# non-triviality of the underlying real vector bundle of the complexification of a real vector bundle

Let $M$ be a given manifold and $\xi$ be a given $k$-dimensional vector bundle over $M$. How to determine whether the underlying real vector bundle of $\xi\otimes\mathbb{C}$, i.e. the Whitney sum $\xi\oplus\xi$, is trivial or not? I want to try different methods as many as possible.

My attempt: To prove $\xi\oplus\xi$ is not a trivial bundle, we only need to prove the Stiefel-Whitney class $$1\neq (w(\xi))^2=(1+\sum_{i=1}^k w_i(\xi))^2=1+\sum_{i=1}^k (w_i(\xi))^2,$$ i.e. there exists $1\leq i\leq k$ such that $$(w_i(\xi))^2\neq 0.$$ But even the cup-product of $$H^*(M;\mathbb{Z}_2)$$ is known, I still do not know how to determine the element $w_i(\xi)$?

Are there any other methods to try?

## 2 Answers

It is a bit unclear what the question is asking exactly, so let me try to give some examples what can happen to a vector bundle $E$ when taking $E\oplus E$ (for simplicity, let me call this Whitney doubling).

Certainly, as already mentioned in the question, the non-vanishing of the Stiefel-Whitney classes is a sufficient condition for non-triviality of $E\oplus E$, and the Stiefel-Whitney classes of $E\oplus E$ are of the form $w_{2i}(E\oplus E)=w_i(E)^{\cup 2}$.

Another possibility to show non-triviality of $E\oplus E$ would be using the Pontryagin classes which are classes in $H^{4k}(M,\mathbb{Z})$. We have $2p(E\oplus E)=2p(E)^{\cup 2}$, and non-vanishing of the Pontryagin classes of $E\oplus E$ will imply non-triviality. In favourable cases, this will be possible by computations in rational cohomology.

However, real vector bundles are not determined by their characteristic classes in general. Generally, classification results can be obtained using obstruction theory (to the extent that homotopy groups of $O(n)$ can be computed). The general formalism is explained e.g. in Hatcher's book on algebraic topology. For $M$ a manifold, the obstruction to lifting a map $M\to BSO(n)^{(k)}$ along $BSO(n)^{(k+1)}\to BSO(n)^{(k)}$ lies in $H^{k+1}(M,\pi_{k-1}BSO(n))$, and the possible lifts are classified by $H^k(M,\pi_{k-1}BSO(n))$. So the vector bundle classification hinges on knowledge of cohomology with coefficients in homotopy groups. Moreover, knowing the induced morphism on homotopy groups $\pi_k(O(n))\to\pi_k(O(2n))$ allows to analyze what happens to vector bundles under Whitney doubling. The following examples can be obtained by obstruction theory:

• As a very simple example, the morphism $O(1)\to O(2)$ is null-homotopic, hence every line bundle $L$ will have $L\oplus L$ trivial (generalizing Ben McKay's answer).

• For $O(2)\to O(4)$, the only nontrivial map induced on homotopy groups is the reduction mod 2 $\mathbb{Z}\cong\pi_1(O(2))\to\pi_1(O(4))\cong\mathbb{Z}/2\mathbb{Z}$. In particular, the Whitney doubling of a rank 2 bundle will be trivial iff the Euler class is even.

• To get some more interesting examples, recall the paper of Kervaire ( Non-parallelizability of the $n$-sphere for $n>7$. Proc.N.A.S. 44 (1958), 280-283.) Lemma 1 states that the induced morphism $\pi_k(O(n))\to\pi_k(O(2n))$ is twice the stabilization morphism for $k<n-1$ (where the stabilization morphism happens to be an isomorphism). In particular, the stable range lifting classes in the obstruction classification will just be multiplied by $2$ under the Whitney doubling. Now let's use this to get examples of bundles $E$ with trivial Stiefel-Whitney classes such that $E\oplus E$ is stably non-trivial. Consider a manifold $M=S^5/(\mathbb{Z}/8\mathbb{Z})$ which has $H^4(M,\mathbb{Z})=\mathbb{Z}/8\mathbb{Z}$. Any such class $\sigma$ gives a rank $5$ bundle on $M$ whose Stiefel-Whitney $w_4$ class will be the reduction mod $2$, and whose Pontryagin class is $2\sigma$. If we choose $\sigma$ to be twice a generator, we get a bundle with trivial Stiefel-Whitney classes and Pontryagin class $2$-torsion. Now employ the result of Kervaire, which states that the morphism $\pi_3(SO(5))\to\pi_3(SO(10))$ induced by Whitney doubling is multiplication by two. This means, that the Whitney doubling will have lifting class four times a generator of $H^4(M,\mathbb{Z})$. Hence the Whitney doubling is stably non-trivial, but has all characteristic classes trivial.

• Better examples exist in higher dimensions. The homomorphism $\pi_{4k}(BO)\to\mathbb{Z}$ defining the $k$-th Pontryagin class is multiplication with $(2k-1)!$ if $k$ is even and multiplication by $2(2k-1)!$ if $k$ is odd (see the paper of Bott and Milnor on parallelizability of spheres). In particular, we can construct bundles over $S^{13}/(\mathbb{Z}/8\mathbb{Z})$ whose only lifting class is twice a generator of $H^{12}$. These bundles will have all characteristic classes trivial, but will have non-trivial Whitney doubling.

The Moebius strip as a bundle $\xi$ over the circle $M=S^1$ has $\xi \oplus \xi$ trivial.