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I need an example of vector bundles $E_1 \to B$, $E_2 \to B$ such that $E_1, E_2$ are nontrivial and $E_1 \oplus E_2 $ is trivial.There aren't much of a bundles which I can prove to be nontrivial. One of them is the nontrivial line bundle over the circle (which turns the fiber upside down). I tried to check if the sum of two copies of those bundles is trivial.

It looks to me that it isn't because the gluing map is just the matrix $\operatorname{diag}(-1, -1)$ so it still turns things upside down and therefore any section must be zero on the intersection. I am not sure of it, though.

So what is the simplest example and does the bundle over the circle work?

Remark: I don't know characteristic classes and I am not allowed to use them anyway.

Invincible
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    You're on the right track with the two Möbius strips, but it's not true that any section must vanish on the intersection. Perhaps it would help to think of the sum as the quotient of $\mathbb{R}\times\mathbb{R}^2\to\mathbb{R}$ by the equivalence$$(x,\vec{v})\sim(x+n,(-1)^n\vec{v})\ \ \ \ \ n\in\mathbb{Z}$$ – Kajelad May 07 '21 at 01:11
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    This may be of interest. – Andrew D. Hwang May 07 '21 at 01:47
  • @ Andrew D. Hwang, it could be but I don't understand it. – Invincible May 07 '21 at 08:59
  • IIRC, that's a picture of of a sum of two Moebius bands being a trivial rank-$2$ bundle over the circle. – Andrew D. Hwang May 12 '21 at 13:59

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Question: "So what is the simplest example and does the bundle over the circle work?"

Answer: An elementary example is the real n-sphere. If $k$ is the real number field and $f:=x_1^2+\cdots + x_n^2-1$ with $A:=k[x_1,..,x_n]/(f)$ it follows there is an exact sequence

$$ 0 \rightarrow Adf \rightarrow A\{dx_1,..,dx_n\} \rightarrow \Omega^1_{A/k}\rightarrow 0$$

and $\Omega^1_{A/k}$ is a non-trivial projective $A$-module in general. Hence

$$\Omega^1_{A/k} \oplus A \cong A^n$$

is such a sum. A similar result holds for $Hom_A(\Omega^1_{A/k},A):=T_A$.

I believe the same result holds for the real $n$-sphere $S^n$ viewed as a differentiable manifold. The tangent bundle $T_{S^n}$ (and cotangent bundle $\Omega^1_{S^n}$) is non-trivial and you get a similar result:

$$T_{S^n}\oplus \mathcal{O}_{S^n} \cong \mathcal{O}_{S^n}^n.$$

https://en.wikipedia.org/wiki/Hairy_ball_theorem

hm2020
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    Unfortunately, I don't know algebraic geometry any further than the definition of variety and Zariski topology. I was talking about smooth situation. – Invincible May 07 '21 at 08:57
  • @Vladislav - the result is true for the real n-sphere viewed as a differentiable manifold. Hence you get an elementary and non-trivial example that is non-algebraic – hm2020 May 07 '21 at 13:51