Mixing
Are there interesting example of pseudo-Riemannian manifold other than spacetime manifold?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
The 4 dimensional spacetime manifold is a typical example of pseudo-Riemannian manifold. Are there other mathematically or physically interesting example of it?
differential-geometry soft-question riemannian-geometry physics motivation
asked Jul 31 at 5:16
marimo
945
add a comment |Â
up vote
4
down vote
favorite
The 4 dimensional spacetime manifold is a typical example of pseudo-Riemannian manifold. Are there other mathematically or physically interesting example of it?
differential-geometry soft-question riemannian-geometry physics motivation
asked Jul 31 at 5:16
marimo
945
It isn't quite what you're asking, but one of the slickest constructions of $n$-dimensional hyperbolic space $mathbbH^n$ (i.e., the hyperboloid model) is as a space-like hypersurface in $(n+1)$-dimensional Minkowski space.
– Branimir Ćaćić
Jul 31 at 8:06
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The 4 dimensional spacetime manifold is a typical example of pseudo-Riemannian manifold. Are there other mathematically or physically interesting example of it?
differential-geometry soft-question riemannian-geometry physics motivation
asked Jul 31 at 5:16
marimo
945
The 4 dimensional spacetime manifold is a typical example of pseudo-Riemannian manifold. Are there other mathematically or physically interesting example of it?
differential-geometry soft-question riemannian-geometry physics motivation
asked Jul 31 at 5:16
marimo
945
asked Jul 31 at 5:16
marimo
945
asked Jul 31 at 5:16
marimo
945
asked Jul 31 at 5:16
marimo
945
945
It isn't quite what you're asking, but one of the slickest constructions of $n$-dimensional hyperbolic space $mathbbH^n$ (i.e., the hyperboloid model) is as a space-like hypersurface in $(n+1)$-dimensional Minkowski space.
– Branimir Ćaćić
Jul 31 at 8:06
add a comment |Â
It isn't quite what you're asking, but one of the slickest constructions of $n$-dimensional hyperbolic space $mathbbH^n$ (i.e., the hyperboloid model) is as a space-like hypersurface in $(n+1)$-dimensional Minkowski space.
– Branimir Ćaćić
Jul 31 at 8:06
It isn't quite what you're asking, but one of the slickest constructions of $n$-dimensional hyperbolic space $mathbbH^n$ (i.e., the hyperboloid model) is as a space-like hypersurface in $(n+1)$-dimensional Minkowski space.
– Branimir Ćaćić
Jul 31 at 8:06
It isn't quite what you're asking, but one of the slickest constructions of $n$-dimensional hyperbolic space $mathbbH^n$ (i.e., the hyperboloid model) is as a space-like hypersurface in $(n+1)$-dimensional Minkowski space.
– Branimir Ćaćić
Jul 31 at 8:06
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
A Lie group $G$ acts on itself both by left translation (group multiplication on the left) and by right translation (group multiplication on the right). Hence, it's natural to look for Riemannian metrics that are bi-invariant, i.e., that make both left translation and right translation by a group element into isometries. However, there's a folkloric result that $G$ admits a bi-invariant Riemannian metric if and only if $G cong H times K$ for $H$ compact and $K$ abelian, which means, for instance, that the non-compact simple Lie group
$$
SL(2,mathbbR) := A in M_2(mathbbR) mid det(A) = 1
$$
does not admit any bi-invariant Riemannian metrics. However, as we'll see below, it does admit a canonical bi-invariant pseudo-Riemannian metric of signature $(+,-,-)$.
Now, in general, if $mathfrakg$ is the Lie algebra of $G$, we can define the Killing form $B : mathfrakg times mathfrakg to mathbbR$ by
$$
forall X, Y in mathfrakg, quad B(X,Y) := -operatornametrace(operatornamead(X)operatornamead(Y)),
$$
where $operatornamead(X) : mathfrakg to mathfrakg$ is defined by $operatornamead(X)Y := [X,Y]$. The Killing form $B$ is bilinear, symmetric, and invariant under the adjoint action of $G$ on $mathfrakg$, so by the identification of $mathfrakg$ with left-invariant vector fields on $G$, it induces a bi-invariant possibly degenerate semi-Riemannian metric on $G$ with the same signature as $B$. If $G$ is semi-simple, then $B$ is actually non-degenerate, and hence induces a completely canonical bi-invariant pseudo-Riemannian metric; moreover, the Killing form $B$ is positive-definite (and hence yields a bi-invariant Riemannian metric) if and only if $G$ is compact. So, for instance, the compact semi-simple Lie group
$$
SU(2) := A in M_2(mathbbC) mid A^ast A = I_2, det(A) = 1
$$
admits a canonical bi-invariant Riemannian metric induced by its Killing form, but the non-compact simple (and hence, semi-simple) Lie group $SL(2,mathbbR)$, which admits no bi-invariant Riemannian metric, nonetheless admits a canonical pseudo-Riemannian metric of signature $(+,-,-)$ induced by its Killing form.
