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Existence of Orthonormal Basis of a Metric in a Manifold
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Definition: A metric $ g$ on a manifold $ M$ is a tensor field of type $ (0,2)$ such that
(1) it is symmetric, i.e. $ g(v,w)=g(w,v)$ for any $ w,v in V_p, pin M$, and
(2) it is non-degenerate, i.e. if $ v_1 in V_p$ such that $g(v_1,v)=0$ for any $ v in V_p$, then $ v=0$.
($V_p$ denotes the tangent space at $p$)
Theorem: Given a metric $ g$, there exists some orthonormal basis $v_1,...,v_n$ for $ T_p$ for each $ p in M$, i.e. $ g(v_mu,v_nu)=pm delta_mu^nu$. Moreover, if $ S,S'$ are two orthonormal basis, then
$$#s in S: g(s,s)=1=#s' in S': g(s',s')=1$$
and
$$#s in S: g(s,s)=-1=#s' in S': g(s',s')=-1$$
For the first part, I am wondering there is anything like the Gram-Schmidt process which can be applied to this case, but since the metric may not be positive definite, I can't see how I can do so.
For the second part, I do not have any idea how to approach it.
Thank you for providing me suggestions.
differential-geometry smooth-manifolds general-relativity semi-riemannian-geometry gram-schmidt
edited 9 hours ago
Sou
2,6142720
asked 18 hours ago
Jerry
364111
add a comment |Â
up vote
1
down vote
favorite
Definition: A metric $ g$ on a manifold $ M$ is a tensor field of type $ (0,2)$ such that
(1) it is symmetric, i.e. $ g(v,w)=g(w,v)$ for any $ w,v in V_p, pin M$, and
(2) it is non-degenerate, i.e. if $ v_1 in V_p$ such that $g(v_1,v)=0$ for any $ v in V_p$, then $ v=0$.
($V_p$ denotes the tangent space at $p$)
Theorem: Given a metric $ g$, there exists some orthonormal basis $v_1,...,v_n$ for $ T_p$ for each $ p in M$, i.e. $ g(v_mu,v_nu)=pm delta_mu^nu$. Moreover, if $ S,S'$ are two orthonormal basis, then
$$#s in S: g(s,s)=1=#s' in S': g(s',s')=1$$
and
$$#s in S: g(s,s)=-1=#s' in S': g(s',s')=-1$$
For the first part, I am wondering there is anything like the Gram-Schmidt process which can be applied to this case, but since the metric may not be positive definite, I can't see how I can do so.
For the second part, I do not have any idea how to approach it.
Thank you for providing me suggestions.
differential-geometry smooth-manifolds general-relativity semi-riemannian-geometry gram-schmidt
edited 9 hours ago
Sou
2,6142720
asked 18 hours ago
Jerry
364111
Sylvester's theorem
– Tom Chalmer
8 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Definition: A metric $ g$ on a manifold $ M$ is a tensor field of type $ (0,2)$ such that
(1) it is symmetric, i.e. $ g(v,w)=g(w,v)$ for any $ w,v in V_p, pin M$, and
(2) it is non-degenerate, i.e. if $ v_1 in V_p$ such that $g(v_1,v)=0$ for any $ v in V_p$, then $ v=0$.
($V_p$ denotes the tangent space at $p$)
Theorem: Given a metric $ g$, there exists some orthonormal basis $v_1,...,v_n$ for $ T_p$ for each $ p in M$, i.e. $ g(v_mu,v_nu)=pm delta_mu^nu$. Moreover, if $ S,S'$ are two orthonormal basis, then
$$#s in S: g(s,s)=1=#s' in S': g(s',s')=1$$
and
$$#s in S: g(s,s)=-1=#s' in S': g(s',s')=-1$$
For the first part, I am wondering there is anything like the Gram-Schmidt process which can be applied to this case, but since the metric may not be positive definite, I can't see how I can do so.
For the second part, I do not have any idea how to approach it.
Thank you for providing me suggestions.
differential-geometry smooth-manifolds general-relativity semi-riemannian-geometry gram-schmidt
edited 9 hours ago
Sou
2,6142720
asked 18 hours ago
Jerry
364111
Definition: A metric $ g$ on a manifold $ M$ is a tensor field of type $ (0,2)$ such that
(1) it is symmetric, i.e. $ g(v,w)=g(w,v)$ for any $ w,v in V_p, pin M$, and
(2) it is non-degenerate, i.e. if $ v_1 in V_p$ such that $g(v_1,v)=0$ for any $ v in V_p$, then $ v=0$.
