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Stream: deprecated: discrete geometry and entanglement

Topic: Commuting diagrams

Eric Forgy (Dec 19 2020 at 05:41):

I made a mistake (imagine that!) :sweat_smile:

I said that the first-order calculus can be summarized by the commuting diagram

image.png

That is wrong. Instead, we need two separate commuting diagrams

20201218_213822.jpg

Eric Forgy (Dec 19 2020 at 05:54):

It seemed natural to me to try to combine things but if I put the two diagrams together and claim the whole thing is commutative, then I force some things to be isomorphisms that aren't isomorphisms.

Eric Forgy (Dec 19 2020 at 09:28):

I think I found a way to extend this to $\tilde\Omega^2$ via

20201219_012348.jpg

The literature, e.g. here, says $\tilde\Omega^2 = \tilde\Omega^1\otimes_A\tilde\Omega^1$, but I am not sure if that is correct :thinking:

I think $\tilde\Omega^2$ is a sub-bimodule of $\tilde\Omega^1\otimes_A\tilde\Omega^1.$

Eric Forgy (Dec 19 2020 at 09:36):

The map $\mu_1:A\otimes_K A\otimes_K A\to A\otimes_K A$ is a bimodule morphism that acts on basis elements by "popping out the middle", i.e.

$e^i\otimes_K e^j\otimes_K e^k \mapsto e^i\otimes_K e^k$

and extends linearly to all of $A^{\otimes_K 3},$ i.e.

$\mu_1(\alpha) = \sum_{i,j,k} \alpha(i,j,k) \mu_1(e^i\otimes_K e^j\otimes_K e^k) = \sum_{i,k} \left[\sum_j \alpha(i,j,k)\right] e^i\otimes_K e^k.$

Eric Forgy (Dec 19 2020 at 09:38):

If $f,g\in A$, then clearly

$\mu_1(f\,\alpha\, g) = f\,\mu_1(\alpha)\, g.$

Eric Forgy (Dec 19 2020 at 10:04):

I think $\tilde\Omega^2$ should be the kernel of $\mu_1\circ m: \tilde\Omega^1\otimes_A\Omega^1\to A\otimes_K A.$

Let $\alpha,\beta\in\tilde\Omega^1$ so that

$\begin{aligned} \alpha\otimes_A\beta &= \sum_{(i,j)}\sum_{(k,l)} \alpha(i,j) \beta(k,l) (e^i\otimes_K e^j) \otimes_A (e^k\otimes_K e^l) \\ &= \sum_{(i,j)}\sum_{(k,l)} \alpha(i,j) \beta(k,l) \delta^{j,k} (e^i\otimes_K e^j) \otimes_A (e^j\otimes_K e^l) \\ &= \sum_{(i,j,k)} \alpha(i,j) \beta(j,k) (e^i\otimes_K e^j) \otimes_A (e^j\otimes_K e^k), \end{aligned}$

where $(i,j,k)$ is a length two path $i\to j\to k$ and

$m\left(\alpha\otimes_A\beta\right)= \sum_{(i,j,k)} \alpha(i,j) \beta(j,k) e^i\otimes_K e^j \otimes_K e^k.$

so that

$\mu_1\circ m\left(\alpha\otimes_A\beta\right)= \sum_{(i,-,k)} \left[\sum_j \alpha(i,j) \beta(j,k)\right] e^i\otimes_K e^k.$

This is zero when

$\sum_j \alpha(i,j) \beta(j,k) = 0.$