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Stream: deprecated: discrete geometry and entanglement

Topic: Bourbaki III 10.10: Universal problem for derivations

Eric Forgy (Dec 14 2020 at 20:29):

image.png

image.png

Eric Forgy (Dec 14 2020 at 20:37):

Note: Corrections and comments are welcome.

Given an algebra (with unit) $A$ and a product $m: A\otimes A\to A$ given by $m(a\otimes a') = aa'$ (product in $A$), let $I$ denote the kernel $\mathsf{ker}(m).$

The universal differential $\tilde d: A\to I$

$\tilde d: x\mapsto 1\otimes x - x\otimes 1$ (Note: Bourbaki has a sign difference, but I will use this convention.)

can be written as a (graded) commutator

$\tilde d: x\mapsto [\tilde G,x],$

where $\tilde G=1\otimes 1.$

Eric Forgy (Dec 14 2020 at 20:46):

$\tilde d$ generates $I$ as a left $A$-module. If $a = \sum_i x_i\otimes y_i\in I$, then $m(a) = 0$ by definition and

$\begin{aligned} a &= \sum_i x_i\otimes y_i \\ &= \sum_i x_i \left(1\otimes y_i - y_i\otimes 1\right) \\ &= \sum_i x_i \tilde d y_i.\end{aligned}$

Eric Forgy (Dec 14 2020 at 20:47):

I love how Bourbaki spells out the proofs explicitly regardless how "obvious" they may be. Even mortals like me can follow :blush:

Eric Forgy (Dec 14 2020 at 20:52):

Similarly,

$\begin{aligned} a &= \sum_i x_i\otimes y_i \\ &= \sum_i \left(x_i\otimes 1 - 1\otimes x_i\right) y_i \\ &= \sum_i - (\tilde dx_i) y_i.\end{aligned}$

Eric Forgy (Dec 14 2020 at 21:00):

In particular, I guess this means

$x\otimes y\in I\implies x\otimes y = x \tilde d y = -(\tilde d x) y.$

Eric Forgy (Dec 14 2020 at 21:12):

If $f: I\to E$ is an $A$-bimodule morphism, then

$d=f\circ\tilde d: A\to E$

is also a derivation.

Eric Forgy (Dec 14 2020 at 21:48):

That makes sense. Given a bimodule morphism $f:I\to E$, you get a derivation $d = f\circ\tilde d: A\to E.$

The opposite direction is less clear to me despite staring at the Bourbaki paragraphs for a while :thinking:

Given a derivation $d: A\to E$, you get a unique bimodule morphism $f: I\to E$ :thinking:

Eric Forgy (Dec 14 2020 at 23:15):

I guess the argument is that since

$(x,y) \mapsto x d y$

is bilinear, there is a unique linear map

$g: A\otimes A\to E$

and we need to show that $g$ restricted to $I$ is a bimodule morphism.

Eric Forgy (Dec 14 2020 at 23:18):

But doesn't that require that $E$ is generated by $d$ as a left $A$-module? :thinking:

Eric Forgy (Dec 14 2020 at 23:23):

It is clear to me that if $x\otimes y\in I,$ then $g(x\otimes y) = x dy\in E,$

but can there be elements of $E$ not of that form? :thinking:

Eric Forgy (Dec 14 2020 at 23:38):

Btw, I think the proof of $A$-bilinearity is cleaner to note that

$\sum_i x_i d(y_i y) = -\sum_i (dx_i) y_i y.$

:nerd:

Eric Forgy (Dec 15 2020 at 00:41):

I think I get it.

$g: A\otimes A\to E$

is unique, so restricted to $I$, we have a unique $A$-bimodule morphism

$g: I\to E$

given by

$g\left(\sum_i x_i\otimes y_i\right) := \sum_i x_i d y_i.$

Since $g$ is an $A$-bimodule morphism, we have

$\begin{aligned} g\left(\sum_i x_i\otimes y_i\right) &= g\left(\sum_i x_i\tilde d y_i\right) \\ &= \sum_i x_i g(\tilde d y_i),\end{aligned}$

so that

$d = g\circ\tilde d.$

Cool :sunglasses:

Eric Forgy (Dec 15 2020 at 00:44):

To summarize, given an $A$-bimodule morphism $g: I\to E$, we get a unique derivation $d: A\to E$ given by

$d := g\circ\tilde d.$

Conversely, given a derivation $d: A\to E$, we get a unique $A$-bimodule morphism $g: I\to E$ given by

$g: \sum_i x_i\otimes y_i \mapsto \sum_i x_i d y_i.$

Eric Forgy (Dec 15 2020 at 00:48):

Yay! I understand the proof! :tada:

Next. Is there an arrow theoretic / functorial way to understand this?

