Mixing
Lie groups in synthetic differential geometry
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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 $Dto Dtimes Dto D$ is a colimit" where the first two arrows are $dmapsto (d,0)$ and $dmapsto (0,d)$ 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 $+: Dtimes Dto R$. 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, $d+d = +(d,d) = fcirc m (d,d) = f(0)$ 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 $X,Yin T_eG$ he defines $Xstar Y(d_1,d_2) := X(d_1)Y(d_2)X(-d_1)Y(-d_2)$ (an "infinitesimal commutator") and says $Xstar Y(0,d) = Xstar Y (d,0) = e$ 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 $X(d_1)Y(d_2)X(d_1)^-1Y(d_2)^-1$ instead ?
differential-geometry lie-groups constructive-mathematics intuitionistic-logic synthetic-differential-geometry
asked Jul 14 at 16:28
Max
10.4k1836
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up vote
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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 $Dto Dtimes Dto D$ is a colimit" where the first two arrows are $dmapsto (d,0)$ and $dmapsto (0,d)$ 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 $+: Dtimes Dto R$. 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, $d+d = +(d,d) = fcirc m (d,d) = f(0)$ 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 $X,Yin T_eG$ he defines $Xstar Y(d_1,d_2) := X(d_1)Y(d_2)X(-d_1)Y(-d_2)$ (an "infinitesimal commutator") and says $Xstar Y(0,d) = Xstar Y (d,0) = e$ 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 $X(d_1)Y(d_2)X(d_1)^-1Y(d_2)^-1$ instead ?
differential-geometry lie-groups constructive-mathematics intuitionistic-logic synthetic-differential-geometry
asked Jul 14 at 16:28
Max
10.4k1836
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 $Dto Dtimes Dto D$ is a colimit" where the first two arrows are $dmapsto (d,0)$ and $dmapsto (0,d)$ 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 $+: Dtimes Dto R$. 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, $d+d = +(d,d) = fcirc m (d,d) = f(0)$ 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 $X,Yin T_eG$ he defines $Xstar Y(d_1,d_2) := X(d_1)Y(d_2)X(-d_1)Y(-d_2)$ (an "infinitesimal commutator") and says $Xstar Y(0,d) = Xstar Y (d,0) = e$ 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 $X(d_1)Y(d_2)X(d_1)^-1Y(d_2)^-1$ instead ?
differential-geometry lie-groups constructive-mathematics intuitionistic-logic synthetic-differential-geometry
asked Jul 14 at 16:28
Max
10.4k1836
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 $Dto Dtimes Dto D$ is a colimit" where the first two arrows are $dmapsto (d,0)$ and $dmapsto (0,d)$ 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 $+: Dtimes Dto R$. 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, $d+d = +(d,d) = fcirc m (d,d) = f(0)$ 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 $X,Yin T_eG$ he defines $Xstar Y(d_1,d_2) := X(d_1)Y(d_2)X(-d_1)Y(-d_2)$ (an "infinitesimal commutator") and says $Xstar Y(0,d) = Xstar Y (d,0) = e$ 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 $X(d_1)Y(d_2)X(d_1)^-1Y(d_2)^-1$ instead ?
differential-geometry lie-groups constructive-mathematics intuitionistic-logic synthetic-differential-geometry
asked Jul 14 at 16:28
Max
10.4k1836
edited Jul 14 at 17:47
asked Jul 14 at 16:28
Max
10.4k1836
asked Jul 14 at 16:28
Max
10.4k1836
asked Jul 14 at 16:28
Max
10.4k1836
10.4k1836
add a comment |Â
add a comment |Â
1 Answer
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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 $Dsubstackto\to\toDtimes Dstackrelmto D$ where this includes the maps $dmapsto (d,0)$ and $dmapsto (0,d)$ and also $dmapsto (0,0)$. This requires an arrow $f : Dtimes D to R$ 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, $(Xstar Y)(d,0)=(Xstar Y)(0,d)=e=(Xstar Y)(0,0)$ validates the condition to apply the colimit's universal property.
