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Stream: deprecated: id my structure

Topic: Quasi-Derivations?

Carlos Zapata-Carratala (Sep 22 2023 at 23:49):

Hi everyone!

Let $R$ be a ring. An additive map $\Delta: R\to R$ is (provisionally) called a quasi-derivation when

$\Delta(a\cdot b) = \Delta(a)\cdot b + a \cdot \Delta(b) + \Delta(a)\cdot \Delta(b)$.

Quasi-derivations do not close under addition, they do not form a $R$-module in the usual way and they do not close under commutator bracket (in fact they don't seem to carry any obvious Lie algebra structure).

Short simple question: has anyone seen these guys before? They appear naturally when defining finite difference operators, and particularly in discrete analogues of directional derivatives.

John Baez (Sep 23 2023 at 09:10):

No, I haven't seen those. That formula reminds me slightly of the Rota-Baxter identity but I don't see an actual connection.

John Baez (Sep 23 2023 at 09:11):

Say we define a finite difference operator on functions $f: \mathbb{R} \to \mathbb{R}$ in the usual way:

$(\Delta(f))(x) = f(x+1) - f(x)$

John Baez (Sep 23 2023 at 09:13):

$\Delta$ is not a derivation but it's a twisted derivation:

$(\Delta(f g))(x) = f(x+1)g(x+1) - f(x) g(x)$
$= (f(x+1) - f(x)) g(x+1) + f(x)(g(x+1) - g(x))$
$= \Delta(f) \alpha(g) + f \Delta(g)$

John Baez (Sep 23 2023 at 09:17):

where $\alpha$ is the automorphism of the ring of functions $f : \mathbb{R} \to \mathbb{R}$ given by

$(\alpha(g))(x) = g(x+1)$

John Baez (Sep 23 2023 at 09:17):

So we say $\Delta$ is a derivation twisted by the automorphism $\alpha$.

John Baez (Sep 23 2023 at 09:18):

Twisted derivations are in turn a special case of derivations on a ring taking values in a bimodule of that ring. I've been thinking about those lately. Given a ring $R$ and a bimodule $M$, a derivation on $R$ with values in $M$ is a function

$\Delta : R \to M$

such that

$\Delta(1) = 0$
$\Delta(r + r') = \Delta(r) + \Delta(r')$
$\Delta(r r') = \Delta(r) r' + r \Delta(r')$

John Baez (Sep 23 2023 at 09:21):

where the last formula the multiplication is defined using the fact that $M$ is a bimodule of $R$. Taking $M = R$ with various interesting bimodule structures we get various kinds of 'twisted' derivations on $R$.

John Baez (Sep 23 2023 at 09:22):

Derivations of this generalized sort are connected to 'Hochschild cohomology'.

Tim Hosgood (Sep 23 2023 at 12:32):

this is an interesting structure for sure! I haven't seen it, so I can't say anything useful about it, but the fact that $\Delta(x^2)=2x\Delta(x)+\Delta(x)^2$ seems to suggest that it's like a derivation "up to higher order differential information", so maybe usual derivations sort of factor through these things?

Tim Hosgood (Sep 23 2023 at 12:32):

(not a very helpful answer but I'm mainly posting so I'll be notified if anybody has an answer!)

Jacques Carette (Sep 24 2023 at 13:41):

You might want to look into work done around Ore Algebras.

JS PL (he/him) (Sep 25 2023 at 21:41):

Carlos Zapata-Carratala said:

Let $R$ be a ring. An additive map $\Delta: R\to R$ is (provisionally) called a quasi-derivation when

$\Delta(a\cdot b) = \Delta(a)\cdot b + a \cdot \Delta(b) + \Delta(a)\cdot \Delta(b)$.

This is called a derivation of weight 1 and it is the differentiation operator for weighted differential algebras

JS PL (he/him) (Sep 25 2023 at 21:43):

So in general, for a base commutative ring $k$ and a $k$-algebra $R$ and $\lambda \in k$, a derivation of weight $\lambda$ on $R$ is a $k$-linear map $D: R \to R$ such that:
$D(a \cdot b) = D(a) \cdot b + a \cdot D(b) + \lambda \cdot D(a) \cdot D(b)$

JS PL (he/him) (Sep 25 2023 at 21:46):

Here are some references:
https://arxiv.org/pdf/2003.03899.pdf
https://arxiv.org/pdf/math/0703780.pdf

Carlos Zapata-Carratala (Sep 25 2023 at 21:46):

JS PL (he/him) said:

So in general, for a base commutative ring $k$ and a $k$-algebra $R$ and $\lambda \in k$, a derivation of weight $\lambda$ on $R$ is a $k$-linear map $D: R \to R$ such that:
$D(a \cdot b) = D(a) \cdot b + a \cdot D(b) + \lambda \cdot D(a) \cdot D(b)$

Precisely! That's the identity that I found for finite difference operators on smooth manifolds (where the weight is the "width" of the finite difference). [Edit: I asked too soon! Just saw the references]

JS PL (he/him) (Sep 25 2023 at 21:47):

yes: I just posted some above (right when you posted apparently haha)

JS PL (he/him) (Sep 25 2023 at 21:50):

John Baez said:

No, I haven't seen those. That formula reminds me slightly of the Rota-Baxter identity but I don't see an actual connection.

Oh yes they are very much related to Rota-Baxter stuff! Briefly, a Rota-Baxter algebra is the "intergration" analogue of a differential algebra. Then a differential Rota-Baxter algebra is one with a differential and an integral operation which satisfy versions of the Fundamental theorems of Calculus in nice ways. In one of the references I posted above, they explain how to construct cofree differential Rota-Baxter algebras of any weight, and how the free Rota-Baxter algebra monad and cofree differential algebra comonad distribute over one another.

JS PL (he/him) (Sep 25 2023 at 21:50):

I like differential algebras and Rota-Baxter algebras very much, and use them in my research lots. So if anyone wants to chat some more about them: happy to do so!

Carlos Zapata-Carratala (Sep 25 2023 at 22:17):

JS PL (he/him) said:

I like differential algebras and Rota-Baxter algebras very much, and use them in my research lots. So if anyone wants to chat some more about them: happy to do so!

I would love to chat! I will need some days to go over those references but I can DM you to set some time maybe towards the end of the week or next week. In the meantime, in case someone is interested where this is coming from, I will be in a Wolfram Institute livestream on Wednesday at 18:00 UTC presenting this exact topic:

https://twitter.com/WolframInst/status/1706403496914718780

https://www.youtube.com/@wolframinstitute

Carlos Zapata-Carratala (Sep 25 2023 at 22:24):

John Baez said:

$\Delta$ is not a derivation but it's a twisted derivation:

$(\Delta(f g))(x) = f(x+1)g(x+1) - f(x) g(x)$
$= (f(x+1) - f(x)) g(x+1) + f(x)(g(x+1) - g(x))$
$= \Delta(f) \alpha(g) + f \Delta(g)$

Very interesting perspective, I was not aware of twisted derivations (thanks for the embedded reference!). Those are the kinds of identities I was working with but used the (trivial) fact that

$g(x+1)=g(x)+\Delta g (x)$

to get the formula I wrote instead. From @JS PL (he/him) 's comments it seems to me that when it comes to finite difference operators there is a choice to regard them as twisted derivations or weigthed derivations. Very interesting stuff, thanks for the reply!