($V_p$ denotes the tangent space at $p$)
Theorem: Given a metric $ g$, there exists some orthonormal basis $v_1,...,v_n$ for $ T_p$ for each $ p in M$, i.e. $ g(v_mu,v_nu)=pm delta_mu^nu$. Moreover, if $ S,S'$ are two orthonormal basis, then
$$#s in S: g(s,s)=1=#s' in S': g(s',s')=1$$
and
$$#s in S: g(s,s)=-1=#s' in S': g(s',s')=-1$$
For the first part, I am wondering there is anything like the Gram-Schmidt process which can be applied to this case, but since the metric may not be positive definite, I can't see how I can do so.
For the second part, I do not have any idea how to approach it.
Thank you for providing me suggestions.
differential-geometry smooth-manifolds general-relativity semi-riemannian-geometry gram-schmidt
edited 9 hours ago
Sou
2,6142720
asked 18 hours ago
Jerry
364111
edited 9 hours ago
Sou
2,6142720
edited 9 hours ago
Sou
2,6142720
edited 9 hours ago
Sou
2,6142720
2,6142720
asked 18 hours ago
Jerry
364111
asked 18 hours ago
Jerry
364111
asked 18 hours ago
Jerry
364111
364111
Sylvester's theorem
– Tom Chalmer
8 hours ago
add a comment |Â
Sylvester's theorem
– Tom Chalmer
8 hours ago
Sylvester's theorem
– Tom Chalmer
8 hours ago
Sylvester's theorem
– Tom Chalmer
8 hours ago
add a comment |Â
2 Answers
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For any symmetric bilinear form (not necessarily nondegenerate) $langle_,_rangle$ on a finite-dimensional vector space $V$ over $mathbbR$, I claim that there exists a basis of $V$, called a good basis, consisting of $$u_1,u_2,ldots,u_p,v_1,v_2,ldots,v_q,w_1,w_2,ldots,w_rin V$$
such that
- $p+q+r=dim_mathbbR(V)$
- $langle u_i,u_jrangle =+delta_i,j$ for $i,j=1,2,ldots,p$,
- $langle v_i,v_jrangle = -delta_i,j$ for $i,j=1,2,ldots,q$,
- $langle u_i,v_jrangle=0$ for $i=1,2,ldots,p$ and $j=1,2,ldots,q$, and
- $langle x,w_krangle=0$ for all $xin V$For each $vin V$.
Here, $delta$ is the Kronecker delta. Thus, in the basis above, the bilinear form is represented by the matrix
$$J_p,q,r:=beginbmatrix +I_ptimes p&0_ptimes q&0_ptimes r\
0_qtimes p&-I_qtimes q&0_qtimes r\
0_rtimes p&0_rtimes q&0_rtimes rendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_alphatimes beta$ is the $alpha$-by-$beta$ zero matrix. For each $xin V$, we write $|x|$ for $sqrtlangle x,xranglebig$, and write $sigma(x)in-1,0,+1$ for the sign of $langle x,xrangle$.
First, let $W$ be the kernel of the bilinear form. That is, $W$ consists of all vectors $zin V$ for which $langle x,zrangle=0$ for all $xin V$. Let $r$ denote the dimension of $W$ over $mathbbR$. We can take $leftw_1,w_2,ldots,w_rright$ to be any basis of $W$. (This also shows that $r$ is independent of the choice of good basis of $V$, as it must be the $mathbbR$-dimension of $W$, a fixed subspace of $V$.) We can from now on assume that $W=0$ (that is, the bilinear form $langle_,_rangle$ is nondegenerate). Otherwise, we study the vector space $V/W$ with the bilinear form $langle!langle_,_rangle!rangle$ defined by
$$langle!langle x+W,y+Wrangle!rangle:=langle x,yrangletext for all x,yin V,.$$
Fix a basis $leftz_1,z_2,ldots,z_nright$ of $V$. Write $[n]:=1,2,ldots,n$. We shall perform the following orthonormalization procedure. First, we look at $|z_1|$. If $|z_1|=0$, then note that, for some index $jin[n]setminus1$, $langle z_1,z_jrangle neq 0$, (as the bilinear form is nondegenerate), and so we can replace $z_1$ by $z_1+tz_j$ where $tinmathbbRsetminus0$ is so chosen that $2,langle z_1,z_jrangle +tlangle z_j,z_jrangle neq 0$. Hence, we may always assume that $|z_1|neq 0$. Dividing $z_1$ by $|z_1|$, we may also assume that $|z_1|=1$. Now, we replace $z_j$ for $j=2,3,ldots,n$ by
$$z_j-langle z_j,z_1rangle z_1,.$$
By doing so, we may assume that each $z_j$ is orthogonal to $z_1$ already.
