Mixing
If $exp(X) in G$, and $X$ is Hermitian, do we have $X in mathfrak g$?
$begingroup$
Let $G$ be the $mathbb R$-points of an algebraic group in $operatorname{GL}_n(mathbb C)$ which is defined over $mathbb R$. Let $mathfrak g$ be the Lie algebra of $G$. Assume $G$ is topologically connected, and closed under the conjugate transpose $ g mapsto g^{ast} = space ^toverline{g}$. Let $$mathfrak h = { X in mathfrak{gl}_n(mathbb C) : space ^toverline{X} = X}$$
be the space of Hermitian matrices, and let $U(n) = {g in operatorname{GL}_n(mathbb C) : gg^{ast} = 1 }$ be the unitary group. Let $K = G cap U(n)$. By the polar decomposition theorem, we have that
$$U(n) times mathfrak h rightarrow operatorname{GL}_n(mathbb C)$$ $$(u,X) mapsto u exp(X) $$
is a homeomorphism. If we set $mathfrak p = mathfrak h cap mathfrak g$, I want to prove the Cartan decomposition for $G$, which states that polar decomposition restricts to a homeomorphism
$$K times mathfrak p rightarrow G$$
For $g in G$, uniquely write $g = k exp(X)$ for $k in U(n)$ and $X$ Hermitian. I know how to show that $k$ actually lies in $K$, and therefore that $exp(X)$ lies in $G$. However, I don't see why this implies $X in mathfrak p$.
I think my difficulty comes down to the following question: if $X$ is Hermitian, and $exp(X)$ is in $G$, do we have $X in mathfrak g$?
differential-geometry lie-groups algebraic-groups
asked Jan 25 at 17:33
D_SD_S
14k61553
$endgroup$
add a comment |
$begingroup$
Let $G$ be the $mathbb R$-points of an algebraic group in $operatorname{GL}_n(mathbb C)$ which is defined over $mathbb R$. Let $mathfrak g$ be the Lie algebra of $G$. Assume $G$ is topologically connected, and closed under the conjugate transpose $ g mapsto g^{ast} = space ^toverline{g}$. Let $$mathfrak h = { X in mathfrak{gl}_n(mathbb C) : space ^toverline{X} = X}$$
be the space of Hermitian matrices, and let $U(n) = {g in operatorname{GL}_n(mathbb C) : gg^{ast} = 1 }$ be the unitary group. Let $K = G cap U(n)$. By the polar decomposition theorem, we have that
$$U(n) times mathfrak h rightarrow operatorname{GL}_n(mathbb C)$$ $$(u,X) mapsto u exp(X) $$
is a homeomorphism. If we set $mathfrak p = mathfrak h cap mathfrak g$, I want to prove the Cartan decomposition for $G$, which states that polar decomposition restricts to a homeomorphism
$$K times mathfrak p rightarrow G$$
For $g in G$, uniquely write $g = k exp(X)$ for $k in U(n)$ and $X$ Hermitian. I know how to show that $k$ actually lies in $K$, and therefore that $exp(X)$ lies in $G$. However, I don't see why this implies $X in mathfrak p$.
I think my difficulty comes down to the following question: if $X$ is Hermitian, and $exp(X)$ is in $G$, do we have $X in mathfrak g$?
differential-geometry lie-groups algebraic-groups
asked Jan 25 at 17:33
D_SD_S
14k61553
$endgroup$
1
$begingroup$
Take $G = I$, the trivial group. Take $X$ to be the diagonal matrix with all entries $2pi i$. Then $exp(X) = 1$. What you need to assume is that $exp(tX) in G$ for all $t in Bbb R$, or at least $t$ near zero.
$endgroup$
– user98602
Jan 25 at 17:52
$begingroup$
Maybe there is some pathology with zero dimensional things. Do you think similar counterexamples arise for nontrivial $G$?
$endgroup$
– D_S
Jan 25 at 18:39
1
$begingroup$
The same example works! Take $G$ to be any subgroup that does not include the diagonal $U(1)$, for instance $SU(n)$.
$endgroup$
– user98602
Jan 25 at 19:00
1
$begingroup$
The suggested counterexample is not Hermitian. One shows that $text{exp}(tX) in G$ for all $t in mathbb{R}$ using that $G$ is algebraic and that $text{exp}(nX) in G$ for all $n in mathbb{Z}$. This implies $X in mathfrak{g}$.
