Mixing
Decomposition of Lie group into factors when one is not simply connected?
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
As far as I am aware, for a group to be the direct product of groups it is sufficient that $Gcong Htimes K$ as a set and $H$ and $K$ are normal in $G$. For Lie groups, normal groups corresspond to ideals of the Lie algebra. For $G$ compact, given an ideal $mathfrakh subset mathfrakg$, we can always find a complementary ideal $mathfrakk$ such that $mathfrakgcongmathfrakkoplusmathfrakh$ by using an appropriate inner product, however, in order for the the group to be a direct product proofs require that $G$ be simply connected. As it seems we can get the product structure on $G$ from using the fact that $exp$ is surjective, I am not sure where the requirement for $G$ to be simply connected comes in... Additionally, if $G$ is not simply connected, is there a condition one can add to ensure the group is still a direct product?
lie-groups lie-algebras
asked Jul 24 at 12:28
R Mary
812213
add a comment |Â
up vote
5
down vote
favorite
As far as I am aware, for a group to be the direct product of groups it is sufficient that $Gcong Htimes K$ as a set and $H$ and $K$ are normal in $G$. For Lie groups, normal groups corresspond to ideals of the Lie algebra. For $G$ compact, given an ideal $mathfrakh subset mathfrakg$, we can always find a complementary ideal $mathfrakk$ such that $mathfrakgcongmathfrakkoplusmathfrakh$ by using an appropriate inner product, however, in order for the the group to be a direct product proofs require that $G$ be simply connected. As it seems we can get the product structure on $G$ from using the fact that $exp$ is surjective, I am not sure where the requirement for $G$ to be simply connected comes in... Additionally, if $G$ is not simply connected, is there a condition one can add to ensure the group is still a direct product?
lie-groups lie-algebras
asked Jul 24 at 12:28
R Mary
812213
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
As far as I am aware, for a group to be the direct product of groups it is sufficient that $Gcong Htimes K$ as a set and $H$ and $K$ are normal in $G$. For Lie groups, normal groups corresspond to ideals of the Lie algebra. For $G$ compact, given an ideal $mathfrakh subset mathfrakg$, we can always find a complementary ideal $mathfrakk$ such that $mathfrakgcongmathfrakkoplusmathfrakh$ by using an appropriate inner product, however, in order for the the group to be a direct product proofs require that $G$ be simply connected. As it seems we can get the product structure on $G$ from using the fact that $exp$ is surjective, I am not sure where the requirement for $G$ to be simply connected comes in... Additionally, if $G$ is not simply connected, is there a condition one can add to ensure the group is still a direct product?
lie-groups lie-algebras
asked Jul 24 at 12:28
R Mary
812213
As far as I am aware, for a group to be the direct product of groups it is sufficient that $Gcong Htimes K$ as a set and $H$ and $K$ are normal in $G$. For Lie groups, normal groups corresspond to ideals of the Lie algebra. For $G$ compact, given an ideal $mathfrakh subset mathfrakg$, we can always find a complementary ideal $mathfrakk$ such that $mathfrakgcongmathfrakkoplusmathfrakh$ by using an appropriate inner product, however, in order for the the group to be a direct product proofs require that $G$ be simply connected. As it seems we can get the product structure on $G$ from using the fact that $exp$ is surjective, I am not sure where the requirement for $G$ to be simply connected comes in... Additionally, if $G$ is not simply connected, is there a condition one can add to ensure the group is still a direct product?
lie-groups lie-algebras
asked Jul 24 at 12:28
R Mary
812213
edited Jul 24 at 15:37
asked Jul 24 at 12:28
R Mary
812213
asked Jul 24 at 12:28
R Mary
812213
asked Jul 24 at 12:28
R Mary
812213
812213
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
First, your original supposition is wrong: If we let $G = mathbbZ_4$ and $H = K = mathbbZ_2$, then $Gcong Htimes K$ as sets (since both are $4$ element sets), but $G$ is not a non-trivial direct product.
In the Lie group setting, it makes more sense to ask that $Gcong Htimes K$ as smooth manifolds. There are still counterexamples:
Let $G' = S^1times SU(2n)$ and $G = G'/pm(1,I)$. We let $H$ be the image of the $S^1$ factor in $G'$ in $G$, and $K$ be the image of the $SU(2n)$ factor.
Because $H$ and $K$ are images of normal subgroups under a surjective map, they are normal subgroups of $G$. Further, we claim that $G$ is diffeomorphic to $Htimes K$.
To make this diffeomorphism, we first let $phi:S^1rightarrow SU(2n)$ be given by $phi(z) = operatornamediag(z,z,z...,z, overlinez, overlinez,...,overlinez)$, where there are $n$ copies of $z$ and $overlinez$. Then $phi$ is a homomorphism, embedding $S^1$ into $SU(2n)$.
