Let $E\to X$ be a principal $U(N)$-bundle over a (nice) topological space $X$. It is well known that vanishing of the Chern classes of $E$ is not a sufficient condition for $E$ to be trivial, the simplest example being probably the nontrivial $U(2)$-bundle over $S^5$. However one may wonder what happens for a specific base $X$, e.g, for $k$-tori. In this case it is completely trivial that any $U(N)$-bundle on $S^1$ is trivial and that an $U(N)$-bundle on $S^1\times S^1$ is trivial precisely when its first Chern class vanishes. A very little obstruction theory then shows that this holds true for $U(N)$-bundles over $S^1\times S^1\times S^1$. Such simple arguments, however do not work for $(S^1)^k$ for $k\geq 4$. Can anything general be said?



Complex vector bundles with trivial Chern classes on k-tori
As the cohomology of $(S^1)^n$ is torsion free every stable bundle on $(S^1)^n$ is
determined by Chern classes (this also follows from the $K$-theory Künneth
formula) so just as for the spheres it is an unstable problem. As for the
unstable problem unless I have miscalculated, if $(S^1)^5\rightarrow S^5$ is a
degree $1$ map, then the pullback of the non-trivial $U(2)$ bundle on $S^5$ with
trivial Chern class is non-trivial. (The proof uses that the $5$'th step in the
Postnikov tower of $\mathrm{BU}(2)$ is a principal fibration $K(\mathbb
Z/2,5)\rightarrow U\rightarrow K(\mathbb Z,4)\times K(\mathbb Z,2)$.)


Some more details of the calculation: The first and second Chern class gives a
map 
$$\mathrm{BU}(2)\rightarrow K((\mathbb Z,4)\times K(\mathbb Z,2)$$
which induces an isomorphism on homotopy groups in degrees up to $4$. As
$\pi\_i(\mathrm{BU}(2))=\pi\_{i-1}(\mathrm{SU}(2))$ for $i>2$ we get that
$\pi\_5(\mathrm{BU}(2))=\pi\_4(S^3)=\mathbb Z/2$. Hence, the next step $U$ in the
Postnikov tower of $\mathrm{BU}(2)$ is the pullback of the path space fibration
of a morphism $K(\mathbb Z,4)\times K(\mathbb Z,2)\rightarrow K(\mathbb Z/2,6)$.
In particular we have a principal fibration 
$$K(\mathbb Z/2,5)\rightarrow
U\rightarrow K(\mathbb Z,4)\times K(\mathbb Z,2).$$
This means that for any space
$X$, the image of $[X,K(\mathbb Z/2,5)]$ in $[X,U]$ is in bijection with the
cokernel of $[X,K(\mathbb Z,3)\times K(\mathbb Z,1)]\rightarrow[X,K(\mathbb
Z/2,5)]$ obtained by applying $[X,-]$ to the looping of the structure map
$K(\mathbb Z,4)\times K(\mathbb Z,2)\rightarrow K(\mathbb Z/2,6)$. As
$H^4(K(\mathbb Z,3),\mathbb Z/2)=0$ the Künneth formula shows that any map
$K(\mathbb Z,3)\times K(\mathbb Z,1)\rightarrow K(\mathbb Z/2,5)$ factors
through the projection $K(\mathbb Z,3)\times K(\mathbb Z,1)\rightarrow K(\mathbb
Z,3)$ and $H^5(K(\mathbb Z,3),\mathbb Z/2)=\mathbb Z/2\mathrm{Sq}^2\rho\iota$ (where
$\iota$ is the canonical class, $\iota\in H^3(K(\mathbb Z,3),\mathbb Z)$ and
$\rho$ is induced by the reduction $\mathbb Z\rightarrow\mathbb Z/2$). Hence,
the map $[X,K(\mathbb Z,3)\times K(\mathbb Z,1)]\rightarrow[X,K(\mathbb
Z/2,5)]$ is either the zero map or given by the composite of the projection to
$H^3(X,\mathbb Z)$, the reduction to $\mathbb Z/2$ coefficients and
$\mathrm{Sq}^2$ (I actually think it is non-zero as otherwise the cohomology of
$H^\ast(\mathrm{BU}(2),\mathbb Z)$ would have $2$-torsion). If we apply it to
$X=(S^1)^5$ we get that $[X,K(\mathbb Z,3)\times K(\mathbb Z,1)]\rightarrow[X,K(\mathbb
Z/2,5)]$ is zero provided that 
$$\mathrm{Sq}^2\colon H^3((S^1)^5,\mathbb
Z/2)\rightarrow H^5((S^1)^5,\mathbb
Z/2)$$
 is zero. However, all Steenrod squares are zero on all of
$H^\*((S^1)^n,\mathbb Z/2)$. Indeed, the Künneth and Cartan formulas reduce this to
$n=1$ where it is obvious.


