# Gluing locally defined continous functions over complex domain

This is a cross-post to the question I asked at MSE over almost a month ago.

Suppose $n, l, m \in \mathbb N$ and $n \ge l > m$. Let $T: \mathbb C \to \mathcal M(n \times l; \mathbb C)$ be continuous and $T(\lambda)$ has constant rank $m$ for every $\lambda \in \mathbb C$. Suppose for every $\lambda \in \mathbb C$, there is an open neighborhood $U(\lambda)$ of $\lambda$ and a locally defined continuous function $h_{\lambda}: U(\lambda) \to \mathcal M(n \times m; \mathbb C)$ where the image $h_{\lambda}(\beta)$ is a basis for $T(\beta)$ for every point $\beta \in U(\lambda)$. Could we construct a globally continuous function $\phi: \mathbb C \to \mathcal M(n \times m; \mathbb C)$ by gluing together these locally defined $h_{\lambda}$'s, such that the image $\phi(x)$, of every point $x \in \mathbb C$, is a basis for $T(x)$?

What I have in mind: Suppose we have two continuous families of $g_1: \mathbb C \supseteq U_1 \to \mathcal M(n \times m; \mathbb C)$ and $g_2 : \mathbb C \supseteq U_2 \to \mathcal M(n \times m; \mathbb C)$ where $U_1$ and $U_2$ are two open disks in $\mathbb C$ with $U_1 \cap U_2 \neq \emptyset$. Then $(g_1^1(x), \dots, g_1^m(x))$ is a basis for $T(x)$ for all $x \in U_1$ and $(g_2^1(y), \dots, g_2^m(y))$ is a basis for $T(y)$ for all $y \in U_2$ where $g_i^j: x \mapsto \mathbf c_j( g_i(x))$ where $\mathbf c_j(\cdot)$ denotes the operation of taking the $j^{th}$ column of a matrix. Now for every $y \in U_2 \cap U_1$, ${g}_1 (y) = g_2(y) S(y)$ for some $S(y) \in GL_m(\mathbb C)$. It follows $S(y) = (g_2^T(y) g_2(y))^{-1} g_2^T(y) g_1(y)$. By Cramer's rule, $S(y)$ is continuous with respect to $y$. That is we have a well defined continuous function on $U_1 \cap U_2$, $S \colon U_1 \cap U_2 \to GL_m(\mathbb C)$. Since $U_1 \cap U_2$ is simply connected, there is a continuous retract $r :\mathbb C \to U_1 \cap U_2$. It follows $S \circ r : \mathbb C \to GL_m(\mathbb C)$ is well defined and continuous with $S \circ r|_{U_1 \cap U_2} = S$. So we can define $g$ by \begin{align*} g(x) = \begin{cases} g_1(x), & \text{ if } x \in U_1 \setminus U_2 \\ g_2(x)(S\circ r)(x) , & \text{ if } x \in U_2. \end{cases} \end{align*} This function is clearly continuous by construction and satisfies the prescribed condition.

I am not sure whether above argument is completely correct. Even if it was correct, my problem to proceed is: after we glue some collection of maps, the domain $U$ would become not so "regular". I am not sure the intersection of $U$ and an open disk would still be simple connected.

• Probably I am missing something, but why aren't the columns of $T(\lambda)$ a basis for its image? Jul 10 '18 at 4:24
• @FanZheng: An obvious mistake. The rank I have in mind is something smaller than $m$. Thanks for pointing out. Jul 10 '18 at 4:40
• Then it looks like the question "is every rank $l$ complex vector bundle over $\mathbb C$ trivial?", which is true because $\mathbb C$ is contractible. Jul 10 '18 at 20:57
• I am aware of the argument to assert the existence. But more interested in the explicit construction by gluing together locally defined maps. Is this doable? Jul 10 '18 at 21:47
• I guess it is precisely because of the combinatorial complications arising from the "gluing local data" argument that full-fledged theories of vector bundles, characteristic classes and sheaf cohomology are born. Jul 11 '18 at 4:59

I am not sure whether this gluing method will work, but as I mentioned in the comment, this problem is equivalent to "whether every rank $m$ (typo in the comment, sorry) complex vector bundle over $\mathbb C$ is trivial". To see this, first construct the trivial bundle of rank $n$ over $\mathbb C$, that is, $\mathbb C\times\mathbb C^n$. For every $x\in\mathbb C$, we have an $m$ dimensional subspace $E_x$ spanned by the columns of the matrix $T(x)$. By the local condition, there is a map $\phi_x:U(x)\times\mathbb C^m\to\cup_{y\in U(x)} E_y$, $(y,v)\mapsto h_x(y)v$. The transition maps $\phi_y^{-1}\circ\phi_x: (U(x)\cap U(y))\times\mathbb C^m\to(U(x)\cap U(y))\times\mathbb C^m$ are continuous, as the second paragraph of the question shows. Therefore $\phi_x^{-1}$ are local trivializations of $E:=\cup_{x\in\mathbb C} E_x$, which is then a complex vector bundle of rank $m$ over $\mathbb C$. The desired global continuous function $\phi$ similarly gives rise to a continuous function $\mathbb C\times\mathbb C^m\to E$ mapping $\{x\}\times\mathbb C^m$ to $E_x$, and hence establishes an isomorphism between $E$ and the trivial bundle $\mathbb C\times\mathbb C^m$.
Now the question reduces to "whether every rank $m$ complex vector bundle over $\mathbb C$ is trivial". The answer is Yes and the reasoning is the following. Since $\mathbb C$ is contractible, the identity map $i$ on $\mathbb C$ is homotopic to the map $j:\mathbb C\to\{0\}$. By Theorem 1.6 of http://pi.math.cornell.edu/~hatcher/VBKT/VB.pdf, $E=i^*E$ is isomorphic to $j^*E=\mathbb C\times E_0$, which is trivial because $E_0$ is isomorphic to $\mathbb C^m$.