Now, let $f:G'rightarrow Htimes K$ be given by $f([z,A]) = (z^2, phi(z)A)$.
This is well defined: $f([-z,-A]) = (z^2, phi(z)A) = f([z,A])$.
This is injective: if $(z^2, zA) = (w^2, wB)$, then $z^2 = w^2$ which forces $z = pm w$. If $z = w$, clearly $A=B$. If $z = -w$, clearly $A = -B$, so $(w,B) = -(z,A)$, so $[w,B] = [z,A]$.
This is surjective: Given $(w,B)in Htimes K$, let $z$ be any square root of $w$. Then $f([z, phi(z^-1)B]) = (w,B)$.
This is smooth: If one precomposes with the map $G'rightarrow Grightarrow Htimes K$, one gets the map $(z,A)mapsto (z^2, zA)$, which is obviously smooth. Since $G'rightarrow G$ is a submersion, $f$ must be smooth as well.
The inverse is smooth: At a point $(w,B)in Htimes K$, we pick a square root of $w$ and call it $z$. Then, locally, the inverse of $f$ at $(w,B)$ looks like $([sqrtw, sqrtw^-1 B])$. Since the square root function, viewed as a map $mathbbCsetminus0rightarrow mathbbC$ is locally smooth (if we keep picking the same branch), this is smooth.
As far as your intuition regarding $exp$, the issue is that we could have $exp(h) = exp(k)neq e$ for some $hin mathfrakhsubseteq mathfrakg$ and $kinmathfrakksubseteq mathfrakg$. For the $G$ example above, the vector $h = pi in mathfrakhoplusmathfrakk$ and $k = pi I$ exponentiate to $(-1, I)$ and $(1,-I)$ in $Htimes K$. But in $G$, these get identified to the same (non-identity) point.
Finally, to answer your last question:
Suppose $G$ is any compact Lie group for which $Z(G) = e$, i.e., $G$ is centerless. If $mathfrakg = mathfrakhoplusmathfrakk$ with both $mathfrakh$ and $mathfrakk$ ideals, then $G$ is a direct product.
(So, for example, $SO(4)/pm I$ must be a direct product, and indeed, it is $SO(3)times SO(3)$.)
Proof: Let $pi: G'rightarrow G $ denote the universal cover of $G$. Using the group theoretic fact that $pi(Z(G'))subseteq Z(G) = e$, it follows that $Z(G')subseteq ker pi$. The reverse inclusion $ker pi subseteq Z(G')$ is always true for the universal cover (see this answer by Jack Lee and the references he gives). Thus, $Z(G') = ker pi$. Of course, writing $G' = Htimes K$ (where $H$ and $K$ are the unique simply connected groups with Lie algebras $mathfrakh$ and $mathfrakk$), it follows that $Z(G') = Z(H)times Z(K)$.
Then $G = G'/ker pi = (Htimes K)/(Z(H)times Z(K)) = (H/Z(H)) times (K/Z(K))$ is a direct product, as claimed. $square$
Edit Your claim in the comment is correct:
Theorem Assume $G$ is a connected compact Lie group with Lie algebra $mathfrakg$, which decomposes as a sum of two ideals $mathfrakhoplus mathfrakk$. Assume further that $H:=exp(mathfrakh)$ and $K:=exp(mathfrakk)$ intersect only at $ein G$. Then $G$ is Lie isomorphic to $Htimes K$.
Proof: Consider $f:Htimes Krightarrow G$ given by $f(h,k) = hk$. This is clearly smooth because the multiplication in $G$ is smooth. This map induces an isomorphism on the Lie algebra level, so it follows from general Lie theory that $f$ is a covering.
However, $f$ is injective: if $f(h,k) = e$, then $hk = e$, so $h = k^-1 in Hcap K =e$. An injective covering map is a diffeomorphism, so we are done.
answered Jul 24 at 14:59
Jason DeVito
29.5k473129
Thank you for such a thorough and informative answer!
– R Mary
Jul 24 at 15:31
Dear Jason, sorry to bother you again after you put such trouble into your answer, but just so I know I have fully understood: the problem here is that even though the lie algebra split $mathfrakgcongmathfrakkoplusmathfrakh$, we don't necessarily have $exp(mathfrakk)cap exp(mathfrakh)=e$. So if $G$ is not simply connected, checking this would be enough i.e. the statement "if the lie algebra of $G$ is the sum ideals $mathfrakh, mathfrakk$ and $exp(mathfrakk)cap exp(mathfrakh)=e$, then $G$ is the direct product of Lie groups" is true?
