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Stream: deprecated: discrete geometry and entanglement

Topic: Invitation (Nonassociativity)

Matteo Capucci (he/him) (Jan 10 2021 at 21:16):

I read your remark on cochains only now. What's up with their associativity? How can they lose it if their ambient algebra has it? :thinking:

Eric Forgy (Jan 10 2021 at 21:24):

It's been a long time since I thought about the argument (like 20 years). :thinking:

Eric Forgy (Jan 10 2021 at 21:40):

Maybe it helps to go back to the universal derivations stuff?

Consider three vertices $V = \{1,2,3\}$ with corresponding basis $e^1,e^2, e^3\in K^V$ together with a complete edge set
${}$
$\tilde E = \{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\} = V\times V$
${}$
with corresponding bases
${}$
$e^{1,1},e^{1,2},e^{1,3},e^{2,1},e^{2,2},e^{2,3},e^{3,1},e^{3,2},e^{3,3}\in K^{\tilde E}.$
${}$
The unit element is
${}$
$1 = e^1+e^2+e^3$
${}$
and the (complete) graph operator is
${}$
$\tilde G = 1\otimes_K 1 = \sum_{(i,j)\in\tilde E} e^{i,j} = e^{1,1}+e^{1,2}+e^{1,3}+e^{2,1}+e^{2,2}+e^{2,3}+e^{3,1}+e^{3,2}+e^{3,3}.$

Eric Forgy (Jan 10 2021 at 21:52):

The universal derivation $\tilde d: A\to\tilde\Omega$ is given by
${}$
$\begin{aligned}\tilde d f & = \sum_{(i,j)\in\tilde E} \left[f(j)-f(i)\right] e^{i,j}\end{aligned}.$

Eric Forgy (Jan 10 2021 at 21:53):

Note that all terms involving a "loop" $e^{i,i}$ vanish from the sum.

Eric Forgy (Jan 10 2021 at 21:57):

Also note that although $\tilde G$ is not an element of $\tilde\Omega = \mathsf{ker}(m)$ (since it contains loops), $\tilde G$ generates $\tilde\Omega$ via $\tilde d.$ The loops vanish due to the commutator.

Eric Forgy (Jan 10 2021 at 22:06):

This is truly a universal derivation (proven in Bourbaki among other places) so any other derivation $d:A\to\Omega$ factors through this one, i.e. for every $d$, there is a unique (up to isomorphism) $\phi:\tilde\Omega\to\Omega$ given by
${}$
$\phi(e^{i,j}) = e^i d e^j$
${}$
where $i\ne j$ such that
${}$
$d = \phi\circ\tilde d$
${}$
and vice versa. $\phi$ amount to basically setting some of the $e^{i,j} = 0.$

Eric Forgy (Jan 10 2021 at 22:15):

Back to this example, it means:
${}$

• $\tilde d e^1 = e^{2,1}+e^{3,1}-e^{1,2}-e^{1,3}$
• $\tilde d e^2 = e^{1,2}+e^{3,2}-e^{2,1}-e^{2,3}$
• $\tilde d e^3 = e^{1,3}+e^{2,3}-e^{3,1}-e^{3,2}$

${}$
Adding the three expressions above demonstrates
${}$
$\tilde d 1 = \tilde d (e^1+e^2+e^3) = 0.$

Eric Forgy (Jan 10 2021 at 22:29):

Multiplying on the left demonstrates:
${}$
$e^{i,j} = e^i\tilde d e^j$
${}$
for $i\ne j$ and $\tilde\Omega$ is spanned by these guys (see Bourbaki) so a general element in $\tilde\Omega$ can be expressed as
${}$
$\alpha = \sum_{(i,j)\in\tilde E, i\ne j} \alpha(i,j) e^{i,j}$
${}$
so you can kind of see why I go back on forth on whether I want to include loops in $\tilde E.$ If I don't include loops, then I can just write
${}$
$\alpha = \sum_{(i,j)\in\tilde E} \alpha(i,j) e^{i,j}$
${}$
but then I can't write $\tilde G = 1\otimes_K 1$, which is kind of nice to be able to do. Either way is fine :shrug:

Eric Forgy (Jan 10 2021 at 22:34):

To see where nonassociativity comes in, I think we need to look at what is $\tilde\Omega^2.$

Eric Forgy (Jan 10 2021 at 22:58):

