The n-Category Café

The minimal triangulation of the projective plane

We are going to construct our graph from the minimal triangulation of $\mathbb{R} P^2$ so let’s have a look at that first. We want to see how the $2$-torsion in the homology of $\mathbb{R} P^2$ can be expressed using this triangulation as we will need that later for the $2$-torsion in the graph magnitude homology.

The real projective plane can be thought of as the two-sphere quotiented out by the antipodal map. The antipodal map acts on the icosahedral triangulation of the two-sphere. So quotienting the icosahedron by the antipodal map gives us a triangulation $T_0$ of $\mathbb{R} P^2$ which is in fact the triangulation with fewest simplices. Here is a picture of it.

I’ve numbered the vertices and we will label each simplex by its vertices, so $(0, 1, 2)$ is a 2-simplex you can see in the triangulation above. The label for a simplex will have the vertices in linear order.

Let’s recall how we get the $2$-torsion element in homology. If we take the boundary $\partial$ of the ten 2-simplices with the orientation drawn above then we get

$$(3, 4)+(4,5)-(3,5) + (3, 4)+(4,5)-(3,5).$$

(We write $-(3,5)$ rather than $(5,3)$ because of the orientation conventions which I will gloss over. You can figure them out if you’re interested/concerned. I think they are right!)

As $\partial^2=0$ this boundary chain is a cycle. To get homology we quotient cycles out by boundaries, so this cycle is trivial in homology, which means

$$2[(3, 4)+(4,5)-(3,5)] = 0 \in \mathrm{H}_1^{\mathrm{simp}}(T_0).$$

Thus $[(3, 4)+(4,5)-(3,5)]$ is a $2$-torsion element. Off the top of my head I can’t think of a nice argument showing that this is non-trivial in homology, but hopefully someone can provide one in the comments! Anyway, we’re going to need this later.

Constructing the graph and the magnitude cycles

Following Kaneta and Yoshinaga, we are now going to use the above triangulation $T_0$ to build our graph $G({T_0})$. We take the simplices of the triangulation as the nodes of the graph and have an edge $\sigma \to \tau$ if $\sigma$ is a facet of $\tau$, remember that a facet is a face of maximal dimension. We then add top and bottom nodes, so $\mathrm{bottom}$ has an arrow to each $0$-simplex and $\mathrm{top}$ has an arrow from each $2$-dimensional simplex. Here is the graph again. You should be able to see the six vertices, fifteen edges and ten faces of the original triangulation.

A more sophisticated way of saying what we’ve done is the following. The vertices in the triangulation form a poset, the face poset, with the ordering is ‘is a face of’. We add top and bottom elements to that poset. We then take the Hasse diagram which is the graph which has the elements of the poset as its nodes and where there is an edge $x\to y$ if $x\le y$ but there is no $z$ with $x\le z\le y$. Clearly this process gives us a graph $G({T})$ from any triangulation $T$ of a manifold.

We can obtain the graph $G({T_0})$ in SageMath with a couple of commands:

triangulation = simplicial_complexes.RealProjectivePlane()
poset = triangulation.face_poset().with_bounds()
graph = poset.hasse_diagram()

To see the above picture you can use the following command:

show(poset.plot())

For what we do next it doesn’t matter if we stick with this directed graph or take the associated undirected graph as we are going to be forced to only consider upward pointing edges.

The magnitude chains for this graph

In the previous post I explained that for a finite graph $G$ the magnitude chain groups are defined as follows.

A chain generator is a tuple of the form $c= (x_0, x_1,\dots, x_{k-1},x_k),$ where each $x_i$ is a node of the graph $G$ and $x_{i-1}\ne x_{i}$. The degree is $\mathrm{deg}(c)=k$ and the length is $\mathrm{len}(c)=\sum \mathrm{d}(x_{i-1}, x_i)$.

The face map $\partial_i$ for $i=1,\dots, k-1$ is defined by:

$$\partial_{i}(x_0,\ldots,x_k) = \begin{cases} (x_0,\ldots,\widehat{x_i},\ldots,x_k) & \text{if}\,\, x_{i-1} \lt x_{i} \lt x_{i+1}, \\ 0 & \text{otherwise}. \end{cases}$$

where $x_{i-1} \lt x_{i}\lt x_{i+1}$ means that $x_i$ lies on a shortest path between $x_{i-1}$ and $x_{i+1}$, i.e., $\text{d}(x_{i-1},x_i)+\text{d}(x_i,x_{i+1})=\text{d}(x_{i-1},x_{i+1})$.

