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Let $M$ be a (smooth) manifold of dimension $n$, and let $\pi : E \to M$ be a (smooth) vector bundle of rank $r$. If I choose a connection on $E$, then I obtain a decomposition of $T E$ as $V E \oplus_E H E$ where $V E \cong E \times_M E \cong \pi^* E$ (right...?) and $H E \cong E \times_M T M \cong \pi^* T M$, and thus a split short exact sequence: $$0 \longrightarrow V E \longrightarrow T E \longrightarrow \pi^* T M \longrightarrow 0$$ Now, take the $(n+r)$-th exterior power of $T^* E$. The decomposition $T E \cong \pi^* E \oplus \pi^* T M$ should then yield an isomorphism as below (right...?): $$\Lambda^{n+r} T^* E \cong \Lambda^r \pi^* E^* \otimes_E \Lambda^n \pi^* T^* M \cong \pi^* \left( \Lambda^r E^* \otimes_M \Lambda^n T^* M \right)$$

It is now clear that any two of the following conditions implies the third:

  1. $\Lambda^{n+r} T^* E$ is a trivial line bundle over $E$.

  2. $\Lambda^r E^*$ is a trivial line bundle over $M$.

  3. $\Lambda^n T^*M$ is a trivial line bundle over $M$.

Condition (2) can be replaced by “$\Lambda^r E$ is a trivial line bundle over $M$” if we pick a metric on $E$. Thus, assuming $M$ is orientable, $E$ is orientable as a manifold if and only if $\Lambda^r E$ is a trivial line bundle.

Question 1. Is the above proof sketch correct? I'm more used to sheaves than vector bundles, and casually commuting $\otimes$ and $\Lambda^\bullet$ through $\pi^*$ makes me feel queasy.

Question 2. Is there a way to do this without invoking a connection or a metric? Given that neither the hypotheses nor the conclusion mention such extra structure, it feels as if there has to be a proof that does not use arbitrary choices.

For example, the same proof shows that $T^* M$ is an orientable manifold (regardless of the orientability of $M$), but the proof via the canonical symplectic form feels more morally correct and does not invoke connections or metrics.

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The argument seems perfectly sound to me. However, if yor goal was to show that any two of the three orientabilities (that of the tree vector bundles $TM\rightarrow M, E\rightarrow M$ and $TE\rightarrow E$) implies the third, then this can be seen directly by looking at atlases and without connections and exterior powers. – Olivier Bégassat May 7 '12 at 22:47
You can also show that $TM$ is always orientable as a manifold directly by computations in local charts. – Olivier Bégassat May 8 '12 at 9:32
You don't really need to the connection since you always get that short exact sequence ($VE = \ker \pi_* : TE \to \pi^* TM$) and you can use a metric on $E$ to split this instead of a connection. – Eric O. Korman May 9 '12 at 13:56
@Eric: How so? I was under the impression that a splitting of this exact sequence is the same thing as a connection... – Zhen Lin May 9 '12 at 14:05
@ZhenLin: In general any short exact sequences $0 \to E_1 \to F \to E_2 \to 0$ of vector bundles split since you can put a metric on the vector bundle $F$ and get $E_2 \simeq E_1^\perp$. I think in this case a connection gives you something stronger since $HE$ will satisfy some additional properties (which aren't necessary for this problem). – Eric O. Korman May 9 '12 at 14:43

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