Matteo Capucci (he/him) (Dec 15 2020 at 09:29):

Maybe that derivations out of A are represented by I?

Eric Forgy (Dec 15 2020 at 19:27):

I made some progress (at least for first order). The nLab has a nice diagram to represent kernels, so this universal derivation can be summarized by the commutative diagram:

image.png

where $m: A\otimes A\to A$ is multiplication in $A$ and $\tilde\Omega^1 = \mathsf{ker}(m)$ and $d= \phi\circ \tilde d: A\to\Omega^1$ is any other derivation.

Eric Forgy (Dec 15 2020 at 19:42):

The morphisms are $A$-bimodule morphisms, so we must be sitting in the category of $A$-bimodules :thinking:

Eric Forgy (Dec 15 2020 at 21:02):

Extending to second order seems formally straighforward to write down:

image.png

The question is: What is $\tilde\Omega^1\otimes\tilde\Omega^1\to\tilde\Omega^1$?

I know what I want it to be, but I don't know how to motivate it. Maybe it is fine to define it and prove it works. That is probably justification enough :thinking:

Eric Forgy (Dec 15 2020 at 21:31):

A guiding principle should be (I believe):

$a = \sum_i x_i\otimes y_i\otimes z_i\in\tilde\Omega^2\implies a = \sum_i x_i d y_i d z_i.$

Eric Forgy (Dec 15 2020 at 22:20):

As an intermediate step, we have a product

$(A\otimes A)\otimes (A\otimes A)\to A\otimes A\otimes A$

written as juxtaposition and given by

$(w\otimes x)(y\otimes z) = w\otimes (xy)\otimes z.$

This is similar to the composition of spans.

Eric Forgy (Dec 15 2020 at 22:34):

If $w\otimes x, y\otimes z\in\tilde\Omega^1$, then their product is zero if $x\otimes y$ is also in $\tilde\Omega^1$ so we don't want that. In fact, we'll see later that we want $x = y$ and we know $x\otimes x\not\in\tilde\Omega^1.$

Eric Forgy (Dec 15 2020 at 22:35):

There is one thing missing before we completely define $\tilde\Omega^1\otimes\tilde\Omega^1\to\tilde\Omega^1.$

Eric Forgy (Dec 15 2020 at 22:45):

Spolier: I think we want $\tilde G^2 = 1\otimes 1\otimes 1 = 0.$

Eric Forgy (Dec 15 2020 at 22:55):

The product $m: \tilde\Omega^1\otimes\tilde\Omega^1\to\tilde\Omega^1$ should be an $A$-bimodule morphism so

$m\left(u(v\otimes w)\otimes(x\otimes y)z\right) = u \left(m\left((v\otimes w)\otimes(x\otimes y)\right)\right) z.$

Eric Forgy (Dec 15 2020 at 23:29):

Eric Forgy said:

Spolier: I think we want $\tilde G^2 = 1\otimes 1\otimes 1 = 0.$

Maybe a better way to say it is we want

$x\otimes y\in\tilde\Omega^1\implies x\otimes 1\otimes y = 0.$

Eric Forgy (Dec 15 2020 at 23:32):

(I think)

Eric Forgy (Dec 15 2020 at 23:40):

Put another way,

$x\otimes y, y\otimes z\in\tilde\Omega^1\implies x\otimes z\not\in\tilde\Omega^1.$

Eric Forgy (Dec 16 2020 at 17:41):

Updated diagram to make that intermediate step more explicit:

image.png

Eric Forgy (Dec 16 2020 at 18:05):

I'm probably thinking about this all wrong.