– R Mary
Jul 24 at 17:47
I am not certain, but I think that's right. That's certainly the condition you need on the group level to guarantee that that $G$ is isomorphic to $Htimes K$. (That is, for an abstract group $G$, if you have normal subgroups $H,Ksubseteq G$ with $Hcap K = e$, then $Gcong Htimes K$). I am not 100% confident about the topological aspects, but I would bet everything works. I'll think about it for a bit and then edit the answer if I can get all the details straight.
– Jason DeVito
Jul 24 at 17:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
First, your original supposition is wrong: If we let $G = mathbbZ_4$ and $H = K = mathbbZ_2$, then $Gcong Htimes K$ as sets (since both are $4$ element sets), but $G$ is not a non-trivial direct product.
In the Lie group setting, it makes more sense to ask that $Gcong Htimes K$ as smooth manifolds. There are still counterexamples:
Let $G' = S^1times SU(2n)$ and $G = G'/pm(1,I)$. We let $H$ be the image of the $S^1$ factor in $G'$ in $G$, and $K$ be the image of the $SU(2n)$ factor.
Because $H$ and $K$ are images of normal subgroups under a surjective map, they are normal subgroups of $G$. Further, we claim that $G$ is diffeomorphic to $Htimes K$.
To make this diffeomorphism, we first let $phi:S^1rightarrow SU(2n)$ be given by $phi(z) = operatornamediag(z,z,z...,z, overlinez, overlinez,...,overlinez)$, where there are $n$ copies of $z$ and $overlinez$. Then $phi$ is a homomorphism, embedding $S^1$ into $SU(2n)$.
Now, let $f:G'rightarrow Htimes K$ be given by $f([z,A]) = (z^2, phi(z)A)$.
This is well defined: $f([-z,-A]) = (z^2, phi(z)A) = f([z,A])$.
This is injective: if $(z^2, zA) = (w^2, wB)$, then $z^2 = w^2$ which forces $z = pm w$. If $z = w$, clearly $A=B$. If $z = -w$, clearly $A = -B$, so $(w,B) = -(z,A)$, so $[w,B] = [z,A]$.
This is surjective: Given $(w,B)in Htimes K$, let $z$ be any square root of $w$. Then $f([z, phi(z^-1)B]) = (w,B)$.
This is smooth: If one precomposes with the map $G'rightarrow Grightarrow Htimes K$, one gets the map $(z,A)mapsto (z^2, zA)$, which is obviously smooth. Since $G'rightarrow G$ is a submersion, $f$ must be smooth as well.
The inverse is smooth: At a point $(w,B)in Htimes K$, we pick a square root of $w$ and call it $z$. Then, locally, the inverse of $f$ at $(w,B)$ looks like $([sqrtw, sqrtw^-1 B])$. Since the square root function, viewed as a map $mathbbCsetminus0rightarrow mathbbC$ is locally smooth (if we keep picking the same branch), this is smooth.
As far as your intuition regarding $exp$, the issue is that we could have $exp(h) = exp(k)neq e$ for some $hin mathfrakhsubseteq mathfrakg$ and $kinmathfrakksubseteq mathfrakg$. For the $G$ example above, the vector $h = pi in mathfrakhoplusmathfrakk$ and $k = pi I$ exponentiate to $(-1, I)$ and $(1,-I)$ in $Htimes K$. But in $G$, these get identified to the same (non-identity) point.
Finally, to answer your last question:
Suppose $G$ is any compact Lie group for which $Z(G) = e$, i.e., $G$ is centerless. If $mathfrakg = mathfrakhoplusmathfrakk$ with both $mathfrakh$ and $mathfrakk$ ideals, then $G$ is a direct product.
(So, for example, $SO(4)/pm I$ must be a direct product, and indeed, it is $SO(3)times SO(3)$.)
Proof: Let $pi: G'rightarrow G $ denote the universal cover of $G$. Using the group theoretic fact that $pi(Z(G'))subseteq Z(G) = e$, it follows that $Z(G')subseteq ker pi$. The reverse inclusion $ker pi subseteq Z(G')$ is always true for the universal cover (see this answer by Jack Lee and the references he gives). Thus, $Z(G') = ker pi$. Of course, writing $G' = Htimes K$ (where $H$ and $K$ are the unique simply connected groups with Lie algebras $mathfrakh$ and $mathfrakk$), it follows that $Z(G') = Z(H)times Z(K)$.
