Mixing
Is this pseudo-Cartan decomposition of $SO(n)$ valid?
$begingroup$
I'm a graduate student in a field of science where we frequently need to optimize a matrix in $U(n)$ or $SO(n)$ (henceforth $SO(n)$ for concreteness) to get an "optimal" orthonormal basis before doing a more complicated second optimization. We perform our first optimization by varying over the Lie algebra. Almost always, the Lie algebra is too large a variational space because the basis is only determined up to orthogonal transformations of special subsets. Accordingly, we eliminate the generators of the unwanted orthogonal transformations. This is usually done without comment, but the one source I've found that attempts to justify that we don't lose any cosets this way uses a matrix decomposition. The hypotheses of the decomposition are not stated, so this is my best attempt to construct them:
Consider $SO(n)$. Given a composition of $n$, let $K$ be the special orthogonal group "applied to" that composition via direct product, e.g., $SO(100)$ has decomposition ${80, 14, 6},$ and $K = SO(80) otimes SO(14) otimes SO(6)$. $K$ is a Lie subgroup of $G$, with Lie subalgebra $mathfrak{k}$. Therefore as vector spaces, $mathfrak{g} = mathfrak{k} oplus mathfrak{p}$. We then have $$G = K exp(mathfrak{p}) = exp(mathfrak{p}) K$$
The original paper references no proof of the decomposition. This decomposition is especially relevant to a current research project of mine, so I would like to give a proper citation. This seems very similar to Cartan decomposition, but in the case where you have more than two invariant subspaces, the condition $[mathfrak{p}, mathfrak{p}] subset mathfrak{k}$ fails, and Cartan decomposition cannot be used directly. While you could do a sequence of Cartan decompositions, this would give a product of exponentials, which isn't what my field uses, and different sequences give exponentials of different subsets of the parent Lie algebra.
This brings me to the following questions:
- Is this matrix decomposition valid as stated?
- If it is valid, is there a reference for it? An ideal reference is clearly applicable to my specific case, proves the theorem, and avoids specialized Lie theoretic machinery.
- If it is not valid, is there some substitute to justify eliminating the "redundant" parameters?
In case these are relevant:
- While the problem should also be expressible in the language of Grassmann manifolds, I'm much less familiar with that language.
- My own background in Lie theory is an undergraduate course through Stillwell's Naive Lie Theory plus odds and ends I've picked up while trying to identify this matrix decomposition.
- Lie theory is not a standard tool in my field. In particular, differential geometry is unheard of.
reference-request lie-groups lie-algebras matrix-decomposition
asked Jan 4 at 17:37
JonathonJonathon
112
$endgroup$
add a comment |
$begingroup$
I'm a graduate student in a field of science where we frequently need to optimize a matrix in $U(n)$ or $SO(n)$ (henceforth $SO(n)$ for concreteness) to get an "optimal" orthonormal basis before doing a more complicated second optimization. We perform our first optimization by varying over the Lie algebra. Almost always, the Lie algebra is too large a variational space because the basis is only determined up to orthogonal transformations of special subsets. Accordingly, we eliminate the generators of the unwanted orthogonal transformations. This is usually done without comment, but the one source I've found that attempts to justify that we don't lose any cosets this way uses a matrix decomposition. The hypotheses of the decomposition are not stated, so this is my best attempt to construct them:
Consider $SO(n)$. Given a composition of $n$, let $K$ be the special orthogonal group "applied to" that composition via direct product, e.g., $SO(100)$ has decomposition ${80, 14, 6},$ and $K = SO(80) otimes SO(14) otimes SO(6)$. $K$ is a Lie subgroup of $G$, with Lie subalgebra $mathfrak{k}$. Therefore as vector spaces, $mathfrak{g} = mathfrak{k} oplus mathfrak{p}$. We then have $$G = K exp(mathfrak{p}) = exp(mathfrak{p}) K$$
The original paper references no proof of the decomposition. This decomposition is especially relevant to a current research project of mine, so I would like to give a proper citation. This seems very similar to Cartan decomposition, but in the case where you have more than two invariant subspaces, the condition $[mathfrak{p}, mathfrak{p}] subset mathfrak{k}$ fails, and Cartan decomposition cannot be used directly. While you could do a sequence of Cartan decompositions, this would give a product of exponentials, which isn't what my field uses, and different sequences give exponentials of different subsets of the parent Lie algebra.
This brings me to the following questions:
- Is this matrix decomposition valid as stated?
- If it is valid, is there a reference for it? An ideal reference is clearly applicable to my specific case, proves the theorem, and avoids specialized Lie theoretic machinery.
- If it is not valid, is there some substitute to justify eliminating the "redundant" parameters?
In case these are relevant:
- While the problem should also be expressible in the language of Grassmann manifolds, I'm much less familiar with that language.
- My own background in Lie theory is an undergraduate course through Stillwell's Naive Lie Theory plus odds and ends I've picked up while trying to identify this matrix decomposition.
