Question about the diffeomorphism group of a Lie group.
$begingroup$
I was reading about the diffeomorphism group of varying Lie groups. Wikipedia states that:
When M = G is a Lie group, there is a natural inclusion of G in its
own diffeomorphism group via left-translation. Let Diff(G) denote the
diffeomorphism group of G, then there is a splitting Diff(G) ≃ G ×
Diff(G, e), where Diff(G, e) is the subgroup of Diff(G) that fixes the
identity element of the group.
I'm pretty new to this and not sure how to interpret that statement. Specifically the group Diff(G,e), my brain wants to interpret this by saying that the group G with the identity e fixed is like the group of isometries of G but that isn't right. Can someone steer me in the right direction?
For example, suppose I'm considering the group $R^{3,1}$. Then the group $Diff(R^{3,1})backsimeq R^{3,1}xDiff(R^{3,1},e)$ my intuition (probably wrong) is to say that the identity fixed here leads to the group of isometries (which is the Poincare group for R^{3,1}. This would leave us with something like:
$$Diff(R^{3,1})backsimeq R^{3,1}XDiff(Poincare)$$
However, I think I'm missing something here, but I'm trying to learn this on my own
lie-groups lie-algebras diffeomorphism
asked Dec 13 '18 at 23:19
R. RankinR. Rankin
333213
$endgroup$
add a comment |
$begingroup$
I was reading about the diffeomorphism group of varying Lie groups. Wikipedia states that:
When M = G is a Lie group, there is a natural inclusion of G in its
own diffeomorphism group via left-translation. Let Diff(G) denote the
diffeomorphism group of G, then there is a splitting Diff(G) ≃ G ×
Diff(G, e), where Diff(G, e) is the subgroup of Diff(G) that fixes the
identity element of the group.
I'm pretty new to this and not sure how to interpret that statement. Specifically the group Diff(G,e), my brain wants to interpret this by saying that the group G with the identity e fixed is like the group of isometries of G but that isn't right. Can someone steer me in the right direction?
For example, suppose I'm considering the group $R^{3,1}$. Then the group $Diff(R^{3,1})backsimeq R^{3,1}xDiff(R^{3,1},e)$ my intuition (probably wrong) is to say that the identity fixed here leads to the group of isometries (which is the Poincare group for R^{3,1}. This would leave us with something like:
$$Diff(R^{3,1})backsimeq R^{3,1}XDiff(Poincare)$$
However, I think I'm missing something here, but I'm trying to learn this on my own
lie-groups lie-algebras diffeomorphism
asked Dec 13 '18 at 23:19
R. RankinR. Rankin
333213
$endgroup$
add a comment |
$begingroup$
I was reading about the diffeomorphism group of varying Lie groups. Wikipedia states that:
When M = G is a Lie group, there is a natural inclusion of G in its
own diffeomorphism group via left-translation. Let Diff(G) denote the
diffeomorphism group of G, then there is a splitting Diff(G) ≃ G ×
Diff(G, e), where Diff(G, e) is the subgroup of Diff(G) that fixes the
identity element of the group.
I'm pretty new to this and not sure how to interpret that statement. Specifically the group Diff(G,e), my brain wants to interpret this by saying that the group G with the identity e fixed is like the group of isometries of G but that isn't right. Can someone steer me in the right direction?
For example, suppose I'm considering the group $R^{3,1}$. Then the group $Diff(R^{3,1})backsimeq R^{3,1}xDiff(R^{3,1},e)$ my intuition (probably wrong) is to say that the identity fixed here leads to the group of isometries (which is the Poincare group for R^{3,1}. This would leave us with something like:
$$Diff(R^{3,1})backsimeq R^{3,1}XDiff(Poincare)$$
However, I think I'm missing something here, but I'm trying to learn this on my own
lie-groups lie-algebras diffeomorphism
asked Dec 13 '18 at 23:19
R. RankinR. Rankin
333213
$endgroup$
I was reading about the diffeomorphism group of varying Lie groups. Wikipedia states that:
When M = G is a Lie group, there is a natural inclusion of G in its
own diffeomorphism group via left-translation. Let Diff(G) denote the
diffeomorphism group of G, then there is a splitting Diff(G) ≃ G ×
Diff(G, e), where Diff(G, e) is the subgroup of Diff(G) that fixes the
identity element of the group.
I'm pretty new to this and not sure how to interpret that statement. Specifically the group Diff(G,e), my brain wants to interpret this by saying that the group G with the identity e fixed is like the group of isometries of G but that isn't right. Can someone steer me in the right direction?