Then $G = G'/ker pi = (Htimes K)/(Z(H)times Z(K)) = (H/Z(H)) times (K/Z(K))$ is a direct product, as claimed. $square$
Edit Your claim in the comment is correct:
Theorem Assume $G$ is a connected compact Lie group with Lie algebra $mathfrakg$, which decomposes as a sum of two ideals $mathfrakhoplus mathfrakk$. Assume further that $H:=exp(mathfrakh)$ and $K:=exp(mathfrakk)$ intersect only at $ein G$. Then $G$ is Lie isomorphic to $Htimes K$.
Proof: Consider $f:Htimes Krightarrow G$ given by $f(h,k) = hk$. This is clearly smooth because the multiplication in $G$ is smooth. This map induces an isomorphism on the Lie algebra level, so it follows from general Lie theory that $f$ is a covering.
However, $f$ is injective: if $f(h,k) = e$, then $hk = e$, so $h = k^-1 in Hcap K =e$. An injective covering map is a diffeomorphism, so we are done.
answered Jul 24 at 14:59
Jason DeVito
29.5k473129
Thank you for such a thorough and informative answer!
– R Mary
Jul 24 at 15:31
Dear Jason, sorry to bother you again after you put such trouble into your answer, but just so I know I have fully understood: the problem here is that even though the lie algebra split $mathfrakgcongmathfrakkoplusmathfrakh$, we don't necessarily have $exp(mathfrakk)cap exp(mathfrakh)=e$. So if $G$ is not simply connected, checking this would be enough i.e. the statement "if the lie algebra of $G$ is the sum ideals $mathfrakh, mathfrakk$ and $exp(mathfrakk)cap exp(mathfrakh)=e$, then $G$ is the direct product of Lie groups" is true?
– R Mary
Jul 24 at 17:47
I am not certain, but I think that's right. That's certainly the condition you need on the group level to guarantee that that $G$ is isomorphic to $Htimes K$. (That is, for an abstract group $G$, if you have normal subgroups $H,Ksubseteq G$ with $Hcap K = e$, then $Gcong Htimes K$). I am not 100% confident about the topological aspects, but I would bet everything works. I'll think about it for a bit and then edit the answer if I can get all the details straight.
– Jason DeVito
Jul 24 at 17:53
add a comment |Â
up vote
3
down vote
accepted
First, your original supposition is wrong: If we let $G = mathbbZ_4$ and $H = K = mathbbZ_2$, then $Gcong Htimes K$ as sets (since both are $4$ element sets), but $G$ is not a non-trivial direct product.
In the Lie group setting, it makes more sense to ask that $Gcong Htimes K$ as smooth manifolds. There are still counterexamples:
Let $G' = S^1times SU(2n)$ and $G = G'/pm(1,I)$. We let $H$ be the image of the $S^1$ factor in $G'$ in $G$, and $K$ be the image of the $SU(2n)$ factor.
Because $H$ and $K$ are images of normal subgroups under a surjective map, they are normal subgroups of $G$. Further, we claim that $G$ is diffeomorphic to $Htimes K$.
To make this diffeomorphism, we first let $phi:S^1rightarrow SU(2n)$ be given by $phi(z) = operatornamediag(z,z,z...,z, overlinez, overlinez,...,overlinez)$, where there are $n$ copies of $z$ and $overlinez$. Then $phi$ is a homomorphism, embedding $S^1$ into $SU(2n)$.
Now, let $f:G'rightarrow Htimes K$ be given by $f([z,A]) = (z^2, phi(z)A)$.
This is well defined: $f([-z,-A]) = (z^2, phi(z)A) = f([z,A])$.
This is injective: if $(z^2, zA) = (w^2, wB)$, then $z^2 = w^2$ which forces $z = pm w$. If $z = w$, clearly $A=B$. If $z = -w$, clearly $A = -B$, so $(w,B) = -(z,A)$, so $[w,B] = [z,A]$.
This is surjective: Given $(w,B)in Htimes K$, let $z$ be any square root of $w$. Then $f([z, phi(z^-1)B]) = (w,B)$.
This is smooth: If one precomposes with the map $G'rightarrow Grightarrow Htimes K$, one gets the map $(z,A)mapsto (z^2, zA)$, which is obviously smooth. Since $G'rightarrow G$ is a submersion, $f$ must be smooth as well.
The inverse is smooth: At a point $(w,B)in Htimes K$, we pick a square root of $w$ and call it $z$. Then, locally, the inverse of $f$ at $(w,B)$ looks like $([sqrtw, sqrtw^-1 B])$. Since the square root function, viewed as a map $mathbbCsetminus0rightarrow mathbbC$ is locally smooth (if we keep picking the same branch), this is smooth.
