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The question I have is pretty straightforward, and its answer could very well be contained in some more complicated question(s) asked previously. Here it is:

Why are deformations typically considered as structures over the ring of dual numbers?

My motivation is rather general: I just want to know more about deformation theory. And this question seems like a place to start. The problem for me is that all resources I have read just jump into the fray with "Consider the ring of dual numbers..." and define first-order deformations straight away.

I am sure that I am just missing something obvious; any nudge in the right direction will be greatly appreciated.

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Minor aside: If you're working mod p, it might be necessary to work mod $p^2$ instead of over mod p dual numbers. More generally, one might consider the spectra of artinian local rings as fat points or "spaces of small wiggles". –  S. Carnahan Jul 27 '10 at 22:50
@Scott: thank you for the tip; that makes sense! @everyone else: thank you very much for the various answers! MO is spectacular: five high-quality answers all written within a few hours of my original posting! It's going to be hard to choose a winner... –  user5395 Jul 28 '10 at 14:31

5 Answers 5

Here are two main ideas which I found helpful when I was learning this stuff (very basic and handwavy, but good starting points to see where more precise and sophisticated statements are going):

  1. see how, for instance, an algebra defined over $k[x]$ can be seen as a continuously-(in some sense)-varying $k$-indexed family of $k$-algebras. In algebraical terms, you recover the elements of the family by localising at the ideal $x=t$ for each specific $t \in k$. Geometrically, think of the Grothendieck-style picture where rings really represent some kind of nice spaces, but by duality, so the maps go the “wrong” way. So a $k[x]$-algebra $R$ corresponds to a ring map $k[x] \to R$, and hence (since $k[x]$ represents the affine line $\mathbb{A}^1(k)$) to a map of spaces $X \to \mathbb{A}^1(k)$. But then we can think of $X$ as the bundle of all its fibers, a family varying over $\mathbb{A}^1(k)$. More generally, a widget defined over $k[x]$ can be thought of a a continuous (in some sense) map from $\mathbb{A}^1(k)$ into the space of widgets.

  2. now, see how just as $k[x]$ represented the affine line, $k[x]/(x^2)$ represents an “infinitessimally short line segment”, or a “walking tangent vector” (other answers include good sketches of what's meant by this and why it's a powerful viewpoint). In particular, if $R$ correpsonds to a space $X$, a map $R \to k[x]/(x^2)$ corresponds to a map from the walking tangent vector into the space $X$, or in other words a point of $X$ and a tangent vector at that point. Algebraically, we get a $k$-point of $R$ (by setting $x=0$), and (under good circumstances) a point of the Zariski tangent space.

So then putting these two together, one can see why a widget defined over the dual numbers should behave like some kind of an infinitessimal deformation: a widget, together with a tangent vector at that point in the space of widgets. (This may be literally true if there's a good moduli space of widgets!)

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If you want to look at Deformation Theory from a complex-analytic point of view ( i.e. in the spirit of Kodaira-Spencer "deformations of complex structures" ), you need to solve the Maurer-Cartan equation

$\bar \partial \varphi + \frac{1}{2}[\varphi, \varphi]=0$,

were $\varphi \in \mathcal{E}^{0,1}(T^{1,0})$. In order to do this, one first look at a solution which is a formal power series

$\varphi(t)=\varphi_1 t + \varphi_2 t^2 + \varphi_3 t^3 +...$

Collecting powers of $t$ we obtain equations

$\bar \partial \varphi_1=0$

$\bar \partial \varphi_2 + \frac{1}{2}[\varphi_1, \varphi_1]=0$


The first equation states that $\varphi_1$ is an harmonic form, that is an element of $\mathcal{H}^1(T^{1,0})$. By Hodge Theorem, this space can be identified with $H^1(X, T_X)$, which is exactly the space parametrizing "first-order" deformations.

