# Projective space, explicit descriptions of isomorphism between homology.

NOTE: This is a followup question to my question here.

Consider $$\mathbb{P}_\mathbb{R}^2$$, i.e. the $$\mathbb{R}$$-projective plane.

Here is a CW complex structure on $$\mathbb{P}_\mathbb{R}^2$$. A $$0$$-cell $$e^0$$, a $$1$$-cell $$e^1$$, and a $$2$$-cell $$e^2$$ which is attached to $$e^1$$ by a degree $$2$$ map.

Here is a $$\Delta$$-complex structure on $$\mathbb{P}_\mathbb{R}^2$$.

Question. What's an explicit description of the isomorphism between $$H_*^{\text{cellular}}(\mathbb{P}_\mathbb{R}^2)$$ and $$H_*^{\text{simplicial}}(\mathbb{P}_\mathbb{R}^2)$$ given the specific cellular structure and specific simplicial structure I have chosen to endow $$\mathbb{P}_\mathbb{R}^2$$ with here?

Daniel McLaury gave the following answer here.

In general, if you want the map from cellular homology to singular homology, just notice that if you have a homology class represented by a cell then that cell's closure can be regarded as a singular simplex, so send the cell to itself.

If you want to talk about simplicial homology here, probably the easiest way of thinking about things is to show that both cellular and simplicial homology are isomorphic to singular homology. That way you don't have to worry about comparing the simplicial structure you've chosen with a cellular structure you've chosen. And the situation above works for both cases.

However, that does not answer my question. I'm interested in an explicit description of the isomorphism between $$H_*^{\text{cellular}}(\mathbb{P}_\mathbb{R}^2)$$ and $$H_*^{\text{simplicial}}(\mathbb{P}_\mathbb{R}^2)$$ given the specific cellular structure and specific simplicial structure I have chosen to endow $$\mathbb{P}_\mathbb{R}^2$$ with here.

• Well, if you insist in that, you have to write explicitly the isomorphisms and that ingeneral depends on subdividing your simplicial structure. It is not going to be fun. – Mariano Suárez-Álvarez Nov 11 '16 at 4:17
• You have not given a valid simplicial decomposition of the projective plane. Look up in eg Munkries how he does it. – Rene Schipperus Nov 11 '16 at 4:25
• @Rene It's a valid $\Delta$-complex structure, as in Hatcher. It's the same idea, relaxing some combinatorial constraints. Simplicial homology works fine, and the generalization is healthier both pedagogically and just for one's general sanity. – user98602 Nov 11 '16 at 7:03
• @Mike Miller I believe Hatcher's $\Delta$-complex business is just something like geometric realization of abstract simplicial sets. – user144221 Nov 11 '16 at 14:37

As for your choice of cellular structure, we have the complex

$$0 \longrightarrow C_2 \longrightarrow C_1 \longrightarrow C_0\longrightarrow 0$$

where the chain modules are given as

$$C_2 = \mathbb Z e_2, \qquad C_1 = \mathbb Z e_1, \qquad C_0 = \mathbb Z e_0,$$

and the maps are given as (always according to your definitions)

$$\partial _2: \mathbb Z e_2 \longrightarrow \mathbb Z e_1$$

$$\partial _2(e_2) = 2 e_1$$

and

$$\partial _1: \mathbb Z e_1 \longrightarrow \mathbb Z e_0$$

$$\partial _1(e_1) = e_0 - e_0 = 0.$$

So, cellular homology turns out to be

$$H_2 = \ker \partial _2 = 0, \qquad H_1 = \frac {\ker \partial _1}{\text {Im }\partial _2} = \frac {\mathbb Z}{2\mathbb Z} = \mathbb Z_2, \qquad H_0 = \mathbb Z ,$$

with representatives $\ [e_1]\$ and $\ [e_0]. \$ Of course $\ 2 [e_1] = 0.$

As for your choice of simplicial structure, we have the complex

$$0 \longrightarrow C'_2 \longrightarrow C'_1 \longrightarrow C'_0\longrightarrow 0$$

where the chain modules are given as

$$C'_2 = \mathbb Z U \oplus \mathbb Z L, \qquad C'_1 = \mathbb Z a\oplus \mathbb Z b\oplus \mathbb Z c, \qquad C'_0 = \mathbb Z v \oplus \mathbb Z w.$$

