I made a number of relevant comments to this question on tigu's previous question.
Here they are again, lightly edited.
No, but this there is a true statement like this. When you describe $V_i$ as the locus where some sections are dependent, there are a lot of subtleties in that description. It is only literally correct if there are "enough" holomorphic sections and if those sections are "generic enough". Otherwise, you have to start talking about keeping tracks of multiplicities and signs.
The details of how you deal with the technicalities will be different depending on whether you are looking at algebraic/holomorphic sections (in this case, your best source is Fulton's "Intersection Theory") or whether you are looking at smooth sections (in this case, Milnor's "Characteristic Classes" has a good reputation, thought I haven't read it). I'll discuss a single example.
Look at $\mathbb{P}^2$ and let $U:= \mathcal{O}(1) \oplus \mathcal{O}(−1)$. The Chern class is $1−h^2$, where $h$ generates $H^2(\mathbb{P}^2, \mathbb{Z})$. The point is that $\mathcal{O}(−1)$ has no nonzero holomorphic sections. So all the holomorphic sections of $U$ are of the form $(0, \sigma)$ for $\sigma$ a section of $\mathcal{O}(1)$. In particular, any two holomorphic sections of $U$ are proportional everywhere on $\mathbb{P}^2$.
If we perturb the above sections so that they are not holomorphic then, I think, there will be are two curves $C$ and $C'$ on which they become dependent. One of these curve should be weighted positively and one negatively, and their degrees cancel in $H^2(\mathbb{P}^2)$. It would be fun to work out an example of this. UPDATE: I worked this out below and, at least for the two sections I tried, I got a single contractible sphere rather than two curves.
Similarly, if you look at one holomorphic section of $U$, it will vanish along an entire line in $\mathbb{P}^2$. If you perturb this section to be smooth, then it will vanish along a codimension $2$ subvariety, with multiplicity $−1$. You might want to read this question for a further understanding of how negative numbers show up as intersection multiplicities. That $-1$ explains that $-h^2$.
The true statement I allude to above should be something like "If $U$ has enough holomorphic sections, and $c_1(U)=0$, then $U$ is trivial." I do not know what the right definition of "enough" is; I would guess "globally generated".
Okay, let's actually work out the examples discussed above. We will describe a point of $\mathbb{P}^2$ using homogenous coordinates $(x:y:z)$. The fiber of $\mathcal{O}(-1)$ over $(x:y:z)$ is the line in $\mathbb{C}^3$ spanned by $(x,y,z)$. The fiber of $\mathcal{O}(1)$ over $(x:y:z)$ is the dual of that, in other words, $\mathrm{Hom}(\mathrm{Span}_{\mathbb{C}} (x,y,z), \mathbb{C})$. It is easy to give global holomorphic sections of $\mathcal{O}(1)$: Just take an element of $\mathrm{Hom}(\mathbb{C}^3, \mathbb{C})$ and restrict it to every fiber of $\mathcal{O}(-1)$. Let $\sigma_1$ be the section obtained by restricting $(x,y,z) \mapsto x$ and let $\sigma_2$ and $\sigma_3$ use the other two coordinate functions.
There are no holomorphic sections of $\mathcal{O}(-1)$, but there are lots of smooth sections. Let
$$\tau_1(x:y:z) = \frac{1}{|x|^2+|y|^2+|z|^2} ( x \overline{x}, y \overline{x}, z \overline{x}).$$
Define $\tau_2$ and $\tau_3$ similarly.
It turns out that $\sigma_1 \oplus \tau_1$ is not sufficiently generic is, but $\sigma_1 \oplus \tau_2$, so let's look at that. The section $\sigma_1$ vanishes whenever $x=0$, a line in $\mathbb{P}^2$. The section $\tau_2$ vanishes when $y=0$. However, if you check closely, you'll see that $\tau_2$ is anti-holomorphic, not holomorphic, in a neighborhood of its vanishing locus. So that vanishing should count negatively, and $\tau_2$ vanishes with multiplicity $-1$ on the line $y=0$. The section $(\sigma_1, \tau_2)$ vanishes at the point $(0:0:1)$, where $x=y=0$, and does so with sign $1 \times (-1) = -1$. This shows that $c_2 = -h^2$.
Similarly, let's look at the pair of sections $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$.
Assuming that $x$ and $y$ are nonzero, the ratio $\sigma_1/\sigma_2$ is $x/y$, while the ratio $\tau_3/\tau_2 = \overline{z}/\overline{y}$. So $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$ are proportional when $x/y = \overline{y}/\overline{z}$, in other words, when $x \overline{z} = |y|^2$. I didn't check carefully, but I believe that $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_1)$ are linearly dependent precisely when $x \overline{z} = |y|^2$, including in the degenerate cases where these coordinates vanish.
The set $\{ (x:y:z) \in \mathbb{P}^2 : x \overline{z} = |y|^2 \}$ is a sphere. I can parameterize it as $(e^{i \theta} (1+\sin \phi) : \cos \phi : e^{i \theta} (1-\sin \phi))$ where $\theta$ is the longitude coordinate, running through $\mathbb{R}/(2 \pi \mathbb{Z})$, and $\phi$ is the lattitude coordinate, living in $[-\pi/2, \pi/2]$. I claim that this sphere is contractible in $\mathbb{P}^2$. To see this, map $S^2 \times [0,1]$ to $\mathbb{P}^2$ by $(\theta, \phi, t) \mapsto (e^{i \theta} (1+\sin \phi) : (1-t) \cos \phi : e^{i \theta} (1-\sin \phi))$. At one end of the homotopy, we have the previously described embedding. At the other end, we have the line segment $\{ (1+r:0:1-r) : r \in [-1, 1] \}$. This line segment is obviously contractible.
