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Affine Transformations isomorphic to Heisenberg group
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I want to show that the lie group $G$ of affine transformations of the form $$ beginbmatrix 1 & c & -fracc^22 \ 0 & 1 & -c \ 0 & 0 & 1 endbmatrix + beginbmatrix a \ b \ c endbmatrix $$ for $a,b,cinmathbbR$ is isomorphic to the Heisenberg group given by matrices of the form $$
beginbmatrix 1 & x & z \ 0 & 1 & y \ 0 & 0 & 1 endbmatrix$$ for $x,y,zinmathbbR$.
My idea was to identify both groups with $mathbbR^3$ and then I hoped to see that the induced group multiplications on $mathbbR^3$ are the same for both groups. But this is not the case (at least the way I choosed my maps). If I identify an element of the Heisenberg group with a vector $(x,y,z)$ then the induced group multiplication is $(x,y,z)cdot (x',y',z')=(x+x',y+y',z+z'+xy')$. But when I identify an element of $G$ with a vector $(a,b,c)$, then the induced multiplication is $(a,b,c)cdot (a',b',c')=(a+a'+b'c+c'fracc^22,b+b'-cc',c+c')$ (assuming I have no mistake in my computation). So this does not work.
Is there a better way to see that those groups are isomorphic?
matrices group-theory manifolds lie-groups group-isomorphism
edited Sep 8 at 17:12
mathcounterexamples.net
25.6k21754
asked Sep 8 at 11:32
user450093
995
add a comment |Â
up vote
11
down vote
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I want to show that the lie group $G$ of affine transformations of the form $$ beginbmatrix 1 & c & -fracc^22 \ 0 & 1 & -c \ 0 & 0 & 1 endbmatrix + beginbmatrix a \ b \ c endbmatrix $$ for $a,b,cinmathbbR$ is isomorphic to the Heisenberg group given by matrices of the form $$
beginbmatrix 1 & x & z \ 0 & 1 & y \ 0 & 0 & 1 endbmatrix$$ for $x,y,zinmathbbR$.
My idea was to identify both groups with $mathbbR^3$ and then I hoped to see that the induced group multiplications on $mathbbR^3$ are the same for both groups. But this is not the case (at least the way I choosed my maps). If I identify an element of the Heisenberg group with a vector $(x,y,z)$ then the induced group multiplication is $(x,y,z)cdot (x',y',z')=(x+x',y+y',z+z'+xy')$. But when I identify an element of $G$ with a vector $(a,b,c)$, then the induced multiplication is $(a,b,c)cdot (a',b',c')=(a+a'+b'c+c'fracc^22,b+b'-cc',c+c')$ (assuming I have no mistake in my computation). So this does not work.
Is there a better way to see that those groups are isomorphic?
matrices group-theory manifolds lie-groups group-isomorphism
edited Sep 8 at 17:12
mathcounterexamples.net
25.6k21754
asked Sep 8 at 11:32
user450093
995
1
I don't know the answer. But I would start by finding the commutator of two elements in $G$, and hopefully that would give an idea of what the homomorphism should be.
– Steve D
Sep 8 at 12:45
I computed the commutators. $[(a,b,c),(a^prime,b^prime, c^prime]= (b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0)$ while $[(x,y,z),(x^prime,y^prime, z^prime] = (0,0,x y^prime - y x^prime)$. However I don't know what identification to do in order to progress...
– mathcounterexamples.net
Sep 8 at 16:06
@user450093 There is a small mistake in your computation. $(a,b,c)cdot (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$.
– mathcounterexamples.net
Sep 8 at 16:26
Ah, thanks for computing the commutator! That suggests $(a,0,0)$ is the center of the first group, and $(0,0,z)$ is the center of the second group. In fact your formula shows a potential isomorphism is $$(a,b,c)mapsto (c, b+c^2/2, a)$$
– Steve D
Sep 8 at 17:08
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
I want to show that the lie group $G$ of affine transformations of the form $$ beginbmatrix 1 & c & -fracc^22 \ 0 & 1 & -c \ 0 & 0 & 1 endbmatrix + beginbmatrix a \ b \ c endbmatrix $$ for $a,b,cinmathbbR$ is isomorphic to the Heisenberg group given by matrices of the form $$
beginbmatrix 1 & x & z \ 0 & 1 & y \ 0 & 0 & 1 endbmatrix$$ for $x,y,zinmathbbR$.
My idea was to identify both groups with $mathbbR^3$ and then I hoped to see that the induced group multiplications on $mathbbR^3$ are the same for both groups. But this is not the case (at least the way I choosed my maps). If I identify an element of the Heisenberg group with a vector $(x,y,z)$ then the induced group multiplication is $(x,y,z)cdot (x',y',z')=(x+x',y+y',z+z'+xy')$. But when I identify an element of $G$ with a vector $(a,b,c)$, then the induced multiplication is $(a,b,c)cdot (a',b',c')=(a+a'+b'c+c'fracc^22,b+b'-cc',c+c')$ (assuming I have no mistake in my computation). So this does not work.
Is there a better way to see that those groups are isomorphic?
matrices group-theory manifolds lie-groups group-isomorphism
edited Sep 8 at 17:12
mathcounterexamples.net
25.6k21754
asked Sep 8 at 11:32
user450093
995
I want to show that the lie group $G$ of affine transformations of the form $$ beginbmatrix 1 & c & -fracc^22 \ 0 & 1 & -c \ 0 & 0 & 1 endbmatrix + beginbmatrix a \ b \ c endbmatrix $$ for $a,b,cinmathbbR$ is isomorphic to the Heisenberg group given by matrices of the form $$
beginbmatrix 1 & x & z \ 0 & 1 & y \ 0 & 0 & 1 endbmatrix$$ for $x,y,zinmathbbR$.
My idea was to identify both groups with $mathbbR^3$ and then I hoped to see that the induced group multiplications on $mathbbR^3$ are the same for both groups. But this is not the case (at least the way I choosed my maps). If I identify an element of the Heisenberg group with a vector $(x,y,z)$ then the induced group multiplication is $(x,y,z)cdot (x',y',z')=(x+x',y+y',z+z'+xy')$. But when I identify an element of $G$ with a vector $(a,b,c)$, then the induced multiplication is $(a,b,c)cdot (a',b',c')=(a+a'+b'c+c'fracc^22,b+b'-cc',c+c')$ (assuming I have no mistake in my computation). So this does not work.
Is there a better way to see that those groups are isomorphic?
matrices group-theory manifolds lie-groups group-isomorphism
matrices group-theory manifolds lie-groups group-isomorphism
edited Sep 8 at 17:12
mathcounterexamples.net
25.6k21754
asked Sep 8 at 11:32
user450093
995
edited Sep 8 at 17:12
mathcounterexamples.net
25.6k21754
asked Sep 8 at 11:32
user450093
995
edited Sep 8 at 17:12
mathcounterexamples.net
25.6k21754
edited Sep 8 at 17:12
mathcounterexamples.net
25.6k21754
edited Sep 8 at 17:12
mathcounterexamples.net
25.6k21754
25.6k21754
asked Sep 8 at 11:32
user450093
995
asked Sep 8 at 11:32
user450093
995
asked Sep 8 at 11:32
user450093
995
995
1
I don't know the answer. But I would start by finding the commutator of two elements in $G$, and hopefully that would give an idea of what the homomorphism should be.