For the latter, Proposition 6.1 states that any function $f :Dtimes D to R$ such that $f(d,0)=f(0,d)$ is of the form $f(d_1,d_2) = g(d_1+d_2)$ where $g:D_2to R$. That is, $R$ sees $Drightrightarrows Dtimes Dto D_2$ as a coequalizer. Microlinearity means that $G$ also sees this as a coequalizer, which is to say: $(d_1,d_2)mapsto X(d_1)X(d_2):Dtimes D to G$ 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 $$X(d)X(-d) = g(d+(-d)) = g(0) = X(0)X(0) = e$$
answered Jul 14 at 20:42
Derek Elkins
14.9k11035
Ok for the second point, that's very helpful ! But for the first one I wrote $+: Dtimes Dto R$, not $Dtimes Dto D$. And the text states that $m: Dtimes Dto D$ is the coequalizer (in the eyes of $R$) of $dmapsto (d,0)$ and $dmapsto (0,d)$ (and actually uses this for the Lie bracket)
– Max
Jul 14 at 21:40
@Max You're right. Edit in a moment.
– Derek Elkins
Jul 14 at 22:21
Great, now it's much clearer ! Thanks for the notes as well, they look great !
– Max
Jul 15 at 8:28
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 $Dsubstackto\to\toDtimes Dstackrelmto D$ where this includes the maps $dmapsto (d,0)$ and $dmapsto (0,d)$ and also $dmapsto (0,0)$. This requires an arrow $f : Dtimes D to R$ 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, $(Xstar Y)(d,0)=(Xstar Y)(0,d)=e=(Xstar Y)(0,0)$ validates the condition to apply the colimit's universal property.
For the latter, Proposition 6.1 states that any function $f :Dtimes D to R$ such that $f(d,0)=f(0,d)$ is of the form $f(d_1,d_2) = g(d_1+d_2)$ where $g:D_2to R$. That is, $R$ sees $Drightrightarrows Dtimes Dto D_2$ as a coequalizer. Microlinearity means that $G$ also sees this as a coequalizer, which is to say: $(d_1,d_2)mapsto X(d_1)X(d_2):Dtimes D to G$ 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 $$X(d)X(-d) = g(d+(-d)) = g(0) = X(0)X(0) = e$$
answered Jul 14 at 20:42
Derek Elkins
14.9k11035
Ok for the second point, that's very helpful ! But for the first one I wrote $+: Dtimes Dto R$, not $Dtimes Dto D$. And the text states that $m: Dtimes Dto D$ is the coequalizer (in the eyes of $R$) of $dmapsto (d,0)$ and $dmapsto (0,d)$ (and actually uses this for the Lie bracket)
– Max
Jul 14 at 21:40
@Max You're right. Edit in a moment.
– Derek Elkins
Jul 14 at 22:21
Great, now it's much clearer ! Thanks for the notes as well, they look great !
– Max
Jul 15 at 8:28
add a comment |Â
up vote
1
down vote
accepted
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 $Dsubstackto\to\toDtimes Dstackrelmto D$ where this includes the maps $dmapsto (d,0)$ and $dmapsto (0,d)$ and also $dmapsto (0,0)$. This requires an arrow $f : Dtimes D to R$ 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, $(Xstar Y)(d,0)=(Xstar Y)(0,d)=e=(Xstar Y)(0,0)$ validates the condition to apply the colimit's universal property.
For the latter, Proposition 6.1 states that any function $f :Dtimes D to R$ such that $f(d,0)=f(0,d)$ is of the form $f(d_1,d_2) = g(d_1+d_2)$ where $g:D_2to R$. That is, $R$ sees $Drightrightarrows Dtimes Dto D_2$ as a coequalizer. Microlinearity means that $G$ also sees this as a coequalizer, which is to say: $(d_1,d_2)mapsto X(d_1)X(d_2):Dtimes D to G$ 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 $$X(d)X(-d) = g(d+(-d)) = g(0) = X(0)X(0) = e$$
answered Jul 14 at 20:42
Derek Elkins
14.9k11035
Ok for the second point, that's very helpful ! But for the first one I wrote $+: Dtimes Dto R$, not $Dtimes Dto D$. And the text states that $m: Dtimes Dto D$ is the coequalizer (in the eyes of $R$) of $dmapsto (d,0)$ and $dmapsto (0,d)$ (and actually uses this for the Lie bracket)
– Max
Jul 14 at 21:40
@Max You're right. Edit in a moment.
– Derek Elkins
Jul 14 at 22:21
Great, now it's much clearer ! Thanks for the notes as well, they look great !
– Max
Jul 15 at 8:28
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 $Dsubstackto\to\toDtimes Dstackrelmto D$ where this includes the maps $dmapsto (d,0)$ and $dmapsto (0,d)$ and also $dmapsto (0,0)$. This requires an arrow $f : Dtimes D to R$ 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, $(Xstar Y)(d,0)=(Xstar Y)(0,d)=e=(Xstar Y)(0,0)$ validates the condition to apply the colimit's universal property.