- Lie theory is not a standard tool in my field. In particular, differential geometry is unheard of.
reference-request lie-groups lie-algebras matrix-decomposition
asked Jan 4 at 17:37
JonathonJonathon
112
$endgroup$
add a comment |
$begingroup$
I'm a graduate student in a field of science where we frequently need to optimize a matrix in $U(n)$ or $SO(n)$ (henceforth $SO(n)$ for concreteness) to get an "optimal" orthonormal basis before doing a more complicated second optimization. We perform our first optimization by varying over the Lie algebra. Almost always, the Lie algebra is too large a variational space because the basis is only determined up to orthogonal transformations of special subsets. Accordingly, we eliminate the generators of the unwanted orthogonal transformations. This is usually done without comment, but the one source I've found that attempts to justify that we don't lose any cosets this way uses a matrix decomposition. The hypotheses of the decomposition are not stated, so this is my best attempt to construct them:
Consider $SO(n)$. Given a composition of $n$, let $K$ be the special orthogonal group "applied to" that composition via direct product, e.g., $SO(100)$ has decomposition ${80, 14, 6},$ and $K = SO(80) otimes SO(14) otimes SO(6)$. $K$ is a Lie subgroup of $G$, with Lie subalgebra $mathfrak{k}$. Therefore as vector spaces, $mathfrak{g} = mathfrak{k} oplus mathfrak{p}$. We then have $$G = K exp(mathfrak{p}) = exp(mathfrak{p}) K$$
The original paper references no proof of the decomposition. This decomposition is especially relevant to a current research project of mine, so I would like to give a proper citation. This seems very similar to Cartan decomposition, but in the case where you have more than two invariant subspaces, the condition $[mathfrak{p}, mathfrak{p}] subset mathfrak{k}$ fails, and Cartan decomposition cannot be used directly. While you could do a sequence of Cartan decompositions, this would give a product of exponentials, which isn't what my field uses, and different sequences give exponentials of different subsets of the parent Lie algebra.
This brings me to the following questions:
- Is this matrix decomposition valid as stated?
- If it is valid, is there a reference for it? An ideal reference is clearly applicable to my specific case, proves the theorem, and avoids specialized Lie theoretic machinery.
- If it is not valid, is there some substitute to justify eliminating the "redundant" parameters?
In case these are relevant:
- While the problem should also be expressible in the language of Grassmann manifolds, I'm much less familiar with that language.
- My own background in Lie theory is an undergraduate course through Stillwell's Naive Lie Theory plus odds and ends I've picked up while trying to identify this matrix decomposition.
- Lie theory is not a standard tool in my field. In particular, differential geometry is unheard of.
reference-request lie-groups lie-algebras matrix-decomposition
asked Jan 4 at 17:37
JonathonJonathon
112
$endgroup$
I'm a graduate student in a field of science where we frequently need to optimize a matrix in $U(n)$ or $SO(n)$ (henceforth $SO(n)$ for concreteness) to get an "optimal" orthonormal basis before doing a more complicated second optimization. We perform our first optimization by varying over the Lie algebra. Almost always, the Lie algebra is too large a variational space because the basis is only determined up to orthogonal transformations of special subsets. Accordingly, we eliminate the generators of the unwanted orthogonal transformations. This is usually done without comment, but the one source I've found that attempts to justify that we don't lose any cosets this way uses a matrix decomposition. The hypotheses of the decomposition are not stated, so this is my best attempt to construct them:
Consider $SO(n)$. Given a composition of $n$, let $K$ be the special orthogonal group "applied to" that composition via direct product, e.g., $SO(100)$ has decomposition ${80, 14, 6},$ and $K = SO(80) otimes SO(14) otimes SO(6)$. $K$ is a Lie subgroup of $G$, with Lie subalgebra $mathfrak{k}$. Therefore as vector spaces, $mathfrak{g} = mathfrak{k} oplus mathfrak{p}$. We then have $$G = K exp(mathfrak{p}) = exp(mathfrak{p}) K$$
The original paper references no proof of the decomposition. This decomposition is especially relevant to a current research project of mine, so I would like to give a proper citation. This seems very similar to Cartan decomposition, but in the case where you have more than two invariant subspaces, the condition $[mathfrak{p}, mathfrak{p}] subset mathfrak{k}$ fails, and Cartan decomposition cannot be used directly. While you could do a sequence of Cartan decompositions, this would give a product of exponentials, which isn't what my field uses, and different sequences give exponentials of different subsets of the parent Lie algebra.
This brings me to the following questions:
- Is this matrix decomposition valid as stated?
- If it is valid, is there a reference for it? An ideal reference is clearly applicable to my specific case, proves the theorem, and avoids specialized Lie theoretic machinery.
- If it is not valid, is there some substitute to justify eliminating the "redundant" parameters?
In case these are relevant:
- While the problem should also be expressible in the language of Grassmann manifolds, I'm much less familiar with that language.
- My own background in Lie theory is an undergraduate course through Stillwell's Naive Lie Theory plus odds and ends I've picked up while trying to identify this matrix decomposition.
- Lie theory is not a standard tool in my field. In particular, differential geometry is unheard of.
reference-request lie-groups lie-algebras matrix-decomposition
reference-request lie-groups lie-algebras matrix-decomposition
asked Jan 4 at 17:37
JonathonJonathon
112
asked Jan 4 at 17:37
JonathonJonathon
112
asked Jan 4 at 17:37
JonathonJonathon
112
asked Jan 4 at 17:37
JonathonJonathon
112
asked Jan 4 at 17:37
JonathonJonathon
112
112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It depends on the precise meaning of your question. Here is what's true:
Proposition. Suppose that $G$ is a compact connected Lie group with the Lie agebra ${mathfrak g}$, $K< G$ is a compact (Lie) subgroup with the Lie algebra ${mathfrak k}$ and ${mathfrak g}={mathfrak p} oplus {mathfrak k}$ is the orthogonal decomposition. Then every element $g$ of $G$ can be written as a product $g=pk=kp'$, where $kin K$ and $p, p'in exp({mathfrak p})$.
Proof. The proof below requires some differential geometry, which (as I can deduce from your question) you are unfamiliar with. Maybe there is a linear algebra proof in the case of $G=SO(n)$, I just do not see one.