– Steve D
Sep 8 at 12:45
I computed the commutators. $[(a,b,c),(a^prime,b^prime, c^prime]= (b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0)$ while $[(x,y,z),(x^prime,y^prime, z^prime] = (0,0,x y^prime - y x^prime)$. However I don't know what identification to do in order to progress...
– mathcounterexamples.net
Sep 8 at 16:06
@user450093 There is a small mistake in your computation. $(a,b,c)cdot (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$.
– mathcounterexamples.net
Sep 8 at 16:26
Ah, thanks for computing the commutator! That suggests $(a,0,0)$ is the center of the first group, and $(0,0,z)$ is the center of the second group. In fact your formula shows a potential isomorphism is $$(a,b,c)mapsto (c, b+c^2/2, a)$$
– Steve D
Sep 8 at 17:08
add a comment |Â
1
I don't know the answer. But I would start by finding the commutator of two elements in $G$, and hopefully that would give an idea of what the homomorphism should be.
– Steve D
Sep 8 at 12:45
I computed the commutators. $[(a,b,c),(a^prime,b^prime, c^prime]= (b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0)$ while $[(x,y,z),(x^prime,y^prime, z^prime] = (0,0,x y^prime - y x^prime)$. However I don't know what identification to do in order to progress...
– mathcounterexamples.net
Sep 8 at 16:06
@user450093 There is a small mistake in your computation. $(a,b,c)cdot (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$.
– mathcounterexamples.net
Sep 8 at 16:26
Ah, thanks for computing the commutator! That suggests $(a,0,0)$ is the center of the first group, and $(0,0,z)$ is the center of the second group. In fact your formula shows a potential isomorphism is $$(a,b,c)mapsto (c, b+c^2/2, a)$$
– Steve D
Sep 8 at 17:08
1
1
I don't know the answer. But I would start by finding the commutator of two elements in $G$, and hopefully that would give an idea of what the homomorphism should be.
– Steve D
Sep 8 at 12:45
I don't know the answer. But I would start by finding the commutator of two elements in $G$, and hopefully that would give an idea of what the homomorphism should be.
– Steve D
Sep 8 at 12:45
I computed the commutators. $[(a,b,c),(a^prime,b^prime, c^prime]= (b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0)$ while $[(x,y,z),(x^prime,y^prime, z^prime] = (0,0,x y^prime - y x^prime)$. However I don't know what identification to do in order to progress...
– mathcounterexamples.net
Sep 8 at 16:06
I computed the commutators. $[(a,b,c),(a^prime,b^prime, c^prime]= (b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0)$ while $[(x,y,z),(x^prime,y^prime, z^prime] = (0,0,x y^prime - y x^prime)$. However I don't know what identification to do in order to progress...
– mathcounterexamples.net
Sep 8 at 16:06
@user450093 There is a small mistake in your computation. $(a,b,c)cdot (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$.
– mathcounterexamples.net
Sep 8 at 16:26
@user450093 There is a small mistake in your computation. $(a,b,c)cdot (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$.
– mathcounterexamples.net
Sep 8 at 16:26
Ah, thanks for computing the commutator! That suggests $(a,0,0)$ is the center of the first group, and $(0,0,z)$ is the center of the second group. In fact your formula shows a potential isomorphism is $$(a,b,c)mapsto (c, b+c^2/2, a)$$
– Steve D
Sep 8 at 17:08
Ah, thanks for computing the commutator! That suggests $(a,0,0)$ is the center of the first group, and $(0,0,z)$ is the center of the second group. In fact your formula shows a potential isomorphism is $$(a,b,c)mapsto (c, b+c^2/2, a)$$
– Steve D
Sep 8 at 17:08
add a comment |Â
3 Answers
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up vote
7
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accepted
One method, that eliminates the appearance of the sticky quadratic term, is to show instead that the Lie algebra $mathfrak g$ of $G$ and the Lie algebra $mathfrak h$ of the Heisenberg group $H$ are isomorphic. This fact, together with the fact that both $G$ and $H$ are connected and simply connected (the manifolds underlying both are diffeomorphic to $Bbb R^3$), implies that $G cong H$ as Lie groups.
Now, $mathfrak g$ consists exactly of the infinitesimal affine transformations of the form $$bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\C, quad textrmwhich we respectively denote $(A, B, C)$,$$ and $mathfrak h$ consists of the matrices
$$pmatrixcdot&X&Z\cdot&cdot&Y\cdot&cdot&cdot, quad textrmwhich we respectively denote $(X, Y, Z)$.$$
Computing Lie brackets gives
$$beginalign[(A, B, C), (A', B', C')] &= (C B' - B C', 0, 0),\
quad [(X, Y, Z), (X', Y', Z')] &= (0, 0, X Y' - Y X'). endalign$$
Comparing components immediately shows that $$phi((A, B, C)) := (C, B, A)$$ is a Lie algebra isomorphism. Alternatively it's a straightforward to check that both Lie algebras admit bases $(E_1, E_2, E_3)$ such that $[E_1, E_2] = E_3$, $[E_2, E_3] = [E_3, E_1] = 0$. Either way we're done, and we've established the existence of an isomorphism without messy computation.
Remark If you prefer to produce an explicit group map, one can exponentiate $phi$ to yield a Lie group isomorphism $tildephi : G to H$ that maps $exp W$ to $exp phi(W)$ for all $W in mathfrak g$. Generically computing matrix exponentials is complicated, but because $mathfrak g$ (and therefore $mathfrak h$) is nilpotent, one only needs finitely many terms in the usual expansion $exp W sim I + W + tfrac12 W^2 + cdots$---the rest are zero. Carrying this out gives that
$$tildephi(A + tfrac12 B C - tfrac16 C^3, B - tfrac12 C^2, C) = (C, B, A + tfrac12 BC).$$ (Notice that triples here represent elements of the groups, as in the statement of the original problem, not of their Lie algebras.) Solving to write $tildephi(a, b, c)$ in terms of a generic element $(a, b, c) in G$ recovers precisely the isomorphism in Steve D's answer: $$color#bf0000boxedtildephi(a, b, c) = (c, b + tfrac12 c^2, a + tfrac16 c^3) .$$
Note that we computed the exponential $exp : mathfrak g to G$, but I only mentioned the formula for the exponential of a matrix (Lie algebra), and $mathfrak g$ is not given as a set of matrices. This is no problem, as we can apply a standard trick: It's straightforward to check that we can identify $mathfrak g$ with a matrix algebra via the map $$(A, B, C) leftrightarrow left(bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\Cright) mapsto pmatrixcdot&cdot&cdot&cdot\A&cdot&C&cdot\B&cdot&cdot&-C\C&cdot&cdot&cdot .$$
answered Sep 8 at 18:44
Travis
56k764138
Unfortunately, I don’t have theory background on Lie group... Therefore the question I’ll ask is quite lumpish. Why is exponentiating a good way to explicit the group isomorphism ?
– mathcounterexamples.net
Sep 9 at 4:20
@mathcounterexamples.net: when you have a Lie algebra (as a subset of matrices), the exp map creates a Lie group, which is the unique simply connected Lie group with that Lie algebra. So a Lie algebra isomorphism then become a Lie group isomorphism. In other words, exp is a functor.
– Steve D
Sep 9 at 11:54
@SteveD The image of $exp$ does not always generate a simply connected Lie group. For example, the map $exp : mathfrakgl(n, Bbb C) to GL(n, Bbb C)$ is surjective, but $GL(n, Bbb C)$ has fundamental group $Bbb Z$.
