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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?

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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.

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    $\begingroup$ Why is the combination of the Plücker relations (for the Grassmannian) and the determinant condition sufficient? $\endgroup$ – Mark Wildon Feb 10 at 12:06
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    $\begingroup$ To be invertible over $\mathbb{Z}$, the determinant should be $\pm 1$. $\endgroup$ – spin yesterday

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