Mixing
Deriving Lie Algebra for a Lie Group
$begingroup$
I'm reading Helgason's DG, Lie Groups, and Symmetric Spaces and at one point he briefly mentions the Lobatchevski half-plane on page 136:
The group $G$ of the mappings $T_{a,b}: x to ax+b$ with $x in mathbb{R}$ and $a>0$ has a Lie algebra $mathfrak{g} = mathbb{R}e_1 + mathbb{R}e_2$ where $[e_1,e_2] = e_2$.
I'm trying to verify the claim that the bracket is what it is by taking $[cdot, cdot] equiv dleft (Adright )_e:mathfrak{gtimes gto g}$ but I'm having trouble verifying this. We can see that
$$
Ad_{a,b} circ T_{c,d} ;; =;; T_{c,; ad-bc+b}
$$
If $gamma, eta$ are curves such that $gamma(0) = eta(0) = e$ and $dot{gamma}(0) =X, ; dot{eta}(0) = Y$ then we obtain
begin{eqnarray*}
[X,Y] & = & ad_X(Y) \
& = & left. frac{d}{dt} right |_{t=0} Ad_{gamma(t)} circ T_{eta(t)} \
& = & left . frac{d}{dt} right |_{t=0} left (c(t), ; a(t)d(t) - b(t)c(t) + b(t) right ) \
& = & left (dot{c}(0), ; dot{a}(0)d(0) + a(0)dot{d}(0) - dot{b}(0)c(0) - b(0)dot{c}(0) + dot{b}(0)right ) \
& = & left (Y_1, Y_2right ) ;; =;; Y
end{eqnarray*}
where I make the short hand $gamma(t) = (a(t), b(t)), ; eta(t) = (c(t), d(t))$ and $X = (X_1, X_2), ; Y = (Y_1,Y_2)$.
On first inspection it looks as if this computation satisfies the bracket, but I'm not satisfied with it since this is done independent of a basis for $mathfrak{g}$. Can anyone point out what I'm missing or what's wrong with this computation/approach?
lie-groups lie-algebras
asked Jan 18 at 23:34
MnifldzMnifldz
6,86011634
$endgroup$
add a comment |
$begingroup$
I'm reading Helgason's DG, Lie Groups, and Symmetric Spaces and at one point he briefly mentions the Lobatchevski half-plane on page 136:
The group $G$ of the mappings $T_{a,b}: x to ax+b$ with $x in mathbb{R}$ and $a>0$ has a Lie algebra $mathfrak{g} = mathbb{R}e_1 + mathbb{R}e_2$ where $[e_1,e_2] = e_2$.
I'm trying to verify the claim that the bracket is what it is by taking $[cdot, cdot] equiv dleft (Adright )_e:mathfrak{gtimes gto g}$ but I'm having trouble verifying this. We can see that
$$
Ad_{a,b} circ T_{c,d} ;; =;; T_{c,; ad-bc+b}
$$
If $gamma, eta$ are curves such that $gamma(0) = eta(0) = e$ and $dot{gamma}(0) =X, ; dot{eta}(0) = Y$ then we obtain
begin{eqnarray*}
[X,Y] & = & ad_X(Y) \
& = & left. frac{d}{dt} right |_{t=0} Ad_{gamma(t)} circ T_{eta(t)} \
& = & left . frac{d}{dt} right |_{t=0} left (c(t), ; a(t)d(t) - b(t)c(t) + b(t) right ) \
& = & left (dot{c}(0), ; dot{a}(0)d(0) + a(0)dot{d}(0) - dot{b}(0)c(0) - b(0)dot{c}(0) + dot{b}(0)right ) \
& = & left (Y_1, Y_2right ) ;; =;; Y
end{eqnarray*}
where I make the short hand $gamma(t) = (a(t), b(t)), ; eta(t) = (c(t), d(t))$ and $X = (X_1, X_2), ; Y = (Y_1,Y_2)$.