– Travis
2 days ago
@mathcounterexamples.net A Lie group isomorphism $tildephi$ and its differential $phi$ are related by $tildephi circ exp = exp circ ,phi$. In this case $phi$ is easier to find than $tildephi$, so we find $phi$ and then apply the above identity to determine $tildephi$ explicitly.
– Travis
2 days ago
Yes sorry, I meant exp followed by universal covering.
– Steve D
2 days ago
add a comment |Â
up vote
4
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Here is an isomorphism:
$$ phi(x,y,z) = left(z-fracx^36, y-fracx^22, xright) $$
Note this has an inverse, namely
$$ phi(a,b,c) = left(c, b+fracc^22, a+fracc^36right)$$
So to show it is an isomorphism, we must show it preserves products. First we have
$$ (x,y,z)cdot(hatx,haty,hatz) = (x+hatx, y+haty, z+hatz+xhaty) $$
and so we have
$$ phi((x,y,z)cdot(hatx,haty,hatz)) = (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)$$
Meanwhile
$$ (a,b,c)ast (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$$
and thus
beginalign*
phi(x,y,z)astphi(hatx,haty,hatz) &=(z+hatz-fracx^36-frachatx^36+x(haty-frachatx^22)-frachatxx^22,y+haty-fracx^22-frachatx^22-xhatx, x+hatx)\
&= (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)\
&= phi((x,y,z)cdot(hatx,haty,hatz))
endalign*
Where did this come from?
Well, if these groups were indeed isomorphic, they should have a 1-dimensional commutator subgroup that was central. Computing the commutators then showed the subgroup generated by $(0,0,1)$ should go to the subgroup generated by $(1,0,0)$. From the comments above, it was clear what should happen to the $x$ and $y$ components; where to send $z$ was just a matter of looking at the difference between the $a$-component of $phi((x,y,z)cdot(hatx,haty,hatz))$ and of $phi(x,y,z)astphi(hatx,haty,hatz)$.
Note that once you know both are 3-dimensional simply-connected Lie groups with a central commutator, general Lie theory shows they must be isomorphic. I don't, however, know of a high-level argument that also shows this explicit isomorphism works. Namely, I don't know how to show $phi$ works without doing the (messy) computations.
answered Sep 8 at 17:57
Steve D
2,7301622
In my answer I show that you can produce exactly the isomorphism you gave here by exponentiating an easier-to-guess isomorphism between the Lie algebras of the two groups. This guarantees that your $phi$ works, but I'm not sure that computing the relevant matrix exponentials to produce an explicit formula for $phi$ is faster verifying directly that it is indeed an isomorphism.
– Travis
Sep 8 at 18:48
Yeah,I don't think you can avoid some computation of you want an explicit isomorphism. Of course, the existence of one is pretty easy.
– Steve D
Sep 8 at 18:54
add a comment |Â
up vote
2
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This is not a solution... just some initial thoughts
The commutator of two elements of $G$ is
$$[(a,b,c),(a^prime,b^prime, c^prime)]
= left(b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0right)=left(left(b^prime + dfrac(c^prime)^22right)c -left(b + dfracc^22right)c^prime,0,0 right)$$
While the commutator of two elements of $H$ (the Heisenberg group) is
$$[(x,y,z),(x^prime,y^prime, z^prime)] = (0,0,x y^prime - y x^prime)$$
This encourages us to map an element $(cdot,b,c) in G$ to $(c,b+dfracc^22, overlinecdot) in H$.
But I don't know how to move forward...
answered Sep 8 at 17:02
mathcounterexamples.net
25.6k21754
The map is clearly bijective. Now just show it preserves products.
– Steve D
Sep 8 at 17:15
@SteveD It seems that making that simple, the inverse of the image is not the image of the inverse unfortunately. $a$ has probably to be transformed too. I image that a way is to compute the powers of the elements in each group. Time for appetizer in France now!
– mathcounterexamples.net
Sep 8 at 17:28
One can still recover a solution from this approximation: We can see that $psi : (a, b, c) mapsto (c, b + fracc^22, a)$ is almost a homomorphism in the sense that $psi([(a, b, c), (a', b', c')]) = [psi((a, b, c)), psi(a', b', c')] + R$, where the remainder $R$ is consists of cubic and higher terms in $a, b, c$. This suggests modifying the last component $a$ by something that vanishes to second order. By expanding the homomorphism condition, one can quickly impose conditions on the possible modifications.
– Travis
Sep 8 at 20:45
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
One method, that eliminates the appearance of the sticky quadratic term, is to show instead that the Lie algebra $mathfrak g$ of $G$ and the Lie algebra $mathfrak h$ of the Heisenberg group $H$ are isomorphic. This fact, together with the fact that both $G$ and $H$ are connected and simply connected (the manifolds underlying both are diffeomorphic to $Bbb R^3$), implies that $G cong H$ as Lie groups.
Now, $mathfrak g$ consists exactly of the infinitesimal affine transformations of the form $$bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\C, quad textrmwhich we respectively denote $(A, B, C)$,$$ and $mathfrak h$ consists of the matrices
$$pmatrixcdot&X&Z\cdot&cdot&Y\cdot&cdot&cdot, quad textrmwhich we respectively denote $(X, Y, Z)$.$$
Computing Lie brackets gives
$$beginalign[(A, B, C), (A', B', C')] &= (C B' - B C', 0, 0),\
quad [(X, Y, Z), (X', Y', Z')] &= (0, 0, X Y' - Y X'). endalign$$
Comparing components immediately shows that $$phi((A, B, C)) := (C, B, A)$$ is a Lie algebra isomorphism. Alternatively it's a straightforward to check that both Lie algebras admit bases $(E_1, E_2, E_3)$ such that $[E_1, E_2] = E_3$, $[E_2, E_3] = [E_3, E_1] = 0$. Either way we're done, and we've established the existence of an isomorphism without messy computation.
Remark If you prefer to produce an explicit group map, one can exponentiate $phi$ to yield a Lie group isomorphism $tildephi : G to H$ that maps $exp W$ to $exp phi(W)$ for all $W in mathfrak g$. Generically computing matrix exponentials is complicated, but because $mathfrak g$ (and therefore $mathfrak h$) is nilpotent, one only needs finitely many terms in the usual expansion $exp W sim I + W + tfrac12 W^2 + cdots$---the rest are zero. Carrying this out gives that
$$tildephi(A + tfrac12 B C - tfrac16 C^3, B - tfrac12 C^2, C) = (C, B, A + tfrac12 BC).$$ (Notice that triples here represent elements of the groups, as in the statement of the original problem, not of their Lie algebras.) Solving to write $tildephi(a, b, c)$ in terms of a generic element $(a, b, c) in G$ recovers precisely the isomorphism in Steve D's answer: $$color#bf0000boxedtildephi(a, b, c) = (c, b + tfrac12 c^2, a + tfrac16 c^3) .$$
Note that we computed the exponential $exp : mathfrak g to G$, but I only mentioned the formula for the exponential of a matrix (Lie algebra), and $mathfrak g$ is not given as a set of matrices. This is no problem, as we can apply a standard trick: It's straightforward to check that we can identify $mathfrak g$ with a matrix algebra via the map $$(A, B, C) leftrightarrow left(bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\Cright) mapsto pmatrixcdot&cdot&cdot&cdot\A&cdot&C&cdot\B&cdot&cdot&-C\C&cdot&cdot&cdot .$$
answered Sep 8 at 18:44
Travis
56k764138
Unfortunately, I don’t have theory background on Lie group... Therefore the question I’ll ask is quite lumpish. Why is exponentiating a good way to explicit the group isomorphism ?
