G} gives a continuous map $\CGr_{\<\mathbb C}(1,n)\rightarrow\Gr_{\<\mathbb C}(d,n)$ taking $(V_1,\:V'_{n-1})$ to a subspace of its second component. Composing, we get a continuous function $\Gr_{\<\mathbb C}(1,n)\rightarrow\Gr_{\<\mathbb C}(d,n)$ taking each $\!1\!$-dimensional subspace $L\subseteq\mathbb C\<^n$ to a $\!d\!$-dimensional subspace $L'$ of $L^\perp.$ This gives a $\!d\!$-dimensional subbundle of the tangent bundle on $\!(n\<{-}1)\!$-dimensional complex projective space, which by \mbox{\cite[Theorem~1.1(ii)]{G+H+S}} is possible if and only if $n$ is even and $d=1$ or $n\,{-}\<2.$ Alternatively we may note that for $L'$ as in the preceding paragraph, the map $L\mapsto L\oplus L'$ takes each $\!1\!$-dimensional subspace $L$ of $\mathbb C\<^n$ to a $\!(d\<{+}1)\!$-dimensional subspace containing $L,$ which, by \mbox{\cite[Theorem~1.5(a)]{CC+ZR}} can only happen if $n$ is even and $d+1=2$ or $n-1,$ i.e., again $d=1$ or $n-2.$\qed \end{proof} \section{The question in positive characteristic.}\label{unproved} I do not know whether Theorem~\ref{main} remains true if the characteristic~$0$ hypothesis is deleted. One could hope to prove such a result using algebraic geometry in place of our topological arguments. Now the analog of \cite[Theorem~1.5(a)]{CC+ZR} with continuous maps of topological spaces replaced by morphisms of algebraic varieties over a general algebraically closed field indeed holds [{\em ibid.,} Theorem~1.5(b)]. However, the map $L\mapsto (L, L^\perp)$ that we also called on is not a morphism of algebraic varieties, so we cannot use it as before to yoke the abovementioned result together with Proposition~\ref{CG>G}. (That map was constructed from a Hermitian inner product, which is not bilinear but sesquilinear. A genuine bilinear form on $\mathbb C\<^n$ cannot be positive definite, so that $L^\perp,$ defined using such a form, will not always be complementary to $L.$ And if one tries to retreat to the case $k=\mathbb R$ and use a real inner product, this will not keep its positive definiteness at non-real points, hence still does not lead to a morphism of varieties $\Gr_k(1,n)\rightarrow \CGr_k(1,n).)$ Indeed, there can be no nontrivial morphism of algebraic varieties $\Gr_k(1,n)\rightarrow \CGr_k(1,n),$ because $\Gr_k(1,n)$ is projective while $\CGr_k(1,n)$ is affine. What we may hope for, instead, is an analog of \mbox{\cite[Theorem~1.5(b)]{CC+ZR}} applying directly to morphisms $\CGr_k(1,n)\rightarrow \Gr_k(d,n).$ We note, however, that \mbox{\cite[Theorem~1.5(b)]{CC+ZR}} has only one exceptional case for $n$ even, the case $d=n\<{-}1,$ in place of the two cases $d=2$ and $d=n\<{-}1$ of \cite[Theorem~1.5(a)]{CC+ZR}. Yet the example of the next section shows that both of the latter cases occur; so the desired result on maps $\CGr_k(1,n)\rightarrow \Gr_k(d,n)$ would have to have a weaker conclusion than that of \mbox{\cite[Theorem~1.5(b)]{CC+ZR}}. \section{A factorization when $n$ is even.}\label{even_n} We shall now see that factorizations of the sort allowed by Theorem~\ref{main} when $n$ is even do occur. Our argument is inspired by the construction of Buchweitz and Leuschke \cite{RB+GL}. \begin{lemma}\label{rkadj} Let $R$ be a commutative integral domain, $n$ a positive integer, and $X$ an \nxn\ matrix over $R$ having determinant~$0.$ Then \\[2pt] {\rm(i)}~$\rank(\adj(X)) \leq 1.$ \\[2pt] {\rm(ii)}~For any alternating \nxn\ matrix $A$ over $R,$ one has $\adj(X)\;A\;\adj(X)\tr = 0.$ \end{lemma}\begin{proof} (i)~If $\rank(X) = n-1,$ this follows from the equation $X\ \adj(X) = \det(X)\