First of all, you need to know what a Riemannian metric is: A Riemannian metric on a differentiable manifold is a choice of inner product on tangent spaces which varies smoothly from point to point. If the manifold happens to be a compact Lie group $G$, one can make such metric biinvariant, i.e. invariant under the multiplication by the elements of $G$ on the left and on the right. In particular, it suffices to prescribe the metric on the Lie algebra. The biinvariance condition then becomes the condition of invariance under the adjoint representation of $G$. In the case $G=SO(n)$, the standard metric on the Lie algebra $o(n)$ is given by
$$
langle A, Brangle = frac{1}{2} tr(AB).
$$
(The $1/2$ factor will become clear below. A simple Lie group such as $SO(n)$ admits a unique, up to a constant multiple, biinvariant metric.) The invariance under the adjoint representation is clear since trace is conjugation-invariant.
A Riemannian metric on a manifold $M$ defines the length of curves $c(t), tin [a,b]$, on $M$ by the familiar formula
$$
L(c)= int_a^b ||c'(t)||dt,
$$
where
$$
||v||= sqrt{langle v, v rangle}
$$
is the norm on the tangent space of $M$ (at the point $c(t)$ in our case).
A curve $c(t)$ on $M$ is called geodesic if it has constant speed and is locally length-minimizing. Equivalently, geodesics are defined as curves with zero acceleration:
$$
c''(t)= nabla_{c'(t)} c'(t)=0,
$$
where $nabla$ is the covariant derivative (the Levi-Civita connection of the metric). This equation, written in local coordinates, is a 2nd order linear ODE. As such, it has unique solution of the initial value problem $c(0)=pin M, c'(0)=vin T_pM$, a tangent vector at $p$. If $M$ happens to be compact, this solution exists for all $t$ in ${mathbb R}$.
Given this, one defines Riemannian exponential maps $exp_p: T_pMto M$ by sending $vin T_pM$ to $c(1)in M$ where $c(t)$ is the unique geodesic on $M$ satisfying $c(0)=p, c'(0)=v$.
If $M$ is compact and connected, each exponential map is surjective: Any two points are connected by a geodesic (this geodesic is nonunique unless $M$ is a single point). Moreover, if $Nsubset M$ is a compact smooth submanifold and $pin M$, there exists a shortest geodesic $c$ from $p$ to $N$; such a geodesic meets $N$ at the right angle. (The latter is a generalization of the familiar geometric fact of Euclidean geometry: Every point $p$ in $E^n$ is connected to an affine subspace $Vsubset E^n$ by a shortest path, namely the unique line segment from $p$ which meets $V$ at the right angle.)
It turns out that if $M$ is a compact Lie group equipped with a biinvariant Riemannian metric, this exponential map equals to the Lie-theoretic exponential map at $p=e$, the neutral element of $G$. This fact is a computation which is the key to the entire proof. (You can find it in Helgasson's book and other places.)
In the case $G=SO(n)$, the factor $1/2$ in the definition of the metric on $G$ is chosen so that $exp_e=exp$, the Riemannian exponential map equals the matrix exponential.
Now, with all these preliminaries out of the way, I can give a proof of the proposition.
I will equip the compact Lie group $G$ with a biinvariant metric. The subgroup $K< G$ is a smooth compact submanifold. Take $gin G$ and let $c: [0,1]to G$ be a shortest geodesic connecting $g$ to $K$ ($c(0)=g, c(1)=k$), this path meets $K$ at $kin K$ and its terminal velocity vector $c'(1)=v$ is orthogonal to the tangent space $T_k(K)$. (Such a path is highly nonunique due to compactness of $G$, which is responsible for the lack of uniqueness in the decomposition $g=kp$ in the proposition.)
If $k=e$ then $g=exp_e(-v)= exp(-v)in exp({mathfrak p})$, since $vperp {mathfrak k}=T_eK$, and, hence, $-vin {mathfrak p}$ as well. In the general case, when $k$ need not be equal to $e$, we apply to $c$ the left translation $L_{k^{-1}}$ by $k^{-1}$. The left translation preserves the metric on $G$, hence, sends geodesics to geodesics, so
$$
L_{k^{-1}} c= gamma,
$$
a geodesic connecting $k^{-1}g$ to $e$. Since $kin K$, $L_{k^{-1}}K= k^{-1}K=K$, hence, $gamma$ meets $K$ orthogonally at the point $e$ (since $c$ was orthogonal to $K$ at $k$). Therefore, denoting $w:= gamma'(1)$, we again obtain
$$
k^{-1}g= exp(-w)in exp({mathfrak p}),
$$
and
$$
g= k exp(-w)= kpin K cdot exp({mathfrak p}).
$$
To obtain a decomposition
$$
g= p'k, p'in exp({mathfrak p}), kin K
$$
one uses the same argument but right multiplication by $k^{-1}$ instead of the left multiplication. qed
It is instructive to understand the proof in the case $G=SU(2)$ and $K=U(1)< G$. Then $G$ is isometric to a round 3-sphere in $R^4$, $K$ corresponds to a great circle in $S^3$, say the intersection of $S^3$ with the plane $x_3=x_4=0$. For points $gin S^3$ which are not on the "dual circle" $S^3cap {x_1=x_2=0}$, the shortest path $c$ from $g$ to $K$ is unique and is an arc of a great circle on $S^3$. However, for each point $g$ on the dual great circle, there is continuum of such arcs which can land anywhere on the circle $K$.
Moreover, one can see that a decomposition of $gin G$,
$$
g=kp, kin K, pin exp({mathfrak p}),
$$
need not be unique (even in the case of "special" compact subgroups you are considering). You can already see this when $n=3=2+1$, $G=SO(3), K=SO(2)$ or $2=1+1$ and
$$
G=SU(2), K=U(1)= U(1)otimes U(1).$$
If a decomposition were unique, the group $G$ would be homeomorphic to a product of $K$ with another topological space, while $SO(3)$ is homeomorphic to the 3-dimensional real projective space and hence does not split as a nontrivial product; same for $SU(2)$.
