Complex Vector Bundles with Trivial Chern Classes on k-Tori – Algebraic Topology

Question

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?

Answer 1

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.