like that of the preceding theorem might look bewildering, and just like with the previously mentioned results, its proof does not seem to explain what is special about these particular complexes. But the following definition, going back to Schild [@Sch], does it, by $7/8$:

A subcomplex of $\partial\Delta^n$ is called [*self-dual*]{} in $\partial\Delta^n$ if it contains precisely one face out of each pair $\Delta^k$, $\Delta^{n-k-1}$ of complementary faces of $\Delta^n$.

Now observe that $K_5$, as well as each $HJ_i$, is self-dual as a subcomplex of $\partial\Delta^4$. Indeed, this is obvious for $HJ_0$. But each $HJ_{i+1}$, can be obtained from $HJ_i$ by exchanging a pair of complementary faces of $\partial\Delta^4$, except that $HJ_5$ is obtained in this way from $HJ_3$ not $HJ_4$. Also, $K_5$ is obtained in this way from $HJ_5$.

It is not hard to check that the seven complexes $K_5$ and the $HJ_i$, are in fact all the self-dual subcomplexes of $\partial\Delta^4$ (up to isomorphism). Note that $K_{3,3}$ is missing in this picture (but see Example \[Halin-Jung revisited\] below).

Similarly one can check that the three complexes $\Delta^2$, $(\Delta^2)^{(0)}*\Delta^0$ and $\Delta^0\sqcup\partial\Delta^2$ in the MacLane–Adkisson theorem are precisely all the self-dual subcomplexes of $\partial\Delta^3$.

This is no coincidence; in fact, 