For the latter, Proposition 6.1 states that any function $f :Dtimes D to R$ such that $f(d,0)=f(0,d)$ is of the form $f(d_1,d_2) = g(d_1+d_2)$ where $g:D_2to R$. That is, $R$ sees $Drightrightarrows Dtimes Dto D_2$ as a coequalizer. Microlinearity means that $G$ also sees this as a coequalizer, which is to say: $(d_1,d_2)mapsto X(d_1)X(d_2):Dtimes D to G$ 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 $$X(d)X(-d) = g(d+(-d)) = g(0) = X(0)X(0) = e$$
answered Jul 14 at 20:42
Derek Elkins
14.9k11035
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 $Dsubstackto\to\toDtimes Dstackrelmto D$ where this includes the maps $dmapsto (d,0)$ and $dmapsto (0,d)$ and also $dmapsto (0,0)$. This requires an arrow $f : Dtimes D to R$ 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, $(Xstar Y)(d,0)=(Xstar Y)(0,d)=e=(Xstar Y)(0,0)$ validates the condition to apply the colimit's universal property.
For the latter, Proposition 6.1 states that any function $f :Dtimes D to R$ such that $f(d,0)=f(0,d)$ is of the form $f(d_1,d_2) = g(d_1+d_2)$ where $g:D_2to R$. That is, $R$ sees $Drightrightarrows Dtimes Dto D_2$ as a coequalizer. Microlinearity means that $G$ also sees this as a coequalizer, which is to say: $(d_1,d_2)mapsto X(d_1)X(d_2):Dtimes D to G$ 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 $$X(d)X(-d) = g(d+(-d)) = g(0) = X(0)X(0) = e$$
answered Jul 14 at 20:42
Derek Elkins
14.9k11035
edited Jul 14 at 22:23
answered Jul 14 at 20:42
Derek Elkins
14.9k11035
answered Jul 14 at 20:42
Derek Elkins
14.9k11035
answered Jul 14 at 20:42
Derek Elkins
14.9k11035
14.9k11035
Ok for the second point, that's very helpful ! But for the first one I wrote $+: Dtimes Dto R$, not $Dtimes Dto D$. And the text states that $m: Dtimes Dto D$ is the coequalizer (in the eyes of $R$) of $dmapsto (d,0)$ and $dmapsto (0,d)$ (and actually uses this for the Lie bracket)
– Max
Jul 14 at 21:40
@Max You're right. Edit in a moment.
– Derek Elkins
Jul 14 at 22:21
Great, now it's much clearer ! Thanks for the notes as well, they look great !
– Max
Jul 15 at 8:28
add a comment |Â
Ok for the second point, that's very helpful ! But for the first one I wrote $+: Dtimes Dto R$, not $Dtimes Dto D$. And the text states that $m: Dtimes Dto D$ is the coequalizer (in the eyes of $R$) of $dmapsto (d,0)$ and $dmapsto (0,d)$ (and actually uses this for the Lie bracket)
– Max
Jul 14 at 21:40
@Max You're right. Edit in a moment.
– Derek Elkins
Jul 14 at 22:21
Great, now it's much clearer ! Thanks for the notes as well, they look great !
– Max
Jul 15 at 8:28
Ok for the second point, that's very helpful ! But for the first one I wrote $+: Dtimes Dto R$, not $Dtimes Dto D$. And the text states that $m: Dtimes Dto D$ is the coequalizer (in the eyes of $R$) of $dmapsto (d,0)$ and $dmapsto (0,d)$ (and actually uses this for the Lie bracket)
– Max
Jul 14 at 21:40
Ok for the second point, that's very helpful ! But for the first one I wrote $+: Dtimes Dto R$, not $Dtimes Dto D$. And the text states that $m: Dtimes Dto D$ is the coequalizer (in the eyes of $R$) of $dmapsto (d,0)$ and $dmapsto (0,d)$ (and actually uses this for the Lie bracket)
– Max
Jul 14 at 21:40
@Max You're right. Edit in a moment.
– Derek Elkins
Jul 14 at 22:21
@Max You're right. Edit in a moment.
– Derek Elkins
Jul 14 at 22:21
Great, now it's much clearer ! Thanks for the notes as well, they look great !
– Max
Jul 15 at 8:28
Great, now it's much clearer ! Thanks for the notes as well, they look great !
– Max
Jul 15 at 8:28
add a comment |Â
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