On first inspection it looks as if this computation satisfies the bracket, but I'm not satisfied with it since this is done independent of a basis for $mathfrak{g}$. Can anyone point out what I'm missing or what's wrong with this computation/approach?
lie-groups lie-algebras
asked Jan 18 at 23:34
MnifldzMnifldz
6,86011634
$endgroup$
$begingroup$
This is a beautiful fake proof! Hint: $Ad_g$ should act on $mathfrak{g},$ not $G$ - you need to do one derivative to define $Ad_g,$ then another to define $ad_X.$
$endgroup$
– Dap
Jan 26 at 11:09
$begingroup$
@Dap I'm not sure why I need a derivative to define $Ad_g$. Isn't this just $Ad_g:G to G$ given by $Ad_g(h) = ghg^{-1}$? I'm not sure in what sense $Ad$ acts on $mathfrak{g}$, isn't $ad_X equiv d(Ad)_e$ given a curve $gamma$ such that $gamma(0) = e$ and $dot{gamma}(0) = X$?
$endgroup$
– Mnifldz
Jan 28 at 8:15
$begingroup$
I suggest checking the definition in the book
$endgroup$
– Dap
Jan 28 at 15:30
add a comment |
$begingroup$
I'm reading Helgason's DG, Lie Groups, and Symmetric Spaces and at one point he briefly mentions the Lobatchevski half-plane on page 136:
The group $G$ of the mappings $T_{a,b}: x to ax+b$ with $x in mathbb{R}$ and $a>0$ has a Lie algebra $mathfrak{g} = mathbb{R}e_1 + mathbb{R}e_2$ where $[e_1,e_2] = e_2$.
I'm trying to verify the claim that the bracket is what it is by taking $[cdot, cdot] equiv dleft (Adright )_e:mathfrak{gtimes gto g}$ but I'm having trouble verifying this. We can see that
$$
Ad_{a,b} circ T_{c,d} ;; =;; T_{c,; ad-bc+b}
$$
If $gamma, eta$ are curves such that $gamma(0) = eta(0) = e$ and $dot{gamma}(0) =X, ; dot{eta}(0) = Y$ then we obtain
begin{eqnarray*}
[X,Y] & = & ad_X(Y) \
& = & left. frac{d}{dt} right |_{t=0} Ad_{gamma(t)} circ T_{eta(t)} \
& = & left . frac{d}{dt} right |_{t=0} left (c(t), ; a(t)d(t) - b(t)c(t) + b(t) right ) \
& = & left (dot{c}(0), ; dot{a}(0)d(0) + a(0)dot{d}(0) - dot{b}(0)c(0) - b(0)dot{c}(0) + dot{b}(0)right ) \
& = & left (Y_1, Y_2right ) ;; =;; Y
end{eqnarray*}
where I make the short hand $gamma(t) = (a(t), b(t)), ; eta(t) = (c(t), d(t))$ and $X = (X_1, X_2), ; Y = (Y_1,Y_2)$.
On first inspection it looks as if this computation satisfies the bracket, but I'm not satisfied with it since this is done independent of a basis for $mathfrak{g}$. Can anyone point out what I'm missing or what's wrong with this computation/approach?
lie-groups lie-algebras
asked Jan 18 at 23:34
MnifldzMnifldz
6,86011634
$endgroup$
I'm reading Helgason's DG, Lie Groups, and Symmetric Spaces and at one point he briefly mentions the Lobatchevski half-plane on page 136:
The group $G$ of the mappings $T_{a,b}: x to ax+b$ with $x in mathbb{R}$ and $a>0$ has a Lie algebra $mathfrak{g} = mathbb{R}e_1 + mathbb{R}e_2$ where $[e_1,e_2] = e_2$.
I'm trying to verify the claim that the bracket is what it is by taking $[cdot, cdot] equiv dleft (Adright )_e:mathfrak{gtimes gto g}$ but I'm having trouble verifying this. We can see that
$$
Ad_{a,b} circ T_{c,d} ;; =;; T_{c,; ad-bc+b}
$$
If $gamma, eta$ are curves such that $gamma(0) = eta(0) = e$ and $dot{gamma}(0) =X, ; dot{eta}(0) = Y$ then we obtain
begin{eqnarray*}
[X,Y] & = & ad_X(Y) \
& = & left. frac{d}{dt} right |_{t=0} Ad_{gamma(t)} circ T_{eta(t)} \
& = & left . frac{d}{dt} right |_{t=0} left (c(t), ; a(t)d(t) - b(t)c(t) + b(t) right ) \
& = & left (dot{c}(0), ; dot{a}(0)d(0) + a(0)dot{d}(0) - dot{b}(0)c(0) - b(0)dot{c}(0) + dot{b}(0)right ) \
& = & left (Y_1, Y_2right ) ;; =;; Y
end{eqnarray*}
where I make the short hand $gamma(t) = (a(t), b(t)), ; eta(t) = (c(t), d(t))$ and $X = (X_1, X_2), ; Y = (Y_1,Y_2)$.