As far as your intuition regarding $exp$, the issue is that we could have $exp(h) = exp(k)neq e$ for some $hin mathfrakhsubseteq mathfrakg$ and $kinmathfrakksubseteq mathfrakg$. For the $G$ example above, the vector $h = pi in mathfrakhoplusmathfrakk$ and $k = pi I$ exponentiate to $(-1, I)$ and $(1,-I)$ in $Htimes K$. But in $G$, these get identified to the same (non-identity) point.
Finally, to answer your last question:
Suppose $G$ is any compact Lie group for which $Z(G) = e$, i.e., $G$ is centerless. If $mathfrakg = mathfrakhoplusmathfrakk$ with both $mathfrakh$ and $mathfrakk$ ideals, then $G$ is a direct product.
(So, for example, $SO(4)/pm I$ must be a direct product, and indeed, it is $SO(3)times SO(3)$.)
Proof: Let $pi: G'rightarrow G $ denote the universal cover of $G$. Using the group theoretic fact that $pi(Z(G'))subseteq Z(G) = e$, it follows that $Z(G')subseteq ker pi$. The reverse inclusion $ker pi subseteq Z(G')$ is always true for the universal cover (see this answer by Jack Lee and the references he gives). Thus, $Z(G') = ker pi$. Of course, writing $G' = Htimes K$ (where $H$ and $K$ are the unique simply connected groups with Lie algebras $mathfrakh$ and $mathfrakk$), it follows that $Z(G') = Z(H)times Z(K)$.
Then $G = G'/ker pi = (Htimes K)/(Z(H)times Z(K)) = (H/Z(H)) times (K/Z(K))$ is a direct product, as claimed. $square$
Edit Your claim in the comment is correct:
Theorem Assume $G$ is a connected compact Lie group with Lie algebra $mathfrakg$, which decomposes as a sum of two ideals $mathfrakhoplus mathfrakk$. Assume further that $H:=exp(mathfrakh)$ and $K:=exp(mathfrakk)$ intersect only at $ein G$. Then $G$ is Lie isomorphic to $Htimes K$.
Proof: Consider $f:Htimes Krightarrow G$ given by $f(h,k) = hk$. This is clearly smooth because the multiplication in $G$ is smooth. This map induces an isomorphism on the Lie algebra level, so it follows from general Lie theory that $f$ is a covering.
However, $f$ is injective: if $f(h,k) = e$, then $hk = e$, so $h = k^-1 in Hcap K =e$. An injective covering map is a diffeomorphism, so we are done.
answered Jul 24 at 14:59
Jason DeVito
29.5k473129
Thank you for such a thorough and informative answer!
– R Mary
Jul 24 at 15:31
Dear Jason, sorry to bother you again after you put such trouble into your answer, but just so I know I have fully understood: the problem here is that even though the lie algebra split $mathfrakgcongmathfrakkoplusmathfrakh$, we don't necessarily have $exp(mathfrakk)cap exp(mathfrakh)=e$. So if $G$ is not simply connected, checking this would be enough i.e. the statement "if the lie algebra of $G$ is the sum ideals $mathfrakh, mathfrakk$ and $exp(mathfrakk)cap exp(mathfrakh)=e$, then $G$ is the direct product of Lie groups" is true?
– R Mary
Jul 24 at 17:47
I am not certain, but I think that's right. That's certainly the condition you need on the group level to guarantee that that $G$ is isomorphic to $Htimes K$. (That is, for an abstract group $G$, if you have normal subgroups $H,Ksubseteq G$ with $Hcap K = e$, then $Gcong Htimes K$). I am not 100% confident about the topological aspects, but I would bet everything works. I'll think about it for a bit and then edit the answer if I can get all the details straight.
– Jason DeVito
Jul 24 at 17:53
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
First, your original supposition is wrong: If we let $G = mathbbZ_4$ and $H = K = mathbbZ_2$, then $Gcong Htimes K$ as sets (since both are $4$ element sets), but $G$ is not a non-trivial direct product.
In the Lie group setting, it makes more sense to ask that $Gcong Htimes K$ as smooth manifolds. There are still counterexamples:
Let $G' = S^1times SU(2n)$ and $G = G'/pm(1,I)$. We let $H$ be the image of the $S^1$ factor in $G'$ in $G$, and $K$ be the image of the $SU(2n)$ factor.
Because $H$ and $K$ are images of normal subgroups under a surjective map, they are normal subgroups of $G$. Further, we claim that $G$ is diffeomorphic to $Htimes K$.
To make this diffeomorphism, we first let $phi:S^1rightarrow SU(2n)$ be given by $phi(z) = operatornamediag(z,z,z...,z, overlinez, overlinez,...,overlinez)$, where there are $n$ copies of $z$ and $overlinez$. Then $phi$ is a homomorphism, embedding $S^1$ into $SU(2n)$.
