# 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?

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.
• To be invertible over $\mathbb{Z}$, the determinant should be $\pm 1$. – spin Feb 16 at 20:48