Stably trivial vector bundles over a torus

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?

• Because it might help: the torus is stably $S^n \vee S^{n-1} \vee S^{n-1}$ so its (real) K theory is $\mathbb{Z}_2^2$. Jul 11 '19 at 0:42
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.
• 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? Jul 11 '19 at 13:57
• @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 $\mathbb{R}^3$). Jul 11 '19 at 14:46