at.algebraic topology - 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 $(S^1)^5rightarrow S^5$ 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 $K(mathbb
Z/2,5)rightarrow Urightarrow K(mathbb Z,4)times K(mathbb Z,2)$.)

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
$$K(mathbb Z/2,5)rightarrow
Urightarrow K(mathbb Z,4)times K(mathbb Z,2).$$
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 $H^5(K(mathbb Z,3),mathbb Z/2)=mathbb Z/2mathrm{Sq}^2rhoiota$ (where
$iota$ is the canonical class, $iotain H^3(K(mathbb Z,3),mathbb Z)$ and
$rho$ is induced by the reduction $mathbb Zrightarrowmathbb Z/2$). 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
$$mathrm{Sq}^2colon H^3((S^1)^5,mathbb
Z/2)rightarrow H^5((S^1)^5,mathbb
Z/2)$$
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.