is easy to see that its $1$-skeleton is the complete graph $K_6$, and for each pair of disjoint circuits in $K_6$, precisely one bounds a $2$-simplex in $\R P^2_\triangle$. Hence $\R P^2_\triangle$ is self-dual as a subcomplex of $\partial\Delta^5$ (compare [@Mat 5.8.5]). By Theorem \[self-dual\] this implies that the projective plane $\R P^2=|\R P^2_\triangle|$ does not embed in $S^3$, embeds in $S^4$, and every embedding of $\R P^2$ in $S^4$ is inequivalent to its reflection. In fact, every embedding of $\R P^2$ in $S^4$ with fundamental group of the complement isomorphic to $\Z/2$ is known to be topologically equivalent to either the standard embedding or its reflection [@La3].

A well-known $9$-vertex triangulation of $\C P^2$ [@KB] (see [@MY], [@BD1], [@MS], [@BD3] for other constructions) is easily seen to be self-dual as a subcomplex of $\partial\Delta^8$. Thus Theorem \[self-dual\] applies again. In this connection we note that every embedding of $\C P^2$ in $S^7$ is known to be equivalent to either the standard embedding or its reflection [@Sk-A3].

In fact, it is known that a combinatorial $n$-manifold with $v$ vertices is self-dual as a subcomplex of $\partial\Delta^{v-1}$ if and only if $2v=3n+6$ [@ArM] (if), [@Da] (only if). Combinatorial $n$-manifolds with $\frac{3n}2+3$ 