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Index (or signature) of a pseudo-riemannian metric on manifold
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Let $(M , g)$ a $n$-dimensional pseudo-riemannian manifold. As $g(p)$ is a bilinear mapping from $T_pM times T_pM$ to $mathbb{R}$, we can get a basis
$$
{left{{left(frac{partial}{partial x_i}right)}_pright}}_{i = 1}^n
$$
on $T_pM$ such that
$$
g(p)left({left(frac{partial}{partial x_i}right)}_p , {left(frac{partial}{partial x_j}right)}_pright) = 0
$$
for each $i , j in {1 , ldots , n}$ such that $i neq j$. It allows us to define the index (or signature) of $g(p)$ because it is independent of the choice of basis in $T_pM$, a fact known classically as Sylvester's law of inertia.
Well, many texts talk about index of $g$ strightly, and define for instance riemannian metrics as pseudo-riemannian metrics which index coincide with the dimension of the manifold (or signature ($n , 0$) in this case).
How can I show that it is independent of the point $p$ in $M$? As it can be observed, I have not used neither $g(p)$ is non-degenerate or symetric nor the fact $T_pM cong {mathbb{R}}^n cong T_qM$, as isomorphism of vector spaces ($p , q in M$), for defining signature of $g(p)$ (for $p in M$ fixed). Must I use one of these statements to show that the index of $g(p)$ and $g(q)$ coincide for each $p , q in M$?
semi-riemannian-geometry
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joseabp91
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0
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favorite
Let $(M , g)$ a $n$-dimensional pseudo-riemannian manifold. As $g(p)$ is a bilinear mapping from $T_pM times T_pM$ to $mathbb{R}$, we can get a basis
$$
{left{{left(frac{partial}{partial x_i}right)}_pright}}_{i = 1}^n
$$
on $T_pM$ such that
$$
g(p)left({left(frac{partial}{partial x_i}right)}_p , {left(frac{partial}{partial x_j}right)}_pright) = 0
$$
for each $i , j in {1 , ldots , n}$ such that $i neq j$. It allows us to define the index (or signature) of $g(p)$ because it is independent of the choice of basis in $T_pM$, a fact known classically as Sylvester's law of inertia.
Well, many texts talk about index of $g$ strightly, and define for instance riemannian metrics as pseudo-riemannian metrics which index coincide with the dimension of the manifold (or signature ($n , 0$) in this case).
How can I show that it is independent of the point $p$ in $M$? As it can be observed, I have not used neither $g(p)$ is non-degenerate or symetric nor the fact $T_pM cong {mathbb{R}}^n cong T_qM$, as isomorphism of vector spaces ($p , q in M$), for defining signature of $g(p)$ (for $p in M$ fixed). Must I use one of these statements to show that the index of $g(p)$ and $g(q)$ coincide for each $p , q in M$?
semi-riemannian-geometry
asked yesterday
joseabp91
1,233411
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(M , g)$ a $n$-dimensional pseudo-riemannian manifold. As $g(p)$ is a bilinear mapping from $T_pM times T_pM$ to $mathbb{R}$, we can get a basis
$$
{left{{left(frac{partial}{partial x_i}right)}_pright}}_{i = 1}^n
$$
on $T_pM$ such that
$$
g(p)left({left(frac{partial}{partial x_i}right)}_p , {left(frac{partial}{partial x_j}right)}_pright) = 0
$$
for each $i , j in {1 , ldots , n}$ such that $i neq j$. It allows us to define the index (or signature) of $g(p)$ because it is independent of the choice of basis in $T_pM$, a fact known classically as Sylvester's law of inertia.
Well, many texts talk about index of $g$ strightly, and define for instance riemannian metrics as pseudo-riemannian metrics which index coincide with the dimension of the manifold (or signature ($n , 0$) in this case).