Now, let $f:G'rightarrow Htimes K$ be given by $f([z,A]) = (z^2, phi(z)A)$.
This is well defined: $f([-z,-A]) = (z^2, phi(z)A) = f([z,A])$.
This is injective: if $(z^2, zA) = (w^2, wB)$, then $z^2 = w^2$ which forces $z = pm w$. If $z = w$, clearly $A=B$. If $z = -w$, clearly $A = -B$, so $(w,B) = -(z,A)$, so $[w,B] = [z,A]$.
This is surjective: Given $(w,B)in Htimes K$, let $z$ be any square root of $w$. Then $f([z, phi(z^-1)B]) = (w,B)$.
This is smooth: If one precomposes with the map $G'rightarrow Grightarrow Htimes K$, one gets the map $(z,A)mapsto (z^2, zA)$, which is obviously smooth. Since $G'rightarrow G$ is a submersion, $f$ must be smooth as well.
The inverse is smooth: At a point $(w,B)in Htimes K$, we pick a square root of $w$ and call it $z$. Then, locally, the inverse of $f$ at $(w,B)$ looks like $([sqrtw, sqrtw^-1 B])$. Since the square root function, viewed as a map $mathbbCsetminus0rightarrow mathbbC$ is locally smooth (if we keep picking the same branch), this is smooth.
As far as your intuition regarding $exp$, the issue is that we could have $exp(h) = exp(k)neq e$ for some $hin mathfrakhsubseteq mathfrakg$ and $kinmathfrakksubseteq mathfrakg$. For the $G$ example above, the vector $h = pi in mathfrakhoplusmathfrakk$ and $k = pi I$ exponentiate to $(-1, I)$ and $(1,-I)$ in $Htimes K$. But in $G$, these get identified to the same (non-identity) point.
Finally, to answer your last question:
Suppose $G$ is any compact Lie group for which $Z(G) = e$, i.e., $G$ is centerless. If $mathfrakg = mathfrakhoplusmathfrakk$ with both $mathfrakh$ and $mathfrakk$ ideals, then $G$ is a direct product.
(So, for example, $SO(4)/pm I$ must be a direct product, and indeed, it is $SO(3)times SO(3)$.)
Proof: Let $pi: G'rightarrow G $ denote the universal cover of $G$. Using the group theoretic fact that $pi(Z(G'))subseteq Z(G) = e$, it follows that $Z(G')subseteq ker pi$. The reverse inclusion $ker pi subseteq Z(G')$ is always true for the universal cover (see this answer by Jack Lee and the references he gives). Thus, $Z(G') = ker pi$. Of course, writing $G' = Htimes K$ (where $H$ and $K$ are the unique simply connected groups with Lie algebras $mathfrakh$ and $mathfrakk$), it follows that $Z(G') = Z(H)times Z(K)$.
Then $G = G'/ker pi = (Htimes K)/(Z(H)times Z(K)) = (H/Z(H)) times (K/Z(K))$ is a direct product, as claimed. $square$
Edit Your claim in the comment is correct:
Theorem Assume $G$ is a connected compact Lie group with Lie algebra $mathfrakg$, which decomposes as a sum of two ideals $mathfrakhoplus mathfrakk$. Assume further that $H:=exp(mathfrakh)$ and $K:=exp(mathfrakk)$ intersect only at $ein G$. Then $G$ is Lie isomorphic to $Htimes K$.
Proof: Consider $f:Htimes Krightarrow G$ given by $f(h,k) = hk$. This is clearly smooth because the multiplication in $G$ is smooth. This map induces an isomorphism on the Lie algebra level, so it follows from general Lie theory that $f$ is a covering.
However, $f$ is injective: if $f(h,k) = e$, then $hk = e$, so $h = k^-1 in Hcap K =e$. An injective covering map is a diffeomorphism, so we are done.
answered Jul 24 at 14:59
Jason DeVito
29.5k473129
First, your original supposition is wrong: If we let $G = mathbbZ_4$ and $H = K = mathbbZ_2$, then $Gcong Htimes K$ as sets (since both are $4$ element sets), but $G$ is not a non-trivial direct product.
In the Lie group setting, it makes more sense to ask that $Gcong Htimes K$ as smooth manifolds. There are still counterexamples:
Let $G' = S^1times SU(2n)$ and $G = G'/pm(1,I)$. We let $H$ be the image of the $S^1$ factor in $G'$ in $G$, and $K$ be the image of the $SU(2n)$ factor.
Because $H$ and $K$ are images of normal subgroups under a surjective map, they are normal subgroups of $G$. Further, we claim that $G$ is diffeomorphic to $Htimes K$.