– mathcounterexamples.net
Sep 9 at 4:20
@mathcounterexamples.net: when you have a Lie algebra (as a subset of matrices), the exp map creates a Lie group, which is the unique simply connected Lie group with that Lie algebra. So a Lie algebra isomorphism then become a Lie group isomorphism. In other words, exp is a functor.
– Steve D
Sep 9 at 11:54
@SteveD The image of $exp$ does not always generate a simply connected Lie group. For example, the map $exp : mathfrakgl(n, Bbb C) to GL(n, Bbb C)$ is surjective, but $GL(n, Bbb C)$ has fundamental group $Bbb Z$.
– Travis
2 days ago
@mathcounterexamples.net A Lie group isomorphism $tildephi$ and its differential $phi$ are related by $tildephi circ exp = exp circ ,phi$. In this case $phi$ is easier to find than $tildephi$, so we find $phi$ and then apply the above identity to determine $tildephi$ explicitly.
– Travis
2 days ago
Yes sorry, I meant exp followed by universal covering.
– Steve D
2 days ago
add a comment |Â
up vote
7
down vote
accepted
One method, that eliminates the appearance of the sticky quadratic term, is to show instead that the Lie algebra $mathfrak g$ of $G$ and the Lie algebra $mathfrak h$ of the Heisenberg group $H$ are isomorphic. This fact, together with the fact that both $G$ and $H$ are connected and simply connected (the manifolds underlying both are diffeomorphic to $Bbb R^3$), implies that $G cong H$ as Lie groups.
Now, $mathfrak g$ consists exactly of the infinitesimal affine transformations of the form $$bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\C, quad textrmwhich we respectively denote $(A, B, C)$,$$ and $mathfrak h$ consists of the matrices
$$pmatrixcdot&X&Z\cdot&cdot&Y\cdot&cdot&cdot, quad textrmwhich we respectively denote $(X, Y, Z)$.$$
Computing Lie brackets gives
$$beginalign[(A, B, C), (A', B', C')] &= (C B' - B C', 0, 0),\
quad [(X, Y, Z), (X', Y', Z')] &= (0, 0, X Y' - Y X'). endalign$$
Comparing components immediately shows that $$phi((A, B, C)) := (C, B, A)$$ is a Lie algebra isomorphism. Alternatively it's a straightforward to check that both Lie algebras admit bases $(E_1, E_2, E_3)$ such that $[E_1, E_2] = E_3$, $[E_2, E_3] = [E_3, E_1] = 0$. Either way we're done, and we've established the existence of an isomorphism without messy computation.
Remark If you prefer to produce an explicit group map, one can exponentiate $phi$ to yield a Lie group isomorphism $tildephi : G to H$ that maps $exp W$ to $exp phi(W)$ for all $W in mathfrak g$. Generically computing matrix exponentials is complicated, but because $mathfrak g$ (and therefore $mathfrak h$) is nilpotent, one only needs finitely many terms in the usual expansion $exp W sim I + W + tfrac12 W^2 + cdots$---the rest are zero. Carrying this out gives that
$$tildephi(A + tfrac12 B C - tfrac16 C^3, B - tfrac12 C^2, C) = (C, B, A + tfrac12 BC).$$ (Notice that triples here represent elements of the groups, as in the statement of the original problem, not of their Lie algebras.) Solving to write $tildephi(a, b, c)$ in terms of a generic element $(a, b, c) in G$ recovers precisely the isomorphism in Steve D's answer: $$color#bf0000boxedtildephi(a, b, c) = (c, b + tfrac12 c^2, a + tfrac16 c^3) .$$
Note that we computed the exponential $exp : mathfrak g to G$, but I only mentioned the formula for the exponential of a matrix (Lie algebra), and $mathfrak g$ is not given as a set of matrices. This is no problem, as we can apply a standard trick: It's straightforward to check that we can identify $mathfrak g$ with a matrix algebra via the map $$(A, B, C) leftrightarrow left(bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\Cright) mapsto pmatrixcdot&cdot&cdot&cdot\A&cdot&C&cdot\B&cdot&cdot&-C\C&cdot&cdot&cdot .$$
answered Sep 8 at 18:44
Travis
56k764138
Unfortunately, I don’t have theory background on Lie group... Therefore the question I’ll ask is quite lumpish. Why is exponentiating a good way to explicit the group isomorphism ?
– mathcounterexamples.net
Sep 9 at 4:20
@mathcounterexamples.net: when you have a Lie algebra (as a subset of matrices), the exp map creates a Lie group, which is the unique simply connected Lie group with that Lie algebra. So a Lie algebra isomorphism then become a Lie group isomorphism. In other words, exp is a functor.
– Steve D
Sep 9 at 11:54
@SteveD The image of $exp$ does not always generate a simply connected Lie group. For example, the map $exp : mathfrakgl(n, Bbb C) to GL(n, Bbb C)$ is surjective, but $GL(n, Bbb C)$ has fundamental group $Bbb Z$.
– Travis
2 days ago
@mathcounterexamples.net A Lie group isomorphism $tildephi$ and its differential $phi$ are related by $tildephi circ exp = exp circ ,phi$. In this case $phi$ is easier to find than $tildephi$, so we find $phi$ and then apply the above identity to determine $tildephi$ explicitly.
– Travis
2 days ago
Yes sorry, I meant exp followed by universal covering.
– Steve D
2 days ago
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
One method, that eliminates the appearance of the sticky quadratic term, is to show instead that the Lie algebra $mathfrak g$ of $G$ and the Lie algebra $mathfrak h$ of the Heisenberg group $H$ are isomorphic. This fact, together with the fact that both $G$ and $H$ are connected and simply connected (the manifolds underlying both are diffeomorphic to $Bbb R^3$), implies that $G cong H$ as Lie groups.
Now, $mathfrak g$ consists exactly of the infinitesimal affine transformations of the form $$bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\C, quad textrmwhich we respectively denote $(A, B, C)$,$$ and $mathfrak h$ consists of the matrices
$$pmatrixcdot&X&Z\cdot&cdot&Y\cdot&cdot&cdot, quad textrmwhich we respectively denote $(X, Y, Z)$.$$
Computing Lie brackets gives
$$beginalign[(A, B, C), (A', B', C')] &= (C B' - B C', 0, 0),\
quad [(X, Y, Z), (X', Y', Z')] &= (0, 0, X Y' - Y X'). endalign$$
Comparing components immediately shows that $$phi((A, B, C)) := (C, B, A)$$ is a Lie algebra isomorphism. Alternatively it's a straightforward to check that both Lie algebras admit bases $(E_1, E_2, E_3)$ such that $[E_1, E_2] = E_3$, $[E_2, E_3] = [E_3, E_1] = 0$. Either way we're done, and we've established the existence of an isomorphism without messy computation.