On first inspection it looks as if this computation satisfies the bracket, but I'm not satisfied with it since this is done independent of a basis for $mathfrak{g}$. Can anyone point out what I'm missing or what's wrong with this computation/approach?
lie-groups lie-algebras
lie-groups lie-algebras
asked Jan 18 at 23:34
MnifldzMnifldz
6,86011634
asked Jan 18 at 23:34
MnifldzMnifldz
6,86011634
edited Jan 19 at 0:47
Mnifldz
asked Jan 18 at 23:34
MnifldzMnifldz
6,86011634
asked Jan 18 at 23:34
MnifldzMnifldz
6,86011634
asked Jan 18 at 23:34
MnifldzMnifldz
6,86011634
6,86011634
$begingroup$
This is a beautiful fake proof! Hint: $Ad_g$ should act on $mathfrak{g},$ not $G$ - you need to do one derivative to define $Ad_g,$ then another to define $ad_X.$
$endgroup$
– Dap
Jan 26 at 11:09
$begingroup$
@Dap I'm not sure why I need a derivative to define $Ad_g$. Isn't this just $Ad_g:G to G$ given by $Ad_g(h) = ghg^{-1}$? I'm not sure in what sense $Ad$ acts on $mathfrak{g}$, isn't $ad_X equiv d(Ad)_e$ given a curve $gamma$ such that $gamma(0) = e$ and $dot{gamma}(0) = X$?
$endgroup$
– Mnifldz
Jan 28 at 8:15
$begingroup$
I suggest checking the definition in the book
$endgroup$
– Dap
Jan 28 at 15:30
add a comment |
$begingroup$
This is a beautiful fake proof! Hint: $Ad_g$ should act on $mathfrak{g},$ not $G$ - you need to do one derivative to define $Ad_g,$ then another to define $ad_X.$
$endgroup$
– Dap
Jan 26 at 11:09
$begingroup$
@Dap I'm not sure why I need a derivative to define $Ad_g$. Isn't this just $Ad_g:G to G$ given by $Ad_g(h) = ghg^{-1}$? I'm not sure in what sense $Ad$ acts on $mathfrak{g}$, isn't $ad_X equiv d(Ad)_e$ given a curve $gamma$ such that $gamma(0) = e$ and $dot{gamma}(0) = X$?
$endgroup$
– Mnifldz
Jan 28 at 8:15
$begingroup$
I suggest checking the definition in the book
$endgroup$
– Dap
Jan 28 at 15:30
$begingroup$
This is a beautiful fake proof! Hint: $Ad_g$ should act on $mathfrak{g},$ not $G$ - you need to do one derivative to define $Ad_g,$ then another to define $ad_X.$
$endgroup$
– Dap
Jan 26 at 11:09
$begingroup$
This is a beautiful fake proof! Hint: $Ad_g$ should act on $mathfrak{g},$ not $G$ - you need to do one derivative to define $Ad_g,$ then another to define $ad_X.$
$endgroup$
– Dap
Jan 26 at 11:09
$begingroup$
@Dap I'm not sure why I need a derivative to define $Ad_g$. Isn't this just $Ad_g:G to G$ given by $Ad_g(h) = ghg^{-1}$? I'm not sure in what sense $Ad$ acts on $mathfrak{g}$, isn't $ad_X equiv d(Ad)_e$ given a curve $gamma$ such that $gamma(0) = e$ and $dot{gamma}(0) = X$?
$endgroup$
– Mnifldz
Jan 28 at 8:15
$begingroup$
@Dap I'm not sure why I need a derivative to define $Ad_g$. Isn't this just $Ad_g:G to G$ given by $Ad_g(h) = ghg^{-1}$? I'm not sure in what sense $Ad$ acts on $mathfrak{g}$, isn't $ad_X equiv d(Ad)_e$ given a curve $gamma$ such that $gamma(0) = e$ and $dot{gamma}(0) = X$?