answered Jul 31 at 8:21
Branimir Ćaćić
9,66421846
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
A Lie group $G$ acts on itself both by left translation (group multiplication on the left) and by right translation (group multiplication on the right). Hence, it's natural to look for Riemannian metrics that are bi-invariant, i.e., that make both left translation and right translation by a group element into isometries. However, there's a folkloric result that $G$ admits a bi-invariant Riemannian metric if and only if $G cong H times K$ for $H$ compact and $K$ abelian, which means, for instance, that the non-compact simple Lie group
$$
SL(2,mathbbR) := A in M_2(mathbbR) mid det(A) = 1
$$
does not admit any bi-invariant Riemannian metrics. However, as we'll see below, it does admit a canonical bi-invariant pseudo-Riemannian metric of signature $(+,-,-)$.
Now, in general, if $mathfrakg$ is the Lie algebra of $G$, we can define the Killing form $B : mathfrakg times mathfrakg to mathbbR$ by
$$
forall X, Y in mathfrakg, quad B(X,Y) := -operatornametrace(operatornamead(X)operatornamead(Y)),
$$
where $operatornamead(X) : mathfrakg to mathfrakg$ is defined by $operatornamead(X)Y := [X,Y]$. The Killing form $B$ is bilinear, symmetric, and invariant under the adjoint action of $G$ on $mathfrakg$, so by the identification of $mathfrakg$ with left-invariant vector fields on $G$, it induces a bi-invariant possibly degenerate semi-Riemannian metric on $G$ with the same signature as $B$. If $G$ is semi-simple, then $B$ is actually non-degenerate, and hence induces a completely canonical bi-invariant pseudo-Riemannian metric; moreover, the Killing form $B$ is positive-definite (and hence yields a bi-invariant Riemannian metric) if and only if $G$ is compact. So, for instance, the compact semi-simple Lie group
$$
SU(2) := A in M_2(mathbbC) mid A^ast A = I_2, det(A) = 1
$$
admits a canonical bi-invariant Riemannian metric induced by its Killing form, but the non-compact simple (and hence, semi-simple) Lie group $SL(2,mathbbR)$, which admits no bi-invariant Riemannian metric, nonetheless admits a canonical pseudo-Riemannian metric of signature $(+,-,-)$ induced by its Killing form.
answered Jul 31 at 8:21
Branimir Ćaćić
9,66421846
add a comment |Â
up vote
1
down vote
A Lie group $G$ acts on itself both by left translation (group multiplication on the left) and by right translation (group multiplication on the right). Hence, it's natural to look for Riemannian metrics that are bi-invariant, i.e., that make both left translation and right translation by a group element into isometries. However, there's a folkloric result that $G$ admits a bi-invariant Riemannian metric if and only if $G cong H times K$ for $H$ compact and $K$ abelian, which means, for instance, that the non-compact simple Lie group
$$
SL(2,mathbbR) := A in M_2(mathbbR) mid det(A) = 1
$$
does not admit any bi-invariant Riemannian metrics. However, as we'll see below, it does admit a canonical bi-invariant pseudo-Riemannian metric of signature $(+,-,-)$.
Now, in general, if $mathfrakg$ is the Lie algebra of $G$, we can define the Killing form $B : mathfrakg times mathfrakg to mathbbR$ by
$$
forall X, Y in mathfrakg, quad B(X,Y) := -operatornametrace(operatornamead(X)operatornamead(Y)),
$$
where $operatornamead(X) : mathfrakg to mathfrakg$ is defined by $operatornamead(X)Y := [X,Y]$. The Killing form $B$ is bilinear, symmetric, and invariant under the adjoint action of $G$ on $mathfrakg$, so by the identification of $mathfrakg$ with left-invariant vector fields on $G$, it induces a bi-invariant possibly degenerate semi-Riemannian metric on $G$ with the same signature as $B$. If $G$ is semi-simple, then $B$ is actually non-degenerate, and hence induces a completely canonical bi-invariant pseudo-Riemannian metric; moreover, the Killing form $B$ is positive-definite (and hence yields a bi-invariant Riemannian metric) if and only if $G$ is compact. So, for instance, the compact semi-simple Lie group
$$
SU(2) := A in M_2(mathbbC) mid A^ast A = I_2, det(A) = 1
$$
admits a canonical bi-invariant Riemannian metric induced by its Killing form, but the non-compact simple (and hence, semi-simple) Lie group $SL(2,mathbbR)$, which admits no bi-invariant Riemannian metric, nonetheless admits a canonical pseudo-Riemannian metric of signature $(+,-,-)$ induced by its Killing form.
answered Jul 31 at 8:21
Branimir Ćaćić
9,66421846
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A Lie group $G$ acts on itself both by left translation (group multiplication on the left) and by right translation (group multiplication on the right). Hence, it's natural to look for Riemannian metrics that are bi-invariant, i.e., that make both left translation and right translation by a group element into isometries. However, there's a folkloric result that $G$ admits a bi-invariant Riemannian metric if and only if $G cong H times K$ for $H$ compact and $K$ abelian, which means, for instance, that the non-compact simple Lie group
$$
SL(2,mathbbR) := A in M_2(mathbbR) mid det(A) = 1
$$
does not admit any bi-invariant Riemannian metrics. However, as we'll see below, it does admit a canonical bi-invariant pseudo-Riemannian metric of signature $(+,-,-)$.