Neither the length nor the endpoints of chains are altered by the face maps, so the magnitude complex splits up into subcomplexes with specified endpoints. So if we define

$$MC^{y,z}_{k,l}(G)=\left\langle (y, x_1,\dots, x_{k-1},z) \,\, |\,\, \text{chain generator of length }\,\,l\right\rangle.$$

Then the magnitude chain complex splits as

$$MC_{\ast,\ast}(G)=\bigoplus_{y,z\in G}\bigoplus_l MC^{y,z}_{\ast,l}(G)$$

We will concentrate on the subcomplex of length-four chains from the bottom element to the top element in our graph (here, four is dimension of $\mathbb{R} P^2$ plus two). Writing $\mathrm{b}$ and $\mathrm{t}$ for the bottom and top elements we consider the magnitude chain complex $MC^{\mathrm{b},\mathrm{t}}_{\ast,4}(G({T_0})$. We will see that the homology of this is isomorphic to $\widetilde{\mathrm{H}}_{\ast+2}(\mathbb{R} P_2)$ and so we get the embedding $\widetilde{\mathrm{H}}_{\ast}(\mathbb{R} P_2)\hookrightarrow MC_{\ast+2,4}(G)$.

Looking at our graph it is easy to see that a length four chain must be of the form

$$(\mathrm{b}, \sigma_1, \dots,\sigma_{k-1},\mathrm{t})$$

where $\sigma_1\subset \dots\subset\sigma_{k-1}$ is a sequence of simplices, each of which is a face of the following one, in other words it is flag of simplices in our original triangulation. Such a flag can be a full flag like $(0)\subset(0,1)\subset(0,1,2)$ in which each is a facet of the following simplex, or it can be a partial flag like $(0)\subset (0,1,2)$.

Looking at the formula for the face maps you can see that given a generator corresponding to a flag of simplices, each facet of the generator corresponds to a flag with one of the simplices removed.

Those of you familiar with combinatorial topology might have spotted the connection with the barycentric subdivision construction. We will look a little more closely at these flags now.

Barycentric subdivision

One of the things you usually learn in a first course on algebraic topology is that if you have a triangulation of a space then you can form a finer triangulation – the barycentric subdivision – in the following way. Take the midpoint of each simplex in the original triangulation and use these as the vertices of the new triangulation, decomposing each of the old $n$-simplices into $(n+1)!$ new $n$-simplices in ‘the obvious fashion.

For instance, if we take a triangle in our triangulation then each $1$-simplex gets split into $2$ new $1$-simplices and the $2$-simplex gets split into $6$ new $2$-simplices.

Each new vertex can be labelled by the $n$-simplex that it is the midpoint of. What then is ‘the obvious fashion’ for creating the new simplices? Well each new $n$-simplex in the subdivision corresponds to a flag of old simplices $\sigma_0\subset \sigma_1\subset \dots \subset \sigma_n$. We can picture the $2$-simplex corresponding to $(0)\subset(0,1)\subset(0,1,2)$ and the $1$-simplex corresponding to $(0)\subset (0,1,2)$ as follows.

It ought to be clear that for the $n$-simplex in the subdivision corresponding to a flag $\sigma_0\subset \sigma_1\subset \dots \subset \sigma_n$, each facet corresponds to a flag where one of the old simplices have been removed. So for $(0)\subset(0,1)\subset(0,1,2)$ the facets correspond to $(0,1)\subset(0,1,2)$, $(0)\subset(0,1,2)$ and $(0)\subset(0,1)$.

So the barycentric subdivision $T'$, as a simplicial complex is isomorphic to the simplicial complex of flags where an $n$-simplex is a flag $\sigma_0\subset \sigma_1\subset \dots \subset \sigma_n$ of simplices of the original triangulation $T$. Well, there’s a subtlety here in that we shouldn’t forget the empty flag! The flags naturally form an augmented simplicial complex, meaning that there is a unique simplex in degree $-1$, the empty flag, which is the facet of every $0$-simplex.