$endgroup$
– Mnifldz
Jan 28 at 8:15
$begingroup$
I suggest checking the definition in the book
$endgroup$
– Dap
Jan 28 at 15:30
$begingroup$
I suggest checking the definition in the book
$endgroup$
– Dap
Jan 28 at 15:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are several ways to do this. The first one is more to my liking, but the second one closest to your computations.
1) One way to compute the Lie Bracket, assuming we know what the Lie bracket on $mathfrak{gl}(n)$ is, is to work through a linear representation. Let $Psi : G to GL(2)$ be the following map:
$$Psi(T_{a,b})=left( begin{array}{cc} a & b \ 0 & 1 end{array} right).$$
This map is an injective morphism, that allows to identify $G$ with its image.
Thus, $mathfrak{g}$ is identifed with the subalgebra
$$mathfrak{g}simeq left{ left(begin{array}{cc} * & * \ 0 & 0 end{array} right) right}, $$
and the Lie bracket in this representation is the usual Lie bracket on $mathfrak{gl}(n)$ - this comes from the fact this is true for the group $GL(n)$, which contains our group.
Put
$$e_1=left(begin{array}{cc} 1 & 0 \ 0 & 0 end{array} right), ; e_2=left(begin{array}{cc} 0 & 1 \ 0 & 0 end{array} right),$$
this is clearly a basis and then
$$[e_1,e_2]=e_1e_2-e_2e_1=e_2,$$
by direct matrix product computations.
2) As noted by Dap in comments, you should be careful with definitions; $Ad$ is a map from $G$ to $Aut(mathfrak{g})$, $Ad_g$ is the derivative of $hmapsto ghg^{-1}$ at identity, and $ad: mathfrak{g}to Der(mathfrak{g})$ can be identified with the derivative of the latter.
Thus, to correct what you wrote,
$$c_{a,b}(T_{c,d})=T_{c,ad-bc+b},$$
where $c_{a,b}$ is $gmapsto T_{a,b} circ g circ T_{1/a,-b/a}$ is the conjugation by $T_{a,b}$.
Let
$$e_1=frac{partial T_{1+t,0}}{partial t} (0), e_2=frac{partial T_{1,t}}{partial t} (0).$$
In particular,
$$c_{a,b}(T_{1+talpha,tbeta})=T_{1+talpha,t(abeta-balpha)},$$
so when $tto 0$, we have computed the derivative
$$Ad_{a,b}(alpha e_1+beta e_2)=alpha e_1+ (abeta-balpha)e_2,$$
that is, the matrix of $Ad_{a,b}$ in the basis $(e_1,e_2)$ is
$$Ad_{a,b}=left( begin{array}{cc} 1& 0 \ -b & a end{array}right)$$
so the endomorphism $ad_{e_1}$ has matrix in the basis $(e_1,e_2)$
$$frac{partial }{partial t} left( begin{array}{cc} 1 & 0 \ 0 & t end{array}right)=left( begin{array}{cc} 0 & 0 \ 0 & 1 end{array}right),$$
so $[e_1,e_2]=ad_{e_1}(e_2)=e_2$, as expected.
answered Jan 29 at 16:01
user120527user120527
1,910315
$endgroup$
$begingroup$
Thank you so much, this was the explanation I was looking for.
$endgroup$
– Mnifldz
Jan 30 at 21:35
add a comment |
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$begingroup$
There are several ways to do this. The first one is more to my liking, but the second one closest to your computations.
1) One way to compute the Lie Bracket, assuming we know what the Lie bracket on $mathfrak{gl}(n)$ is, is to work through a linear representation. Let $Psi : G to GL(2)$ be the following map:
$$Psi(T_{a,b})=left( begin{array}{cc} a & b \ 0 & 1 end{array} right).$$
This map is an injective morphism, that allows to identify $G$ with its image.
Thus, $mathfrak{g}$ is identifed with the subalgebra
$$mathfrak{g}simeq left{ left(begin{array}{cc} * & * \ 0 & 0 end{array} right) right}, $$
and the Lie bracket in this representation is the usual Lie bracket on $mathfrak{gl}(n)$ - this comes from the fact this is true for the group $GL(n)$, which contains our group.