Now, in general, if $mathfrakg$ is the Lie algebra of $G$, we can define the Killing form $B : mathfrakg times mathfrakg to mathbbR$ by
$$
forall X, Y in mathfrakg, quad B(X,Y) := -operatornametrace(operatornamead(X)operatornamead(Y)),
$$
where $operatornamead(X) : mathfrakg to mathfrakg$ is defined by $operatornamead(X)Y := [X,Y]$. The Killing form $B$ is bilinear, symmetric, and invariant under the adjoint action of $G$ on $mathfrakg$, so by the identification of $mathfrakg$ with left-invariant vector fields on $G$, it induces a bi-invariant possibly degenerate semi-Riemannian metric on $G$ with the same signature as $B$. If $G$ is semi-simple, then $B$ is actually non-degenerate, and hence induces a completely canonical bi-invariant pseudo-Riemannian metric; moreover, the Killing form $B$ is positive-definite (and hence yields a bi-invariant Riemannian metric) if and only if $G$ is compact. So, for instance, the compact semi-simple Lie group
$$
SU(2) := A in M_2(mathbbC) mid A^ast A = I_2, det(A) = 1
$$
admits a canonical bi-invariant Riemannian metric induced by its Killing form, but the non-compact simple (and hence, semi-simple) Lie group $SL(2,mathbbR)$, which admits no bi-invariant Riemannian metric, nonetheless admits a canonical pseudo-Riemannian metric of signature $(+,-,-)$ induced by its Killing form.
answered Jul 31 at 8:21
Branimir Ćaćić
9,66421846
A Lie group $G$ acts on itself both by left translation (group multiplication on the left) and by right translation (group multiplication on the right). Hence, it's natural to look for Riemannian metrics that are bi-invariant, i.e., that make both left translation and right translation by a group element into isometries. However, there's a folkloric result that $G$ admits a bi-invariant Riemannian metric if and only if $G cong H times K$ for $H$ compact and $K$ abelian, which means, for instance, that the non-compact simple Lie group
$$
SL(2,mathbbR) := A in M_2(mathbbR) mid det(A) = 1
$$
does not admit any bi-invariant Riemannian metrics. However, as we'll see below, it does admit a canonical bi-invariant pseudo-Riemannian metric of signature $(+,-,-)$.
Now, in general, if $mathfrakg$ is the Lie algebra of $G$, we can define the Killing form $B : mathfrakg times mathfrakg to mathbbR$ by
$$
forall X, Y in mathfrakg, quad B(X,Y) := -operatornametrace(operatornamead(X)operatornamead(Y)),
$$
where $operatornamead(X) : mathfrakg to mathfrakg$ is defined by $operatornamead(X)Y := [X,Y]$. The Killing form $B$ is bilinear, symmetric, and invariant under the adjoint action of $G$ on $mathfrakg$, so by the identification of $mathfrakg$ with left-invariant vector fields on $G$, it induces a bi-invariant possibly degenerate semi-Riemannian metric on $G$ with the same signature as $B$. If $G$ is semi-simple, then $B$ is actually non-degenerate, and hence induces a completely canonical bi-invariant pseudo-Riemannian metric; moreover, the Killing form $B$ is positive-definite (and hence yields a bi-invariant Riemannian metric) if and only if $G$ is compact. So, for instance, the compact semi-simple Lie group
$$
SU(2) := A in M_2(mathbbC) mid A^ast A = I_2, det(A) = 1
$$
admits a canonical bi-invariant Riemannian metric induced by its Killing form, but the non-compact simple (and hence, semi-simple) Lie group $SL(2,mathbbR)$, which admits no bi-invariant Riemannian metric, nonetheless admits a canonical pseudo-Riemannian metric of signature $(+,-,-)$ induced by its Killing form.
answered Jul 31 at 8:21
Branimir Ćaćić
9,66421846
edited Jul 31 at 9:42
answered Jul 31 at 8:21
Branimir Ćaćić
9,66421846
answered Jul 31 at 8:21
Branimir Ćaćić
9,66421846
answered Jul 31 at 8:21
Branimir Ćaćić
9,66421846
9,66421846
add a comment |Â
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%2f2867682%2fare-there-interesting-example-of-pseudo-riemannian-manifold-other-than-spacetime%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
It isn't quite what you're asking, but one of the slickest constructions of $n$-dimensional hyperbolic space $mathbbH^n$ (i.e., the hyperboloid model) is as a space-like hypersurface in $(n+1)$-dimensional Minkowski space.
– Branimir Ćaćić
Jul 31 at 8:06