For example, suppose I'm considering the group $R^{3,1}$. Then the group $Diff(R^{3,1})backsimeq R^{3,1}xDiff(R^{3,1},e)$ my intuition (probably wrong) is to say that the identity fixed here leads to the group of isometries (which is the Poincare group for R^{3,1}. This would leave us with something like:
$$Diff(R^{3,1})backsimeq R^{3,1}XDiff(Poincare)$$
However, I think I'm missing something here, but I'm trying to learn this on my own
lie-groups lie-algebras diffeomorphism
lie-groups lie-algebras diffeomorphism
asked Dec 13 '18 at 23:19
R. RankinR. Rankin
333213
asked Dec 13 '18 at 23:19
R. RankinR. Rankin
333213
edited Dec 14 '18 at 1:56
R. Rankin
asked Dec 13 '18 at 23:19
R. RankinR. Rankin
333213
asked Dec 13 '18 at 23:19
R. RankinR. Rankin
333213
asked Dec 13 '18 at 23:19
R. RankinR. Rankin
333213
333213
add a comment |
add a comment |
1 Answer
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$begingroup$
The group of diffeomorphisms fixing a point is not signifcantly simpler or smaller than the diffeomorphism group itself. Given any manifold $M$ and a point $x_0in M$, you get a map $Diff(M)to M$ by sending a diffeomorphism $f$ to $f(x_0)in M$. Assuming for exmaple that $M$ is connected, this map is surjective, and the preimage of ${x_0}$ is the subgroup of all diffeomorphisms fixing the point $x_0$. This is still infinite dimensional and shares many properties with the full diffeomorphism group.
In the special case of a Lie group, it is natural to choose $x_0=e$ and you get a natural map $Gto Diff(G)$ splitting the evaluation at $e$ by sending $g$ to left translation $lambda_g$ by $g$. The splitting you refer to, then simply sends a diffeomorphism $f$ to $(f(e),lambda_{f(e)^{-1}}circ f)$ and you can easily compute what the multiplication looks like in this picture. (You get a structure of a semi-direct product, but that's not very illuminating.)
Your example sounds a bit strange. If you want to view $mathbb R^{3,1}$ as a goup, you have to take addition as a operation. Then the statment boils down to the fact that any diffeomorphism of $mathbb R^{3,1}$ can be written as a composition of a translation and a diffeomorphism fixing the origin. But as far as I am aware of, there is no relation to isometries.
answered Dec 16 '18 at 12:04
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
Thank you very much! In truth I am interested in the manifold $SU(2)XU(1)$ In the context of being a spacetime manifold. So it would go $SU(2)XU(1)XDiff(SU(2)XU(1), e)$
$endgroup$
– R. Rankin
Dec 16 '18 at 12:38
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The group of diffeomorphisms fixing a point is not signifcantly simpler or smaller than the diffeomorphism group itself. Given any manifold $M$ and a point $x_0in M$, you get a map $Diff(M)to M$ by sending a diffeomorphism $f$ to $f(x_0)in M$. Assuming for exmaple that $M$ is connected, this map is surjective, and the preimage of ${x_0}$ is the subgroup of all diffeomorphisms fixing the point $x_0$. This is still infinite dimensional and shares many properties with the full diffeomorphism group.
In the special case of a Lie group, it is natural to choose $x_0=e$ and you get a natural map $Gto Diff(G)$ splitting the evaluation at $e$ by sending $g$ to left translation $lambda_g$ by $g$. The splitting you refer to, then simply sends a diffeomorphism $f$ to $(f(e),lambda_{f(e)^{-1}}circ f)$ and you can easily compute what the multiplication looks like in this picture. (You get a structure of a semi-direct product, but that's not very illuminating.)
Your example sounds a bit strange. If you want to view $mathbb R^{3,1}$ as a goup, you have to take addition as a operation. Then the statment boils down to the fact that any diffeomorphism of $mathbb R^{3,1}$ can be written as a composition of a translation and a diffeomorphism fixing the origin. But as far as I am aware of, there is no relation to isometries.
answered Dec 16 '18 at 12:04
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
Thank you very much! In truth I am interested in the manifold $SU(2)XU(1)$ In the context of being a spacetime manifold. So it would go $SU(2)XU(1)XDiff(SU(2)XU(1), e)$
$endgroup$
– R. Rankin
Dec 16 '18 at 12:38
add a comment |
$begingroup$
The group of diffeomorphisms fixing a point is not signifcantly simpler or smaller than the diffeomorphism group itself. Given any manifold $M$ and a point $x_0in M$, you get a map $Diff(M)to M$ by sending a diffeomorphism $f$ to $f(x_0)in M$. Assuming for exmaple that $M$ is connected, this map is surjective, and the preimage of ${x_0}$ is the subgroup of all diffeomorphisms fixing the point $x_0$. This is still infinite dimensional and shares many properties with the full diffeomorphism group.
In the special case of a Lie group, it is natural to choose $x_0=e$ and you get a natural map $Gto Diff(G)$ splitting the evaluation at $e$ by sending $g$ to left translation $lambda_g$ by $g$. The splitting you refer to, then simply sends a diffeomorphism $f$ to $(f(e),lambda_{f(e)^{-1}}circ f)$ and you can easily compute what the multiplication looks like in this picture. (You get a structure of a semi-direct product, but that's not very illuminating.)
