Given a smooth vector bundle $\pi: E\to M$ it is a standard result that the exterior vector bundle $\pi':\wedge E\to M$ where $\wedge E= \sqcup_{p}\wedge E_p$ is also a smooth vector bundle. My question is how do we identify which vector bundles are exterior vector bundles?
Suppose $E\to M$ is a smooth vector bundle whose fibres are isomorphic to the exterior algebra of a fixed vector space. Then can we say that $E\to M$ is the exterior bundle of some smooth vector bundle?
Requiring the fibers to be isomorphic as vector spaces to an exterior algebra is merely a condition on the dimension of the fiber, namely that the said dimension is a power of $2$. Certainly there are bundles of rank $2^n$ that are not exterior bundles, such as $T (S^{2^n})$.
As pointed out by Mikhail Katz, a finite-dimensional real vector space $V$ is isomorphic to the exterior algebra of another finite-dimensional real vector space $W$ if and only if $\dim V = 2^{\dim W}$. It follows that a necessary condition for a vector bundle $E \to M$ to be isomorphic to the exterior algebra of a vector bundle $F$ is that $\operatorname{rank}E = 2^{\operatorname{rank}F}$, but this is far from sufficient.
Just as $\bigwedge W \cong \bigoplus_{k=0}^n\bigwedge^kW$ where $n = \dim W$, we have $\bigwedge F \cong \bigoplus_{k=0}^n\bigwedge^kF$ where $n = \operatorname{rank}F$ - note that $\bigwedge^kF$ is a rank $\binom{n}{k}$ bundle. When $k = 0$, the bundle $\bigwedge^0F$ is isomorphic to $\varepsilon^1$, the trivial line bundle, so if $E \cong\bigwedge F$, then $E$ admits a nowhere-zero section. This is one way to see that the bundles $T(S^{2^n})$ mentioned in Mikhail Katz's answer are not isomorphic to exterior algebra bundles (by the Poincaré-Hopf Theorem, every vector field on an even-dimensional sphere has at least one zero).
When $n = 1$, we have $\bigwedge F \cong \bigwedge^0F\oplus\bigwedge^1 F \cong \varepsilon^1\oplus F$. So if $E$ is a rank two bundle isomorphic to the exterior algebra of the rank one bundle $F$, then $E \cong \varepsilon^1\oplus F$. Conversely, if $E \cong \varepsilon^1\oplus F$, then $E \cong \bigwedge F$. That is, a rank two bundle $E$ is isomorphic to the exterior algebra of a bundle if and only if $E$ admits a nowhere-zero section.
When $n = 2$, we have $\bigwedge F \cong \bigwedge^0F\oplus\bigwedge^1 F\oplus\bigwedge^2F \cong \varepsilon^1\oplus F\oplus\det(F)$. Again, if a rank four bundle $E$ can be decomposed in this way, then it will be isomorphic to $\bigwedge F$, but that isn't a useful characterisation. However, if we suppose $\det F$ is trivial, then $\bigwedge F \cong\varepsilon^2\oplus F$ with $F$ orientable. This leads to the following characterisation: a rank four bundle $E$ is isomorphic to the exterior algebra of an orientable bundle if and only if $E$ is orientable and admits two linearly independent sections. For example, the tangent bundle of the four manifold in this answer is isomorphic to the exterior algebra of a vector bundle
For any $n \geq 2$, one can show that $\bigwedge F$ is always orientable using the fact that for $k \geq 1$ we have $w_1(\bigwedge^kF) = \binom{n-1}{k-1}w_1(F)$ where $w_1$ denotes the first Stiefel-Whitney class. So if $E$ is a rank $2^n > 2$ bundle isomorphic to the exterior algebra of another vector bundle, it is orientable. One could use characteristic classes to obtain other necessary conditions. The isomorphism $\bigwedge^{n-k}F \cong (\bigwedge^kF)^*\otimes\det F$ could be useful if one were to pursue this further.