Let $V_1$ denote the orthogonal complement of $z_1$. That is, $V_1$ consists of all vectors $x$ in $V$ such that $langle x,z_1rangle=0$. Clearly, $V_1$ is an $(n-1)$-dimensional $mathbbR$-subspace of $V$. By the paragraph above, $V_1$ is the $mathbbR$-span of $z_2,z_3,ldots,z_n$, and $V_1$ inherits the bilinear form $langle_,_rangle_1$ from $V$ (by simply restricting $langle _,_rangle$ onto $V_1times V_1$). We can then repeat the paragraph above for $V_1$, noting that $langle_,_rangle_1$ is nondegenerate. Hence, by induction, you may assume that the vectors $z_2,z_3,ldots,z_n$ are orthogonal, and each $|z_j|$ is equal to $1$ for $j=2,3,ldots,n$.
Then, we let $u_1,u_2,ldots,u_p$ to be the vectors $z_j$ with $sigma(z_j)=+1$ ($jin[n]$). Likewise, $v_1,v_2,ldots,v_q$ are the vectors $z_j$ with $sigma(z_j)=-1$ ($jin[n]$). Note that $p$ and $q$ are also independent of the choice of good bases. However, it is easier to show that, if $S$ is an $n$-by-$n$ real symmetric and $A$ is an $n$-by-$n$ invertible real matrix, then $A^top,S,A$ and $S$ have the same number of positive eigenvalues (with multiplicities), and the same number of negative eigenvalues (with multiplicities). The triple $(p,q,r)$ is called the signature of a symmetric bilinear form.
answered 15 hours ago
Batominovski
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Suppose we have a Semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the map $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form. We can find an orthonormal basis for $T_pM$ for any $p in M$ using induction as shown in this lemma (lemma 24 of Barrett O'neill's Semi-Riemannian Geometry, p.50).
For the second question, it's enough to do that in the level of vector space. Suppose $V$ is a $n$-dimensional real vector space endowed with a non-degenerate, symmetric bilinear form $g : V times V to BbbR$. Let $e_1,dots,e_n$ is an arbitrary orthonormal basis for $V$.
Let $k leq n$ be the number of negative values in $g(e_i,e_i)=pm1 : i=1,dots ,n$. The case $k=0$ is trivial. If $k>0$, then $V$ will have a subspaces which $g$ is negative definite, e.g. the span of one of the basis elements. Let $W$ be the subspace of maximal dimension on which $g$ is negative definite. If we can show that $k=textdim W$ then we're done since this number $k$ is independent of basis.
By rearranging the basis $e_1,dots,e_k,e_k+1,dots,e_n$, we have $g(e_i,e_i)=-1$ for $1leq i leq k$ and $g(e_j,e_j)=1$ for $k+1leq j leq n$. Since $g$ negatives definite on $X = textspan (e_1,dots,e_k)$ then $k=textdim X leq textdim W$. To show $k geq textdim W$, define a map $T : W to X$ as follow; for any $w = sum_i=1^n w^i e_i in W$
$$
T(w) = sum_i=1^k w^ie_i.
$$
You can check directly that $T$ is injective and therefore $textdim W = 0 + textdim Im T leq textdim X=k$. Therefore the number $k$ is fixed by $g$, independent of basis.
answered 15 hours ago
Sou
2,6142720
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
14 hours ago
@Jerry Ah. I think this is not gonna work, since $|cdot|$ may be zero on some places on $U$. It is defined as $|cdot| = sqrt$. I will delete this answer.
– Sou
14 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
14 hours ago
I just leave it that way.