Your example sounds a bit strange. If you want to view $mathbb R^{3,1}$ as a goup, you have to take addition as a operation. Then the statment boils down to the fact that any diffeomorphism of $mathbb R^{3,1}$ can be written as a composition of a translation and a diffeomorphism fixing the origin. But as far as I am aware of, there is no relation to isometries.
answered Dec 16 '18 at 12:04
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
Thank you very much! In truth I am interested in the manifold $SU(2)XU(1)$ In the context of being a spacetime manifold. So it would go $SU(2)XU(1)XDiff(SU(2)XU(1), e)$
$endgroup$
– R. Rankin
Dec 16 '18 at 12:38
add a comment |
$begingroup$
The group of diffeomorphisms fixing a point is not signifcantly simpler or smaller than the diffeomorphism group itself. Given any manifold $M$ and a point $x_0in M$, you get a map $Diff(M)to M$ by sending a diffeomorphism $f$ to $f(x_0)in M$. Assuming for exmaple that $M$ is connected, this map is surjective, and the preimage of ${x_0}$ is the subgroup of all diffeomorphisms fixing the point $x_0$. This is still infinite dimensional and shares many properties with the full diffeomorphism group.
In the special case of a Lie group, it is natural to choose $x_0=e$ and you get a natural map $Gto Diff(G)$ splitting the evaluation at $e$ by sending $g$ to left translation $lambda_g$ by $g$. The splitting you refer to, then simply sends a diffeomorphism $f$ to $(f(e),lambda_{f(e)^{-1}}circ f)$ and you can easily compute what the multiplication looks like in this picture. (You get a structure of a semi-direct product, but that's not very illuminating.)
Your example sounds a bit strange. If you want to view $mathbb R^{3,1}$ as a goup, you have to take addition as a operation. Then the statment boils down to the fact that any diffeomorphism of $mathbb R^{3,1}$ can be written as a composition of a translation and a diffeomorphism fixing the origin. But as far as I am aware of, there is no relation to isometries.
answered Dec 16 '18 at 12:04
Andreas CapAndreas Cap
11.1k923
$endgroup$
The group of diffeomorphisms fixing a point is not signifcantly simpler or smaller than the diffeomorphism group itself. Given any manifold $M$ and a point $x_0in M$, you get a map $Diff(M)to M$ by sending a diffeomorphism $f$ to $f(x_0)in M$. Assuming for exmaple that $M$ is connected, this map is surjective, and the preimage of ${x_0}$ is the subgroup of all diffeomorphisms fixing the point $x_0$. This is still infinite dimensional and shares many properties with the full diffeomorphism group.
In the special case of a Lie group, it is natural to choose $x_0=e$ and you get a natural map $Gto Diff(G)$ splitting the evaluation at $e$ by sending $g$ to left translation $lambda_g$ by $g$. The splitting you refer to, then simply sends a diffeomorphism $f$ to $(f(e),lambda_{f(e)^{-1}}circ f)$ and you can easily compute what the multiplication looks like in this picture. (You get a structure of a semi-direct product, but that's not very illuminating.)
Your example sounds a bit strange. If you want to view $mathbb R^{3,1}$ as a goup, you have to take addition as a operation. Then the statment boils down to the fact that any diffeomorphism of $mathbb R^{3,1}$ can be written as a composition of a translation and a diffeomorphism fixing the origin. But as far as I am aware of, there is no relation to isometries.
answered Dec 16 '18 at 12:04
Andreas CapAndreas Cap
11.1k923
answered Dec 16 '18 at 12:04
Andreas CapAndreas Cap
11.1k923
answered Dec 16 '18 at 12:04
Andreas CapAndreas Cap
11.1k923
answered Dec 16 '18 at 12:04
Andreas CapAndreas Cap
11.1k923
11.1k923
$begingroup$
Thank you very much! In truth I am interested in the manifold $SU(2)XU(1)$ In the context of being a spacetime manifold. So it would go $SU(2)XU(1)XDiff(SU(2)XU(1), e)$
$endgroup$
– R. Rankin
Dec 16 '18 at 12:38
add a comment |
$begingroup$
Thank you very much! In truth I am interested in the manifold $SU(2)XU(1)$ In the context of being a spacetime manifold. So it would go $SU(2)XU(1)XDiff(SU(2)XU(1), e)$
$endgroup$
– R. Rankin
Dec 16 '18 at 12:38
$begingroup$
Thank you very much! In truth I am interested in the manifold $SU(2)XU(1)$ In the context of being a spacetime manifold. So it would go $SU(2)XU(1)XDiff(SU(2)XU(1), e)$
$endgroup$
– R. Rankin
Dec 16 '18 at 12:38
$begingroup$
Thank you very much! In truth I am interested in the manifold $SU(2)XU(1)$ In the context of being a spacetime manifold. So it would go $SU(2)XU(1)XDiff(SU(2)XU(1), e)$
$endgroup$
– R. Rankin
Dec 16 '18 at 12:38
add a comment |
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