$endgroup$
– Oven
Jan 26 at 21:30
add a comment |
$begingroup$
Let $G$ be the $mathbb R$-points of an algebraic group in $operatorname{GL}_n(mathbb C)$ which is defined over $mathbb R$. Let $mathfrak g$ be the Lie algebra of $G$. Assume $G$ is topologically connected, and closed under the conjugate transpose $ g mapsto g^{ast} = space ^toverline{g}$. Let $$mathfrak h = { X in mathfrak{gl}_n(mathbb C) : space ^toverline{X} = X}$$
be the space of Hermitian matrices, and let $U(n) = {g in operatorname{GL}_n(mathbb C) : gg^{ast} = 1 }$ be the unitary group. Let $K = G cap U(n)$. By the polar decomposition theorem, we have that
$$U(n) times mathfrak h rightarrow operatorname{GL}_n(mathbb C)$$ $$(u,X) mapsto u exp(X) $$
is a homeomorphism. If we set $mathfrak p = mathfrak h cap mathfrak g$, I want to prove the Cartan decomposition for $G$, which states that polar decomposition restricts to a homeomorphism
$$K times mathfrak p rightarrow G$$
For $g in G$, uniquely write $g = k exp(X)$ for $k in U(n)$ and $X$ Hermitian. I know how to show that $k$ actually lies in $K$, and therefore that $exp(X)$ lies in $G$. However, I don't see why this implies $X in mathfrak p$.
I think my difficulty comes down to the following question: if $X$ is Hermitian, and $exp(X)$ is in $G$, do we have $X in mathfrak g$?
differential-geometry lie-groups algebraic-groups
asked Jan 25 at 17:33
D_SD_S
14k61553
$endgroup$
Let $G$ be the $mathbb R$-points of an algebraic group in $operatorname{GL}_n(mathbb C)$ which is defined over $mathbb R$. Let $mathfrak g$ be the Lie algebra of $G$. Assume $G$ is topologically connected, and closed under the conjugate transpose $ g mapsto g^{ast} = space ^toverline{g}$. Let $$mathfrak h = { X in mathfrak{gl}_n(mathbb C) : space ^toverline{X} = X}$$
be the space of Hermitian matrices, and let $U(n) = {g in operatorname{GL}_n(mathbb C) : gg^{ast} = 1 }$ be the unitary group. Let $K = G cap U(n)$. By the polar decomposition theorem, we have that
$$U(n) times mathfrak h rightarrow operatorname{GL}_n(mathbb C)$$ $$(u,X) mapsto u exp(X) $$
is a homeomorphism. If we set $mathfrak p = mathfrak h cap mathfrak g$, I want to prove the Cartan decomposition for $G$, which states that polar decomposition restricts to a homeomorphism
$$K times mathfrak p rightarrow G$$
For $g in G$, uniquely write $g = k exp(X)$ for $k in U(n)$ and $X$ Hermitian. I know how to show that $k$ actually lies in $K$, and therefore that $exp(X)$ lies in $G$. However, I don't see why this implies $X in mathfrak p$.
I think my difficulty comes down to the following question: if $X$ is Hermitian, and $exp(X)$ is in $G$, do we have $X in mathfrak g$?
differential-geometry lie-groups algebraic-groups
differential-geometry lie-groups algebraic-groups
asked Jan 25 at 17:33
D_SD_S
14k61553
asked Jan 25 at 17:33
D_SD_S
14k61553
asked Jan 25 at 17:33
D_SD_S
14k61553
asked Jan 25 at 17:33
D_SD_S
14k61553
asked Jan 25 at 17:33
D_SD_S
14k61553
14k61553
1
$begingroup$
Take $G = I$, the trivial group. Take $X$ to be the diagonal matrix with all entries $2pi i$. Then $exp(X) = 1$. What you need to assume is that $exp(tX) in G$ for all $t in Bbb R$, or at least $t$ near zero.
$endgroup$
– user98602
Jan 25 at 17:52
$begingroup$
Maybe there is some pathology with zero dimensional things. Do you think similar counterexamples arise for nontrivial $G$?
$endgroup$
– D_S
Jan 25 at 18:39
1
$begingroup$
The same example works! Take $G$ to be any subgroup that does not include the diagonal $U(1)$, for instance $SU(n)$.
$endgroup$
– user98602
Jan 25 at 19:00
1
$begingroup$
The suggested counterexample is not Hermitian. One shows that $text{exp}(tX) in G$ for all $t in mathbb{R}$ using that $G$ is algebraic and that $text{exp}(nX) in G$ for all $n in mathbb{Z}$. This implies $X in mathfrak{g}$.
