The n-Category Café

Nice! Some casual readers may wonder:

“But what if the path $\gamma_1$ is a loop?”

The point is that in this case, the 1-chain it defined by this path is a cycle, and if this is orthogonal to all cycles it’s orthogonal to itself, hence it vanishes. So this case is easy.

So, now we’re assuming without loss of generality that our surface is connected, and choosing two vertices $v,w$ on the graph embedded in this surface, and seeking two ‘substantially distinct’ paths from $v$ to $w$.

It seems helpful to find paths that aren’t ridiculously redundant. After all, if a path from $v$ to $w$ uses all the edges of our graph, no other path can be substantially distinct from it.

Let’s say a path is irredundant if it doesn’t go through the same vertex twice. Given any path from $v$ to $w$, we can always make it irredundant as follows. Suppose it goes through the vertex $x$ twice. Then it’s of the form $\gamma \delta \epsilon$ where $\delta$ starts and ends at $x$. Then $\gamma \epsilon$ is another, strictly shorter, path from $v$ to $w$. By repeating this move we must eventually reach an irredundant path from $v$ to $w$.

(Note that if we apply this procedure to a loop that starts and ends at $v = w$, we’ll eventually reach the trivial loop, with no edges at all, that starts an ends at this vertex.)

An irredundant path can never go over the same edge twice, in either direction. That is, it can never contain an edge $e$ twice, nor can it contain can it edge $e$ and its inverse $e^{-1}$, nor can it contain $e^{-1}$ twice

Thus, by working with irredundant paths we can fix what seems like a little hole in Greg’s result. He wrote:

$$\langle c, \gamma_1 \rangle = \langle \gamma_1, \gamma_1 \rangle - \langle \gamma_2, \gamma_1 \rangle$$

The first term on the RHS is the number of edges in $\gamma_1$ [….]

That’s true when $\gamma_1$ is irredundant, but not in general. Luckily we can always make $\gamma_1$ and $\gamma_2$ irredundant before running the argument, so this shouldn’t be a serious problem!

An irredundant path can still be absurdly circuitous: for example, it can visit every vertex of our graph, or even every edge. But, it’s a step toward finding a path from $v$ to $w$ that’s efficient enough that we can find another substantially distinct path.

Thanks for pointing out that hole in my argument! I wasn’t thinking clearly.

Here’s a slightly different approach that might cope better with arbitrary paths. Given our path $\gamma_1$, we just need another path $\gamma_2$ with the same endpoints such that:

$$\langle \gamma_1, \gamma_2 \rangle \neq \langle \gamma_1, \gamma_1 \rangle$$

Then we can join $\gamma_1$ and $\gamma_2^{-1}$ to make a loop that’s not orthogonal to $\gamma_1$.

Consider some edge $e_i$ of $\gamma_1$, and suppose it starts and ends on vertices $v_{i,1}$ and $v_{i,2}$. There are two obvious alternative paths from $v_{i,1}$ to $v_{i,2}$, given by the two faces incident on $e_i$: you just go around either face, in the opposite direction to $e_i$, using all the other edges of that face except $e_i$.

We want to construct the path $\gamma_2$ by taking $\gamma_1$ and substituting one of these “detours” in place of one of its edges, $e_i$. Generically, you’d expect the inner product of $\gamma_1$ and $e_i$ to differ from the inner product of $\gamma_1$ and the detour around at least one of the faces. Of course we can’t appeal to that … but my hunch is that the only paths where the inner products remain the same for every possible such substitution would have to be loops.

Does that sound plausible? Provable?

Hmm, my last suggestion might not be all that helpful, since requiring the edge and the detour to have different inner products with the path is really just the same as requiring the path to have a non-zero inner product with the loop around the whole face that’s being used to construct the detour. So it’s really just saying that if a path has a non-zero inner product with the loop around some face, here’s how to construct another loop with which the path has a non-zero inner product!

I like the idea of Ilya Bogdanov’s proof, but I don’t see how to make it rigorous. If a path $p$ is not simple it contains a loop $c$, but we don’t always have $\langle c, p \rangle = |c|$, and indeed we can have $\langle c, p \rangle = 0.$

There are various kinds of counterexamples: here’s one. Suppose $p = a b c b^{-1} c^{-1}$ where $b, c$ are loos that start and end at the same point but share no edges, and $a$ is path sharing no edges with $b$ or $c$. Then $p$ is not simple, and it contains a loop $c$, but $\langle p, c \rangle = 0$.

Adam wrote:

I haven’t thought deeply about the case of a triangulation, but shouldn’t the 1-chains have a Hodge decomposition similar to how 1-forms do in cohomology?

Yes! And this works for any graph drawn on a surface whose complement consists of open disks, not just graphs coming from triangulations.

Thanks for bringing up Hodge theory. I had tried to solve my problem using Hodge theory, but I wasn’t able to use the standard manipulations involving $\partal, \partial^*$ and the inner product to settle the question I was really interested in:

Question: Suppose we have a graph $\Gamma$ embedded in a surface $S$ such that $S - \Gamma$ is a union of open disks whose closures are embedded closed disks. Any path in $\Gamma$ gives a 1-chain $c \in C^1(\Gamma,\mathbb{R})$. If this 1-chain is orthogonal to all 1-cycles, must it vanish?

The condition about ‘embedded closed disks’ can’t be omitted from this statement — that’s what makes it tricky. But Ilya Bogdanov seems to have proved a stronger version that avoids mentioning the surface $S$ altogether:

Stronger Claim: Suppose $\Gamma$ is a graph having no edge whose removal would increase the number of connected components. Any path in $\Gamma$ gives a 1-chain $c \in C^1(\Gamma,\mathbb{R})$. If this 1-chain is orthogonal to all 1-cycles, it vanishes.

Is there a nice proof of this using Hodge theory?

Hodge theory seems like a bad idea (or perhaps I am interpreting it too literally) because this seems to be all about the integral structure on the chain groups. Indeed, the statement (or rather, the generalization of it which seems most natural to me) is false with rational coefficients!

Consider the tetrahedron, with vertices $v_1$, $v_2$, $v_3$ and $v_4$. Write $e_{ij}$ for the edge $v_i \to v_j$. Then the chain

$$\tfrac{1}{2} e{14} + \tfrac{1}{4}(e{12}+e{24}+e{13}+e{14})$has boundary$v1-v4$and is orthogonal to every cycle. So the question is to understand the finite group$C0/\partial \mathrm{Ker}(\partial^{\ast}: C1 \to C2)$, which one can think of as the "integral failure of Hodge theory", and see that$u-v$is not in it for any vertices$u$and$v$.