# Barycentric subdivision in convex sets: is Rotman's definition incorrect?

In Rotman's An Introduction to Algebraic Topology, 4th printing, pp. 113, he provided a definition of barycentric subdivision in a convex set $E$:

Let $E$ be a convex set. Then barycentric subdivision is a homomorphism $\operatorname{Sd}_n:S_n(E)\to S_n(E)$ defined inductively on generators $\tau:\Delta^n\to E$ as follows:

(i) If $n=0$, then $\operatorname{Sd}_0(\tau)=\tau$;
(ii) If $n>0$, then $\operatorname{Sd}_n(\tau)=\tau(b_n).\operatorname{Sd}_{n-1}(\partial\tau)$, where $b_n$ is the barycenter of $\Delta^n$.

Here $b.\sigma$ is the cone construction. This definition does not really affect the subsequent arguments: the barycentric subdivision in a general space $X$ is immediately defined as

$$\operatorname{Sd}_n(\sigma)=\sigma_\sharp\operatorname{Sd}_n(\delta^n)$$ (here $\sigma:\Delta^n\to X$ is a $n$-simplex and $\delta^n$ is the identity map of $\Delta^n$).

Thus we only need the definition of $\operatorname{Sd}_n(\delta^n)$ to be correct, which is nevertheless the case. However, the aforementioned definition in convex sets does not make sense to me and seems wrong in its own regard, as Rotman mentioned

It is easy to see that both definitions of $\operatorname{Sd}_n(\sigma)$ agree when $X$ is convex.

This is obviously not true. Consider the unit circle on complex plane $\sigma:I\to\mathbb C$, $t\mapsto e^{2\pi it}$. By the definition in convex sets, the barycentric subdivision of $\sigma$ (viewed as a $1$-cycle) is $0$, while by the definition in general spaces it is a non-zero linear combination of two semicircles.

In fact, in other sources the barycentric subdivision in convex/affine set is only defined on the free abelian group generated by affine simplicies, i.e. the linear chain $LC_n(E)$ (the notation in Hatcher's Algebraic Topology). In this case, all arguments run smoothly and the definition coincides with that for general $X$.

So is this a typo in the book? Is Rotman's definition incorrect?

• The unit circle is not a convex subset of the plane. – Cheerful Parsnip Aug 25 '18 at 3:53
• @CheerfulParsnip From what I cite, it seems the definition only requires the underlying space (in this case, the complex plane) to be convex. – Cave Johnson Aug 25 '18 at 4:09
• @CheerfulParsnip And even if we ask the image of $\sigma$ to be convex, it still appears incorrect. One may replace the circle example by a path going straight from $a$ to $b$ and then coming back. – Cave Johnson Aug 25 '18 at 4:13
• okay, I see what you mean. I read your post too quickly. I think it's probably a mistake in Rotman's book. – Cheerful Parsnip Aug 25 '18 at 5:07