Put
$$e_1=left(begin{array}{cc} 1 & 0 \ 0 & 0 end{array} right), ; e_2=left(begin{array}{cc} 0 & 1 \ 0 & 0 end{array} right),$$
this is clearly a basis and then
$$[e_1,e_2]=e_1e_2-e_2e_1=e_2,$$
by direct matrix product computations.
2) As noted by Dap in comments, you should be careful with definitions; $Ad$ is a map from $G$ to $Aut(mathfrak{g})$, $Ad_g$ is the derivative of $hmapsto ghg^{-1}$ at identity, and $ad: mathfrak{g}to Der(mathfrak{g})$ can be identified with the derivative of the latter.
Thus, to correct what you wrote,
$$c_{a,b}(T_{c,d})=T_{c,ad-bc+b},$$
where $c_{a,b}$ is $gmapsto T_{a,b} circ g circ T_{1/a,-b/a}$ is the conjugation by $T_{a,b}$.
Let
$$e_1=frac{partial T_{1+t,0}}{partial t} (0), e_2=frac{partial T_{1,t}}{partial t} (0).$$
In particular,
$$c_{a,b}(T_{1+talpha,tbeta})=T_{1+talpha,t(abeta-balpha)},$$
so when $tto 0$, we have computed the derivative
$$Ad_{a,b}(alpha e_1+beta e_2)=alpha e_1+ (abeta-balpha)e_2,$$
that is, the matrix of $Ad_{a,b}$ in the basis $(e_1,e_2)$ is
$$Ad_{a,b}=left( begin{array}{cc} 1& 0 \ -b & a end{array}right)$$
so the endomorphism $ad_{e_1}$ has matrix in the basis $(e_1,e_2)$
$$frac{partial }{partial t} left( begin{array}{cc} 1 & 0 \ 0 & t end{array}right)=left( begin{array}{cc} 0 & 0 \ 0 & 1 end{array}right),$$
so $[e_1,e_2]=ad_{e_1}(e_2)=e_2$, as expected.
answered Jan 29 at 16:01
user120527user120527
1,910315
$endgroup$
$begingroup$
Thank you so much, this was the explanation I was looking for.
$endgroup$
– Mnifldz
Jan 30 at 21:35
add a comment |
$begingroup$
There are several ways to do this. The first one is more to my liking, but the second one closest to your computations.
1) One way to compute the Lie Bracket, assuming we know what the Lie bracket on $mathfrak{gl}(n)$ is, is to work through a linear representation. Let $Psi : G to GL(2)$ be the following map:
$$Psi(T_{a,b})=left( begin{array}{cc} a & b \ 0 & 1 end{array} right).$$
This map is an injective morphism, that allows to identify $G$ with its image.
Thus, $mathfrak{g}$ is identifed with the subalgebra
$$mathfrak{g}simeq left{ left(begin{array}{cc} * & * \ 0 & 0 end{array} right) right}, $$
and the Lie bracket in this representation is the usual Lie bracket on $mathfrak{gl}(n)$ - this comes from the fact this is true for the group $GL(n)$, which contains our group.
Put
$$e_1=left(begin{array}{cc} 1 & 0 \ 0 & 0 end{array} right), ; e_2=left(begin{array}{cc} 0 & 1 \ 0 & 0 end{array} right),$$
this is clearly a basis and then
$$[e_1,e_2]=e_1e_2-e_2e_1=e_2,$$
by direct matrix product computations.
2) As noted by Dap in comments, you should be careful with definitions; $Ad$ is a map from $G$ to $Aut(mathfrak{g})$, $Ad_g$ is the derivative of $hmapsto ghg^{-1}$ at identity, and $ad: mathfrak{g}to Der(mathfrak{g})$ can be identified with the derivative of the latter.
Thus, to correct what you wrote,
$$c_{a,b}(T_{c,d})=T_{c,ad-bc+b},$$
where $c_{a,b}$ is $gmapsto T_{a,b} circ g circ T_{1/a,-b/a}$ is the conjugation by $T_{a,b}$.