In this regard, the situation is quite different from the Cartan decomposition of a semisimple Lie group.
answered Jan 14 at 2:26
Moishe CohenMoishe Cohen
46.3k342106
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061880%2fis-this-pseudo-cartan-decomposition-of-son-valid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It depends on the precise meaning of your question. Here is what's true:
Proposition. Suppose that $G$ is a compact connected Lie group with the Lie agebra ${mathfrak g}$, $K< G$ is a compact (Lie) subgroup with the Lie algebra ${mathfrak k}$ and ${mathfrak g}={mathfrak p} oplus {mathfrak k}$ is the orthogonal decomposition. Then every element $g$ of $G$ can be written as a product $g=pk=kp'$, where $kin K$ and $p, p'in exp({mathfrak p})$.
Proof. The proof below requires some differential geometry, which (as I can deduce from your question) you are unfamiliar with. Maybe there is a linear algebra proof in the case of $G=SO(n)$, I just do not see one.
First of all, you need to know what a Riemannian metric is: A Riemannian metric on a differentiable manifold is a choice of inner product on tangent spaces which varies smoothly from point to point. If the manifold happens to be a compact Lie group $G$, one can make such metric biinvariant, i.e. invariant under the multiplication by the elements of $G$ on the left and on the right. In particular, it suffices to prescribe the metric on the Lie algebra. The biinvariance condition then becomes the condition of invariance under the adjoint representation of $G$. In the case $G=SO(n)$, the standard metric on the Lie algebra $o(n)$ is given by
$$
langle A, Brangle = frac{1}{2} tr(AB).
$$
(The $1/2$ factor will become clear below. A simple Lie group such as $SO(n)$ admits a unique, up to a constant multiple, biinvariant metric.) The invariance under the adjoint representation is clear since trace is conjugation-invariant.
A Riemannian metric on a manifold $M$ defines the length of curves $c(t), tin [a,b]$, on $M$ by the familiar formula
$$
L(c)= int_a^b ||c'(t)||dt,
$$
where
$$
||v||= sqrt{langle v, v rangle}
$$
is the norm on the tangent space of $M$ (at the point $c(t)$ in our case).
A curve $c(t)$ on $M$ is called geodesic if it has constant speed and is locally length-minimizing. Equivalently, geodesics are defined as curves with zero acceleration:
$$
c''(t)= nabla_{c'(t)} c'(t)=0,
$$
where $nabla$ is the covariant derivative (the Levi-Civita connection of the metric). This equation, written in local coordinates, is a 2nd order linear ODE. As such, it has unique solution of the initial value problem $c(0)=pin M, c'(0)=vin T_pM$, a tangent vector at $p$. If $M$ happens to be compact, this solution exists for all $t$ in ${mathbb R}$.
Given this, one defines Riemannian exponential maps $exp_p: T_pMto M$ by sending $vin T_pM$ to $c(1)in M$ where $c(t)$ is the unique geodesic on $M$ satisfying $c(0)=p, c'(0)=v$.
If $M$ is compact and connected, each exponential map is surjective: Any two points are connected by a geodesic (this geodesic is nonunique unless $M$ is a single point). Moreover, if $Nsubset M$ is a compact smooth submanifold and $pin M$, there exists a shortest geodesic $c$ from $p$ to $N$; such a geodesic meets $N$ at the right angle. (The latter is a generalization of the familiar geometric fact of Euclidean geometry: Every point $p$ in $E^n$ is connected to an affine subspace $Vsubset E^n$ by a shortest path, namely the unique line segment from $p$ which meets $V$ at the right angle.)
It turns out that if $M$ is a compact Lie group equipped with a biinvariant Riemannian metric, this exponential map equals to the Lie-theoretic exponential map at $p=e$, the neutral element of $G$. This fact is a computation which is the key to the entire proof. (You can find it in Helgasson's book and other places.)
In the case $G=SO(n)$, the factor $1/2$ in the definition of the metric on $G$ is chosen so that $exp_e=exp$, the Riemannian exponential map equals the matrix exponential.
Now, with all these preliminaries out of the way, I can give a proof of the proposition.
I will equip the compact Lie group $G$ with a biinvariant metric. The subgroup $K< G$ is a smooth compact submanifold. Take $gin G$ and let $c: [0,1]to G$ be a shortest geodesic connecting $g$ to $K$ ($c(0)=g, c(1)=k$), this path meets $K$ at $kin K$ and its terminal velocity vector $c'(1)=v$ is orthogonal to the tangent space $T_k(K)$. (Such a path is highly nonunique due to compactness of $G$, which is responsible for the lack of uniqueness in the decomposition $g=kp$ in the proposition.)
If $k=e$ then $g=exp_e(-v)= exp(-v)in exp({mathfrak p})$, since $vperp {mathfrak k}=T_eK$, and, hence, $-vin {mathfrak p}$ as well. In the general case, when $k$ need not be equal to $e$, we apply to $c$ the left translation $L_{k^{-1}}$ by $k^{-1}$. The left translation preserves the metric on $G$, hence, sends geodesics to geodesics, so
$$
L_{k^{-1}} c= gamma,
$$
a geodesic connecting $k^{-1}g$ to $e$. Since $kin K$, $L_{k^{-1}}K= k^{-1}K=K$, hence, $gamma$ meets $K$ orthogonally at the point $e$ (since $c$ was orthogonal to $K$ at $k$). Therefore, denoting $w:= gamma'(1)$, we again obtain
$$
k^{-1}g= exp(-w)in exp({mathfrak p}),
$$
and
$$
g= k exp(-w)= kpin K cdot exp({mathfrak p}).
$$
To obtain a decomposition
$$
g= p'k, p'in exp({mathfrak p}), kin K
$$
one uses the same argument but right multiplication by $k^{-1}$ instead of the left multiplication. qed
It is instructive to understand the proof in the case $G=SU(2)$ and $K=U(1)< G$. Then $G$ is isometric to a round 3-sphere in $R^4$, $K$ corresponds to a great circle in $S^3$, say the intersection of $S^3$ with the plane $x_3=x_4=0$. For points $gin S^3$ which are not on the "dual circle" $S^3cap {x_1=x_2=0}$, the shortest path $c$ from $g$ to $K$ is unique and is an arc of a great circle on $S^3$. However, for each point $g$ on the dual great circle, there is continuum of such arcs which can land anywhere on the circle $K$.
