Homomorphism induced by the second exterior power of a linear map Consider the map from $M(n, \mathbb Z) \rightarrow M(\binom{n}{2}, \mathbb Z)$ taking a matrix A to its second compound, i.e, $\bigwedge^2 A$. 
Restricting this map to the invertible matrices we get a homomorphism of groups from $\mathrm{GL}(n, \mathbb Z)$ to $\mathrm{GL}(\binom{n}{2}, \mathbb Z)$. 

How can we determine if a given matrix $B \in \mathrm{GL}(\binom{n}{2}, \mathbb Z)$ is contained in the image of this map? 

 A: First, let us discuss the same question over an algebraically closed field (e.g., over $\overline{\mathbb{Q}}$). Let $V$ be a vector space of dimension $n$. The question is to understand the image of the homomorphism
$$
\lambda \colon \mathrm{GL}(V) \to \mathrm{GL}(\wedge^2V).
$$
Note that $\mathrm{GL}(\wedge^2V)$ acts naturally on the projective space $\mathbb{P}(\wedge^2V)$, which contains as a subvariety the Grassmannian
$$
\mathrm{Gr}(2,V) \subset \mathbb{P}(\wedge^2V).
$$
Clearly, it is preserved by the action of $\mathrm{GL}(V)$. The converse is also true for $n > 4$, i.e., if $g \in \mathrm{GL}(\wedge^2V)$ is such that 
$$
g(\mathrm{Gr}(2,V)) \subset \mathrm{Gr}(2,V), 
$$
then $g \in \mathrm{Im}(\lambda)$. This follows immediately from the isomorphism
$$
\mathrm{Aut}(\mathrm{Gr}(2,V)) \cong \mathrm{PGL}(V).
$$
Over $\mathbb{Z}$, I guess, the equality 
$$
\det(\wedge^2g) = \det(g)^{n-1}
$$
gives an extra constraint; so besides preserving the Grassmannian one should impose the condition that the determinant is $(n-1)$-st power.