The second equation states that you can extend the first order deformation to a second-order one (i.e., you can solve the Maurer-Cartan equation modulo $(t^3)$ ) if and only if the 2-cocycle $[\varphi_1, \varphi_1]$ is a coboundary. So the class of $[\varphi_1, \varphi_1]$ in $H^2(X, T_X)$ is the "primary obstruction" to your deformation problem.

In this way, you can try to solve modulo higher and higher powers of $t$. If all the higher order obstructions vanish and the series defining $\varphi(t)$ converges, you obtain a "genuine" deformation, namely a deformation over a small disk.

Now it should be clear that, in order to generalize this in the algebraic framework, you need a substitute for the step "solve the Maurer-Cartan equation modulo $(t^k)$ ". This substitute is roughly speaking obtained by considering deformations over Spec $k[\epsilon]/(\epsilon^k)$.

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The dual numbers are generated over $R$ by 1 and an element $d$ such that $d^2 = 0$. This $d$ can be interpreted as a first-order differential or an infinitesimal tangent vector. For if $f \in R[x]$ then you may transfer $f$ to the dual numbers and verify that calculating $f(x+d) - f(x)$ gives you $d$ times the first derivative of $f$ at $x \in R$.

In light of that, it shouldn't be surprising that dual numbers may be used to study first-order deformations.

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Your question might be

Why are infinitesimal deformations typically considered as structures over the ring of dual numbers?

A first order (or infinitesimal) deformation of an algebraic structure is a one parameter family of such algebraic structures with axioms satisfied only up to order 1. Dual numbers are just a technical way to deal with identities up to order 1. Namely, if you call the parameter $t$... then computing up to first order is same as computing modulo $t^2$.

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Just a small comment on the terminology. Infinitesimal and first-order deformations are not the same thing; in fact, an infinitesimal deformation is a deformation over Spec($A$), where $A$ is an Artinian local $k$-algebra, whereas a first-order deformation corresponds to the special case $A=k[\epsilon]/(\epsilon^2)$. –  Francesco Polizzi Jul 28 '10 at 9:12
And def's over the dual numbers (i.e. first order) are important because when calculating, one tends to do these first, and then calculate the obstruction to extending to higher order. –  Ketil Tveiten Jul 28 '10 at 17:40

Such first-order infinitesimal deformations allow one to compute the Zariski tangent space when good moduli space exists. You can find some nice motivational remarks in the first chapter of Hartshorne's Deformation Theory. See esp. the two paragraphs preceding the Exercises 1.1.

You may find it helpful to first understand analogous applications of dual numbers in simpler contexts. Below is an excerpt from one of my old sci.math posts which may prove useful in this regard.

What is the factor ring R[x]/(x^2) ?

It is known as the algebra of dual numbers over R, for R a commutative ring. It and its higher order analogs R[x]/(x^n) prove useful when studying derivations. E.g. they permit easy transfer of properties of homomorphisms to derivations -- see section 8.15 in Jacobson, Basic Algebra II. They yield algebraic models of tangent spaces.

They've been applied in many contexts, e.g. deformation theory [2], numerical analysis [3] (along with Levi-Civita fields), where they're viewed simply as truncated Taylor / power series, and in Synthetic Differential Geometry (SDG) [1], another rigorization of inifinitesimals based on work of Lawvere and Kock. SDG employs nilpotent infinitesimals, unlike Abe Robinson's nonstandard analysis which has invertible infinitesimals, hence infinities.

1 Bell, J. L. Infinitesimals. Synthese 75 (1988) #3, 285--315.

2 Szendroi, B. The unbearable lightness of deformation theory, a tutorial introduction.

3 M. Berz, Differential Algebraic Techniques,
in "Handbook of Accelerator Physics and Engineering, M. Tigner, A.Chao (Eds.)" (World Scientific, 1998)

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@Bill: thank you for your answer and the list of references. I should point out that reference [2] (the first "[2]") is a broken link.... –  user5395 Jul 28 '10 at 14:34
@altgr Thanks for pointing out the link rot - I've corrected it. ALas, I don't always have the time to check for link rot when excerpting old posts. –  Bill Dubuque Jul 28 '10 at 15:22

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