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In many places on the internet, you can find that a derivation is defined as a linear operator on a ring or algebra, satisfying the Leibniz rule.

I'm trying to understand the algebraic meaning of this point of view. My fundamental doubt is: Why do we want to define derivations on rings or algebras of functions in the first place? Couldn't we define derivations on vector spaces of functions?

Example: in the basic case of the ring of R->R functions (where both sum and product are defined pointwise) the derivation is of course the derivative. But it's not obvious to me why couldn't we have a structure of functions with pointwise sum where the product is not even defined, and still define the derivative there.

[ EDIT: To clarify this point: Let's take the R-vector space $C^\infty(R^d)$ for example. I'd want to get the sub-vector space of all linear operators on this space, which only contains the directional derivatives (defined in the usual way, as $ \lim_{\epsilon \rightarrow 0} ((f(x+\epsilon v) - f(x)) / \epsilon ) $.

I think (I'm not 100% sure) that if you consider $C^\infty(R^d)$ as an algebra, and consider all linear operators satisfying the Leibniz rule, you do in fact retrieve only the directional derivatives defined above.

If I'm wrong, nevermind. If I'm right, my question (roughly) is why is it true, considering that in the "usual" definition of directional derivative, pointwise product of functions isn't even used. ]

In other places, you learn that derivations are really infinitesimal versions of automorphisms. I haven't yet found a proof that this view implies the Leibniz rule, but I'd like to find a reference, maybe it would help. But still, if I understand correctly, we are starting from automorphisms of algebras, so there's a product.

So in summary: What's the deep algebraic reason why having a product is important to define a derivation as we intuitively know it?

Micoloth
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  • You should avoid the “the derivation is” replace it by an example of derivation is given by the usual differentiation, in order to avoid confusion. The whole point of derivation is to mimic endomorphism that satisfy similar results as the usual differentiation, i.e. $d(f\cdot g) = d(f) \cdot g + \epsilon f \cdot d(g)$ with $\epsilon$ a sign (usually 1 in the normal case). You have a narrow link between these endomorphisms and the tangent space of a differential manifold. However why do we take a ring? Because we need the addition, the multiplication (and in the graded case the $-$) – julio_es_sui_glace Apr 26 '24 at 09:15
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    Another very abstract motivation could be the definition of dg algebras and dg-categories, that are very powerful tools, and also homological algebra in general – julio_es_sui_glace Apr 26 '24 at 09:18
  • Well.. To be fair, I didn't say “the derivation is”. I said "a derivation is defined as", which is precisely what it says here for example https://en.wikipedia.org/wiki/Derivation_(differential_algebra) – Micoloth Apr 26 '24 at 09:18

  • the definition of dg algebras and dg-categories

    This sounds very interesting.. Can you offer any reference? :)

     – Micoloth Apr 26 '24 at 09:19
  • “the derivation is of course the derivative. ” I was mentioning this part, they are other derivations on the space of smooth maps from $\mathbb{R}$ to $\mathbb{R}$ – julio_es_sui_glace Apr 26 '24 at 09:21
  • For dg stuff, you should first study homological algebra, it would be to big of a step to first go to the dg world – julio_es_sui_glace Apr 26 '24 at 09:24
  • Ah.. I though you can prove that there are not, and the Leibniz rule was enough to effectively define a unique operator, which is in fact the Derivative.

    I thought this based on various answers such as this one, and others that I can't find anymore: https://math.stackexchange.com/questions/2000414/are-all-derivations-of-real-valued-functions-derivatives If there are other derivations, can you give me some examples?

     – Micoloth Apr 26 '24 at 09:29
  • Let us continue this discussion in chat. – julio_es_sui_glace Apr 26 '24 at 09:30

1 Answers1

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On an algebra $A$, a derivation $D$ is a linear endomorphism of $A$ which satisfies the Leibniz rule: $D(ab)=aD(b) + D(a)b$. The product is thus essential from the definition. We can remove linearity and move from an algebra to a ring. But how would you define a derivation if there is no longer a product? Or rather, what would you want to add to being an endomorphism of a group?

Emmanuel Royer
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  • I might have been unclear, let me explain better:

    "What do you want to add ? " ->

    Let's take the R-vector space $C^\infty(R^d)$ for example. I'd want to get the sub-vector space of all linear operators on this space, which only contains the directional derivatives (defined in the usual way, as $ \lim_{\epsilon \rightarrow 0} (f(x+\epsilon v) - f(x)) / \epsilon ) $. )

     – Micoloth Apr 26 '24 at 10:09
  • I think (I'm not 100% sure) that if you consider $C^\infty(R^d)$ as an algebra with pointwise product, and add the Leibniz rule, you do in fact get the space containing only the directional derivatives defined above.

    If it's wrong, nevermind.

    If it's true, my question (roughly) is why is it true, considering that in the "basic" definition of directional derivative, pointwise product of functions isn't even used.

    My question really is where does the product come up from.

     – Micoloth Apr 26 '24 at 10:10
  • But really, ideally I don't want an analytic proof on $C^\infty(R^d)$, I want an algebraic reason... – Micoloth Apr 26 '24 at 10:15