Eric Forgy directed space

Warning: Under Construction

Given a directed graph GG, a directed space is the space dual to the graded differential algebra (DGA)

Ω(G)= r0Ω r(G)\Omega(G) = \bigoplus_{r\ge 0} \Omega^r(G)

generated by GG.

The space Ω 0(G)\Omega^0(G) is spanned by the complete set of vertex projections

{π v:G 0G 0|vG 0}\{\pi_v:G_0\to G_0|v\in G_0\}


(1) vG 0π v=1.\sum_{v\in G_0} \pi_v = 1.

The commutative product on Ω 0(G)\Omega^0(G) is given simply by composition

(2)π vπ v=δ v,vπ v.\pi_v \pi_{v'} = \delta_{v,v'} \pi_v.

The space Ω 1(G)\Omega^1(G) is spanned by elements

{π e|eG 1}\{\pi_e | e\in G_1\}

with right multiplication Ω 1(G)×Ω 0(G)Ω 1(G)\Omega^1(G)\times\Omega^0(G)\to\Omega^1(G) given by

(3)π eπ v=δ t(e),vπ e\pi_e\pi_v = \delta_{t(e),v} \pi_e

and left multiplication Ω 0(G)×Ω 1(G)Ω 1(G)\Omega^0(G)\times\Omega^1(G)\to\Omega^1(G) given by

(4)π vπ e=δ v,s(e)π e.\pi_v\pi_e = \delta_{v,s(e)} \pi_e.

Generating a Discrete Space

The derivation

d:Ω 0(G)Ω 1(G)d: \Omega^0(G)\to \Omega^1(G)

applied to vertex projections results in,

(5) vG 0dπ v=0\sum_{v\in G_0} d\pi_v = 0


(6)d(π vπ v)=(dπ v)π v+π v(dπ v)=δ v,vdπ v,d(\pi_v \pi_{v'}) = (d \pi_v) \pi_{v'} + \pi_v (d \pi_{v'}) = \delta_{v,v'} d\pi_v,

which, in particular, provides the graded commutative relation

(7)[π v,dπ v]=dπ v.[\pi_v,d\pi_v] = d\pi_v.

Since {π e|eG 1}\{\pi_e|e\in G_1\} spans Ω 1(G)\Omega^1(G), it follows that

(8)dπ v = eG 1α eπ e,α eR = et 1(v)α e (t)π e+ es 1(v)α e (s)π efrom (3),(4), and (7), \begin{aligned} d\pi_v &= \sum_{e\in G_1} \alpha_e \pi_e, \alpha_e\in R \\ &= \sum_{e\in t^{-1}(v)} \alpha^{(t)}_e \pi_e + \sum_{e\in s^{-1}(v)} \alpha^{(s)}_e \pi_e\quad\text{from (3),(4), and (7),} \end{aligned}

which says dπ vd\pi_v is a weighted sum of edges directed into vv and edges directed out of vv.

Multiplying (6) by π v\pi_v on the left and π v\pi_{v'} on the right, results in

(9)π v(dπ v)π v=(δ v,v1)π v(dπ v)π v\pi_v (d\pi_{v'}) \pi_{v'} = (\delta_{v,v'} - 1) \pi_v (d\pi_v) \pi_{v'}

Case: v=vv = v'

When v=vv = v', combining (8) and (9) gives

(10)π v(dπ v)π v= es 1(v)t 1(v)(α e (t)+α e (s))π e=0, \pi_v (d\pi_v) \pi_v = \sum_{e\in s^{-1}(v)\cap t^{-1}(v)} (\alpha^{(t)}_e + \alpha^{(s)}_e) \pi_e = 0,

which implies that for loops

α e (s)=α e (t).\alpha^{(s)}_e = -\alpha^{(t)}_e.

Case: vvv \ne v'

When vvv \ne v', (8) gives

(11)π v(dπ v)π v= es 1(v)t 1(v)α e (s)π e \pi_v (d\pi_v) \pi_{v'} = \sum_{e\in s^{-1}(v)\cap t^{-1}(v')} \alpha^{(s)}_e \pi_e


(12)π v(dπ v)π v= es 1(v)t 1(v)α e (t)π e \pi_v (d\pi_{v'}) \pi_{v'} = \sum_{e\in s^{-1}(v)\cap t^{-1}(v')} \alpha^{(t)}_e \pi_e

which, when combined with (9), implies that

α e (s)=α e (t).\alpha^{(s)}_e = -\alpha^{(t)}_e.

Therefore, (8) may be rewritten, in general, as

(13)dπ v= et 1(v)α eπ e es 1(v)α eπ e. d\pi_v = \sum_{e\in t^{-1}(v)} \alpha_e \pi_e - \sum_{e\in s^{-1}(v)} \alpha_e \pi_e.

and (12) may be rewritten as

(14)π v(dπ v)π v= es 1(v)t 1(v)α eπ e. \pi_v (d\pi_{v'}) \pi_{v'} = \sum_{e\in s^{-1}(v)\cap t^{-1}(v')} \alpha_e \pi_e.

Question: Could it be that α e\alpha_e is related to the Leinster measure?

Note also that (setting α e=1\alpha_e = 1 and assuming no loops)

(15) v,vπ v(dπ v)π v= v(dπ v)π v= eπ e. \sum_{v,v'} \pi_v (d\pi_{v'}) \pi_{v'} = \sum_{v'} (d\pi_{v'}) \pi_{v'} = \sum_{e} \pi_e.


(16) eπ e= v(dπ v)π v, \sum_{e} \pi_e = \sum_{v} (d\pi_{v}) \pi_{v},

which allows us to convert a sum over all edges to a sum over all nodes.

category: maths

Revised on October 3, 2010 at 14:12:29 by Eric Forgy