It is not too tricky to show that
${}$

• $\tilde d e^{1,2} = e^{3,1,2} - e^{1,3,2} + e^{1,2,3}$
• $\tilde d e^{2,1} = e^{3,2,1} - e^{2,3,1} + e^{2,1,3}$
• $\tilde d e^{2,3} = e^{1,2,3} - e^{2,1,3} + e^{2,3,1}$
• $\tilde d e^{3,2} = e^{1,3,2} - e^{3,1,2} + e^{3,2,1}$
• $\tilde d e^{3,1} = e^{2,3,1} - e^{3,2,1} + e^{3,1,2}$
• $\tilde d e^{1,3} = e^{2,1,3} - e^{1,2,3} + e^{1,3,2}$

${}$
which follows from the definition
${}$
$\begin{aligned}\tilde d e^{i,j} = \tilde d (e^i\tilde d e^j) = (\tilde d e^i)(\tilde d e^j) = \sum_{k\in V} (e^{k,i,j} - e^{i,k,j} + e^{i,j,k})\end{aligned}$

Eric Forgy (Jan 10 2021 at 23:04):

On notation, recall that
${}$
$e^{i,j} = e^i\otimes_K e^j,\quad i\ne j$
${}$
are bases for $\tilde\Omega.$

It is tempting to let
${}$
$e^{i,j,k} = e^i\otimes_K e^j\otimes_K e^k,$
${}$
but it is better to think of these as
${}$
$e^{i,j,k} = e^{i,j}\otimes_A e^{j,k} = (e^i\otimes_K e^j)\otimes_A (e^j\otimes_K e^k)\in\tilde\Omega\otimes_A\tilde\Omega,$
${}$
i.e. these are directed 2-paths (or secretly 2-simplices).

Eric Forgy (Jan 10 2021 at 23:11):

There is, however, an obvious map:
${}$
$\tilde\Omega^{\otimes A n} \to A^{\otimes_K (n+1)}$
${}$
with
${}$
$e^{i,j,k} = e^{i,j}\otimes_A e^{j,k} \mapsto e^i\otimes_K e^j\otimes_K e^k$
${}$
and I actually make use of this when I define diamonds.

Eric Forgy (Jan 10 2021 at 23:33):

Pausing for a second, what I've shown above (in gory tedious detail) is that the universal derivation
${}$
$\tilde d:A\to\tilde\Omega$
${}$
maps $0$-paths to linear combinations of $1$-paths. This can be extended to a derivation
${}$
$\tilde d:\tilde\Omega\to\tilde\Omega\otimes_A\tilde\Omega$
${}$
that maps $1$-paths to linear combinations of $2$-paths.

Eric Forgy (Jan 10 2021 at 23:40):

As long as you are dealing with complete graphs / universal derivations, we have
${}$
$\tilde d^2 = 0,\quad \tilde d(f\, g) = (\tilde d f)\, g + f\,(\tilde d g).$

Eric Forgy (Jan 10 2021 at 23:43):

(btw, I think everything I said in my last few comments are just an application of well know stuff to a simple complex involving 3 vertices)

Eric Forgy (Jan 11 2021 at 00:05):

I think the more traditional approach with simplices assumes a total order on vertices and keeps only the paths with the right order so that the universal
${}$

• $\tilde d e^1 = e^{2,1}+e^{3,1}-e^{1,2}-e^{1,3}$
• $\tilde d e^2 = e^{1,2}+e^{3,2}-e^{2,1}-e^{2,3}$
• $\tilde d e^3 = e^{1,3}+e^{2,3}-e^{3,1}-e^{3,2}$

${}$
becomes
${}$

• $d e^1 = -e^{1,2}-e^{1,3}$
• $d e^2 = +e^{1,2}-e^{2,3}$
• $d e^3 = +e^{1,3}+e^{2,3}$

${}$
i.e. $\phi(e^{i,j}) = 0$ if $i>j$ so that $\Omega$ is spanned by $e^{1,2}, e^{2,3}, e^{1,3}\in\Omega$ and the universal
${}$

• $\tilde d e^{1,2} = e^{3,1,2} - e^{1,3,2} + e^{1,2,3}$
• $\tilde d e^{2,1} = e^{3,2,1} - e^{2,3,1} + e^{2,1,3}$
• $\tilde d e^{2,3} = e^{1,2,3} - e^{2,1,3} + e^{2,3,1}$
• $\tilde d e^{3,2} = e^{1,3,2} - e^{3,1,2} + e^{3,2,1}$
• $\tilde d e^{3,1} = e^{2,3,1} - e^{3,2,1} + e^{3,1,2}$
• $\tilde d e^{1,3} = e^{2,1,3} - e^{1,2,3} + e^{1,3,2}$

