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It is not difficult to see that there are non-trivial stably trivial bundles of rank $2$ for a closed surface $\Sigma$ of genus $\neq 1$ using that $T \Sigma$ is non-trivial, but what happens in the genus $1$ case? Are there stably trivial real vector bundle over $T^2$ which are non-trivial?

In case the answer is affirmative, I'm actually interested in the following situation. Suppose we have an orientable real vector bundle $E \rightarrow T^2$ of rank $2$ such that $E \oplus \epsilon^2$ is trivial, where $\epsilon^2 \rightarrow T^2$ is the trivial real vector bundle of rank $2$. Can we conclude in this case that $E$ is also trivial?

Thank you in advance.

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    $\begingroup$ Because it might help: the torus is stably $S^n \vee S^{n-1} \vee S^{n-1}$ so its (real) K theory is $\mathbb{Z}_2^2$. $\endgroup$ Jul 11 '19 at 0:42
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I solved the question, I'll leave the answer for reference.

The answer is yes for the first question and no for the second. Consider the tangent bundle of $T^2 \times S^2$, which is trivial. $T^2 \times S^2$ contains tori of nonzero self-intersection number. Pick one such torus, say $Y$, and let $E \rightarrow Y$ be the restriction of the tangent bundle of $T^2 \times S^2$ to $Y$. This is a rank $4$ trivial vector bundle over a torus, and it decomposes as $E = TY \oplus NY$, where $TY$ is the tangent bundle and $NY$ the normal bundle. Since a torus is parallelizable, $TY$ is trivial, so we are in the situation of the second question above. However, $NY$ is not trivial, since its Euler class coincides with its self-intersection number, which is nonzero.

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    $\begingroup$ Oh also I just noticed that I don't understand why the tangent bundle to $T^2 \times S^2$ is trivial. Is the explanation comment-sized? $\endgroup$
    – Todd N
    Jul 11 '19 at 13:57
  • $\begingroup$ @Todd N: Right! Thanks, I've fixed it. As for why $T^2 \times S^2$ is parallelizable: its tangent bundle is isomorphic to $\pi_1^*TT^2 \oplus \pi_2^*TS^2 = \pi_2^*TS^2 \oplus \epsilon^2$, since the tangent bundle of a torus is trivial. But $\pi_2^*TS^2 \oplus \epsilon^1 = \pi_2^*(TS^2 \oplus \epsilon^1)$ is trivial since $TS^2 \oplus \epsilon^1 \rightarrow S^2$ is the trivial bundle (think of it as the tangent-normal bundle decomposition of $T\mathbb{R}^3|_{S^2}$ for the usual embedding of $S^2$ into $\mathbb{R}^3$). $\endgroup$
    – Charlie
    Jul 11 '19 at 14:46
  • $\begingroup$ Makes sense, thanks for the explanation. $\endgroup$
    – Todd N
    Jul 11 '19 at 15:02

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