Moreover, one can see that a decomposition of $gin G$,
$$
g=kp, kin K, pin exp({mathfrak p}),
$$
need not be unique (even in the case of "special" compact subgroups you are considering). You can already see this when $n=3=2+1$, $G=SO(3), K=SO(2)$ or $2=1+1$ and
$$
G=SU(2), K=U(1)= U(1)otimes U(1).$$
If a decomposition were unique, the group $G$ would be homeomorphic to a product of $K$ with another topological space, while $SO(3)$ is homeomorphic to the 3-dimensional real projective space and hence does not split as a nontrivial product; same for $SU(2)$.
In this regard, the situation is quite different from the Cartan decomposition of a semisimple Lie group.
answered Jan 14 at 2:26
Moishe CohenMoishe Cohen
46.3k342106
$endgroup$
add a comment |
$begingroup$
It depends on the precise meaning of your question. Here is what's true:
Proposition. Suppose that $G$ is a compact connected Lie group with the Lie agebra ${mathfrak g}$, $K< G$ is a compact (Lie) subgroup with the Lie algebra ${mathfrak k}$ and ${mathfrak g}={mathfrak p} oplus {mathfrak k}$ is the orthogonal decomposition. Then every element $g$ of $G$ can be written as a product $g=pk=kp'$, where $kin K$ and $p, p'in exp({mathfrak p})$.
Proof. The proof below requires some differential geometry, which (as I can deduce from your question) you are unfamiliar with. Maybe there is a linear algebra proof in the case of $G=SO(n)$, I just do not see one.
First of all, you need to know what a Riemannian metric is: A Riemannian metric on a differentiable manifold is a choice of inner product on tangent spaces which varies smoothly from point to point. If the manifold happens to be a compact Lie group $G$, one can make such metric biinvariant, i.e. invariant under the multiplication by the elements of $G$ on the left and on the right. In particular, it suffices to prescribe the metric on the Lie algebra. The biinvariance condition then becomes the condition of invariance under the adjoint representation of $G$. In the case $G=SO(n)$, the standard metric on the Lie algebra $o(n)$ is given by
$$
langle A, Brangle = frac{1}{2} tr(AB).
$$
(The $1/2$ factor will become clear below. A simple Lie group such as $SO(n)$ admits a unique, up to a constant multiple, biinvariant metric.) The invariance under the adjoint representation is clear since trace is conjugation-invariant.
A Riemannian metric on a manifold $M$ defines the length of curves $c(t), tin [a,b]$, on $M$ by the familiar formula
$$
L(c)= int_a^b ||c'(t)||dt,
$$
where
$$
||v||= sqrt{langle v, v rangle}
$$
is the norm on the tangent space of $M$ (at the point $c(t)$ in our case).
A curve $c(t)$ on $M$ is called geodesic if it has constant speed and is locally length-minimizing. Equivalently, geodesics are defined as curves with zero acceleration:
$$
c''(t)= nabla_{c'(t)} c'(t)=0,
$$
where $nabla$ is the covariant derivative (the Levi-Civita connection of the metric). This equation, written in local coordinates, is a 2nd order linear ODE. As such, it has unique solution of the initial value problem $c(0)=pin M, c'(0)=vin T_pM$, a tangent vector at $p$. If $M$ happens to be compact, this solution exists for all $t$ in ${mathbb R}$.
Given this, one defines Riemannian exponential maps $exp_p: T_pMto M$ by sending $vin T_pM$ to $c(1)in M$ where $c(t)$ is the unique geodesic on $M$ satisfying $c(0)=p, c'(0)=v$.
If $M$ is compact and connected, each exponential map is surjective: Any two points are connected by a geodesic (this geodesic is nonunique unless $M$ is a single point). Moreover, if $Nsubset M$ is a compact smooth submanifold and $pin M$, there exists a shortest geodesic $c$ from $p$ to $N$; such a geodesic meets $N$ at the right angle. (The latter is a generalization of the familiar geometric fact of Euclidean geometry: Every point $p$ in $E^n$ is connected to an affine subspace $Vsubset E^n$ by a shortest path, namely the unique line segment from $p$ which meets $V$ at the right angle.)
It turns out that if $M$ is a compact Lie group equipped with a biinvariant Riemannian metric, this exponential map equals to the Lie-theoretic exponential map at $p=e$, the neutral element of $G$. This fact is a computation which is the key to the entire proof. (You can find it in Helgasson's book and other places.)
In the case $G=SO(n)$, the factor $1/2$ in the definition of the metric on $G$ is chosen so that $exp_e=exp$, the Riemannian exponential map equals the matrix exponential.
Now, with all these preliminaries out of the way, I can give a proof of the proposition.
