# Synthetic differential geometry

I'm reading Serre's "Lie algebra and Lie groups" now, and I found a description of Lie algebra of an algebraic matrix group, given for a free algebra (free as a module) $k'$ with a basis $\lbrace 1, \varepsilon\rbrace$, $\varepsilon^2 = 0$ as all $X \in M(n,k)$ such that $1 + \varepsilon X \in G(k')$, where $G(k')$ is a set of zeros of a number of polinomials.

$k'$ struck me as being extremely close to how smooth infinitesimals are defined in http://arxiv.org/abs/0805.3307 (a link I found on Wiki)

I'm a grad student who recently switched from applied math, so I know very little about algebra, differential geometry etc.

Can you please clarify how is such a free algebra related to smooth infinitesimals, and when were the latter invented, and how close is synthetic differential geometry to this construction? I currently don't know much about synthetic differential geometry, but I know it uses smooth infinitesimals to define differential, tangent space etc. differently from how they are usually defined.

Sorry if the questions are not clear enough :) I want to dive into the interesting subject, I just need a push in the right direction. It amazes me how the example fits the concept of a Lie algebra as being an 'infinitesimal Lie group' when you think of $\varepsilon$ as an infinitesimal, but I don't know what to make out of it, or even if I am right in my intuition.

• If you recently switched from applied math and don't know a lot about algebra I don't know if Serre is the best reference for you... what are you reading Serre for? – Qiaochu Yuan Nov 21 '10 at 22:11
• I'm reading Serre to make sense of Lie groups and algebra course in my university :) I basically understand it, but it's hard to wrap my head around without putting various bits of theory in perspective so I picked up the book and started studying various chapters to sort of get the feeling of what is the general line of reasoning that leads to Cartan subalgebras, for example. – Alexei Averchenko Nov 22 '10 at 4:30

Your intuition is right. This $\varepsilon^2 = 0$ stuff is basically an algebraic way to extract differential information, but it's certainly not as sophisticated as the full edifice of synthetic differential geometry because for polynomials everything is trivial. The basic observation here is that $f(a + \varepsilon b) = f(a) + b f'(a) \varepsilon$ and from here everything else follows.
Recall that if $f_1, ... f_k$ are a family of polynomials, then a $k$-homomorphism $k[x_1, ... x_n]/(f_1, ... f_k) \to k$ is the same thing as a $k$-point $(p_1, ... p_n)$ on the variety $f_1 = ... = f_k = 0$. The appropriate generalization here is that a $k$-homomorphism $k[x_1, ... x_n]/(f_1, ... f_k) \to k[\varepsilon]/\varepsilon^2$ is the same thing as a "generalized point" $(p_1 + \varepsilon q_1, ... p_n + \varepsilon q_n)$ on the variety. If you write down what it means for $f_1, ... f_k$ to vanish on this point, the condition you get is that $p = (p_1, ... p_n)$ must be a point in the ordinary sense and $(q_1, ... q_n)$ must be orthogonal to the gradients of all the $f_i$; but this is equivalent to being a tangent vector at $p$.