– Sou
14 hours ago
@Jerry I have corrected my answer. I think this okay now. I'm sorry i misread your question. If you mean orthonormal basis just for a tangent space, then it's done in lemma 24 of barrett o'neill's (as linked above). My answer is kind of overkill since it's about construction of local orthonormal frame.
– Sou
9 hours ago
add a comment |Â
2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
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up vote
0
down vote
For any symmetric bilinear form (not necessarily nondegenerate) $langle_,_rangle$ on a finite-dimensional vector space $V$ over $mathbbR$, I claim that there exists a basis of $V$, called a good basis, consisting of $$u_1,u_2,ldots,u_p,v_1,v_2,ldots,v_q,w_1,w_2,ldots,w_rin V$$
such that
- $p+q+r=dim_mathbbR(V)$
- $langle u_i,u_jrangle =+delta_i,j$ for $i,j=1,2,ldots,p$,
- $langle v_i,v_jrangle = -delta_i,j$ for $i,j=1,2,ldots,q$,
- $langle u_i,v_jrangle=0$ for $i=1,2,ldots,p$ and $j=1,2,ldots,q$, and
- $langle x,w_krangle=0$ for all $xin V$For each $vin V$.
Here, $delta$ is the Kronecker delta. Thus, in the basis above, the bilinear form is represented by the matrix
$$J_p,q,r:=beginbmatrix +I_ptimes p&0_ptimes q&0_ptimes r\
0_qtimes p&-I_qtimes q&0_qtimes r\
0_rtimes p&0_rtimes q&0_rtimes rendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_alphatimes beta$ is the $alpha$-by-$beta$ zero matrix. For each $xin V$, we write $|x|$ for $sqrtlangle x,xranglebig$, and write $sigma(x)in-1,0,+1$ for the sign of $langle x,xrangle$.
First, let $W$ be the kernel of the bilinear form. That is, $W$ consists of all vectors $zin V$ for which $langle x,zrangle=0$ for all $xin V$. Let $r$ denote the dimension of $W$ over $mathbbR$. We can take $leftw_1,w_2,ldots,w_rright$ to be any basis of $W$. (This also shows that $r$ is independent of the choice of good basis of $V$, as it must be the $mathbbR$-dimension of $W$, a fixed subspace of $V$.) We can from now on assume that $W=0$ (that is, the bilinear form $langle_,_rangle$ is nondegenerate). Otherwise, we study the vector space $V/W$ with the bilinear form $langle!langle_,_rangle!rangle$ defined by
$$langle!langle x+W,y+Wrangle!rangle:=langle x,yrangletext for all x,yin V,.$$
Fix a basis $leftz_1,z_2,ldots,z_nright$ of $V$. Write $[n]:=1,2,ldots,n$. We shall perform the following orthonormalization procedure. First, we look at $|z_1|$. If $|z_1|=0$, then note that, for some index $jin[n]setminus1$, $langle z_1,z_jrangle neq 0$, (as the bilinear form is nondegenerate), and so we can replace $z_1$ by $z_1+tz_j$ where $tinmathbbRsetminus0$ is so chosen that $2,langle z_1,z_jrangle +tlangle z_j,z_jrangle neq 0$. Hence, we may always assume that $|z_1|neq 0$. Dividing $z_1$ by $|z_1|$, we may also assume that $|z_1|=1$. Now, we replace $z_j$ for $j=2,3,ldots,n$ by
$$z_j-langle z_j,z_1rangle z_1,.$$
By doing so, we may assume that each $z_j$ is orthogonal to $z_1$ already.
Let $V_1$ denote the orthogonal complement of $z_1$. That is, $V_1$ consists of all vectors $x$ in $V$ such that $langle x,z_1rangle=0$. Clearly, $V_1$ is an $(n-1)$-dimensional $mathbbR$-subspace of $V$. By the paragraph above, $V_1$ is the $mathbbR$-span of $z_2,z_3,ldots,z_n$, and $V_1$ inherits the bilinear form $langle_,_rangle_1$ from $V$ (by simply restricting $langle _,_rangle$ onto $V_1times V_1$). We can then repeat the paragraph above for $V_1$, noting that $langle_,_rangle_1$ is nondegenerate. Hence, by induction, you may assume that the vectors $z_2,z_3,ldots,z_n$ are orthogonal, and each $|z_j|$ is equal to $1$ for $j=2,3,ldots,n$.