I will equip the compact Lie group $G$ with a biinvariant metric. The subgroup $K< G$ is a smooth compact submanifold. Take $gin G$ and let $c: [0,1]to G$ be a shortest geodesic connecting $g$ to $K$ ($c(0)=g, c(1)=k$), this path meets $K$ at $kin K$ and its terminal velocity vector $c'(1)=v$ is orthogonal to the tangent space $T_k(K)$. (Such a path is highly nonunique due to compactness of $G$, which is responsible for the lack of uniqueness in the decomposition $g=kp$ in the proposition.)
If $k=e$ then $g=exp_e(-v)= exp(-v)in exp({mathfrak p})$, since $vperp {mathfrak k}=T_eK$, and, hence, $-vin {mathfrak p}$ as well. In the general case, when $k$ need not be equal to $e$, we apply to $c$ the left translation $L_{k^{-1}}$ by $k^{-1}$. The left translation preserves the metric on $G$, hence, sends geodesics to geodesics, so
$$
L_{k^{-1}} c= gamma,
$$
a geodesic connecting $k^{-1}g$ to $e$. Since $kin K$, $L_{k^{-1}}K= k^{-1}K=K$, hence, $gamma$ meets $K$ orthogonally at the point $e$ (since $c$ was orthogonal to $K$ at $k$). Therefore, denoting $w:= gamma'(1)$, we again obtain
$$
k^{-1}g= exp(-w)in exp({mathfrak p}),
$$
and
$$
g= k exp(-w)= kpin K cdot exp({mathfrak p}).
$$
To obtain a decomposition
$$
g= p'k, p'in exp({mathfrak p}), kin K
$$
one uses the same argument but right multiplication by $k^{-1}$ instead of the left multiplication. qed
It is instructive to understand the proof in the case $G=SU(2)$ and $K=U(1)< G$. Then $G$ is isometric to a round 3-sphere in $R^4$, $K$ corresponds to a great circle in $S^3$, say the intersection of $S^3$ with the plane $x_3=x_4=0$. For points $gin S^3$ which are not on the "dual circle" $S^3cap {x_1=x_2=0}$, the shortest path $c$ from $g$ to $K$ is unique and is an arc of a great circle on $S^3$. However, for each point $g$ on the dual great circle, there is continuum of such arcs which can land anywhere on the circle $K$.
Moreover, one can see that a decomposition of $gin G$,
$$
g=kp, kin K, pin exp({mathfrak p}),
$$
need not be unique (even in the case of "special" compact subgroups you are considering). You can already see this when $n=3=2+1$, $G=SO(3), K=SO(2)$ or $2=1+1$ and
$$
G=SU(2), K=U(1)= U(1)otimes U(1).$$
If a decomposition were unique, the group $G$ would be homeomorphic to a product of $K$ with another topological space, while $SO(3)$ is homeomorphic to the 3-dimensional real projective space and hence does not split as a nontrivial product; same for $SU(2)$.
In this regard, the situation is quite different from the Cartan decomposition of a semisimple Lie group.
answered Jan 14 at 2:26
Moishe CohenMoishe Cohen
46.3k342106
$endgroup$
add a comment |
$begingroup$
It depends on the precise meaning of your question. Here is what's true:
Proposition. Suppose that $G$ is a compact connected Lie group with the Lie agebra ${mathfrak g}$, $K< G$ is a compact (Lie) subgroup with the Lie algebra ${mathfrak k}$ and ${mathfrak g}={mathfrak p} oplus {mathfrak k}$ is the orthogonal decomposition. Then every element $g$ of $G$ can be written as a product $g=pk=kp'$, where $kin K$ and $p, p'in exp({mathfrak p})$.
Proof. The proof below requires some differential geometry, which (as I can deduce from your question) you are unfamiliar with. Maybe there is a linear algebra proof in the case of $G=SO(n)$, I just do not see one.
First of all, you need to know what a Riemannian metric is: A Riemannian metric on a differentiable manifold is a choice of inner product on tangent spaces which varies smoothly from point to point. If the manifold happens to be a compact Lie group $G$, one can make such metric biinvariant, i.e. invariant under the multiplication by the elements of $G$ on the left and on the right. In particular, it suffices to prescribe the metric on the Lie algebra. The biinvariance condition then becomes the condition of invariance under the adjoint representation of $G$. In the case $G=SO(n)$, the standard metric on the Lie algebra $o(n)$ is given by
$$
langle A, Brangle = frac{1}{2} tr(AB).
$$
(The $1/2$ factor will become clear below. A simple Lie group such as $SO(n)$ admits a unique, up to a constant multiple, biinvariant metric.) The invariance under the adjoint representation is clear since trace is conjugation-invariant.
A Riemannian metric on a manifold $M$ defines the length of curves $c(t), tin [a,b]$, on $M$ by the familiar formula
$$
L(c)= int_a^b ||c'(t)||dt,
$$
where
$$
||v||= sqrt{langle v, v rangle}
$$
is the norm on the tangent space of $M$ (at the point $c(t)$ in our case).
A curve $c(t)$ on $M$ is called geodesic if it has constant speed and is locally length-minimizing. Equivalently, geodesics are defined as curves with zero acceleration:
$$
c''(t)= nabla_{c'(t)} c'(t)=0,
$$
where $nabla$ is the covariant derivative (the Levi-Civita connection of the metric). This equation, written in local coordinates, is a 2nd order linear ODE. As such, it has unique solution of the initial value problem $c(0)=pin M, c'(0)=vin T_pM$, a tangent vector at $p$. If $M$ happens to be compact, this solution exists for all $t$ in ${mathbb R}$.
Given this, one defines Riemannian exponential maps $exp_p: T_pMto M$ by sending $vin T_pM$ to $c(1)in M$ where $c(t)$ is the unique geodesic on $M$ satisfying $c(0)=p, c'(0)=v$.
