This is a follow-up to the question https://math.stackexchange.com/questions/3845080/linearly-dependent-vector-fields-that-are-not-spanned-by-fewer-continuous-vect, which attracted very little attention.
Consider the vector fields $w_1(p)=(a,b,0,0)$, $w_2(p)=(0,0,a,b)$, $w_3(p)=(c,0,d,0)$, $w_4(p)=(0,c,0,d)$ in $\mathbb{C}^4$, where the coordinates of a point are given by $p=(a,b,c,d) \in \mathbb{C}^4$. It is easy to see that these fields are linearly dependent at each point (in fact, they are also $\mathbb{C}[a,b,c,d]$-linearly dependent when seen as elements of a module).
However, I suspect that there are no three continuous vector fields $v_1, v_2, v_3$ such that $w_i(p) \in \langle v_1(p), v_2(p), v_3(p) \rangle$ for all $p \in \mathbb{C}^4$ and all $i \in \{1,2,3,4\}$. Note that I allow for $v_1, v_2, v_3$ to vanish at some points, if this is needed. Are there general techniques to prove such things?
I've tried -as in the linked question- studying the vector bundle that they generate outside the singularities. Notice that the span of $w_1, w_2, w_3, w_4$ outside of the planes $\Pi_1:=\{p \in \mathbb{C}^4: a=b=0\}$ and $\Pi_2:=\{p \in \mathbb{C}^4: c=d=0\}$ is $3$-dimensional, so if we restrict the vector fields to the real numbers they form a vector bundle of rank $3$ over $\mathbb{R}^4 \setminus (\Pi_1 \cup \Pi_2)$. You can then circle the holes of the intersection of the $3$-sphere with this space with a torus and I had hoped that the restricted bundle over this torus would be non-trivial. Unfortunately, it is trivial, here is a trivialization that I extended to all of $\mathbb{R}^4 \setminus (\Pi_1 \cup \Pi_2)$: $v_1=(c,0,d,0)$, $v_2=(0,c,0,d)$, $v_3=(ad, bd, -ac, -bc)$. Indeed, in $\mathbb{R}^4 \setminus (\Pi_1 \cup \Pi_2)$ we have $(c^2 + d^2)w_1 = acv_1+bcv_2 +dv_3$ and $(c^2+d^2)w_2 = adv_1+bdv_2 -cv_3$.
So is there a better way of doing these things? Or do these vectors actually exist in this case and I missed an easy decomposition?