Remark If you prefer to produce an explicit group map, one can exponentiate $phi$ to yield a Lie group isomorphism $tildephi : G to H$ that maps $exp W$ to $exp phi(W)$ for all $W in mathfrak g$. Generically computing matrix exponentials is complicated, but because $mathfrak g$ (and therefore $mathfrak h$) is nilpotent, one only needs finitely many terms in the usual expansion $exp W sim I + W + tfrac12 W^2 + cdots$---the rest are zero. Carrying this out gives that
$$tildephi(A + tfrac12 B C - tfrac16 C^3, B - tfrac12 C^2, C) = (C, B, A + tfrac12 BC).$$ (Notice that triples here represent elements of the groups, as in the statement of the original problem, not of their Lie algebras.) Solving to write $tildephi(a, b, c)$ in terms of a generic element $(a, b, c) in G$ recovers precisely the isomorphism in Steve D's answer: $$color#bf0000boxedtildephi(a, b, c) = (c, b + tfrac12 c^2, a + tfrac16 c^3) .$$
Note that we computed the exponential $exp : mathfrak g to G$, but I only mentioned the formula for the exponential of a matrix (Lie algebra), and $mathfrak g$ is not given as a set of matrices. This is no problem, as we can apply a standard trick: It's straightforward to check that we can identify $mathfrak g$ with a matrix algebra via the map $$(A, B, C) leftrightarrow left(bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\Cright) mapsto pmatrixcdot&cdot&cdot&cdot\A&cdot&C&cdot\B&cdot&cdot&-C\C&cdot&cdot&cdot .$$
answered Sep 8 at 18:44
Travis
56k764138
One method, that eliminates the appearance of the sticky quadratic term, is to show instead that the Lie algebra $mathfrak g$ of $G$ and the Lie algebra $mathfrak h$ of the Heisenberg group $H$ are isomorphic. This fact, together with the fact that both $G$ and $H$ are connected and simply connected (the manifolds underlying both are diffeomorphic to $Bbb R^3$), implies that $G cong H$ as Lie groups.
Now, $mathfrak g$ consists exactly of the infinitesimal affine transformations of the form $$bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\C, quad textrmwhich we respectively denote $(A, B, C)$,$$ and $mathfrak h$ consists of the matrices
$$pmatrixcdot&X&Z\cdot&cdot&Y\cdot&cdot&cdot, quad textrmwhich we respectively denote $(X, Y, Z)$.$$
Computing Lie brackets gives
$$beginalign[(A, B, C), (A', B', C')] &= (C B' - B C', 0, 0),\
quad [(X, Y, Z), (X', Y', Z')] &= (0, 0, X Y' - Y X'). endalign$$
Comparing components immediately shows that $$phi((A, B, C)) := (C, B, A)$$ is a Lie algebra isomorphism. Alternatively it's a straightforward to check that both Lie algebras admit bases $(E_1, E_2, E_3)$ such that $[E_1, E_2] = E_3$, $[E_2, E_3] = [E_3, E_1] = 0$. Either way we're done, and we've established the existence of an isomorphism without messy computation.
Remark If you prefer to produce an explicit group map, one can exponentiate $phi$ to yield a Lie group isomorphism $tildephi : G to H$ that maps $exp W$ to $exp phi(W)$ for all $W in mathfrak g$. Generically computing matrix exponentials is complicated, but because $mathfrak g$ (and therefore $mathfrak h$) is nilpotent, one only needs finitely many terms in the usual expansion $exp W sim I + W + tfrac12 W^2 + cdots$---the rest are zero. Carrying this out gives that
$$tildephi(A + tfrac12 B C - tfrac16 C^3, B - tfrac12 C^2, C) = (C, B, A + tfrac12 BC).$$ (Notice that triples here represent elements of the groups, as in the statement of the original problem, not of their Lie algebras.) Solving to write $tildephi(a, b, c)$ in terms of a generic element $(a, b, c) in G$ recovers precisely the isomorphism in Steve D's answer: $$color#bf0000boxedtildephi(a, b, c) = (c, b + tfrac12 c^2, a + tfrac16 c^3) .$$
Note that we computed the exponential $exp : mathfrak g to G$, but I only mentioned the formula for the exponential of a matrix (Lie algebra), and $mathfrak g$ is not given as a set of matrices. This is no problem, as we can apply a standard trick: It's straightforward to check that we can identify $mathfrak g$ with a matrix algebra via the map $$(A, B, C) leftrightarrow left(bf u mapsto pmatrixcdot&C&cdot\cdot&cdot&-C\ cdot&cdot&cdot bf u + pmatrixA\B\Cright) mapsto pmatrixcdot&cdot&cdot&cdot\A&cdot&C&cdot\B&cdot&cdot&-C\C&cdot&cdot&cdot .$$
answered Sep 8 at 18:44
Travis
56k764138
edited 16 hours ago
answered Sep 8 at 18:44
Travis
56k764138
answered Sep 8 at 18:44
Travis
56k764138
answered Sep 8 at 18:44
Travis
56k764138
56k764138
Unfortunately, I don’t have theory background on Lie group... Therefore the question I’ll ask is quite lumpish. Why is exponentiating a good way to explicit the group isomorphism ?
– mathcounterexamples.net
Sep 9 at 4:20
@mathcounterexamples.net: when you have a Lie algebra (as a subset of matrices), the exp map creates a Lie group, which is the unique simply connected Lie group with that Lie algebra. So a Lie algebra isomorphism then become a Lie group isomorphism. In other words, exp is a functor.
– Steve D
Sep 9 at 11:54
@SteveD The image of $exp$ does not always generate a simply connected Lie group. For example, the map $exp : mathfrakgl(n, Bbb C) to GL(n, Bbb C)$ is surjective, but $GL(n, Bbb C)$ has fundamental group $Bbb Z$.
– Travis
2 days ago
@mathcounterexamples.net A Lie group isomorphism $tildephi$ and its differential $phi$ are related by $tildephi circ exp = exp circ ,phi$. In this case $phi$ is easier to find than $tildephi$, so we find $phi$ and then apply the above identity to determine $tildephi$ explicitly.
– Travis
2 days ago
Yes sorry, I meant exp followed by universal covering.
– Steve D
2 days ago
add a comment |Â
Unfortunately, I don’t have theory background on Lie group... Therefore the question I’ll ask is quite lumpish. Why is exponentiating a good way to explicit the group isomorphism ?
– mathcounterexamples.net
Sep 9 at 4:20
@mathcounterexamples.net: when you have a Lie algebra (as a subset of matrices), the exp map creates a Lie group, which is the unique simply connected Lie group with that Lie algebra. So a Lie algebra isomorphism then become a Lie group isomorphism. In other words, exp is a functor.
– Steve D
Sep 9 at 11:54
@SteveD The image of $exp$ does not always generate a simply connected Lie group. For example, the map $exp : mathfrakgl(n, Bbb C) to GL(n, Bbb C)$ is surjective, but $GL(n, Bbb C)$ has fundamental group $Bbb Z$.
– Travis
2 days ago
@mathcounterexamples.net A Lie group isomorphism $tildephi$ and its differential $phi$ are related by $tildephi circ exp = exp circ ,phi$. In this case $phi$ is easier to find than $tildephi$, so we find $phi$ and then apply the above identity to determine $tildephi$ explicitly.
– Travis
2 days ago
Yes sorry, I meant exp followed by universal covering.
– Steve D
2 days ago
Unfortunately, I don’t have theory background on Lie group... Therefore the question I’ll ask is quite lumpish. Why is exponentiating a good way to explicit the group isomorphism ?
– mathcounterexamples.net
Sep 9 at 4:20
Unfortunately, I don’t have theory background on Lie group... Therefore the question I’ll ask is quite lumpish. Why is exponentiating a good way to explicit the group isomorphism ?
– mathcounterexamples.net
Sep 9 at 4:20
@mathcounterexamples.net: when you have a Lie algebra (as a subset of matrices), the exp map creates a Lie group, which is the unique simply connected Lie group with that Lie algebra. So a Lie algebra isomorphism then become a Lie group isomorphism. In other words, exp is a functor.
