# Definition of $T$ operator in proof of Excision Theorem on Homology Theory

I had some problems with the proof of Excision Theorem. I use two books at the same time, one is (Algebraic Topology, Hatcher) and the other one is "An Introduction of Algebraic Topology", Rotman. In the first book the main work is done in Proposition $2.21$ in four steps. At the second step is defined an operator $T$, that work as chain homotopy between the identity and the barycentric subdivision operator $S$ on $LC_n(Y)$, with $Y$ convex. Hatcher defines $T$ operator inductively as $\ T_n\lambda=b_{\lambda}(\lambda-T_{n-1}\partial_n \lambda)$. Is proved that the formula $\partial_{n+1} T_n +T_{n-1} \partial_n=1-S$ holds at degree $n$ using the same formula at degree $n-1$.

The same fact in Rotman is proved in Lemma $6.13$ using $T$ operator defined as $T_n \gamma =b.(\gamma -Sd \gamma -T_{n-1}\partial_n \gamma)$, where $b.$ is the cone operator $b_{\lambda}$ used by Hatcher, and $Sd_n$ operator is $S$ in the notation above. Also in this case the formula $\partial_{n+1} T_n +T_{n-1} \partial_n=1-Sd_n$ holds using the properties of cone operator and the formula $\partial_n(\gamma -Sd_n \gamma-T_{n-1} \partial_n \gamma)=\partial_n \gamma - Sd_{n-1}\partial_n-(1-Sd_{n-1}-T_{n-2} \partial_{n-1}) \partial_n \gamma$
that holds by induction and properties of $Sd$ operator.

The two definitions of $T$ are not the same and differ by an $Sd_n$ term that appears in Rotman definition and in Hatcher doesn't.

$1)$ Can someone explain me the reasons (algebraic and geometric) of this difference?

$2)$ Hatcher and other authors say that the geometric interpretation of $T$ operator is a subdivision of $\Delta^n \times I$ obtained by joining simplices in $\Delta^n \times\{0\} \cup (\partial \Delta^n) \times I$ to the barycenter of $\Delta^n \times \{ 1\}$. I don't understand this explanation, can someone be more clear?

## 2 Answers

The original nonsense response is deleted. Below is a letter from Allen Hatcher himself addressing the issue. I wish to apologize for my earlier misunderstanding and stupid remarks.

Let me try to answer the two questions asked by Lorban in the math.stackexchange posting that you linked to. The first question asks about the difference between the formula for T that I give and the formula of Rotman which has an extra term. Both formulas are inductive, assuming one has already defined T for simplices of lower dimension. My formula is

$$T\lambda = b_\lambda(\lambda - T\partial\lambda) \tag{\ast}$$

The right side of this equation can be rewritten as

$$b_\lambda(\lambda) - b_\lambda(T\partial\lambda) \tag{\ast\ast}$$

As explained in my book, the way to understand this geometrically is to think of the simplex $\Delta^n$ as the projection of a prism $\Delta^n \times I$ that has $\lambda$ as its base face and $b_\lambda$ as the barycenter of its top face. Then the first term $b_\lambda(\lambda)$ in the formula $(\ast\ast)$ is the cone on the base of the prism, with the tip of the cone being $b_\lambda$ in the top face, and the second term $b_\lambda(T\partial\lambda)$ is the cone on the sides of the prism, which have already been subdivided by induction.

Rotman's formula is

$$T\gamma = b.(\gamma - Sd\gamma - T\partial\gamma) = b.\gamma -b.Sd\gamma - b.T\partial\gamma$$

Here $b$ is to be interpreted as the center point of the whole prism, halfway between the top and bottom of the prism, and the three terms in the formula represent the cone on the bottom (this is $b.\gamma$), the cone on the top (this is $b.Sd\gamma$) and the cone on the sides (this is $b.T\partial\gamma$).

I'm not sure what the reason is for doing it Rotman's way, coning to the center of the whole prism rather than to the center of the top. There are other books that do the same thing as Rotman. It seems simpler to do it the way I did because the formulas have fewer terms.