If $M$ is compact and connected, each exponential map is surjective: Any two points are connected by a geodesic (this geodesic is nonunique unless $M$ is a single point). Moreover, if $Nsubset M$ is a compact smooth submanifold and $pin M$, there exists a shortest geodesic $c$ from $p$ to $N$; such a geodesic meets $N$ at the right angle. (The latter is a generalization of the familiar geometric fact of Euclidean geometry: Every point $p$ in $E^n$ is connected to an affine subspace $Vsubset E^n$ by a shortest path, namely the unique line segment from $p$ which meets $V$ at the right angle.)
It turns out that if $M$ is a compact Lie group equipped with a biinvariant Riemannian metric, this exponential map equals to the Lie-theoretic exponential map at $p=e$, the neutral element of $G$. This fact is a computation which is the key to the entire proof. (You can find it in Helgasson's book and other places.)
In the case $G=SO(n)$, the factor $1/2$ in the definition of the metric on $G$ is chosen so that $exp_e=exp$, the Riemannian exponential map equals the matrix exponential.
Now, with all these preliminaries out of the way, I can give a proof of the proposition.
I will equip the compact Lie group $G$ with a biinvariant metric. The subgroup $K< G$ is a smooth compact submanifold. Take $gin G$ and let $c: [0,1]to G$ be a shortest geodesic connecting $g$ to $K$ ($c(0)=g, c(1)=k$), this path meets $K$ at $kin K$ and its terminal velocity vector $c'(1)=v$ is orthogonal to the tangent space $T_k(K)$. (Such a path is highly nonunique due to compactness of $G$, which is responsible for the lack of uniqueness in the decomposition $g=kp$ in the proposition.)
If $k=e$ then $g=exp_e(-v)= exp(-v)in exp({mathfrak p})$, since $vperp {mathfrak k}=T_eK$, and, hence, $-vin {mathfrak p}$ as well. In the general case, when $k$ need not be equal to $e$, we apply to $c$ the left translation $L_{k^{-1}}$ by $k^{-1}$. The left translation preserves the metric on $G$, hence, sends geodesics to geodesics, so
$$
L_{k^{-1}} c= gamma,
$$
a geodesic connecting $k^{-1}g$ to $e$. Since $kin K$, $L_{k^{-1}}K= k^{-1}K=K$, hence, $gamma$ meets $K$ orthogonally at the point $e$ (since $c$ was orthogonal to $K$ at $k$). Therefore, denoting $w:= gamma'(1)$, we again obtain
$$
k^{-1}g= exp(-w)in exp({mathfrak p}),
$$
and
$$
g= k exp(-w)= kpin K cdot exp({mathfrak p}).
$$
To obtain a decomposition
$$
g= p'k, p'in exp({mathfrak p}), kin K
$$
one uses the same argument but right multiplication by $k^{-1}$ instead of the left multiplication. qed
It is instructive to understand the proof in the case $G=SU(2)$ and $K=U(1)< G$. Then $G$ is isometric to a round 3-sphere in $R^4$, $K$ corresponds to a great circle in $S^3$, say the intersection of $S^3$ with the plane $x_3=x_4=0$. For points $gin S^3$ which are not on the "dual circle" $S^3cap {x_1=x_2=0}$, the shortest path $c$ from $g$ to $K$ is unique and is an arc of a great circle on $S^3$. However, for each point $g$ on the dual great circle, there is continuum of such arcs which can land anywhere on the circle $K$.
Moreover, one can see that a decomposition of $gin G$,
$$
g=kp, kin K, pin exp({mathfrak p}),
$$
need not be unique (even in the case of "special" compact subgroups you are considering). You can already see this when $n=3=2+1$, $G=SO(3), K=SO(2)$ or $2=1+1$ and
$$
G=SU(2), K=U(1)= U(1)otimes U(1).$$
If a decomposition were unique, the group $G$ would be homeomorphic to a product of $K$ with another topological space, while $SO(3)$ is homeomorphic to the 3-dimensional real projective space and hence does not split as a nontrivial product; same for $SU(2)$.
In this regard, the situation is quite different from the Cartan decomposition of a semisimple Lie group.
answered Jan 14 at 2:26
Moishe CohenMoishe Cohen
46.3k342106
$endgroup$
It depends on the precise meaning of your question. Here is what's true:
Proposition. Suppose that $G$ is a compact connected Lie group with the Lie agebra ${mathfrak g}$, $K< G$ is a compact (Lie) subgroup with the Lie algebra ${mathfrak k}$ and ${mathfrak g}={mathfrak p} oplus {mathfrak k}$ is the orthogonal decomposition. Then every element $g$ of $G$ can be written as a product $g=pk=kp'$, where $kin K$ and $p, p'in exp({mathfrak p})$.
Proof. The proof below requires some differential geometry, which (as I can deduce from your question) you are unfamiliar with. Maybe there is a linear algebra proof in the case of $G=SO(n)$, I just do not see one.
First of all, you need to know what a Riemannian metric is: A Riemannian metric on a differentiable manifold is a choice of inner product on tangent spaces which varies smoothly from point to point. If the manifold happens to be a compact Lie group $G$, one can make such metric biinvariant, i.e. invariant under the multiplication by the elements of $G$ on the left and on the right. In particular, it suffices to prescribe the metric on the Lie algebra. The biinvariance condition then becomes the condition of invariance under the adjoint representation of $G$. In the case $G=SO(n)$, the standard metric on the Lie algebra $o(n)$ is given by
$$
langle A, Brangle = frac{1}{2} tr(AB).
$$
(The $1/2$ factor will become clear below. A simple Lie group such as $SO(n)$ admits a unique, up to a constant multiple, biinvariant metric.) The invariance under the adjoint representation is clear since trace is conjugation-invariant.