For the maps, we have to choose an orientation for the $2$-"simplexes" $U\$ and $\ L. \$ I choose counter-clockwise for $U\$ and clockwise for $\ L: \$ this choice makes everything simpler, but is not necessary. The boundary maps are

$$\partial '_2: \mathbb Z U \oplus \mathbb Z L \longrightarrow \mathbb Z a\oplus \mathbb Z b\oplus \mathbb Z c$$

$$\partial '_2(U;0) = (-a;b;c), \qquad \partial '_2(0;L) = (a;-b;c).$$

You can see this on your picture, for my choice of orientations.

$$\partial '_1: \mathbb Z a\oplus \mathbb Z b\oplus \mathbb Z c \longrightarrow \mathbb Z v \oplus \mathbb Z w$$

$$\partial '_1 (a;0;0) = (-v;w), \qquad \partial '_1 (0;b;0) = (-v;w) \qquad \partial '_1 (0;0;c) = (0;0).$$

The last follows from the fact that $\ \partial '_1 c = v-v.$

So, again:

$$\ker \partial '_2 = \{(hU;kL) \in C'_2 / \partial '_2 (hU;kL) = (0;0;0) \},$$

that is

$$-h+k = 0, \qquad h - k = 0, \qquad h+k = 0.$$

This is trivially seen to imply $\ h=k=0.$

As we are here, note that $\ \partial '_2 (U;L) = (0;0;2c).$

$$\ker \partial '_1 = \{(la;mb;nc) \in C'_1 / \partial '_1 (la;mb;nc) = (-(l+m)v;(l+m)w) = (0;0) \},$$

so

$$l+m = 0 \iff l=-m$$ and $$\ker \partial '_1 = \{(-m;m;n) \ \forall m, n \in \mathbb z \}.$$

$$\text {Im} \partial '_2 = \{(-h+k;h-k; h+k)\ h,k \in \mathbb Z \}.$$

Notice that $h-k\$ and $h+k \$ have the same parity, so this is not all of the kernel of $\partial '_1: \$ we can see that if $m$ and $n$ do not have the same parity, they are not in the image, while if they do, we can find $h$ and $k:$

$$h-k = m \qquad h+k = n, \iff h = k + m \qquad 2k + m = n \iff 2k = n - m.$$

So

$$\frac {\ker (\partial '_1 )}{\text {Im} (\partial '_2)} = \mathbb Z_2.$$

As for $H_0$, of course Im$(\partial '_1 ) = \{(-sv;sw), \ s \in \mathbb Z\},\$ which is $\ \mathbb Z.$

As representatives for the homology classes, we can take $(0;0;c)\$ (it is a cycle, but, since $0$ and $1$ do not have the same parity, not a boundary,) and $(v;0).$

So, we are ready for our explicit chain complexes homomorphism which induces an isomorphism in homology: $$f_* : C_* \longrightarrow C'_*$$

$$f_2 : e_2 \longrightarrow (U;L)$$

$$f_1 : e_1 \longrightarrow (0;0;c)$$

$$f_1 : e_0 \longrightarrow (v;0)$$

Check commutativity: $f_1 \circ \partial _2 (e_2) = f_1 (2 e_1) = (0;0;2c); \ \partial' _2 \circ f_2 (e_2) = \partial' _2 (U;L) = (0;0;2c). \$ Seen before. $f_0 \circ \partial _1 (e_1) = f_0 (0) = (0;0); \ \partial' _1 \circ f_1 (e_1) = \partial' _1 (0;0;c) = (0;0). \$ This map induces an isomorphism in homology because $\ f_*([e_1]) = [(0;0;c)]\$ and $\ f_* ([e_0]) = [(v,0)], \ i. e. \$ sends representatives to representatives.