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Stream: deprecated: discrete geometry and entanglement

Topic: Free functors

Eric Forgy (Dec 13 2020 at 22:29):

Note: Corrections or comments are welcome.

A directed graph is a functor $\mathcal{G}: X\to\mathsf{Set},$ where $X$ is a category with two objects $X^0,X^1$ and two morphisms $s,t:X^1\to X^0.$ The set $\mathcal{G}^0 = \mathcal{G}(X^0)$ is the set of vertices and the set $\mathcal{G}^1 = \mathcal{G}(X^1)$ is the set of directed edges. If $\epsilon\in\mathcal{G}^1$, it is a directed edge from the source node $s(\epsilon)$ to the target node $t(\epsilon).$

The category $\mathsf{DiGraph}$ is the functor category $[X,\mathsf{Set}].$

Composing a graph $\mathcal{G}$ with a free functor $\mathcal{F}_{\mathsf{Vect}}: \mathsf{Set}\to\mathsf{Vect}$ gives a span

$V^0\overset{s}{\leftarrow} V^1\overset{t}{\rightarrow} V^0$

where $V^0 = \mathcal{F}_{\mathsf{Vect}}(\mathcal{G}^0)$ is a vector space of nodes and $V^1 = \mathcal{F}_{\mathsf{Vect}}(\mathcal{G}^1)$ is a vector space of directed edges.

Eric Forgy (Dec 14 2020 at 00:05):

An algebra is a monoid in $\mathsf{Vect}$ so we can consider a free functor

$\mathcal{F}_{\mathsf{Alg}}: \mathsf{Vect}\to\mathsf{Alg},$

where $\mathsf{Alg} = \mathsf{Mon(Vect)}$ so that the composite

$\mathcal{F}_{\mathsf{Alg}}\circ\mathcal{F}_{\mathsf{Vect}}\circ\mathcal{G}: X\to\mathsf{Alg}$

is a span of algebras

$A^0\overset{s}{\leftarrow} A^1\overset{t}{\rightarrow} A^0,$

where $A^0 = \mathcal{F}_{\mathsf{Alg}}(V^0)$ is an algebra of nodes and $A^1=\mathcal{F}_{\mathsf{Alg}}(V^1)$ is an algebra of directed edges.

Eric Forgy (Dec 14 2020 at 02:36):

(I'm shaky on this one.)

We can consider another free functor

$\mathcal{F}_{\mathsf{Span}_1}: \mathsf{Alg}\to\mathsf{Span_1(Alg)}$

so that the composite

$\mathcal{F}_{\mathsf{Span_1}}\circ\mathcal{F}_{\mathsf{Alg}}\circ\mathcal{F}_{\mathsf{Vect}}\circ\mathcal{G}: X\to\mathsf{Span_1(Alg)}$

consists of two objects $A^0,A^1$ and morphisms consists of composable spans

$A^0\overset{s}{\leftarrow} A^1\overset{t}{\rightarrow} A^0.$

Matteo Capucci (he/him) (Dec 14 2020 at 12:15):

Whats up with the last functor? It seems it does nothing!

Eric Forgy (Dec 14 2020 at 18:51):

Yeah. I had something else written there and edited it and now I am confused myself. That is why I prefaced it "I am shaky on this one" :) I am not sure about that last step.

Eric Forgy (Jan 09 2021 at 21:44):

Since I started this journey, things have changed a little bit (thank goodness!).

For one, I am no longer using the traditional definition of a directed graph as a functor $X\to\mathsf{Set}$ as described above. Instead, now I think of a directed graph as an endomorphism in $\mathsf{Span(Set)}$, which is captured nicely via a functor
${}$
$G:\mathbb{N}\to\mathsf{Span(Set)}$
${}$
with
${}$
$G(\bullet) = V,\quad G(1) = E:V\to V.$
${}$
Given a span of sets
${}$
$V\leftarrow E\rightarrow V$
${}$
we can construct a cospan of algebras
${}$
$K^V\leftarrow K^E\rightarrow K^V$
${}$
or (I think) a span of vector spaces
${}$
$K[V]\leftarrow K[E]\rightarrow K[V].$
${}$
Both can be thought of as functors, i.e.
${}$
$K^-: \mathsf{Set\to Alg}_K,\quad V\mapsto K^V$
${}$
and
${}$
$K[-]: \mathsf{Set\to Vect}_K,\quad V\mapsto K[V].$
${}$
Since I'm trying to do this incrementally, it probably makes sense to think about $K[V]$ a bit since $K^V$ is dual to $K[V]$ :thinking:
${}$
So I think I'll spend some time thinking about
${}$
$\mathbb{N}\to\mathsf{Span(Set)\to Span(Vect}_K).$