3 \times 3}$ does not admit continuous map $p \mapsto x_p$ to elements of its kernels, but we have that $A(0)=0$, and therefore the dimension of the kernel changes. (It is $2$ outside $p=0$ and $3$ at $p=0$.)

Even if $\Omega=\mathbb{R}$, if we don't assume that the kernels have constant dimension, we cannot guarantee the existence of such a continuous choice of vectors. For instance, take the map \begin{align*} f: \mathbb{R} &\to \mathbb{R}^{2 \times 2} \\ x &\mapsto \begin{pmatrix} f(x) & 0 \\ 0 & g(x) \end{pmatrix}, \end{align*} where $f$ is a smooth function which is zero for $x \geq 0$ and nonzero elsewhere and $g$ is a smooth function which is zero for $x \leq 0$ and nonzero elsewhere. Suppose that there was such a $p \mapsto x_p$. Then, by taking $p \mapsto x_p/\Vert x_p\Vert,$ we would have a continuous map on $\mathbb{R}$ with image on $\{(0,1), (0,-1), (1,0), (-1,0)\}$ consisting of two different points, which is a contradiction, since $\mathbb{R}$ is connected.

Just to be clear, if we assume that the kernels have constant dimension, then the result is true for $\Omega=\mathbb{R}^l$, since it is contractible.

Aloizio Macedo
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  • Great answer! Do you know if there is also a counterexample for $\Omega=\mathbb R$? – Claire Aug 23 '19 at 00:14
  • @Chiara If you make the assumption that the kernels of $A_p$ are of constant rank and that the domain $\Omega$ is contractible, I think the result is true. This is due to the fact that, if I'm not mistaken, the kernels will form a bundle over the domain $\Omega$, which is trivializable if the domain is contractible. If you want, I could expand the answer. – Aloizio Macedo Aug 23 '19 at 15:32
  • Yes, that would be great! – Claire Aug 23 '19 at 15:51
  • @Chiara I expanded the answer. Please, see if it is helpful. – Aloizio Macedo Aug 23 '19 at 18:25
  • Did i understand correctly that you use the $C^r$- version of this lemma to show that $\mathcal{E}$ is a subbundle of $\Omega\times\mathbb R^n$? (And i think the $l_i(x)$ change after applying GS but it doesn't really matter.) Also the last thing i am interested in is if there is a counterexample for $\Omega=\mathbb R$ but maybe i ask this in another question. – Claire Aug 23 '19 at 23:29
  • @Chiara Yes, for $\Omega \times \mathbb{R}^{n \times n}$. And you were right about the $l_i$. A counterexample for $\mathbb{R}$ can be seen as follows: pick $x \mapsto A(x)=\begin{pmatrix}f(x) & 0 \ 0 & g(x)\end{pmatrix}$, where $f$ is a smooth function which is $0$ for $x \geq 0$ and nonzero for $x<0$ and $g$ is a smooth function which is $0$ for $x \leq 0$ and nonzero for $x > 0$. Then you can see that there is no way to choose a smooth $p \mapsto x_p$ due to the break at $0$. I expanded the answer to give this counterexample. – Aloizio Macedo Aug 24 '19 at 02:46
  • Thank you very much i learned alot! I still think it should be $\Omega\times\mathbb R^n$ though. – Claire Aug 24 '19 at 03:54
  • @Chiara Yes, you're right! – Aloizio Macedo Aug 24 '19 at 12:50