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Suppose $f: \mathbb C \to \mathcal M(n \times n; \mathbb C)$ is a continuous function such that for each $\alpha \in \mathbb C$, $f(\alpha)$ has constant rank $r < n$. The question is whether we can choose a continuous function $g: \mathbb C \to \mathcal M(n \times r; \mathbb C)$ such that $g(\alpha)$ is a basis for the kernel $\ker f(\alpha)$.

There is a similar question asked here. The domain of the parameter was not specified and there is an answer there by considering the domain of parameter to be $\mathbb R$, i.e., $f: \mathbb R \to \mathcal M(n \times n; \mathbb R)$. I am particularly interested in the case of $\mathbb C$ and could not see directly to adapt the answer. Could someone shed some light on this? Thanks.

p.s。： I asked the author of the answer but got no reply. And it was also mentioned this can be solved by a result of vector bundles. If possible, I would also be interested in the setup in that language.

user1101010
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  • I am willing to say its more or less clear that it can be done. Depends how much detail you want to write it down in. You could try going via the row space. The row spaces will be a continuously-varying family of $r$-dimensional subspaces. The kernel is the orthogonal complement. So you can certainly "get your hands on" a continuously varying family of subspaces. Then I would guess its a standard face about grassmannians that this can be done – SBK Jun 11 '18 at 20:14
  • Wait a minute, $g(\alpha)$ is a matrix. What do you mean that you want it to be a basis? I guess that the column space of $g(\alpha)$ is the kernel of $f(\alpha)$? – SBK Jun 11 '18 at 20:17
  • @T_M: Yes. For each $\alpha$, $g(\alpha)$ is basis of $\ker f(\alpha)$. – user1101010 Jun 11 '18 at 20:18
  • @T_M: The answer I referenced gives a continuous local construction over $\mathbb R$. I think over $\mathbb C$, it would be the same. The problem is how to make it global. – user1101010 Jun 11 '18 at 20:19
  • But the referenced answer explains why local construction suffices. What do you think goes wrong over C? – SBK Jun 11 '18 at 20:25
  • @T_M: That part works for $\mathbb R$, yes. But for $\mathbb C$, how do we 'glue' those sets of basis continuously? Could you explain your idea? – user1101010 Jun 11 '18 at 20:30
  • Let us continue this discussion in chat. – user1101010 Jun 11 '18 at 20:32
  • @T_M: Will appreciate if you can write your idea as an answer. The problem for me is: if we have two open sets $U_1$ and $U_2$ with $U_1 \cap U_2 \neq \emptyset$, then we pick $x_0 \in U_1 \cap U_2$, it is true we can have invertible matrix. But how do we define the function? In $\mathbb R$, we have an ordering. But in $\mathbb C$, we don't have this ordering. So when do we switch basis over $\mathbb C$? – user1101010 Jun 11 '18 at 22:55

1 Answers1

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Here is a sketch using $G$-principal bundles : a space $X$ is a $G$-principal bundle if there is a free transitive action $G$ on $X$. The idea is that for such space, we obtain a map $p : X \to Y$ (where $Y = X/G$, which is a manifold), with fibers isomorphic to $G$ but without an explicit bijection. So locally, it's a product with $G$ but not globally. For example, the Klein bottle is a $S^1$-principal bundle over $S^1$ which is not a global product $S^1 \times S^1$.

Claim : a $G$-bundle is trivial (meaning that $X \cong Y \times G$ and the action is multiplication of the left factor) if and only if there is a section, that is a map $\sigma : Y \to X$ with $p \circ \sigma = id_{X/G}$.

Proposition : There exists a space $BG$ so that isomorphism classes of $G$-bundles over $Y$ are classified by $[Y,BG]$. Here $[A,B]$ is the homotopy classes of map $A \to B$.

In particular, if $Y$ is contractible and $X$ a $G$-principal bundle over $Y$ there is always a section.

Back to your problem :

We take $Y= \Bbb C$ and $f : Y \to R$, where $R$ is the space of matrices of rank $r$. There is a $\rm{GL}_r$-bundle over $R$, called $E_r$. The fiber is as follows : over $A \in R$, the fiber consists of all the bases of $\ker(A)$, abstractly $\rm{GL}(\ker(A))$. This is a $\rm{GL}_r$-principal bundle, and a section of this bundle is exactly the data for each $A \in R$ of a basis of $\ker(A)$. Well, this bundle is certainly not trivial but we don't care : we look at the pullback of $E_r$ by $f$. By definition, fiber of the pullback $f^*E_r$ coincide with fibers of the original bundle, meaning that we have a bundle over $\Bbb C$ so that the fiber over $\alpha$ is given by all basis of $\ker(f(\alpha))$. By the discussion before, a section always exists over $Y$.

Remark : It would be interesting to see an explicit solution, and also to find a non-trivial family with $Y = \Bbb C^*$ instead.

Nicolas Hemelsoet
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  • Thanks for the answer. For the real case, the answer I mentioned managed to 'glue' together the kernel continuously. But for complex field, I could not see how this gluing process can be explicitly described. I can understand the idea of your proof but feel like we are using a sledge hammer for this problem. – user1101010 Jun 11 '18 at 19:59
  • @iris2017 : I am definitely using a sledge hammer. The tricky part is to write down explicit trivialization of $E$. It should not be impossible but it doesn't look really easy. – Nicolas Hemelsoet Jun 11 '18 at 20:02
  • Could you explain what is your definition of $GL(\ker (A))$ and $C^*$ in the remark? Thanks. – user1101010 Jun 11 '18 at 20:13
  • @iris2017 : $GL(ker(A))$ is the group of invertible linear map $f : \ker(A) \to \ker(A)$. $\Bbb C^* = \Bbb C \backslash {0}$, but I think also (I'm not 100% sure) that any map $f : \Bbb C^* \to R$ also has a continuous basis of $\ker(f)$. But in theory there should exists maps $S^2 \to R$ without such continuous basis. – Nicolas Hemelsoet Jun 11 '18 at 20:19
  • Hi Nicolas. In another thought of the problem, I am wondering on what subset of $\mathbb C$, we can specify the image of $g$ at different points $z \in \mathbb C$? We can certainly specify the image at one point by right multiplying a general linear map? Can we do this for a finite subset of $\mathbb C$? – user1101010 Jul 16 '18 at 18:13