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I have a question which is not really precise, unfortunately.

Let $A$ be a Poisson $n$-algebra, i.e. a graded commutative algebra with a Lie bracket of degree $n-1$ s.t. the bracket is a biderivation w.r.t. the commutative product. Add a new square-zero variable $\varepsilon$ to $A$, possibly of positive degree, to get $A[\varepsilon]$. Clearly this is a graded commutative algebra, and there is also a shifted Lie bracket $[a+\varepsilon b, a'+\varepsilon b'] := [a,a'] + \varepsilon ([a,b'] \pm [a',b])$.

If $A$ has some geometric meaning (and this is the imprecise part), does $A[\varepsilon]$ also has some geometric meaning? I'm open to various notions of "geometric meaning". I found this appearing in an algebro-topological setting, and I'm wondering if there is a bigger picture I don't know about. This may be something very simple. My best guess is that $A[\varepsilon]$ is twice the dimension of $A$, so perhaps this is related to some (co)tangent bundle...?

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    $\begingroup$ If $\mathfrak{g}$ is a dg Lie algebra controlling a formal moduli problem $X$, then $\mathfrak{g}[\epsilon]$ controls the formal moduli problem $\mathrm{T} X$ (the tangent bundle). $\endgroup$ Sep 19, 2018 at 15:32
  • $\begingroup$ @Pavel Thanks! That's already a good point of view. $\endgroup$ Sep 19, 2018 at 17:43

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I can sort of give an answer in the context of derived geometry.

So let $A$,$B$ derived rings (i.e. dg-rings, simplicial rings, ${\mathbb{E}_\infty }$-rings).

For a $B$-module $M$, we can define the a square-zero extension ${B \oplus M}$. In your specific case $M$ would be $B\epsilon$.

Let $f:Y = SpecB \to X = SpecA$ be a map of affine derived schemes and $Y\left[ M \right] = Spec\left( {B \oplus M} \right)$ the affine derived scheme corresponding to the square-zero extension of $B$ by $M$.

Then we can define a space of derivations of Y over X via a pullback diagram $\begin{array}{*{20}{c}} {De{r_X}\left( {Y;M} \right)}& \to &{Map{s_X}\left( {Y\left[ M \right],Y} \right)} \\ \downarrow &{}& \downarrow \\ {\left\{ {i{d_Y}} \right\}}& \to &{Map{s_X}\left( {Y,Y} \right)} \end{array}$

Then there exists a $B$-module ${\mathbb{L}_{Y/X}}$, called the cotangent complex, which acts as a sort of universal derivation. i.e. $De{r_X}\left( {Y;M} \right) \simeq Map{s_B}\left( {{\mathbb{L}_{Y/X}},M} \right)$

The contangent complex plays a very important role in derived geometry, pretty much an analogous role to the cotangent bundle in usual geometry.

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  • $\begingroup$ If I unpack everything, you're saying that morphisms $B \to A[\varepsilon]$ are pairs consisting of a morphism and a derivation, right? $\endgroup$ Sep 19, 2018 at 17:46
  • $\begingroup$ I think I can clarify my objection now. I think you are taking the point of view of what happens when we vary the square-zero extension, $A \oplus M$, and the answer is basically that this is represented by the cotangent complex. But here I have a fixed square-zero extension, and I want to know what this particular one means. $\endgroup$ Sep 20, 2018 at 15:16
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    $\begingroup$ M is just an arbitrary target. i.e Derivations $B \to M$ would be defined via maps $B \to B \oplus M$, and the cotangent complex is essentially a universal derivation in the sense we have a factorization $B \to {\mathbb{L}_{Y/X}} \to M$. So the particular case, $M = B\varepsilon $ is what characterizes derivations from B to itself, $B \to B$. $\endgroup$
    – JJJ
    Sep 23, 2018 at 17:53
  • $\begingroup$ I understand that the module is arbitrary in your answer, but my question is about the particular module I wrote... $\endgroup$ Sep 23, 2018 at 18:47
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    $\begingroup$ Yeah I was saying the particular module you wrote characterizes derivations from the ring to itself. $\endgroup$
    – JJJ
    Sep 23, 2018 at 19:46

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