How can I show that it is independent of the point $p$ in $M$? As it can be observed, I have not used neither $g(p)$ is non-degenerate or symetric nor the fact $T_pM cong {mathbb{R}}^n cong T_qM$, as isomorphism of vector spaces ($p , q in M$), for defining signature of $g(p)$ (for $p in M$ fixed). Must I use one of these statements to show that the index of $g(p)$ and $g(q)$ coincide for each $p , q in M$?
semi-riemannian-geometry
asked yesterday
joseabp91
1,233411
Let $(M , g)$ a $n$-dimensional pseudo-riemannian manifold. As $g(p)$ is a bilinear mapping from $T_pM times T_pM$ to $mathbb{R}$, we can get a basis
$$
{left{{left(frac{partial}{partial x_i}right)}_pright}}_{i = 1}^n
$$
on $T_pM$ such that
$$
g(p)left({left(frac{partial}{partial x_i}right)}_p , {left(frac{partial}{partial x_j}right)}_pright) = 0
$$
for each $i , j in {1 , ldots , n}$ such that $i neq j$. It allows us to define the index (or signature) of $g(p)$ because it is independent of the choice of basis in $T_pM$, a fact known classically as Sylvester's law of inertia.
Well, many texts talk about index of $g$ strightly, and define for instance riemannian metrics as pseudo-riemannian metrics which index coincide with the dimension of the manifold (or signature ($n , 0$) in this case).
How can I show that it is independent of the point $p$ in $M$? As it can be observed, I have not used neither $g(p)$ is non-degenerate or symetric nor the fact $T_pM cong {mathbb{R}}^n cong T_qM$, as isomorphism of vector spaces ($p , q in M$), for defining signature of $g(p)$ (for $p in M$ fixed). Must I use one of these statements to show that the index of $g(p)$ and $g(q)$ coincide for each $p , q in M$?
semi-riemannian-geometry
semi-riemannian-geometry
asked yesterday
joseabp91
1,233411
asked yesterday
joseabp91
1,233411
asked yesterday
joseabp91
1,233411
asked yesterday
joseabp91
1,233411
asked yesterday
joseabp91
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1 Answer
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Let $v_1, dots, v_n$ be a basis for $T_pM$ such that $g(p)(v_i, v_i) > 0$ for $i = 1, dots, k$, and $g(p)(v_i, v_i) < 0$ for $i = k + 1, dots, n$. We can extend this basis to a basis of local smooth vector fields $V_1, dots, V_n$ for $TM|_U$ where $U$ is an open subset of $p$.
Now consider the functions $f_i : U to mathbb{R}$ given by $f_i(q) = g(q)(V_i|_q, V_i|_q)$. Note that $f_i$ is smooth and $f_i(p) = g(p)(V_i|_p, V_i|_p) = g(p)(v_i, v_i)$ which has a sign. For $i = 1, dots, k$, let $U_i = f^{-1}((0, infty))$ and for $i = k + 1, dots, n$ let $U_i = f^{-1}((-infty, 0))$. Then on $U' = U_1capdotscap U_kcap U_{k+1}capdotscap U_n$ each $f_i$ has a sign. For any $q in U'$, $f_i(q) > 0$ for $i = 1, dots, k$ and $f_i(q) < 0$ for $i = k + 1, dots, n$, so $g(q)$ has the same index of $g(p)$. That is, the index of $g$ is a locally constant integer-valued function and is therefore constant on each connected component of $M$.
answered yesterday
Michael Albanese
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1 Answer
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1 Answer
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up vote
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Let $v_1, dots, v_n$ be a basis for $T_pM$ such that $g(p)(v_i, v_i) > 0$ for $i = 1, dots, k$, and $g(p)(v_i, v_i) < 0$ for $i = k + 1, dots, n$. We can extend this basis to a basis of local smooth vector fields $V_1, dots, V_n$ for $TM|_U$ where $U$ is an open subset of $p$.
Now consider the functions $f_i : U to mathbb{R}$ given by $f_i(q) = g(q)(V_i|_q, V_i|_q)$. Note that $f_i$ is smooth and $f_i(p) = g(p)(V_i|_p, V_i|_p) = g(p)(v_i, v_i)$ which has a sign. For $i = 1, dots, k$, let $U_i = f^{-1}((0, infty))$ and for $i = k + 1, dots, n$ let $U_i = f^{-1}((-infty, 0))$. Then on $U' = U_1capdotscap U_kcap U_{k+1}capdotscap U_n$ each $f_i$ has a sign. For any $q in U'$, $f_i(q) > 0$ for $i = 1, dots, k$ and $f_i(q) < 0$ for $i = k + 1, dots, n$, so $g(q)$ has the same index of $g(p)$. That is, the index of $g$ is a locally constant integer-valued function and is therefore constant on each connected component of $M$.
answered yesterday
Michael Albanese
62.5k1598299
add a comment |
up vote
0
down vote
Let $v_1, dots, v_n$ be a basis for $T_pM$ such that $g(p)(v_i, v_i) > 0$ for $i = 1, dots, k$, and $g(p)(v_i, v_i) < 0$ for $i = k + 1, dots, n$. We can extend this basis to a basis of local smooth vector fields $V_1, dots, V_n$ for $TM|_U$ where $U$ is an open subset of $p$.