Let
$$e_1=frac{partial T_{1+t,0}}{partial t} (0), e_2=frac{partial T_{1,t}}{partial t} (0).$$
In particular,
$$c_{a,b}(T_{1+talpha,tbeta})=T_{1+talpha,t(abeta-balpha)},$$
so when $tto 0$, we have computed the derivative
$$Ad_{a,b}(alpha e_1+beta e_2)=alpha e_1+ (abeta-balpha)e_2,$$
that is, the matrix of $Ad_{a,b}$ in the basis $(e_1,e_2)$ is
$$Ad_{a,b}=left( begin{array}{cc} 1& 0 \ -b & a end{array}right)$$
so the endomorphism $ad_{e_1}$ has matrix in the basis $(e_1,e_2)$
$$frac{partial }{partial t} left( begin{array}{cc} 1 & 0 \ 0 & t end{array}right)=left( begin{array}{cc} 0 & 0 \ 0 & 1 end{array}right),$$
so $[e_1,e_2]=ad_{e_1}(e_2)=e_2$, as expected.
answered Jan 29 at 16:01
user120527user120527
1,910315
$endgroup$
$begingroup$
Thank you so much, this was the explanation I was looking for.
$endgroup$
– Mnifldz
Jan 30 at 21:35
add a comment |
$begingroup$
There are several ways to do this. The first one is more to my liking, but the second one closest to your computations.
1) One way to compute the Lie Bracket, assuming we know what the Lie bracket on $mathfrak{gl}(n)$ is, is to work through a linear representation. Let $Psi : G to GL(2)$ be the following map:
$$Psi(T_{a,b})=left( begin{array}{cc} a & b \ 0 & 1 end{array} right).$$
This map is an injective morphism, that allows to identify $G$ with its image.
Thus, $mathfrak{g}$ is identifed with the subalgebra
$$mathfrak{g}simeq left{ left(begin{array}{cc} * & * \ 0 & 0 end{array} right) right}, $$
and the Lie bracket in this representation is the usual Lie bracket on $mathfrak{gl}(n)$ - this comes from the fact this is true for the group $GL(n)$, which contains our group.
Put
$$e_1=left(begin{array}{cc} 1 & 0 \ 0 & 0 end{array} right), ; e_2=left(begin{array}{cc} 0 & 1 \ 0 & 0 end{array} right),$$
this is clearly a basis and then
$$[e_1,e_2]=e_1e_2-e_2e_1=e_2,$$
by direct matrix product computations.
2) As noted by Dap in comments, you should be careful with definitions; $Ad$ is a map from $G$ to $Aut(mathfrak{g})$, $Ad_g$ is the derivative of $hmapsto ghg^{-1}$ at identity, and $ad: mathfrak{g}to Der(mathfrak{g})$ can be identified with the derivative of the latter.
Thus, to correct what you wrote,
$$c_{a,b}(T_{c,d})=T_{c,ad-bc+b},$$
where $c_{a,b}$ is $gmapsto T_{a,b} circ g circ T_{1/a,-b/a}$ is the conjugation by $T_{a,b}$.
Let
$$e_1=frac{partial T_{1+t,0}}{partial t} (0), e_2=frac{partial T_{1,t}}{partial t} (0).$$
In particular,
$$c_{a,b}(T_{1+talpha,tbeta})=T_{1+talpha,t(abeta-balpha)},$$
so when $tto 0$, we have computed the derivative
$$Ad_{a,b}(alpha e_1+beta e_2)=alpha e_1+ (abeta-balpha)e_2,$$
that is, the matrix of $Ad_{a,b}$ in the basis $(e_1,e_2)$ is
$$Ad_{a,b}=left( begin{array}{cc} 1& 0 \ -b & a end{array}right)$$
so the endomorphism $ad_{e_1}$ has matrix in the basis $(e_1,e_2)$
$$frac{partial }{partial t} left( begin{array}{cc} 1 & 0 \ 0 & t end{array}right)=left( begin{array}{cc} 0 & 0 \ 0 & 1 end{array}right),$$
so $[e_1,e_2]=ad_{e_1}(e_2)=e_2$, as expected.
answered Jan 29 at 16:01
user120527user120527
1,910315
$endgroup$
There are several ways to do this. The first one is more to my liking, but the second one closest to your computations.