$endgroup$
– Oven
Jan 26 at 21:30
add a comment |
1
$begingroup$
Take $G = I$, the trivial group. Take $X$ to be the diagonal matrix with all entries $2pi i$. Then $exp(X) = 1$. What you need to assume is that $exp(tX) in G$ for all $t in Bbb R$, or at least $t$ near zero.
$endgroup$
– user98602
Jan 25 at 17:52
$begingroup$
Maybe there is some pathology with zero dimensional things. Do you think similar counterexamples arise for nontrivial $G$?
$endgroup$
– D_S
Jan 25 at 18:39
1
$begingroup$
The same example works! Take $G$ to be any subgroup that does not include the diagonal $U(1)$, for instance $SU(n)$.
$endgroup$
– user98602
Jan 25 at 19:00
1
$begingroup$
The suggested counterexample is not Hermitian. One shows that $text{exp}(tX) in G$ for all $t in mathbb{R}$ using that $G$ is algebraic and that $text{exp}(nX) in G$ for all $n in mathbb{Z}$. This implies $X in mathfrak{g}$.
$endgroup$
– Oven
Jan 26 at 21:30
1
1
$begingroup$
Take $G = I$, the trivial group. Take $X$ to be the diagonal matrix with all entries $2pi i$. Then $exp(X) = 1$. What you need to assume is that $exp(tX) in G$ for all $t in Bbb R$, or at least $t$ near zero.
$endgroup$
– user98602
Jan 25 at 17:52
$begingroup$
Take $G = I$, the trivial group. Take $X$ to be the diagonal matrix with all entries $2pi i$. Then $exp(X) = 1$. What you need to assume is that $exp(tX) in G$ for all $t in Bbb R$, or at least $t$ near zero.
$endgroup$
– user98602
Jan 25 at 17:52
$begingroup$
Maybe there is some pathology with zero dimensional things. Do you think similar counterexamples arise for nontrivial $G$?
$endgroup$
– D_S
Jan 25 at 18:39
$begingroup$
Maybe there is some pathology with zero dimensional things. Do you think similar counterexamples arise for nontrivial $G$?
$endgroup$
– D_S
Jan 25 at 18:39
1
1
$begingroup$
The same example works! Take $G$ to be any subgroup that does not include the diagonal $U(1)$, for instance $SU(n)$.
$endgroup$
– user98602
Jan 25 at 19:00
$begingroup$
The same example works! Take $G$ to be any subgroup that does not include the diagonal $U(1)$, for instance $SU(n)$.
$endgroup$
– user98602
Jan 25 at 19:00
1
1
$begingroup$
The suggested counterexample is not Hermitian. One shows that $text{exp}(tX) in G$ for all $t in mathbb{R}$ using that $G$ is algebraic and that $text{exp}(nX) in G$ for all $n in mathbb{Z}$. This implies $X in mathfrak{g}$.
$endgroup$
– Oven
Jan 26 at 21:30
$begingroup$
The suggested counterexample is not Hermitian. One shows that $text{exp}(tX) in G$ for all $t in mathbb{R}$ using that $G$ is algebraic and that $text{exp}(nX) in G$ for all $n in mathbb{Z}$. This implies $X in mathfrak{g}$.
$endgroup$
– Oven
Jan 26 at 21:30
add a comment |
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1
$begingroup$
Take $G = I$, the trivial group. Take $X$ to be the diagonal matrix with all entries $2pi i$. Then $exp(X) = 1$. What you need to assume is that $exp(tX) in G$ for all $t in Bbb R$, or at least $t$ near zero.
$endgroup$
– user98602
Jan 25 at 17:52
$begingroup$
Maybe there is some pathology with zero dimensional things. Do you think similar counterexamples arise for nontrivial $G$?
$endgroup$
– D_S
Jan 25 at 18:39
1
$begingroup$
The same example works! Take $G$ to be any subgroup that does not include the diagonal $U(1)$, for instance $SU(n)$.
$endgroup$
– user98602
Jan 25 at 19:00
1
$begingroup$
The suggested counterexample is not Hermitian. One shows that $text{exp}(tX) in G$ for all $t in mathbb{R}$ using that $G$ is algebraic and that $text{exp}(nX) in G$ for all $n in mathbb{Z}$. This implies $X in mathfrak{g}$.
$endgroup$
– Oven
Jan 26 at 21:30