To make this diffeomorphism, we first let $phi:S^1rightarrow SU(2n)$ be given by $phi(z) = operatornamediag(z,z,z...,z, overlinez, overlinez,...,overlinez)$, where there are $n$ copies of $z$ and $overlinez$. Then $phi$ is a homomorphism, embedding $S^1$ into $SU(2n)$.
Now, let $f:G'rightarrow Htimes K$ be given by $f([z,A]) = (z^2, phi(z)A)$.
This is well defined: $f([-z,-A]) = (z^2, phi(z)A) = f([z,A])$.
This is injective: if $(z^2, zA) = (w^2, wB)$, then $z^2 = w^2$ which forces $z = pm w$. If $z = w$, clearly $A=B$. If $z = -w$, clearly $A = -B$, so $(w,B) = -(z,A)$, so $[w,B] = [z,A]$.
This is surjective: Given $(w,B)in Htimes K$, let $z$ be any square root of $w$. Then $f([z, phi(z^-1)B]) = (w,B)$.
This is smooth: If one precomposes with the map $G'rightarrow Grightarrow Htimes K$, one gets the map $(z,A)mapsto (z^2, zA)$, which is obviously smooth. Since $G'rightarrow G$ is a submersion, $f$ must be smooth as well.
The inverse is smooth: At a point $(w,B)in Htimes K$, we pick a square root of $w$ and call it $z$. Then, locally, the inverse of $f$ at $(w,B)$ looks like $([sqrtw, sqrtw^-1 B])$. Since the square root function, viewed as a map $mathbbCsetminus0rightarrow mathbbC$ is locally smooth (if we keep picking the same branch), this is smooth.
As far as your intuition regarding $exp$, the issue is that we could have $exp(h) = exp(k)neq e$ for some $hin mathfrakhsubseteq mathfrakg$ and $kinmathfrakksubseteq mathfrakg$. For the $G$ example above, the vector $h = pi in mathfrakhoplusmathfrakk$ and $k = pi I$ exponentiate to $(-1, I)$ and $(1,-I)$ in $Htimes K$. But in $G$, these get identified to the same (non-identity) point.
Finally, to answer your last question:
Suppose $G$ is any compact Lie group for which $Z(G) = e$, i.e., $G$ is centerless. If $mathfrakg = mathfrakhoplusmathfrakk$ with both $mathfrakh$ and $mathfrakk$ ideals, then $G$ is a direct product.
(So, for example, $SO(4)/pm I$ must be a direct product, and indeed, it is $SO(3)times SO(3)$.)
Proof: Let $pi: G'rightarrow G $ denote the universal cover of $G$. Using the group theoretic fact that $pi(Z(G'))subseteq Z(G) = e$, it follows that $Z(G')subseteq ker pi$. The reverse inclusion $ker pi subseteq Z(G')$ is always true for the universal cover (see this answer by Jack Lee and the references he gives). Thus, $Z(G') = ker pi$. Of course, writing $G' = Htimes K$ (where $H$ and $K$ are the unique simply connected groups with Lie algebras $mathfrakh$ and $mathfrakk$), it follows that $Z(G') = Z(H)times Z(K)$.
Then $G = G'/ker pi = (Htimes K)/(Z(H)times Z(K)) = (H/Z(H)) times (K/Z(K))$ is a direct product, as claimed. $square$
Edit Your claim in the comment is correct:
Theorem Assume $G$ is a connected compact Lie group with Lie algebra $mathfrakg$, which decomposes as a sum of two ideals $mathfrakhoplus mathfrakk$. Assume further that $H:=exp(mathfrakh)$ and $K:=exp(mathfrakk)$ intersect only at $ein G$. Then $G$ is Lie isomorphic to $Htimes K$.
Proof: Consider $f:Htimes Krightarrow G$ given by $f(h,k) = hk$. This is clearly smooth because the multiplication in $G$ is smooth. This map induces an isomorphism on the Lie algebra level, so it follows from general Lie theory that $f$ is a covering.
However, $f$ is injective: if $f(h,k) = e$, then $hk = e$, so $h = k^-1 in Hcap K =e$. An injective covering map is a diffeomorphism, so we are done.
answered Jul 24 at 14:59
Jason DeVito
29.5k473129
edited Jul 24 at 20:11
answered Jul 24 at 14:59
Jason DeVito
29.5k473129
answered Jul 24 at 14:59
Jason DeVito
29.5k473129
answered Jul 24 at 14:59
Jason DeVito
29.5k473129
29.5k473129
Thank you for such a thorough and informative answer!