So the space where $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$ are dependent has homology class $0$, showing that $c_1=0$.
Question: "Sorry but I have a couple of questions: Why is c(1)c(2)=c2(V) and why is c1(V)=c(1)+c(2)? And also, when we write c using these t's, they sort of serve to indicate what term corresponds to which ci right? @hm2020"
Answer: when $r=2$ the formula follows from your relation $c(V \otimes L)=\prod_j (1 + c_1(L_j') + c_1(L))$: Let $c(i):=c_1(L_i)$ and $c:=c_1(L)$. You get the calculation
$$c_t(V\otimes L)=(1+(c(1)+c)t)(1+(c(2)+c)t)= $$
$$...+(c(1)c(2)+(c(1)+c(2))c+c^2)t^2=$$
$$\cdots + (c_2(V)+c_1(V)c_1(L)+\binom{2}{2}c_1(L)^2)t^2=$$
$$c_0(V\otimes L) +c_1(V\otimes L)t+c_2(V\otimes L)t^2.$$
Note: You get
$$c_t(V)=(1+c(1)t)(1+c(2)t)= $$
$$1+(c(1)+c(2))t+c(1)c(2)t^2=c_0(V)+c_1(V)t+c_2(V)t^2$$
hence
$$c_1(V)=c_1(L_1)+c_1(L_2)\text{ and }c_2(V)=c_1(L_1)c_1(L_2).$$
Here you view the elements as "living" in a commutative ring and multiply (they all live in the even cohomology ring $H^{2*}(X)$ which is commutative) You find this explained in Hartshorne, Appendix A.
Note: Formulas for Chern classes
$c_i(E\otimes F)$ where $E,F$ have "Chern roots" $a_i,b_j$ are expressed in terms of polynomials in the roots $a_i,b_j$. The coefficient of the $t^i$ term "lives" in the group $H^{2i}(X)$. Whenever you have a theory of Chern classes $c_i(E) \in H^{2i}(X)$ living in the even part of a cohomology ring $H^*(X)$ you may perform such calculations. The experession
$$c_t(E):=c_0(E)+c_1(E)t+\cdots + c_r(E)t^r$$
"lives" in
$$H^0(X)\oplus H^2(X)t\oplus \cdots \oplus H^{2r}(X)t^r.$$
For any rank $r$ locally trivial sheaf $E$ there is the complete flag bundle $F(E)$ of $E$ and an injection
$$H^*(X) \subseteq H^*(F(E)),$$
hence you may perform all calculations in $H^*(F(E))$. In $H^*(F(E))$
you have the equality
$$c_r(E)=c_r(L_1)+ \cdots + c_r(L_r)$$
where $L_i$ are invertible sheaves. You get the formula
$$c_t(E)=c_t(L_1)\cdots c_t(L_r)$$
in $H^{2*}(X)[t]$: It holds in $H^{2*}(F(E))[t]$ but since $c_t(E) \in H^{2*}(X)[t]$ the equality holds here.
Best Answer
This is an interesting question! The 'if' part is true and I think the 'only if' part is also (but this would need verification).
Suppose that $E$ is a $SU(n)$-bundle. Given any $SU(n)$-connection form on $E$, its curvature form $\Omega$ is a basic $\mathfrak{su}(n)$-valued 2-form on the principal $SU(n)$-bundle associated to $E$. As such, it can also be understood as a $\mathfrak{su}(n)$-valued closed 2-form on $M$. According to Chern-Weil theory, the first Chern class is given by the cohomology class of $\frac{i}{2\pi} \mathrm{tr}\,\Omega$, which vanishes since the matrices in $\mathfrak{su}(n)$ are traceless.
Suppose that $E$ is $U(n)$-bundle with vanishing first Chern class. The structure group of $E$ can be reduced to $SU(n)$ if and only if there exists a $SU(n)$-principal subbundle in the $U(n)$-principal bundle $P$ associated to $E$, which is equivalent to the existence of an 'equivariant' function $f : P \to U(n)/SU(n)$ (with the canonical left $U(n)$-action on the quotient). In turn, this is equivalent to the existence of a section of the associated fiber bundle $P(U(n)/SU(n))$. Since $SU(n)$ is a normal subgroup of $U(n)$, we deduce that $Q = P(U(n)/SU(n))$ is a principal bundle, so it admits a section if and only if it is trivial. In fact, $U(n)/SU(n)$ being isomorphic to $U(1)$, it is a $U(1)$-principal bundle. Letting $U(1)$ act on $\mathbb{C}$ in the standard way, we can form the complex line bundle $F = Q(\mathbb{C})$. Thus, the reduction of the structure group of $E$ from $U(n)$ to $SU(n)$ is equivalent to the line bundle $F$ being trivial, that is, as you pointed out, equivalent to $F$ having a vanishing first Chern class.
I think, but I am not quite sure, that looking at how a connection form on $P$ induces a connection form on $Q$, we can infer that the vanishing of the first Chern class of $E$ implies the vanishing of the first Chern class of $F$. If true, the reduction to $SU(n)$ would be guaranteed.