When are bundles of odd and even differential forms isomorphic? 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.
 A: 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.
A: I will explain that as long as $n>2$ the real vector bundles $\Omega^{even}$ and $\Omega^{odd}$ over $M$ are isomorphic.
If $n>2$ then $dim(\Omega^{even}) = dim(\Omega^{odd}) = 2^{n-1} > n$ and so $\Omega^{even}$ and $\Omega^{odd}$ are isomorphic if and only if they are stably isomorphic. This follows from obstruction theory, applied to the map $BO(2^{n-1}) \to BO$. We are therefore left with analysing the real $K$-theory class
$$\Omega^{even} - \Omega^{odd} \in KO^0(M).$$
As Meier's comment points out, the complex version of this would be the complex $K$-theory Euler class of $TM$, and we can take inspiration from that construction as follows.
For a real vector space $V$ consider the chain complex of vector bundles
$$V \times \Lambda^0 V \overset{(v,w) \mapsto (v, v \wedge w)}\longrightarrow V \times \Lambda^1 V \overset{(v,w) \mapsto (v, v \wedge w)}\longrightarrow V \times \Lambda^2 V \longrightarrow \cdots,$$
which is exact over every point other than $0 \in V$. This defines (see e.g. Definition 9.23 of Spin Geometry) a compactly-supported real $K$-theory class
$$v_V \in KO^0_c(V).$$
This construction can be done fibrewise to any vector bundle, so in particular gives a class
$$v_{TM} \in KO^0_c(TM),$$
which we can consider as a class in $\widetilde{KO}^0(Th(TM))$, the reduced real $K$-theory of the Thom space of $TM$. Pulling $v_{TM}$ back along the zero-section $s_0 : M \to Th(TM)$ gives the complex of vector bundles
$$\Omega^0 \overset{0}\longrightarrow \Omega^1 \overset{0}\longrightarrow \Omega^2 \overset{0}\longrightarrow \cdots$$
over $M$, which represents the class $\Omega^{even} - \Omega^{odd} \in KO^0(M)$ that we are interested in.
On the other hand, the inclusion of a tangent fibre $S^n \to Th(TM)$ is $n$-connected, so there is a factorisation up to homotopy
$$s_0 : M \overset{q}\longrightarrow S^n \longrightarrow Th(TM)$$
(the map $q$ has degree $\chi(M)$, but that will not matter for the argument). We obtain the equation
$$\Omega^{even} - \Omega^{odd} = q^*(v_{\mathbb{R}^n}) \in KO^0(M).\tag{1}\label{eq}$$
We now turn to determining the class $v_{\mathbb{R}^n} \in KO^0_c(\mathbb{R}^n) = \widetilde{KO}^0(S^n) = KO^{-n}$. It is easy to see from the construction in terms of exterior powers that
$$v_{\mathbb{R}^n} = (v_{\mathbb{R}^1})^n \in KO^{-n},$$
and not hard to see that $v_{\mathbb{R}^1} = \eta \in KO^{-1} = \mathbb{Z}/2\{\eta\}$. It follows that $v_{\mathbb{R}^n}=0$ for $n > 2$, as $\eta^3 \in KO^{-3} = 0$, and combining this with \eqref{eq} proves the claimed result.
Addendum. It is interesting to see what happens for $n=2$. As explained in the question, over an orientable surface the Euler class distinguishes $\Omega^{even}$ and $\Omega^{odd}$, but one can still wonder about them as stable vector bundles. The analysis above, including the fact that $q$ has mod 2 degree $\chi(M)$, shows that
$$\Omega^{even} - \Omega^{odd} = \chi(M) \cdot p^*(\eta^2) \in KO^0(M),$$
where $p : M \to S^2$ is the map that collapses the complement of a ball. As $\eta^2$ has order 2, this shows that these bundles are stably isomorphic if $\chi(M)$ is even.
On the other hand taking $M = \mathbb{RP}^2$ we have $\Omega^{even} = \mathbb{R} \oplus L$, where $L$ is the tautological line bundle, and $\Omega^{odd} = T\mathbb{RP}^2 \cong_{stable} L^{\oplus 3} - \mathbb{R}$. These are not stably isomorphic, as they have different second Stiefel--Whitney classes: this implies that $\chi(\mathbb{RP}^2) \cdot p^*(\eta^2) = p^*(\eta^2) \neq 0 \in KO^0(\mathbb{RP}^2)$, which is indeed the case.