– R Mary
Jul 24 at 15:31
Dear Jason, sorry to bother you again after you put such trouble into your answer, but just so I know I have fully understood: the problem here is that even though the lie algebra split $mathfrakgcongmathfrakkoplusmathfrakh$, we don't necessarily have $exp(mathfrakk)cap exp(mathfrakh)=e$. So if $G$ is not simply connected, checking this would be enough i.e. the statement "if the lie algebra of $G$ is the sum ideals $mathfrakh, mathfrakk$ and $exp(mathfrakk)cap exp(mathfrakh)=e$, then $G$ is the direct product of Lie groups" is true?
– R Mary
Jul 24 at 17:47
I am not certain, but I think that's right. That's certainly the condition you need on the group level to guarantee that that $G$ is isomorphic to $Htimes K$. (That is, for an abstract group $G$, if you have normal subgroups $H,Ksubseteq G$ with $Hcap K = e$, then $Gcong Htimes K$). I am not 100% confident about the topological aspects, but I would bet everything works. I'll think about it for a bit and then edit the answer if I can get all the details straight.
– Jason DeVito
Jul 24 at 17:53
add a comment |Â
Thank you for such a thorough and informative answer!
– R Mary
Jul 24 at 15:31
Dear Jason, sorry to bother you again after you put such trouble into your answer, but just so I know I have fully understood: the problem here is that even though the lie algebra split $mathfrakgcongmathfrakkoplusmathfrakh$, we don't necessarily have $exp(mathfrakk)cap exp(mathfrakh)=e$. So if $G$ is not simply connected, checking this would be enough i.e. the statement "if the lie algebra of $G$ is the sum ideals $mathfrakh, mathfrakk$ and $exp(mathfrakk)cap exp(mathfrakh)=e$, then $G$ is the direct product of Lie groups" is true?
– R Mary
Jul 24 at 17:47
I am not certain, but I think that's right. That's certainly the condition you need on the group level to guarantee that that $G$ is isomorphic to $Htimes K$. (That is, for an abstract group $G$, if you have normal subgroups $H,Ksubseteq G$ with $Hcap K = e$, then $Gcong Htimes K$). I am not 100% confident about the topological aspects, but I would bet everything works. I'll think about it for a bit and then edit the answer if I can get all the details straight.
– Jason DeVito
Jul 24 at 17:53
Thank you for such a thorough and informative answer!
– R Mary
Jul 24 at 15:31
Thank you for such a thorough and informative answer!
– R Mary
Jul 24 at 15:31
Dear Jason, sorry to bother you again after you put such trouble into your answer, but just so I know I have fully understood: the problem here is that even though the lie algebra split $mathfrakgcongmathfrakkoplusmathfrakh$, we don't necessarily have $exp(mathfrakk)cap exp(mathfrakh)=e$. So if $G$ is not simply connected, checking this would be enough i.e. the statement "if the lie algebra of $G$ is the sum ideals $mathfrakh, mathfrakk$ and $exp(mathfrakk)cap exp(mathfrakh)=e$, then $G$ is the direct product of Lie groups" is true?
– R Mary
Jul 24 at 17:47
Dear Jason, sorry to bother you again after you put such trouble into your answer, but just so I know I have fully understood: the problem here is that even though the lie algebra split $mathfrakgcongmathfrakkoplusmathfrakh$, we don't necessarily have $exp(mathfrakk)cap exp(mathfrakh)=e$. So if $G$ is not simply connected, checking this would be enough i.e. the statement "if the lie algebra of $G$ is the sum ideals $mathfrakh, mathfrakk$ and $exp(mathfrakk)cap exp(mathfrakh)=e$, then $G$ is the direct product of Lie groups" is true?
– R Mary
Jul 24 at 17:47
I am not certain, but I think that's right. That's certainly the condition you need on the group level to guarantee that that $G$ is isomorphic to $Htimes K$. (That is, for an abstract group $G$, if you have normal subgroups $H,Ksubseteq G$ with $Hcap K = e$, then $Gcong Htimes K$). I am not 100% confident about the topological aspects, but I would bet everything works. I'll think about it for a bit and then edit the answer if I can get all the details straight.
– Jason DeVito
Jul 24 at 17:53
I am not certain, but I think that's right. That's certainly the condition you need on the group level to guarantee that that $G$ is isomorphic to $Htimes K$. (That is, for an abstract group $G$, if you have normal subgroups $H,Ksubseteq G$ with $Hcap K = e$, then $Gcong Htimes K$). I am not 100% confident about the topological aspects, but I would bet everything works. I'll think about it for a bit and then edit the answer if I can get all the details straight.
– Jason DeVito
Jul 24 at 17:53
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861278%2fdecomposition-of-lie-group-into-factors-when-one-is-not-simply-connected%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password