– Steve D
Sep 9 at 11:54
@mathcounterexamples.net: when you have a Lie algebra (as a subset of matrices), the exp map creates a Lie group, which is the unique simply connected Lie group with that Lie algebra. So a Lie algebra isomorphism then become a Lie group isomorphism. In other words, exp is a functor.
– Steve D
Sep 9 at 11:54
@SteveD The image of $exp$ does not always generate a simply connected Lie group. For example, the map $exp : mathfrakgl(n, Bbb C) to GL(n, Bbb C)$ is surjective, but $GL(n, Bbb C)$ has fundamental group $Bbb Z$.
– Travis
2 days ago
@SteveD The image of $exp$ does not always generate a simply connected Lie group. For example, the map $exp : mathfrakgl(n, Bbb C) to GL(n, Bbb C)$ is surjective, but $GL(n, Bbb C)$ has fundamental group $Bbb Z$.
– Travis
2 days ago
@mathcounterexamples.net A Lie group isomorphism $tildephi$ and its differential $phi$ are related by $tildephi circ exp = exp circ ,phi$. In this case $phi$ is easier to find than $tildephi$, so we find $phi$ and then apply the above identity to determine $tildephi$ explicitly.
– Travis
2 days ago
@mathcounterexamples.net A Lie group isomorphism $tildephi$ and its differential $phi$ are related by $tildephi circ exp = exp circ ,phi$. In this case $phi$ is easier to find than $tildephi$, so we find $phi$ and then apply the above identity to determine $tildephi$ explicitly.
– Travis
2 days ago
Yes sorry, I meant exp followed by universal covering.
– Steve D
2 days ago
Yes sorry, I meant exp followed by universal covering.
– Steve D
2 days ago
add a comment |Â
up vote
4
down vote
Here is an isomorphism:
$$ phi(x,y,z) = left(z-fracx^36, y-fracx^22, xright) $$
Note this has an inverse, namely
$$ phi(a,b,c) = left(c, b+fracc^22, a+fracc^36right)$$
So to show it is an isomorphism, we must show it preserves products. First we have
$$ (x,y,z)cdot(hatx,haty,hatz) = (x+hatx, y+haty, z+hatz+xhaty) $$
and so we have
$$ phi((x,y,z)cdot(hatx,haty,hatz)) = (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)$$
Meanwhile
$$ (a,b,c)ast (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$$
and thus
beginalign*
phi(x,y,z)astphi(hatx,haty,hatz) &=(z+hatz-fracx^36-frachatx^36+x(haty-frachatx^22)-frachatxx^22,y+haty-fracx^22-frachatx^22-xhatx, x+hatx)\
&= (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)\
&= phi((x,y,z)cdot(hatx,haty,hatz))
endalign*
Where did this come from?
Well, if these groups were indeed isomorphic, they should have a 1-dimensional commutator subgroup that was central. Computing the commutators then showed the subgroup generated by $(0,0,1)$ should go to the subgroup generated by $(1,0,0)$. From the comments above, it was clear what should happen to the $x$ and $y$ components; where to send $z$ was just a matter of looking at the difference between the $a$-component of $phi((x,y,z)cdot(hatx,haty,hatz))$ and of $phi(x,y,z)astphi(hatx,haty,hatz)$.
Note that once you know both are 3-dimensional simply-connected Lie groups with a central commutator, general Lie theory shows they must be isomorphic. I don't, however, know of a high-level argument that also shows this explicit isomorphism works. Namely, I don't know how to show $phi$ works without doing the (messy) computations.
answered Sep 8 at 17:57
Steve D
2,7301622
In my answer I show that you can produce exactly the isomorphism you gave here by exponentiating an easier-to-guess isomorphism between the Lie algebras of the two groups. This guarantees that your $phi$ works, but I'm not sure that computing the relevant matrix exponentials to produce an explicit formula for $phi$ is faster verifying directly that it is indeed an isomorphism.
– Travis
Sep 8 at 18:48
Yeah,I don't think you can avoid some computation of you want an explicit isomorphism. Of course, the existence of one is pretty easy.
– Steve D
Sep 8 at 18:54
add a comment |Â
up vote
4
down vote
Here is an isomorphism:
$$ phi(x,y,z) = left(z-fracx^36, y-fracx^22, xright) $$
Note this has an inverse, namely
$$ phi(a,b,c) = left(c, b+fracc^22, a+fracc^36right)$$
So to show it is an isomorphism, we must show it preserves products. First we have
$$ (x,y,z)cdot(hatx,haty,hatz) = (x+hatx, y+haty, z+hatz+xhaty) $$
and so we have
$$ phi((x,y,z)cdot(hatx,haty,hatz)) = (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)$$
Meanwhile
$$ (a,b,c)ast (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$$
and thus
beginalign*
phi(x,y,z)astphi(hatx,haty,hatz) &=(z+hatz-fracx^36-frachatx^36+x(haty-frachatx^22)-frachatxx^22,y+haty-fracx^22-frachatx^22-xhatx, x+hatx)\
&= (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)\
&= phi((x,y,z)cdot(hatx,haty,hatz))
endalign*
Where did this come from?
Well, if these groups were indeed isomorphic, they should have a 1-dimensional commutator subgroup that was central. Computing the commutators then showed the subgroup generated by $(0,0,1)$ should go to the subgroup generated by $(1,0,0)$. From the comments above, it was clear what should happen to the $x$ and $y$ components; where to send $z$ was just a matter of looking at the difference between the $a$-component of $phi((x,y,z)cdot(hatx,haty,hatz))$ and of $phi(x,y,z)astphi(hatx,haty,hatz)$.
Note that once you know both are 3-dimensional simply-connected Lie groups with a central commutator, general Lie theory shows they must be isomorphic. I don't, however, know of a high-level argument that also shows this explicit isomorphism works. Namely, I don't know how to show $phi$ works without doing the (messy) computations.
answered Sep 8 at 17:57
Steve D
2,7301622
In my answer I show that you can produce exactly the isomorphism you gave here by exponentiating an easier-to-guess isomorphism between the Lie algebras of the two groups. This guarantees that your $phi$ works, but I'm not sure that computing the relevant matrix exponentials to produce an explicit formula for $phi$ is faster verifying directly that it is indeed an isomorphism.
– Travis
Sep 8 at 18:48
Yeah,I don't think you can avoid some computation of you want an explicit isomorphism. Of course, the existence of one is pretty easy.
– Steve D
Sep 8 at 18:54
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Here is an isomorphism:
$$ phi(x,y,z) = left(z-fracx^36, y-fracx^22, xright) $$
Note this has an inverse, namely
$$ phi(a,b,c) = left(c, b+fracc^22, a+fracc^36right)$$
So to show it is an isomorphism, we must show it preserves products. First we have
$$ (x,y,z)cdot(hatx,haty,hatz) = (x+hatx, y+haty, z+hatz+xhaty) $$
and so we have
$$ phi((x,y,z)cdot(hatx,haty,hatz)) = (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)$$
Meanwhile
$$ (a,b,c)ast (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$$
and thus
beginalign*
phi(x,y,z)astphi(hatx,haty,hatz) &=(z+hatz-fracx^36-frachatx^36+x(haty-frachatx^22)-frachatxx^22,y+haty-fracx^22-frachatx^22-xhatx, x+hatx)\
&= (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)\
&= phi((x,y,z)cdot(hatx,haty,hatz))
endalign*
Where did this come from?