Then, we let $u_1,u_2,ldots,u_p$ to be the vectors $z_j$ with $sigma(z_j)=+1$ ($jin[n]$). Likewise, $v_1,v_2,ldots,v_q$ are the vectors $z_j$ with $sigma(z_j)=-1$ ($jin[n]$). Note that $p$ and $q$ are also independent of the choice of good bases. However, it is easier to show that, if $S$ is an $n$-by-$n$ real symmetric and $A$ is an $n$-by-$n$ invertible real matrix, then $A^top,S,A$ and $S$ have the same number of positive eigenvalues (with multiplicities), and the same number of negative eigenvalues (with multiplicities). The triple $(p,q,r)$ is called the signature of a symmetric bilinear form.
answered 15 hours ago
Batominovski
22.2k22675
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0
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For any symmetric bilinear form (not necessarily nondegenerate) $langle_,_rangle$ on a finite-dimensional vector space $V$ over $mathbbR$, I claim that there exists a basis of $V$, called a good basis, consisting of $$u_1,u_2,ldots,u_p,v_1,v_2,ldots,v_q,w_1,w_2,ldots,w_rin V$$
such that
- $p+q+r=dim_mathbbR(V)$
- $langle u_i,u_jrangle =+delta_i,j$ for $i,j=1,2,ldots,p$,
- $langle v_i,v_jrangle = -delta_i,j$ for $i,j=1,2,ldots,q$,
- $langle u_i,v_jrangle=0$ for $i=1,2,ldots,p$ and $j=1,2,ldots,q$, and
- $langle x,w_krangle=0$ for all $xin V$For each $vin V$.
Here, $delta$ is the Kronecker delta. Thus, in the basis above, the bilinear form is represented by the matrix
$$J_p,q,r:=beginbmatrix +I_ptimes p&0_ptimes q&0_ptimes r\
0_qtimes p&-I_qtimes q&0_qtimes r\
0_rtimes p&0_rtimes q&0_rtimes rendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_alphatimes beta$ is the $alpha$-by-$beta$ zero matrix. For each $xin V$, we write $|x|$ for $sqrtlangle x,xranglebig$, and write $sigma(x)in-1,0,+1$ for the sign of $langle x,xrangle$.
First, let $W$ be the kernel of the bilinear form. That is, $W$ consists of all vectors $zin V$ for which $langle x,zrangle=0$ for all $xin V$. Let $r$ denote the dimension of $W$ over $mathbbR$. We can take $leftw_1,w_2,ldots,w_rright$ to be any basis of $W$. (This also shows that $r$ is independent of the choice of good basis of $V$, as it must be the $mathbbR$-dimension of $W$, a fixed subspace of $V$.) We can from now on assume that $W=0$ (that is, the bilinear form $langle_,_rangle$ is nondegenerate). Otherwise, we study the vector space $V/W$ with the bilinear form $langle!langle_,_rangle!rangle$ defined by
$$langle!langle x+W,y+Wrangle!rangle:=langle x,yrangletext for all x,yin V,.$$
Fix a basis $leftz_1,z_2,ldots,z_nright$ of $V$. Write $[n]:=1,2,ldots,n$. We shall perform the following orthonormalization procedure. First, we look at $|z_1|$. If $|z_1|=0$, then note that, for some index $jin[n]setminus1$, $langle z_1,z_jrangle neq 0$, (as the bilinear form is nondegenerate), and so we can replace $z_1$ by $z_1+tz_j$ where $tinmathbbRsetminus0$ is so chosen that $2,langle z_1,z_jrangle +tlangle z_j,z_jrangle neq 0$. Hence, we may always assume that $|z_1|neq 0$. Dividing $z_1$ by $|z_1|$, we may also assume that $|z_1|=1$. Now, we replace $z_j$ for $j=2,3,ldots,n$ by
$$z_j-langle z_j,z_1rangle z_1,.$$
By doing so, we may assume that each $z_j$ is orthogonal to $z_1$ already.
