I'm reading Mike Shulman's Synthetic Differential geometry (a small article for the "pizza seminar" it seems); and there's a part about Lie groups that I have trouble understanding.

Actually there are two statements that he makes with respect to the Lie bracket to be defined on $T_eG$ that I don't understand.

The first one is "$R$ thinks $D\to D\times D\to D$ is a colimit" where the first two arrows are $d\mapsto (d,0)$ and $d\mapsto (0,d)$ and the last arrow is multiplication $m(d_1,d_2) = d_1d_2$ ($R$ is the "real" line and $D$ the set of nilsquares of $R$). Now I (think I) understand what that means but it looks as if it were completely wrong. Indeed consider $+: D\times D\to R$. Clearly $+(d,0) = d = +(0,d)$, so $+$ coequalizes the first two arrows; but $+$ doesn't factor through $m$, does it ?

It would mean that, say, $d+d = +(d,d) = f\circ m (d,d) = f(0)$ so $2d = 2d'$ for all nilsquares, so all nilsquares would be $0$ (we've assumed $2$ was invertible), which we know isn't true.

So did I misunderstand something or is the statement wrong ?

The second thing I don't understand is that, for $X,Y\in T_eG$ he defines $X\star Y(d_1,d_2) := X(d_1)Y(d_2)X(-d_1)Y(-d_2)$ (an "infinitesimal commutator") and says $X\star Y(0,d) = X\star Y (d,0) = e$ for all $d$. But I don't see why $X(d)X(-d)$ should be $e$.

Is there something I'm not seeing or should it be something like $X(d_1)Y(d_2)X(d_1)^{-1}Y(d_2)^{-1}$ instead ?


In Kostecki's notes demonstrating the existence of a Lie bracket in this situation (Proposition 3.9 of that paper), it is noted that $R$ sees the following diagram as a colimit $D\substack{\to\\\to\\\to}D\times D\stackrel{m}{\to} D$ where this includes the maps $d\mapsto (d,0)$ and $d\mapsto (0,d)$ and also $d\mapsto (0,0)$. This requires an arrow $f : D\times D \to R$ to satisfy $f(d,0)=f(0,d)=f(0,0)$ before the universal property can be applied. This is probably the colimit Shulmann actually intended, but I think you are correct that as written it is erroneous. Certainly, $(X\star Y)(d,0)=(X\star Y)(0,d)=e=(X\star Y)(0,0)$ validates the condition to apply the colimit's universal property.

For the latter, Proposition 6.1 states that any function $f :D\times D \to R$ such that $f(d,0)=f(0,d)$ is of the form $f(d_1,d_2) = g(d_1+d_2)$ where $g:D_2\to R$. That is, $R$ sees $D\rightrightarrows D\times D\to D_2$ as a coequalizer. Microlinearity means that $G$ also sees this as a coequalizer, which is to say: $(d_1,d_2)\mapsto X(d_1)X(d_2):D\times D \to G$ looks like it is of the form $(d_1,d_2)\mapsto g(d_1+d_2)$ to $G$ because $X(d)X(0)=X(d)=X(0)X(d)$. So $$X(d)X(-d) = g(d+(-d)) = g(0) = X(0)X(0) = e$$

  • $\begingroup$ Ok for the second point, that's very helpful ! But for the first one I wrote $+: D\times D\to R$, not $D\times D\to D$. And the text states that $m: D\times D\to D$ is the coequalizer (in the eyes of $R$) of $d\mapsto (d,0)$ and $d\mapsto (0,d)$ (and actually uses this for the Lie bracket) $\endgroup$ – Maxime Ramzi Jul 14 '18 at 21:40
  • $\begingroup$ @Max You're right. Edit in a moment. $\endgroup$ – Derek Elkins left SE Jul 14 '18 at 22:21
  • $\begingroup$ Great, now it's much clearer ! Thanks for the notes as well, they look great ! $\endgroup$ – Maxime Ramzi Jul 15 '18 at 8:28

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