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.

  • $\begingroup$ Probably I am missing something, but why aren't the columns of $T(\lambda)$ a basis for its image? $\endgroup$
    – Fan Zheng
    Jul 10, 2018 at 4:24
  • 1
    $\begingroup$ @FanZheng: An obvious mistake. The rank I have in mind is something smaller than $m$. Thanks for pointing out. $\endgroup$ Jul 10, 2018 at 4:40
  • $\begingroup$ 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. $\endgroup$
    – Fan Zheng
    Jul 10, 2018 at 20:57
  • $\begingroup$ 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? $\endgroup$ Jul 10, 2018 at 21:47
  • $\begingroup$ 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. $\endgroup$
    – Fan Zheng
    Jul 11, 2018 at 4:59

1 Answer 1


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 $\mathbb C^m$.

PS The OP asks the question in the continuous category (on which this answer is based) but in the comments the OP mentions the smooth category. It's an exercise left to the reader (and the OP) to carry the above argument over to the smooth category.

PPS The cited book (by Allen Hatcher, who also happens to be a MO user) is a wonderful introduction to vector bundles (and beyond).


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