A Riemannian metric on a manifold $M$ defines the length of curves $c(t), tin [a,b]$, on $M$ by the familiar formula
$$
L(c)= int_a^b ||c'(t)||dt,
$$
where
$$
||v||= sqrt{langle v, v rangle}
$$
is the norm on the tangent space of $M$ (at the point $c(t)$ in our case).
A curve $c(t)$ on $M$ is called geodesic if it has constant speed and is locally length-minimizing. Equivalently, geodesics are defined as curves with zero acceleration:
$$
c''(t)= nabla_{c'(t)} c'(t)=0,
$$
where $nabla$ is the covariant derivative (the Levi-Civita connection of the metric). This equation, written in local coordinates, is a 2nd order linear ODE. As such, it has unique solution of the initial value problem $c(0)=pin M, c'(0)=vin T_pM$, a tangent vector at $p$. If $M$ happens to be compact, this solution exists for all $t$ in ${mathbb R}$.
Given this, one defines Riemannian exponential maps $exp_p: T_pMto M$ by sending $vin T_pM$ to $c(1)in M$ where $c(t)$ is the unique geodesic on $M$ satisfying $c(0)=p, c'(0)=v$.
If $M$ is compact and connected, each exponential map is surjective: Any two points are connected by a geodesic (this geodesic is nonunique unless $M$ is a single point). Moreover, if $Nsubset M$ is a compact smooth submanifold and $pin M$, there exists a shortest geodesic $c$ from $p$ to $N$; such a geodesic meets $N$ at the right angle. (The latter is a generalization of the familiar geometric fact of Euclidean geometry: Every point $p$ in $E^n$ is connected to an affine subspace $Vsubset E^n$ by a shortest path, namely the unique line segment from $p$ which meets $V$ at the right angle.)
It turns out that if $M$ is a compact Lie group equipped with a biinvariant Riemannian metric, this exponential map equals to the Lie-theoretic exponential map at $p=e$, the neutral element of $G$. This fact is a computation which is the key to the entire proof. (You can find it in Helgasson's book and other places.)
In the case $G=SO(n)$, the factor $1/2$ in the definition of the metric on $G$ is chosen so that $exp_e=exp$, the Riemannian exponential map equals the matrix exponential.
Now, with all these preliminaries out of the way, I can give a proof of the proposition.
I will equip the compact Lie group $G$ with a biinvariant metric. The subgroup $K< G$ is a smooth compact submanifold. Take $gin G$ and let $c: [0,1]to G$ be a shortest geodesic connecting $g$ to $K$ ($c(0)=g, c(1)=k$), this path meets $K$ at $kin K$ and its terminal velocity vector $c'(1)=v$ is orthogonal to the tangent space $T_k(K)$. (Such a path is highly nonunique due to compactness of $G$, which is responsible for the lack of uniqueness in the decomposition $g=kp$ in the proposition.)
If $k=e$ then $g=exp_e(-v)= exp(-v)in exp({mathfrak p})$, since $vperp {mathfrak k}=T_eK$, and, hence, $-vin {mathfrak p}$ as well. In the general case, when $k$ need not be equal to $e$, we apply to $c$ the left translation $L_{k^{-1}}$ by $k^{-1}$. The left translation preserves the metric on $G$, hence, sends geodesics to geodesics, so
$$
L_{k^{-1}} c= gamma,
$$
a geodesic connecting $k^{-1}g$ to $e$. Since $kin K$, $L_{k^{-1}}K= k^{-1}K=K$, hence, $gamma$ meets $K$ orthogonally at the point $e$ (since $c$ was orthogonal to $K$ at $k$). Therefore, denoting $w:= gamma'(1)$, we again obtain
$$
k^{-1}g= exp(-w)in exp({mathfrak p}),
$$
and
$$
g= k exp(-w)= kpin K cdot exp({mathfrak p}).
$$
To obtain a decomposition
$$
g= p'k, p'in exp({mathfrak p}), kin K
$$
one uses the same argument but right multiplication by $k^{-1}$ instead of the left multiplication. qed
It is instructive to understand the proof in the case $G=SU(2)$ and $K=U(1)< G$. Then $G$ is isometric to a round 3-sphere in $R^4$, $K$ corresponds to a great circle in $S^3$, say the intersection of $S^3$ with the plane $x_3=x_4=0$. For points $gin S^3$ which are not on the "dual circle" $S^3cap {x_1=x_2=0}$, the shortest path $c$ from $g$ to $K$ is unique and is an arc of a great circle on $S^3$. However, for each point $g$ on the dual great circle, there is continuum of such arcs which can land anywhere on the circle $K$.
Moreover, one can see that a decomposition of $gin G$,
$$
g=kp, kin K, pin exp({mathfrak p}),
$$
need not be unique (even in the case of "special" compact subgroups you are considering). You can already see this when $n=3=2+1$, $G=SO(3), K=SO(2)$ or $2=1+1$ and
$$
G=SU(2), K=U(1)= U(1)otimes U(1).$$
If a decomposition were unique, the group $G$ would be homeomorphic to a product of $K$ with another topological space, while $SO(3)$ is homeomorphic to the 3-dimensional real projective space and hence does not split as a nontrivial product; same for $SU(2)$.
In this regard, the situation is quite different from the Cartan decomposition of a semisimple Lie group.
answered Jan 14 at 2:26
Moishe CohenMoishe Cohen
46.3k342106
edited Jan 14 at 17:18
answered Jan 14 at 2:26
Moishe CohenMoishe Cohen
46.3k342106
answered Jan 14 at 2:26
Moishe CohenMoishe Cohen
46.3k342106
answered Jan 14 at 2:26
Moishe CohenMoishe Cohen
46.3k342106
46.3k342106
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061880%2fis-this-pseudo-cartan-decomposition-of-son-valid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