${}$
becomes
${}$

• $d e^{1,2} = e^{1,2,3}$
• $d e^{2,3} = e^{1,2,3}$
• $d e^{1,3} = -e^{1,2,3}$

${}$
so that $\Omega\otimes_A\Omega$ is one dimensional and spanned by $e^{1,2,3}.$

Eric Forgy (Jan 11 2021 at 00:17):

You're right. I can't think of why it should be nonassociative. Proof by exhaustion :joy:

Eric Forgy (Jan 11 2021 at 00:42):

I haven't thought about the simplicial stuff since grad school, but it seems the total ordering can be given a physical interpretation as giving each $n$-simplex a kind of temporal direction so the first vertex can be thought of as an event that happens before the last vertex :thinking:

Eric Forgy (Jan 11 2021 at 00:51):

i.e. traversing an edge involves moving forward in time. This automatically rules out cycles, loops, etc.

Eric Forgy (Jan 11 2021 at 00:52):

This is a common property with diamonds so maybe totally ordered simplices and diamonds are not so different after all :thinking:

Eric Forgy (Jan 11 2021 at 07:15):

Maybe this is where non-associativity comes in...

Given $\alpha,\beta\in\Omega$, what is $\alpha\,\beta\in\Omega\otimes_A\Omega$?

Eric Forgy (Jan 11 2021 at 07:25):

The only thing I can think of that it could be is:
${}$
$\begin{aligned} \alpha\,\beta &= \left[(\alpha(1,2)\,e^{1,2}+\alpha(2,3)\,e^{2,3}+\alpha(1,3)\,e^{1,3}\right]\,\left[(\beta(1,2)\,e^{1,2}+\beta(2,3)\,e^{2,3}+\beta(1,3)\,e^{1,3}\right] \\ &= \alpha(1,2)\,\beta(2,3)\,e^{1,2,3}.\end{aligned}$
${}$
This product would be associative, but I think this might be antisymmetrized so that
${}$
$\alpha\wedge\beta = \left[\alpha(1,2)\,\beta(2,3)-\beta(1,2)\,\alpha(2,3)\right]\,e^{1,2,3}.$
${}$
This product, although looking more like the continuum wedge product, would not be associative :thinking:

Eric Forgy (Jan 11 2021 at 07:28):

The associative product of simplicial cochains
${}$
$\alpha\,\beta = \alpha(1,2)\,\beta(2,3)\,e^{1,2,3}$
${}$
does not, in any obvious way that I can see, approximate wedge product.

Eric Forgy (Jan 11 2021 at 07:37):

However, if you put two simplicial 2-cochains together (forming a 2-diamond), you get
${}$
$\begin{aligned}\alpha\,\beta = \left[\alpha(1,2)\,\beta(2,3)-\alpha(1,2')\,\beta(2',3)\right] (e^{1,2,3} - e^{1,2',3}).\end{aligned}$
${}$
This is associative and approximates the continuum wedge product.

Eric Forgy (Jan 11 2021 at 07:39):

Diamonds for the win :blush: :large_blue_diamond:

Eric Forgy (Jan 11 2021 at 07:42):

Note: 1-diamonds are directed edges and 2-diamonds are "directed squares", but $n$-diamonds for $n>2$ are richer than just $n$-cubes (although directed $n$-cubes are $n$-diamonds).

Eric Forgy (Jan 11 2021 at 07:46):

Note^2: You cannot form a diamond with just 3 vertices. If there is not a fourth vertex present allowing you to construct a 2-diamond from 3 existing vertices, then the 2-simplex formed from the 3 vertices is equal to zero in discrete differential geometry.

Eric Forgy (Jan 11 2021 at 07:49):

i.e. the space of 2-diamonds $\Omega^2$ is spanned by pairs of 2-simplices sharing the same start and end vertex, e.g.
${}$
$e^{i,j,k}-e^{i,j',k}.$

Eric Forgy (Jan 11 2021 at 08:29):

Wilson talks about this a bit (Wilson as in Sullivan-Wilson with a competing formulation to ours) in

• DIFFERENTIAL FORMS, FLUIDS, AND FINITE MODELS

Eric Forgy (Jan 11 2021 at 08:40):

The product defined in this paper :point_of_information: is graded commutative and nonassociative.

Eric Forgy (Jan 11 2021 at 08:42):

On the bottom of page 4:

There is also a graded commutative nonassociative product on $C^\bullet$ described easily in terms of the elementary cochains $a,b$ as follows