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Suppose that $P(t)$ is a one-parameter family of rank 2 self-adjoint projections on $\mathbb{C}^n$ that vary analytically in the real parameter $t \in [0,1]$. I claim that there must exist a vector $x \in \mathbb{C}^n$ such that $P(t)x \neq 0$ for all $t$.

In other words, I am hoping to prove that $\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t) \neq \mathbb{C}^n$. This seems like it is essentially a space-filling curve type argument, and hence the requirement that $P(t)$ be differentiable is probably important (in my example, $P(t)$ is analytic in $t$). Does anyone know a reference that would provide a simple proof of this claim?

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    $\begingroup$ If $P$ is $C^1$ then isn't it just Sard's theorem ? $\endgroup$ – Simon Henry Jul 8 '14 at 15:41
  • $\begingroup$ Maybe it's easier to prove after blowing up at the origin? $\endgroup$ – Marco Golla Jul 8 '14 at 16:53
  • $\begingroup$ Applying Sard's theorem is an interesting idea, but I don't see how it can be directly applied to $P(t)$. What function and what critical values did you have in mind? $\endgroup$ – Brian Lins Jul 8 '14 at 17:10
  • $\begingroup$ Very close to mathoverflow.net/questions/139593 $\endgroup$ – David E Speyer Jul 9 '14 at 20:09
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Let $Q(t):\mathbb C^n\to\mathbb C^n$ be the orthogonal projection onto $\text{ker}(P(t))$. Then $t\mapsto Q(t)$ is as differentiable as $P$ was. Now the mapping $(t,x)\mapsto Q(t)(x)$ has rank at most $2n-2+1<2n$. Apply Sard's theorem: The set of regular values is Lebesque nearly everything. Take a regular value $y$. It cannot be in the image, and thus is in no kernel of $P(t)$ for any $t$.

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I assume $P$ varies analytically as in the body of the question, I do not know how to tackle the case where $P$ is just differentiable, I also read rank 2 as real rank though of course the problem is even easier if the rank is complex.

The map $\operatorname {ker}:A\rightarrow\operatorname {Gr}_{2n-2}\Bbb R^{2n}$ where $A\subseteq End(\Bbb R^{2n})$ are the endomorphisms of rank $2$ is $C^\infty$. Clearly the rank of $D(\operatorname {ker}\circ P)$ is less than or equal to 1. Consider the tautological bundle over the Grassmanian $T$ and the projection map $g:T\rightarrow \Bbb R^{2n}$. Since $T$ is locally trivial and second countable, we have $T$ is a countable union of spaces diffeomorphic to $\operatorname {Gr}_{n-2}\Bbb R^{2n}\times \Bbb R^{2n-2}=M_n$. Consider $f:\Bbb R\times\Bbb R^{2n-2} \rightarrow M_n$, $f_n(x,y)= (\operatorname {ker}\circ P(x),y)$. The rank of $D(g\circ f_n)$ is less than or equal to $2n-1$, so by Sard's theorem we have $\operatorname{img}(g\circ f_n)$ is measure zero in $\Bbb R^{2n}$. So $\bigcup_n\operatorname{img}(g\circ f_n)=\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t)$ is a countable union of measure zero sets hence its complement is non-empty.

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