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 $M$?




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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.



When are bundles of odd and even differential forms isomorphic?
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.


