Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $S^11\times S^1$ 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?

share|cite|improve this question
Whoever has Borges as epigraph on his homepage can't be all bad :-) Great quotation, Domenico! – Georges Elencwajg Aug 11 '10 at 7:35

1 Answer 1

up vote 12 down vote accepted

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.

share|cite|improve this answer
Thanks. I knew the stable answer by Kunneth, but I was really interested in the unstable one. Also had $(S^1)^5$ as my favourite candidate, but I had been unable to prove that the pullback bundle you consider was nontrivial. Could you add a few details of the proof? Thanks. – domenico fiorenza Aug 11 '10 at 15:08
Thanks a lot for the details. Actually in a paper I'm writing I'm interested only in a very specific bundle on a 3-torus, and in that case I knew the answer, but as a side remark it was nice to say what happened for higher dimensions. Now, what's absolutely great of MO is that I can not limit myself to an obscure acknowlkedgement such as "We thank Torsten Ekedahl for having shown us a counterexample in dimension 5", but I can address the reader to this web page here! By the way, any idea of what happens in dimension 4? – domenico fiorenza Aug 11 '10 at 17:41
Dimension $5$ is the first dimension where things can happen. In general if $X$ is of dimension $\leq n$ and $\mathrm{BU}(2)\to U_n$ is the $n$'th Postnikov approximation, then $[X,\mathrm{BU}(2)]\to[X,U_n]$ is a bijection. In particular for $n=4$ we have $U_4=K(\mathbb Z,4)\times K(\mathbb Z,2)$ which means that a $U(2)$ bundle is determined by its Chern classes. (Implicitly I use the same thing for $U=U_5$ so that I only need to look at $U$ instead of $U(2)$.) – Torsten Ekedahl Aug 11 '10 at 17:49
Crystal clear. Thanks a lot! – domenico fiorenza Aug 11 '10 at 18:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.