Now consider the functions $f_i : U to mathbb{R}$ given by $f_i(q) = g(q)(V_i|_q, V_i|_q)$. Note that $f_i$ is smooth and $f_i(p) = g(p)(V_i|_p, V_i|_p) = g(p)(v_i, v_i)$ which has a sign. For $i = 1, dots, k$, let $U_i = f^{-1}((0, infty))$ and for $i = k + 1, dots, n$ let $U_i = f^{-1}((-infty, 0))$. Then on $U' = U_1capdotscap U_kcap U_{k+1}capdotscap U_n$ each $f_i$ has a sign. For any $q in U'$, $f_i(q) > 0$ for $i = 1, dots, k$ and $f_i(q) < 0$ for $i = k + 1, dots, n$, so $g(q)$ has the same index of $g(p)$. That is, the index of $g$ is a locally constant integer-valued function and is therefore constant on each connected component of $M$.
answered yesterday
Michael Albanese
62.5k1598299
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $v_1, dots, v_n$ be a basis for $T_pM$ such that $g(p)(v_i, v_i) > 0$ for $i = 1, dots, k$, and $g(p)(v_i, v_i) < 0$ for $i = k + 1, dots, n$. We can extend this basis to a basis of local smooth vector fields $V_1, dots, V_n$ for $TM|_U$ where $U$ is an open subset of $p$.
Now consider the functions $f_i : U to mathbb{R}$ given by $f_i(q) = g(q)(V_i|_q, V_i|_q)$. Note that $f_i$ is smooth and $f_i(p) = g(p)(V_i|_p, V_i|_p) = g(p)(v_i, v_i)$ which has a sign. For $i = 1, dots, k$, let $U_i = f^{-1}((0, infty))$ and for $i = k + 1, dots, n$ let $U_i = f^{-1}((-infty, 0))$. Then on $U' = U_1capdotscap U_kcap U_{k+1}capdotscap U_n$ each $f_i$ has a sign. For any $q in U'$, $f_i(q) > 0$ for $i = 1, dots, k$ and $f_i(q) < 0$ for $i = k + 1, dots, n$, so $g(q)$ has the same index of $g(p)$. That is, the index of $g$ is a locally constant integer-valued function and is therefore constant on each connected component of $M$.
answered yesterday
Michael Albanese
62.5k1598299
Let $v_1, dots, v_n$ be a basis for $T_pM$ such that $g(p)(v_i, v_i) > 0$ for $i = 1, dots, k$, and $g(p)(v_i, v_i) < 0$ for $i = k + 1, dots, n$. We can extend this basis to a basis of local smooth vector fields $V_1, dots, V_n$ for $TM|_U$ where $U$ is an open subset of $p$.
Now consider the functions $f_i : U to mathbb{R}$ given by $f_i(q) = g(q)(V_i|_q, V_i|_q)$. Note that $f_i$ is smooth and $f_i(p) = g(p)(V_i|_p, V_i|_p) = g(p)(v_i, v_i)$ which has a sign. For $i = 1, dots, k$, let $U_i = f^{-1}((0, infty))$ and for $i = k + 1, dots, n$ let $U_i = f^{-1}((-infty, 0))$. Then on $U' = U_1capdotscap U_kcap U_{k+1}capdotscap U_n$ each $f_i$ has a sign. For any $q in U'$, $f_i(q) > 0$ for $i = 1, dots, k$ and $f_i(q) < 0$ for $i = k + 1, dots, n$, so $g(q)$ has the same index of $g(p)$. That is, the index of $g$ is a locally constant integer-valued function and is therefore constant on each connected component of $M$.
answered yesterday
Michael Albanese
62.5k1598299
answered yesterday
Michael Albanese
62.5k1598299
answered yesterday
Michael Albanese
62.5k1598299
answered yesterday
Michael Albanese
62.5k1598299
62.5k1598299
add a comment |
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