Let be $M$ be a smooth manifold (possibly with boundary) and assume that the complexified tangent bundle $TM \otimes \mathbb{C}$ is trivial. Does this imply that $TM$ is stably trivial? This seems to be true for all two dimensional manifolds. Note that the triviality of $TM \otimes \mathbb{C}$ implies that $TM \oplus TM$ is trivial and hence all the characteristic classes of $TM$ are 2-torsion.
Does the answer to the above question depend on whether $M$ has boundary or is closed?
Best Answer
Let $M = \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$.
Since $\dim M = 4$, we have $TM\otimes_{\mathbb{R}}\mathbb{C} \cong E\oplus\varepsilon^2_{\mathbb{C}}$ where $E$ is a complex rank two bundle - in general, if $V \to X$ is a complex vector bundle with $\dim X = 2r$ or $2r + 1$ and $\operatorname{rank}_{\mathbb{C}}V = r + d$, then $V \cong V_0\oplus\varepsilon^d_{\mathbb{C}}$ with $\operatorname{rank}_{\mathbb{C}}V_0 = r$. As with all complexifications, $\overline{TM\otimes_{\mathbb{R}}\mathbb{C}} \cong TM\otimes_{\mathbb{R}}\mathbb{C}$ so $c_1(E) = c_1(TM\otimes_{\mathbb{R}}\mathbb{C}) = c_1(\overline{TM\otimes_{\mathbb{R}}\mathbb{C}}) = -c_1(TM\otimes_{\mathbb{R}}\mathbb{C}) = -c_1(E)$; that is, $c_1(E)$ is two-torsion. As $H^2(M; \mathbb{Z}) \cong \mathbb{Z}\oplus\mathbb{Z}$ is torsion-free, $c_1(E) = 0$. By the answers to this question, the bundle $E$ is determined up to isomorphism by $c_2(E)$. As $c_2(E) = c_2(TM\otimes_{\mathbb{R}}\mathbb{C}) = -p_1(TM)$ and $0 = \sigma(M) = \frac{1}{3}\langle p_1(TM), [M]\rangle$, we see that $c_2(E) = 0$ so $E$ is trivial, and therefore $TM\otimes_{\mathbb{R}}\mathbb{C}$ is trivial. On the other hand $TM$ is not stably trivial since $w_2(TM) \neq 0$.