Someone called user32240 attempted an answer to Lorban's question, but the answer contains a fundamental mistake. In the middle of the answer there is an equation

$$[a,b]−T(b−a)=(a,b)−(b,b)+(a,a)=(2a−b,a)$$

The first equality here is correct, but the second one is wrong. First of all, there should really be square brackets in these formulas to indicate simplices:

$$[a,b]−T([b]−[a])=[a,b]−[b,b]+[a,a]$$

This is a linear combination of three 1-simplices, two of which are degenerate, $[a,a]$ and [$b,b]$. It makes no sense to try to combine these three 1-simplices into a single 1-simplex $[2a-b,a]$. These are not vectors that we perform vector addition on, they are just combinatorial objects, simplices. The whole point of homology theory is to take formal linear combinations of simplices, not to treat simplices as vectors and add them.

This mistake messes up many of the later formulas as well.

To compute the correct formula for $T([a,b])$ let us denote the barycenter of $[a,b]$ by $c$ (this is $b_\lambda$ in my earlier notation). Then we get

$$T([a,b]) = c([a,b] − T([b]−[a])) = [c,a,b]−[c,b,b]+[c,a,a]$$

This is a linear combination of three 2-simplices, all of which are actually degenerate since $c$ is the barycenter of $[a,b]$. Geometrically, one can see the three 2-simplices in the prism $\Delta^1 \times I$ (which is just a rectangle) by labeling the two vertices on the left edge by $a$, the two vertices on the right edge by $b$, and the midpoint of the top edge by $c$. There are edges in the interior of the rectangle connecting $c$ to the two ends of the bottom edge, and these two edges subdivide the rectangle into three 2-simplices, $[c,a,b]$, $[c,b,b]$, and $[c,a,a]$.

I hope this long explanation clarifies things.

• Thanks for your answer. As you said, Hatcher's interpretation feels very ad hoc and I still think chain homotopies have a geometric meaning that we are missing. I will award the bounty (better than having 50 points go to waste!), but the $\Delta^n \times I$ still remains a mystery... – Emilio Ferrucci Sep 4 '12 at 11:50
• Thank you for your comment and you references, I'll look those books for more ideas and interpretations. – Lorban Sep 4 '12 at 23:13
• @Lorban: Is it polite to ask for your email? I have an email which is quite relevant and should be helpful to you. – Bombyx mori Sep 5 '12 at 9:36
• stack1-1-2-3-5-8-13@live.com Thank you for your courtesy! – Lorban Sep 6 '12 at 10:38
• send you the email(re-typed here). If you wish you can contact hatcher himself in future, as obviously this would be more helpful. – Bombyx mori Sep 6 '12 at 19:40

I am following Hatcher. I don't think answering so late makes much sense, but still I present my interpretation for your consideration. We need an operator T that satisfies $\partial_{n+1} T_n +T_{n-1} \partial_n=1-S$.

Now for simplicity and motivation we look at a $2$-simplex as illustrated in Hatcher. Whenever we need to define a chain homotopy like $T$ we think of a way to decompose $\Delta^n \times I$ to some $n+1$ simplices. $T$ describes this decomposition. The decomposition should be such that, boundary $\partial T \lambda$ of the $3$-simplices when summed up in a certain way (to be thought of next) cancel at all the faces except the ones remaining in the linear combination of $2$-simplices appearing in $(1-S)\lambda-T\partial\lambda$. This hints (though vaguely) that the decomposition should give us some $3$-simplices with the edges and vertices as those appearing in the barycentric division of the $2$-simplex. Thus the decomposition (as in the illustration in Hatcher) is in some sense not that out of the blue. Now we try to define $T$ on $\lambda$ for the $2$-simplex case. We want $T$ to be the map which projects a particular linear combination of the $3$-simplices to the $2$-simplex followed by composition with $\lambda$. Given the way $T\partial\lambda$ looks from Hatcher's calculation we see that $b_\lambda(\lambda - T\partial\lambda)$ is the one possible linear combination (by a brute brute force calculation), and $T$ is the map which projects this sum into the $2$-simplex followed by composition with $\lambda$ (Here $b_\lambda$ is to be identified with the barycenter of $\Delta_{n} \times \{1\}$) . But here we have assumed the definition of $T$ on $1$-simplices. In order to verify that we would have followed same the line of reasoning.We then have a form for $T$ for the $2$-simplex (verified for higher $n$, anyhow considering a 3-simplex for geometric motivation is unyielding due to obvious limitations), which works for higher n when defined as in Hatcher.

P.S. Apologies for poor presentation and vague answer

• You can use LaTeX here pretty much as usual. Just enclose the formulas in dollar signs. – Martin May 29 '13 at 20:16
• Thnaks Martin ! – deb May 29 '13 at 20:33