Let $V_1$ denote the orthogonal complement of $z_1$. That is, $V_1$ consists of all vectors $x$ in $V$ such that $langle x,z_1rangle=0$. Clearly, $V_1$ is an $(n-1)$-dimensional $mathbbR$-subspace of $V$. By the paragraph above, $V_1$ is the $mathbbR$-span of $z_2,z_3,ldots,z_n$, and $V_1$ inherits the bilinear form $langle_,_rangle_1$ from $V$ (by simply restricting $langle _,_rangle$ onto $V_1times V_1$). We can then repeat the paragraph above for $V_1$, noting that $langle_,_rangle_1$ is nondegenerate. Hence, by induction, you may assume that the vectors $z_2,z_3,ldots,z_n$ are orthogonal, and each $|z_j|$ is equal to $1$ for $j=2,3,ldots,n$.
Then, we let $u_1,u_2,ldots,u_p$ to be the vectors $z_j$ with $sigma(z_j)=+1$ ($jin[n]$). Likewise, $v_1,v_2,ldots,v_q$ are the vectors $z_j$ with $sigma(z_j)=-1$ ($jin[n]$). Note that $p$ and $q$ are also independent of the choice of good bases. However, it is easier to show that, if $S$ is an $n$-by-$n$ real symmetric and $A$ is an $n$-by-$n$ invertible real matrix, then $A^top,S,A$ and $S$ have the same number of positive eigenvalues (with multiplicities), and the same number of negative eigenvalues (with multiplicities). The triple $(p,q,r)$ is called the signature of a symmetric bilinear form.
answered 15 hours ago
Batominovski
22.2k22675
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For any symmetric bilinear form (not necessarily nondegenerate) $langle_,_rangle$ on a finite-dimensional vector space $V$ over $mathbbR$, I claim that there exists a basis of $V$, called a good basis, consisting of $$u_1,u_2,ldots,u_p,v_1,v_2,ldots,v_q,w_1,w_2,ldots,w_rin V$$
such that
- $p+q+r=dim_mathbbR(V)$
- $langle u_i,u_jrangle =+delta_i,j$ for $i,j=1,2,ldots,p$,
- $langle v_i,v_jrangle = -delta_i,j$ for $i,j=1,2,ldots,q$,
- $langle u_i,v_jrangle=0$ for $i=1,2,ldots,p$ and $j=1,2,ldots,q$, and
- $langle x,w_krangle=0$ for all $xin V$For each $vin V$.
Here, $delta$ is the Kronecker delta. Thus, in the basis above, the bilinear form is represented by the matrix
$$J_p,q,r:=beginbmatrix +I_ptimes p&0_ptimes q&0_ptimes r\
0_qtimes p&-I_qtimes q&0_qtimes r\
0_rtimes p&0_rtimes q&0_rtimes rendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_alphatimes beta$ is the $alpha$-by-$beta$ zero matrix. For each $xin V$, we write $|x|$ for $sqrtlangle x,xranglebig$, and write $sigma(x)in-1,0,+1$ for the sign of $langle x,xrangle$.
First, let $W$ be the kernel of the bilinear form. That is, $W$ consists of all vectors $zin V$ for which $langle x,zrangle=0$ for all $xin V$. Let $r$ denote the dimension of $W$ over $mathbbR$. We can take $leftw_1,w_2,ldots,w_rright$ to be any basis of $W$. (This also shows that $r$ is independent of the choice of good basis of $V$, as it must be the $mathbbR$-dimension of $W$, a fixed subspace of $V$.) We can from now on assume that $W=0$ (that is, the bilinear form $langle_,_rangle$ is nondegenerate). Otherwise, we study the vector space $V/W$ with the bilinear form $langle!langle_,_rangle!rangle$ defined by
$$langle!langle x+W,y+Wrangle!rangle:=langle x,yrangletext for all x,yin V,.$$
Fix a basis $leftz_1,z_2,ldots,z_nright$ of $V$. Write $[n]:=1,2,ldots,n$. We shall perform the following orthonormalization procedure. First, we look at $|z_1|$. If $|z_1|=0$, then note that, for some index $jin[n]setminus1$, $langle z_1,z_jrangle neq 0$, (as the bilinear form is nondegenerate), and so we can replace $z_1$ by $z_1+tz_j$ where $tinmathbbRsetminus0$ is so chosen that $2,langle z_1,z_jrangle +tlangle z_j,z_jrangle neq 0$. Hence, we may always assume that $|z_1|neq 0$. Dividing $z_1$ by $|z_1|$, we may also assume that $|z_1|=1$. Now, we replace $z_j$ for $j=2,3,ldots,n$ by
$$z_j-langle z_j,z_1rangle z_1,.$$
By doing so, we may assume that each $z_j$ is orthogonal to $z_1$ already.