So the augmented chain complex $\widetilde{\mathrm{C}}_\ast^{\mathrm{simp}}(T')$, which is obtained from the usual chain complex by sticking a unique generator in degree $-1$, is isomorphic to the complex of flags. The homology of the augmented chain complex gives the reduced homology $\widetilde{\mathrm{H}}_\ast^{\mathrm{simp}}(T')\cong \widetilde{\mathrm{H}}_\ast(M)$ which in practice means you kill off a copy of $\mathrm{Z}$ in degree zero from the usual homology.

The important thing is that we are seeing here precisely the same structure that we saw in the magnitude chain group $MC_{\ast, l}^{b,t}(G)$. We should now make precise.

Synthesis of the two sides: the theorem

I’ve hopefully given the impression that a complex of flags of simplices is isomorphic to both the augmented chain complex of the barycentric subdivision of a triangulation and to a subcomplex of the magnitude chain complex of the graph of the triangulation. Let’s give a proper statement of this now. Kaneta and Yoshinaga have a more general statement in their paper, involving ranked posets rather than just triangulations, but for the purpose of finding torsion, the following will suffice.

Theorem. Suppose that $T$ is a finite triangulation of a $m$-manifold $M$, and that $T'$ is its barycentric subdivision, with $G({T})$ the graph obtained as above from the poset structure, then the isomorphism of (augmented) chain groups for $k\ge -1$

$$\begin{aligned} \widetilde{\mathrm{C}}^{\mathrm{simp}}_k (T')&\xrightarrow{\sim} MC^{\mathrm{b},\mathrm{t}}_{k+2,m+2}(G({T})); \\ \sigma_0\subset \sigma_1\subset \dots \subset \sigma_k &\mapsto (b, \sigma_0,\dots, \sigma_k,t) \end{aligned}$$

commutes up to sign with the differentials. Thus this induces an isomorphism of homology groups

$$\widetilde{\mathrm{H}}^{\mathrm{simp}}_\ast (T')\xrightarrow{\simeq} MH^{b,t}_{\ast+2,m+2}(G({T})).$$

As a corollary we get that the homology of $M$ embeds in the magnitude homology of the graph.

Corollary. With $M$, $T$, $T'$ and $G({T})$ as in the above theorem, the homology of $M$ embeds in the magnitude homology of the graph $G({T})$ via the following sequence of isomorphisms and embeddings:

$$\widetilde{\mathrm{H}}_\ast(M) \cong \widetilde{\mathrm{H}}_\ast^{\mathrm{simp}}(T) \xrightarrow{\simeq} \widetilde{\mathrm{H}}^{\mathrm{simp}}_\ast (T') \xrightarrow{\simeq} MH^{b,t}_{\ast+2,m+2}(G({T})) \hookrightarrow MH_{\ast+2,m+2}(G({T})).$$

The payoff: a torsion element in the homology of our graph

We saw earlier on in this post that

$$[(3, 4)+(4,5)-(3,5)] \in \mathrm{H}_1^{\mathrm{simp}}(T_0).$$

is the non-trivial $2$-torsion element. We can follow this element through the sequence of maps above. We just need to note that the map $1$-chains on $T$ to $1$-chains on $T'$ rewrites each edge as the (signed) sum of its two half-edges, namely

$$\begin{aligned} \mathrm{C}_1^{\mathrm{simp}}(T) &\to \mathrm{C}_1^{\mathrm{simp}}(T') \\ (a,b)&\mapsto \bigl((a) \subset (a,b)\bigr) - \bigl((b)\subset (a,b)\bigr). \end{aligned}$$

Then we get our non-trivial $2$-torsion element in $MH_{3,4}(G({T_0}))$ to be

$$\begin{split} [(\mathrm{b}, (3), (3,4), \mathrm{t}) &-(\mathrm{b}, (4), (3,4), \mathrm{t}) +(\mathrm{b}, (4), (4,5), \mathrm{t})\\ &-(\mathrm{b}, (5), (4,5), \mathrm{t}) +(\mathrm{b}, (5), (3,5), \mathrm{t}) -(\mathrm{b}, (3), (3,5), \mathrm{t})]. \end{split}$$

Thus we have a graph with torsion in its magnitude homology groups!