# Does there exist a continuous path between two sets of oriented basis for a vector space out of a collection of subspaces?

This is cross post to the question at MSE.

Let $$V_1, V_2, \dots, V_n$$ be a collection of vector subspaces in $$\mathbb R^n$$. For each $$j=1, \dots, n$$, $$\dim(V_j) = m$$ with $$2 \le m < n$$. We also have the condition: for any collection of $$\lceil{\frac n m}\rceil$$ vector spaces from $$\{V_1, \dots, V_n\}$$, then $$V_{k_1}+ \dots+ V_{k_{\lceil \frac n m \rceil}} = \mathbb R^n$$. Suppose we construct a basis $$U = \{u_1, \dots, u_n\}$$ of $$\mathbb R^n$$ in the manner: $$u_j \in V_j$$ for each $$j$$. Now suppose we construct another basis $$W = \{w_1, \dots, w_n\}$$ in the same manner, i.e., $$w_j \in V_j$$ for each $$j$$. I am wondering whether $$U$$ is connected with $$W$$ in the sense: there is a path $$\gamma = \gamma_1 \times \gamma_2 \times \dots \times \gamma_n$$, where each $$\gamma_j: [0,1] \to V_j$$ is a continuous path connecting $$v_j$$ and $$w_j$$ in $$V_j$$ and for each $$t$$: $$\gamma(t)$$ forms a basis for $$\mathbb R^n$$. We assume the basis $$\{v_j\}$$ and $$\{w_j\}$$ have the same orientation.

The basis can be identified by $$GL_n(\mathbb R)_+$$ or $$GL_n(\mathbb R)_-$$ and we know they are connected. But is there a way to guarantee on the path, each column vector only varies in the corresponding subspace?

An example of $$V_1, \dots, V_n$$: suppose $$n=5$$, $$m=2$$. The construction I have in mind is: $$V_i = \text{span} ( (1, a_i, 0, 0, 0), (0, 0, 1, a_i, a_i^2))$$. As long as $$a_1 \neq \dots \neq a_5 \neq 0$$, any three subspace would span $$\mathbb R^5$$. For other cases, we can use similar idea.

• For the variation of the problem with $\mathbb{R}$ replaced with $\mathbb{C}$ (not sure it helps): consider the set $S$ of all tuples $v_1,...,v_n$, where $v_i\in V_i$. Clearly, $S$ is a vector space, and so we can identify it with a linear subspace of $\mathbb{C}^{n^2}$. Then $\det(v_1,...,v_n)$ is a polynomial of a degree at most $n$ on $S$, who is linearly isomorphic to $\mathbb{C}^{mn}$. Since the zero-set of an analytic function does not separate $\mathbb{C}^{mn}$, it follows that the bases for a connected set. – erz Nov 18 '18 at 13:06