Let $V_1$ denote the orthogonal complement of $z_1$. That is, $V_1$ consists of all vectors $x$ in $V$ such that $langle x,z_1rangle=0$. Clearly, $V_1$ is an $(n-1)$-dimensional $mathbbR$-subspace of $V$. By the paragraph above, $V_1$ is the $mathbbR$-span of $z_2,z_3,ldots,z_n$, and $V_1$ inherits the bilinear form $langle_,_rangle_1$ from $V$ (by simply restricting $langle _,_rangle$ onto $V_1times V_1$). We can then repeat the paragraph above for $V_1$, noting that $langle_,_rangle_1$ is nondegenerate. Hence, by induction, you may assume that the vectors $z_2,z_3,ldots,z_n$ are orthogonal, and each $|z_j|$ is equal to $1$ for $j=2,3,ldots,n$.
Then, we let $u_1,u_2,ldots,u_p$ to be the vectors $z_j$ with $sigma(z_j)=+1$ ($jin[n]$). Likewise, $v_1,v_2,ldots,v_q$ are the vectors $z_j$ with $sigma(z_j)=-1$ ($jin[n]$). Note that $p$ and $q$ are also independent of the choice of good bases. However, it is easier to show that, if $S$ is an $n$-by-$n$ real symmetric and $A$ is an $n$-by-$n$ invertible real matrix, then $A^top,S,A$ and $S$ have the same number of positive eigenvalues (with multiplicities), and the same number of negative eigenvalues (with multiplicities). The triple $(p,q,r)$ is called the signature of a symmetric bilinear form.
answered 15 hours ago
Batominovski
22.2k22675
For any symmetric bilinear form (not necessarily nondegenerate) $langle_,_rangle$ on a finite-dimensional vector space $V$ over $mathbbR$, I claim that there exists a basis of $V$, called a good basis, consisting of $$u_1,u_2,ldots,u_p,v_1,v_2,ldots,v_q,w_1,w_2,ldots,w_rin V$$
such that
- $p+q+r=dim_mathbbR(V)$
- $langle u_i,u_jrangle =+delta_i,j$ for $i,j=1,2,ldots,p$,
- $langle v_i,v_jrangle = -delta_i,j$ for $i,j=1,2,ldots,q$,
- $langle u_i,v_jrangle=0$ for $i=1,2,ldots,p$ and $j=1,2,ldots,q$, and
- $langle x,w_krangle=0$ for all $xin V$For each $vin V$.
Here, $delta$ is the Kronecker delta. Thus, in the basis above, the bilinear form is represented by the matrix
$$J_p,q,r:=beginbmatrix +I_ptimes p&0_ptimes q&0_ptimes r\
0_qtimes p&-I_qtimes q&0_qtimes r\
0_rtimes p&0_rtimes q&0_rtimes rendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_alphatimes beta$ is the $alpha$-by-$beta$ zero matrix. For each $xin V$, we write $|x|$ for $sqrtlangle x,xranglebig$, and write $sigma(x)in-1,0,+1$ for the sign of $langle x,xrangle$.
First, let $W$ be the kernel of the bilinear form. That is, $W$ consists of all vectors $zin V$ for which $langle x,zrangle=0$ for all $xin V$. Let $r$ denote the dimension of $W$ over $mathbbR$. We can take $leftw_1,w_2,ldots,w_rright$ to be any basis of $W$. (This also shows that $r$ is independent of the choice of good basis of $V$, as it must be the $mathbbR$-dimension of $W$, a fixed subspace of $V$.) We can from now on assume that $W=0$ (that is, the bilinear form $langle_,_rangle$ is nondegenerate). Otherwise, we study the vector space $V/W$ with the bilinear form $langle!langle_,_rangle!rangle$ defined by
$$langle!langle x+W,y+Wrangle!rangle:=langle x,yrangletext for all x,yin V,.$$
Fix a basis $leftz_1,z_2,ldots,z_nright$ of $V$. Write $[n]:=1,2,ldots,n$. We shall perform the following orthonormalization procedure. First, we look at $|z_1|$. If $|z_1|=0$, then note that, for some index $jin[n]setminus1$, $langle z_1,z_jrangle neq 0$, (as the bilinear form is nondegenerate), and so we can replace $z_1$ by $z_1+tz_j$ where $tinmathbbRsetminus0$ is so chosen that $2,langle z_1,z_jrangle +tlangle z_j,z_jrangle neq 0$. Hence, we may always assume that $|z_1|neq 0$. Dividing $z_1$ by $|z_1|$, we may also assume that $|z_1|=1$. Now, we replace $z_j$ for $j=2,3,ldots,n$ by
$$z_j-langle z_j,z_1rangle z_1,.$$
By doing so, we may assume that each $z_j$ is orthogonal to $z_1$ already.
