TITLE: When are bundles of odd and even differential forms isomorphic?
QUESTION [9 upvotes]: Let $M$ be a compact oriented $n$-manifold. Denote $\Omega^k := {\bigwedge}^k T^*M$ the vector bundle of differential $k$-forms, and let $\Omega^{\text{odd}} := \bigoplus_{\text{$k$ odd}} \Omega^k$ and $\Omega^{\text{even}} := \bigoplus_{\text{$k$ even}} \Omega^k$ be the bundles of odd and even differential forms, respectively.
When are $\Omega^{\text{odd}}$ and $\Omega^{\text{even}}$ isomorphic as vector bundles over $M$?

A few observations:

If $n$ is odd they are isomorphic, $\Omega^{\text{odd}} \simeq \Omega^{\text{even}}$, as can be seen e.g. by introducing a Riemannian metric $g$ on $M$, and observing that the Hodge star $\star: \Omega^k \to \Omega^{n - k}$ is an isomorphism for eack $k$ and hence also between $\Omega^{\text{odd}}$ and $\Omega^{\text{even}}$.

More generally, if $\chi(M) = 0$, then $\Omega^{\text{odd}} \simeq \Omega^{\text{even}}$. Proof: there is a non-vanishing section $s$ of $\Omega^1 \simeq TM$. Write $\Omega^1 = \mathbb{R}s \oplus V$ for some vector bundle $V$. Then $$\Omega^{\text{odd}} = s \wedge \Omega^{\text{even}}(V) \oplus \Omega^{\text{odd}}(V) \simeq \Omega(V) \simeq s \wedge \Omega^{\text{odd}}(V) \oplus \Omega^{\text{even}}(V) 
 = \Omega^{\text{even}},$$
where $\Omega(V)$ is the full exterior bundle of $V$, and $\Omega^{\text{odd/even}}(V)$ are the odd/even parts of $\Omega(V)$.

However, if $n = 2$ they are not isomorphic unless $M$ is the torus: the Euler class of $\Omega^{\text{odd}} = \Omega^1$ is then non-zero, while $\Omega^{\text{even}} = \Omega^0 \oplus \Omega^2 \simeq \mathbb{R}^2$ has trivial Euler class.

If $n = 2d$, $\Omega^{\text{even}}$ and $\Omega^{\text{odd}}$ can be seen to have identical Chern classes $c_i$, except possibly the top one $c_d$. (I define the Chern class of a real vector bundle to be the Chern class of its complexification.) Namely, if $SM \subset TM$ denotes the unit sphere bundle, and $\pi: SM \to M$ is the footpoint projection, then the pullbacks $\pi^*\Omega^{\text{odd}} \simeq \pi^*\Omega^{\text{even}}$ are isomorphic. (Proof: since there is a non-vanishing tautological section $\tau(x, v) := g_x(v, \bullet)$ of $\Omega^1$, we may repeat the argument in the second point.) By Gysin sequence, $\pi^*$ is an isomorphism on $H^\bullet(M)$ for $\bullet \leq n - 1$, so $c_i(\Omega^{\text{odd}}) = c_i(\Omega^{\text{even}})$ for $i \leq d - 1$, and $c_d(\Omega^{\text{odd}}) - c_d(\Omega^{\text{even}})$ is a multiple of $\chi(M)$ in $H^{2d}(M, \mathbb{Z})$.

If $n = 4$, my computations suggest that $\Omega^{\text{even}}$ and $\Omega^{\text{odd}}$ have equal all Chern, and Stiefel–Whitney characteristic classes. One needs to use that $\Omega^2 = \Omega^1 \wedge \Omega^1$ and the formula for $c_2$ of a wedge product. In particular, $c_2(\Omega^{\text{odd}}) - c_2(\Omega^{\text{even}}) = 0$ is the zero multiple of the Euler characteristic.

My guess after all of this is that $\Omega^{\text{even}} \not \simeq \Omega^{\text{odd}}$ if $n$ is even and $\chi(M) \neq 0$, but I was not able to prove this.

REPLY [4 votes]: Strengthening your fourth bullet point, one can see all the Chern classes are equal using the complex splitting principle. So one cannot use Chern classes to distinguish them.
Indeed, I will prove the stronger statement: Let $V$ be a complex vector bundle of even rank $n$. Then the differences between the Chern classes of $\bigoplus_{i=0}^{\lfloor n/2\rfloor } \wedge^{2i} V$ and $\bigoplus_{i=0}^{\lfloor (n-1)/2 \rfloor} \wedge^{2i+1} V$ is divisible by $c_n(V)$ and thus, in particular, vanishes in degrees $<2n$.
This implies the Chern classes are equal in this setting by taking $V$ to be the complexified cotangent bundle.
To prove this, it suffices by the splitting principle to handle the case when $V$ is a direct sum of the $n$ universal complex line bundles on $(\mathbb C\mathbb P^{\infty})^n$. The Cherm classes of $\bigoplus_{i=0}^{\lfloor n/2\rfloor } \wedge^{2i} V$ and $\bigoplus_{i=0}^{\lfloor (n-1)/2 \rfloor} \wedge^{2i+1} V$ are then polynomials in the Chern classes of these line bundles. If we restrict to $(\mathbb C\mathbb P^{\infty})^{n-1}$ by trivializing one of the line bundles, then the pullbacks of these polynomials must be equal, as the pullbacks of the vector bundles are isomorphic. This pullback is equivalent to quotienting the cohomology ring by the first Chern class of that line bundle.
It follows that the difference between the two polynomials is divisible by the first Chern class of each universal line bundle, hence (since the polynomial ring is a UFD), by their product, which is the $n$th Chern class, as desired.