# Why is the subdivision an extension?

Bredon defines, in page 224, the "subdivision" operator by induction on the affine chains of a simplex as

$$\Upsilon(\sigma)= \underline{\sigma}(\Upsilon\big(\partial \sigma)\big) \quad \text{if deg(\sigma)>0}$$ and $$\Upsilon(\sigma)=\sigma \quad \text{if deg(\sigma)=0},$$ where $\underline{\sigma}$ is the barycentric division.

He then proceeds to extend the definition of the operator to any chain $\sigma$ in any topological space. He makes $$\Upsilon(\sigma):=\sigma_{\Delta}(\Upsilon \iota_p),$$ where $\iota_p : \Delta_p \rightarrow \Delta_p$ is the identity map. He now argues as follows:

Of course, one must check that these coincide with the previous definitions when $\sigma$ is affine, but this is obvious because $\Upsilon$ (...) was defined on affine simplices using only affine operations.

I don't understand the justification, and even less what he means by "affine operation". What does he mean?

Clarification of notation:

What $\underline{\sigma}$ means:

Let an affine simplex with image on $\Delta_q$ be denoted by $\sigma=[v_0,v_1,...,v_p]$, Given $v \in \Delta_q$, we make the cone on $\sigma$ from $v$: $$v \sigma:=[v,v_0,...,v_p].$$

The barycenter of the affine simplex $\sigma$ is $$\underline{\sigma}:=(\sum_{i=0}^pv_i)/(p+1),$$ where $[v_0,...,v_p]=\sigma$.

• Can you explain the notation some? I think I must be interpreting the definition of $\Upsilon$ horribly wrong. For instance, it would seem that by induction $\Upsilon(\sigma)$ is always in degree $0$, since $\partial\sigma$ has degree lower than that of $\sigma$ so by induction $\Upsilon(\partial\sigma)$ has degree $0$, and the barycentric subdivision of a degree $0$ chain is degree $0$. – Eric Wofsey Dec 30 '15 at 4:54
• @EricWofsey Sure! I'll clarify. Thanks for helping. – Aloizio Macedo Dec 30 '15 at 5:18
• @EricWofsey Made the edit, hope it clarifies. What happens is that $\partial \sigma$ will have lower degree, $\Upsilon$ will preserve degree but $\underline{\sigma}$ will increase degree. Hence degree is unchanged. – Aloizio Macedo Dec 30 '15 at 5:34

Any affine map preserves barycenters. That is, if $T:\Delta^n\to\Delta^m$ is an affine map and $\tau$ is an affine simplex in $\Delta^n$, then $T(\underline{\tau})=\underline{T(\tau)}$. It follows by induction on degree that $T_*(\Upsilon(\tau))=\Upsilon(T_*(\tau))$, where $T_*$ is the induced map on affine chains. (More generally, an "affine operation" is an operation that takes a finite tuple of points $(x_0,\dots,x_n)$ to $\sum t_ix_i$ for some $t_i$ such that $\sum t_i=1$. Barycenters are the special case when $t_i=1/(n+1)$ for all $i$. Affine maps can be defined as exactly the maps that preserve all affine operations.)
Now, writing $T=\sigma$ and choosing $\tau=\iota_p$, we get that $$\sigma_*(\Upsilon(\iota_p))=\Upsilon(\sigma_*(\iota_p)).$$ But $\sigma_*(\iota_p)$ is none other than $\sigma$ itself considered as an affine simplex. So in the equation above, the left-hand side is the new definition of $\Upsilon(\sigma)$ (for arbitrary chains), and the right-hand side is the old definition (for affine chains). Thus the two definitions agree.
• First thing I'll do when I wake up is read this. It is kind of late ($5$ am) and I need to get some sleep. Thank you very much for helping me, again. – Aloizio Macedo Dec 30 '15 at 7:04