Well, if these groups were indeed isomorphic, they should have a 1-dimensional commutator subgroup that was central. Computing the commutators then showed the subgroup generated by $(0,0,1)$ should go to the subgroup generated by $(1,0,0)$. From the comments above, it was clear what should happen to the $x$ and $y$ components; where to send $z$ was just a matter of looking at the difference between the $a$-component of $phi((x,y,z)cdot(hatx,haty,hatz))$ and of $phi(x,y,z)astphi(hatx,haty,hatz)$.
Note that once you know both are 3-dimensional simply-connected Lie groups with a central commutator, general Lie theory shows they must be isomorphic. I don't, however, know of a high-level argument that also shows this explicit isomorphism works. Namely, I don't know how to show $phi$ works without doing the (messy) computations.
answered Sep 8 at 17:57
Steve D
2,7301622
Here is an isomorphism:
$$ phi(x,y,z) = left(z-fracx^36, y-fracx^22, xright) $$
Note this has an inverse, namely
$$ phi(a,b,c) = left(c, b+fracc^22, a+fracc^36right)$$
So to show it is an isomorphism, we must show it preserves products. First we have
$$ (x,y,z)cdot(hatx,haty,hatz) = (x+hatx, y+haty, z+hatz+xhaty) $$
and so we have
$$ phi((x,y,z)cdot(hatx,haty,hatz)) = (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)$$
Meanwhile
$$ (a,b,c)ast (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$$
and thus
beginalign*
phi(x,y,z)astphi(hatx,haty,hatz) &=(z+hatz-fracx^36-frachatx^36+x(haty-frachatx^22)-frachatxx^22,y+haty-fracx^22-frachatx^22-xhatx, x+hatx)\
&= (z+hatz+xhaty-frac(x+hatx)^36, y+haty-frac(x+hatx)^22, x+hatx)\
&= phi((x,y,z)cdot(hatx,haty,hatz))
endalign*
Where did this come from?
Well, if these groups were indeed isomorphic, they should have a 1-dimensional commutator subgroup that was central. Computing the commutators then showed the subgroup generated by $(0,0,1)$ should go to the subgroup generated by $(1,0,0)$. From the comments above, it was clear what should happen to the $x$ and $y$ components; where to send $z$ was just a matter of looking at the difference between the $a$-component of $phi((x,y,z)cdot(hatx,haty,hatz))$ and of $phi(x,y,z)astphi(hatx,haty,hatz)$.
Note that once you know both are 3-dimensional simply-connected Lie groups with a central commutator, general Lie theory shows they must be isomorphic. I don't, however, know of a high-level argument that also shows this explicit isomorphism works. Namely, I don't know how to show $phi$ works without doing the (messy) computations.
answered Sep 8 at 17:57
Steve D
2,7301622
edited Sep 8 at 18:09
answered Sep 8 at 17:57
Steve D
2,7301622
answered Sep 8 at 17:57
Steve D
2,7301622
answered Sep 8 at 17:57
Steve D
2,7301622
2,7301622
In my answer I show that you can produce exactly the isomorphism you gave here by exponentiating an easier-to-guess isomorphism between the Lie algebras of the two groups. This guarantees that your $phi$ works, but I'm not sure that computing the relevant matrix exponentials to produce an explicit formula for $phi$ is faster verifying directly that it is indeed an isomorphism.
– Travis
Sep 8 at 18:48
Yeah,I don't think you can avoid some computation of you want an explicit isomorphism. Of course, the existence of one is pretty easy.
– Steve D
Sep 8 at 18:54
add a comment |Â
In my answer I show that you can produce exactly the isomorphism you gave here by exponentiating an easier-to-guess isomorphism between the Lie algebras of the two groups. This guarantees that your $phi$ works, but I'm not sure that computing the relevant matrix exponentials to produce an explicit formula for $phi$ is faster verifying directly that it is indeed an isomorphism.
– Travis
Sep 8 at 18:48
Yeah,I don't think you can avoid some computation of you want an explicit isomorphism. Of course, the existence of one is pretty easy.
– Steve D
Sep 8 at 18:54
In my answer I show that you can produce exactly the isomorphism you gave here by exponentiating an easier-to-guess isomorphism between the Lie algebras of the two groups. This guarantees that your $phi$ works, but I'm not sure that computing the relevant matrix exponentials to produce an explicit formula for $phi$ is faster verifying directly that it is indeed an isomorphism.
– Travis
Sep 8 at 18:48
In my answer I show that you can produce exactly the isomorphism you gave here by exponentiating an easier-to-guess isomorphism between the Lie algebras of the two groups. This guarantees that your $phi$ works, but I'm not sure that computing the relevant matrix exponentials to produce an explicit formula for $phi$ is faster verifying directly that it is indeed an isomorphism.
– Travis
Sep 8 at 18:48
Yeah,I don't think you can avoid some computation of you want an explicit isomorphism. Of course, the existence of one is pretty easy.
– Steve D
Sep 8 at 18:54
Yeah,I don't think you can avoid some computation of you want an explicit isomorphism. Of course, the existence of one is pretty easy.
– Steve D
Sep 8 at 18:54
add a comment |Â
up vote
2
down vote
This is not a solution... just some initial thoughts
The commutator of two elements of $G$ is
$$[(a,b,c),(a^prime,b^prime, c^prime)]
= left(b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0right)=left(left(b^prime + dfrac(c^prime)^22right)c -left(b + dfracc^22right)c^prime,0,0 right)$$
While the commutator of two elements of $H$ (the Heisenberg group) is
$$[(x,y,z),(x^prime,y^prime, z^prime)] = (0,0,x y^prime - y x^prime)$$
This encourages us to map an element $(cdot,b,c) in G$ to $(c,b+dfracc^22, overlinecdot) in H$.
But I don't know how to move forward...
answered Sep 8 at 17:02
mathcounterexamples.net
25.6k21754
The map is clearly bijective. Now just show it preserves products.
– Steve D
Sep 8 at 17:15
@SteveD It seems that making that simple, the inverse of the image is not the image of the inverse unfortunately. $a$ has probably to be transformed too. I image that a way is to compute the powers of the elements in each group. Time for appetizer in France now!
– mathcounterexamples.net
Sep 8 at 17:28
One can still recover a solution from this approximation: We can see that $psi : (a, b, c) mapsto (c, b + fracc^22, a)$ is almost a homomorphism in the sense that $psi([(a, b, c), (a', b', c')]) = [psi((a, b, c)), psi(a', b', c')] + R$, where the remainder $R$ is consists of cubic and higher terms in $a, b, c$. This suggests modifying the last component $a$ by something that vanishes to second order. By expanding the homomorphism condition, one can quickly impose conditions on the possible modifications.
– Travis
Sep 8 at 20:45
add a comment |Â
up vote
2
down vote
This is not a solution... just some initial thoughts
The commutator of two elements of $G$ is
$$[(a,b,c),(a^prime,b^prime, c^prime)]
= left(b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0right)=left(left(b^prime + dfrac(c^prime)^22right)c -left(b + dfracc^22right)c^prime,0,0 right)$$
While the commutator of two elements of $H$ (the Heisenberg group) is
$$[(x,y,z),(x^prime,y^prime, z^prime)] = (0,0,x y^prime - y x^prime)$$
This encourages us to map an element $(cdot,b,c) in G$ to $(c,b+dfracc^22, overlinecdot) in H$.