1) One way to compute the Lie Bracket, assuming we know what the Lie bracket on $mathfrak{gl}(n)$ is, is to work through a linear representation. Let $Psi : G to GL(2)$ be the following map:
$$Psi(T_{a,b})=left( begin{array}{cc} a & b \ 0 & 1 end{array} right).$$
This map is an injective morphism, that allows to identify $G$ with its image.
Thus, $mathfrak{g}$ is identifed with the subalgebra
$$mathfrak{g}simeq left{ left(begin{array}{cc} * & * \ 0 & 0 end{array} right) right}, $$
and the Lie bracket in this representation is the usual Lie bracket on $mathfrak{gl}(n)$ - this comes from the fact this is true for the group $GL(n)$, which contains our group.
Put
$$e_1=left(begin{array}{cc} 1 & 0 \ 0 & 0 end{array} right), ; e_2=left(begin{array}{cc} 0 & 1 \ 0 & 0 end{array} right),$$
this is clearly a basis and then
$$[e_1,e_2]=e_1e_2-e_2e_1=e_2,$$
by direct matrix product computations.
2) As noted by Dap in comments, you should be careful with definitions; $Ad$ is a map from $G$ to $Aut(mathfrak{g})$, $Ad_g$ is the derivative of $hmapsto ghg^{-1}$ at identity, and $ad: mathfrak{g}to Der(mathfrak{g})$ can be identified with the derivative of the latter.
Thus, to correct what you wrote,
$$c_{a,b}(T_{c,d})=T_{c,ad-bc+b},$$
where $c_{a,b}$ is $gmapsto T_{a,b} circ g circ T_{1/a,-b/a}$ is the conjugation by $T_{a,b}$.
Let
$$e_1=frac{partial T_{1+t,0}}{partial t} (0), e_2=frac{partial T_{1,t}}{partial t} (0).$$
In particular,
$$c_{a,b}(T_{1+talpha,tbeta})=T_{1+talpha,t(abeta-balpha)},$$
so when $tto 0$, we have computed the derivative
$$Ad_{a,b}(alpha e_1+beta e_2)=alpha e_1+ (abeta-balpha)e_2,$$
that is, the matrix of $Ad_{a,b}$ in the basis $(e_1,e_2)$ is
$$Ad_{a,b}=left( begin{array}{cc} 1& 0 \ -b & a end{array}right)$$
so the endomorphism $ad_{e_1}$ has matrix in the basis $(e_1,e_2)$
$$frac{partial }{partial t} left( begin{array}{cc} 1 & 0 \ 0 & t end{array}right)=left( begin{array}{cc} 0 & 0 \ 0 & 1 end{array}right),$$
so $[e_1,e_2]=ad_{e_1}(e_2)=e_2$, as expected.
answered Jan 29 at 16:01
user120527user120527
1,910315
answered Jan 29 at 16:01
user120527user120527
1,910315
answered Jan 29 at 16:01
user120527user120527
1,910315
answered Jan 29 at 16:01
user120527user120527
1,910315
1,910315
$begingroup$
Thank you so much, this was the explanation I was looking for.
$endgroup$
– Mnifldz
Jan 30 at 21:35
add a comment |
$begingroup$
Thank you so much, this was the explanation I was looking for.
$endgroup$
– Mnifldz
Jan 30 at 21:35
$begingroup$
Thank you so much, this was the explanation I was looking for.
$endgroup$
– Mnifldz
Jan 30 at 21:35
$begingroup$
Thank you so much, this was the explanation I was looking for.
$endgroup$
– Mnifldz
Jan 30 at 21:35
add a comment |
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$begingroup$
This is a beautiful fake proof! Hint: $Ad_g$ should act on $mathfrak{g},$ not $G$ - you need to do one derivative to define $Ad_g,$ then another to define $ad_X.$
$endgroup$
– Dap
Jan 26 at 11:09
$begingroup$
@Dap I'm not sure why I need a derivative to define $Ad_g$. Isn't this just $Ad_g:G to G$ given by $Ad_g(h) = ghg^{-1}$? I'm not sure in what sense $Ad$ acts on $mathfrak{g}$, isn't $ad_X equiv d(Ad)_e$ given a curve $gamma$ such that $gamma(0) = e$ and $dot{gamma}(0) = X$?
$endgroup$
– Mnifldz
Jan 28 at 8:15
$begingroup$
I suggest checking the definition in the book
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– Dap
Jan 28 at 15:30