1
$\begingroup$

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 ?

$\endgroup$
2
$\begingroup$

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 $$X(d)X(-d) = g(d+(-d)) = g(0) = X(0)X(0) = e$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.