But I don't know how to move forward...
answered Sep 8 at 17:02
mathcounterexamples.net
25.6k21754
The map is clearly bijective. Now just show it preserves products.
– Steve D
Sep 8 at 17:15
@SteveD It seems that making that simple, the inverse of the image is not the image of the inverse unfortunately. $a$ has probably to be transformed too. I image that a way is to compute the powers of the elements in each group. Time for appetizer in France now!
– mathcounterexamples.net
Sep 8 at 17:28
One can still recover a solution from this approximation: We can see that $psi : (a, b, c) mapsto (c, b + fracc^22, a)$ is almost a homomorphism in the sense that $psi([(a, b, c), (a', b', c')]) = [psi((a, b, c)), psi(a', b', c')] + R$, where the remainder $R$ is consists of cubic and higher terms in $a, b, c$. This suggests modifying the last component $a$ by something that vanishes to second order. By expanding the homomorphism condition, one can quickly impose conditions on the possible modifications.
– Travis
Sep 8 at 20:45
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is not a solution... just some initial thoughts
The commutator of two elements of $G$ is
$$[(a,b,c),(a^prime,b^prime, c^prime)]
= left(b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0right)=left(left(b^prime + dfrac(c^prime)^22right)c -left(b + dfracc^22right)c^prime,0,0 right)$$
While the commutator of two elements of $H$ (the Heisenberg group) is
$$[(x,y,z),(x^prime,y^prime, z^prime)] = (0,0,x y^prime - y x^prime)$$
This encourages us to map an element $(cdot,b,c) in G$ to $(c,b+dfracc^22, overlinecdot) in H$.
But I don't know how to move forward...
answered Sep 8 at 17:02
mathcounterexamples.net
25.6k21754
This is not a solution... just some initial thoughts
The commutator of two elements of $G$ is
$$[(a,b,c),(a^prime,b^prime, c^prime)]
= left(b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0right)=left(left(b^prime + dfrac(c^prime)^22right)c -left(b + dfracc^22right)c^prime,0,0 right)$$
While the commutator of two elements of $H$ (the Heisenberg group) is
$$[(x,y,z),(x^prime,y^prime, z^prime)] = (0,0,x y^prime - y x^prime)$$
This encourages us to map an element $(cdot,b,c) in G$ to $(c,b+dfracc^22, overlinecdot) in H$.
But I don't know how to move forward...
answered Sep 8 at 17:02
mathcounterexamples.net
25.6k21754
edited Sep 8 at 17:09
answered Sep 8 at 17:02
mathcounterexamples.net
25.6k21754
answered Sep 8 at 17:02
mathcounterexamples.net
25.6k21754
answered Sep 8 at 17:02
mathcounterexamples.net
25.6k21754
25.6k21754
The map is clearly bijective. Now just show it preserves products.
– Steve D
Sep 8 at 17:15
@SteveD It seems that making that simple, the inverse of the image is not the image of the inverse unfortunately. $a$ has probably to be transformed too. I image that a way is to compute the powers of the elements in each group. Time for appetizer in France now!
– mathcounterexamples.net
Sep 8 at 17:28
One can still recover a solution from this approximation: We can see that $psi : (a, b, c) mapsto (c, b + fracc^22, a)$ is almost a homomorphism in the sense that $psi([(a, b, c), (a', b', c')]) = [psi((a, b, c)), psi(a', b', c')] + R$, where the remainder $R$ is consists of cubic and higher terms in $a, b, c$. This suggests modifying the last component $a$ by something that vanishes to second order. By expanding the homomorphism condition, one can quickly impose conditions on the possible modifications.
– Travis
Sep 8 at 20:45
add a comment |Â
The map is clearly bijective. Now just show it preserves products.
– Steve D
Sep 8 at 17:15
@SteveD It seems that making that simple, the inverse of the image is not the image of the inverse unfortunately. $a$ has probably to be transformed too. I image that a way is to compute the powers of the elements in each group. Time for appetizer in France now!
– mathcounterexamples.net
Sep 8 at 17:28
One can still recover a solution from this approximation: We can see that $psi : (a, b, c) mapsto (c, b + fracc^22, a)$ is almost a homomorphism in the sense that $psi([(a, b, c), (a', b', c')]) = [psi((a, b, c)), psi(a', b', c')] + R$, where the remainder $R$ is consists of cubic and higher terms in $a, b, c$. This suggests modifying the last component $a$ by something that vanishes to second order. By expanding the homomorphism condition, one can quickly impose conditions on the possible modifications.
– Travis
Sep 8 at 20:45
The map is clearly bijective. Now just show it preserves products.
– Steve D
Sep 8 at 17:15
The map is clearly bijective. Now just show it preserves products.
– Steve D
Sep 8 at 17:15
@SteveD It seems that making that simple, the inverse of the image is not the image of the inverse unfortunately. $a$ has probably to be transformed too. I image that a way is to compute the powers of the elements in each group. Time for appetizer in France now!
– mathcounterexamples.net
Sep 8 at 17:28
@SteveD It seems that making that simple, the inverse of the image is not the image of the inverse unfortunately. $a$ has probably to be transformed too. I image that a way is to compute the powers of the elements in each group. Time for appetizer in France now!
– mathcounterexamples.net
Sep 8 at 17:28
One can still recover a solution from this approximation: We can see that $psi : (a, b, c) mapsto (c, b + fracc^22, a)$ is almost a homomorphism in the sense that $psi([(a, b, c), (a', b', c')]) = [psi((a, b, c)), psi(a', b', c')] + R$, where the remainder $R$ is consists of cubic and higher terms in $a, b, c$. This suggests modifying the last component $a$ by something that vanishes to second order. By expanding the homomorphism condition, one can quickly impose conditions on the possible modifications.
– Travis
Sep 8 at 20:45
One can still recover a solution from this approximation: We can see that $psi : (a, b, c) mapsto (c, b + fracc^22, a)$ is almost a homomorphism in the sense that $psi([(a, b, c), (a', b', c')]) = [psi((a, b, c)), psi(a', b', c')] + R$, where the remainder $R$ is consists of cubic and higher terms in $a, b, c$. This suggests modifying the last component $a$ by something that vanishes to second order. By expanding the homomorphism condition, one can quickly impose conditions on the possible modifications.
– Travis
Sep 8 at 20:45
add a comment |Â
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1
I don't know the answer. But I would start by finding the commutator of two elements in $G$, and hopefully that would give an idea of what the homomorphism should be.
– Steve D
Sep 8 at 12:45
I computed the commutators. $[(a,b,c),(a^prime,b^prime, c^prime]= (b^prime c-bc^prime-c^prime dfracc^22 + cdfrac(c^prime)^22,0,0)$ while $[(x,y,z),(x^prime,y^prime, z^prime] = (0,0,x y^prime - y x^prime)$. However I don't know what identification to do in order to progress...
– mathcounterexamples.net
Sep 8 at 16:06
@user450093 There is a small mistake in your computation. $(a,b,c)cdot (a',b',c')=(a+a'+b'c-c'fracc^22,b+b'-cc',c+c')$.
– mathcounterexamples.net
Sep 8 at 16:26
Ah, thanks for computing the commutator! That suggests $(a,0,0)$ is the center of the first group, and $(0,0,z)$ is the center of the second group. In fact your formula shows a potential isomorphism is $$(a,b,c)mapsto (c, b+c^2/2, a)$$
– Steve D
Sep 8 at 17:08