Let $V_1$ denote the orthogonal complement of $z_1$. That is, $V_1$ consists of all vectors $x$ in $V$ such that $langle x,z_1rangle=0$. Clearly, $V_1$ is an $(n-1)$-dimensional $mathbbR$-subspace of $V$. By the paragraph above, $V_1$ is the $mathbbR$-span of $z_2,z_3,ldots,z_n$, and $V_1$ inherits the bilinear form $langle_,_rangle_1$ from $V$ (by simply restricting $langle _,_rangle$ onto $V_1times V_1$). We can then repeat the paragraph above for $V_1$, noting that $langle_,_rangle_1$ is nondegenerate. Hence, by induction, you may assume that the vectors $z_2,z_3,ldots,z_n$ are orthogonal, and each $|z_j|$ is equal to $1$ for $j=2,3,ldots,n$.
Then, we let $u_1,u_2,ldots,u_p$ to be the vectors $z_j$ with $sigma(z_j)=+1$ ($jin[n]$). Likewise, $v_1,v_2,ldots,v_q$ are the vectors $z_j$ with $sigma(z_j)=-1$ ($jin[n]$). Note that $p$ and $q$ are also independent of the choice of good bases. However, it is easier to show that, if $S$ is an $n$-by-$n$ real symmetric and $A$ is an $n$-by-$n$ invertible real matrix, then $A^top,S,A$ and $S$ have the same number of positive eigenvalues (with multiplicities), and the same number of negative eigenvalues (with multiplicities). The triple $(p,q,r)$ is called the signature of a symmetric bilinear form.
answered 15 hours ago
Batominovski
22.2k22675
edited 15 hours ago
answered 15 hours ago
Batominovski
22.2k22675
answered 15 hours ago
Batominovski
22.2k22675
answered 15 hours ago
Batominovski
22.2k22675
22.2k22675
add a comment |Â
add a comment |Â
up vote
0
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Suppose we have a Semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the map $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form. We can find an orthonormal basis for $T_pM$ for any $p in M$ using induction as shown in this lemma (lemma 24 of Barrett O'neill's Semi-Riemannian Geometry, p.50).
For the second question, it's enough to do that in the level of vector space. Suppose $V$ is a $n$-dimensional real vector space endowed with a non-degenerate, symmetric bilinear form $g : V times V to BbbR$. Let $e_1,dots,e_n$ is an arbitrary orthonormal basis for $V$.
Let $k leq n$ be the number of negative values in $g(e_i,e_i)=pm1 : i=1,dots ,n$. The case $k=0$ is trivial. If $k>0$, then $V$ will have a subspaces which $g$ is negative definite, e.g. the span of one of the basis elements. Let $W$ be the subspace of maximal dimension on which $g$ is negative definite. If we can show that $k=textdim W$ then we're done since this number $k$ is independent of basis.
By rearranging the basis $e_1,dots,e_k,e_k+1,dots,e_n$, we have $g(e_i,e_i)=-1$ for $1leq i leq k$ and $g(e_j,e_j)=1$ for $k+1leq j leq n$. Since $g$ negatives definite on $X = textspan (e_1,dots,e_k)$ then $k=textdim X leq textdim W$. To show $k geq textdim W$, define a map $T : W to X$ as follow; for any $w = sum_i=1^n w^i e_i in W$
$$
T(w) = sum_i=1^k w^ie_i.
$$
You can check directly that $T$ is injective and therefore $textdim W = 0 + textdim Im T leq textdim X=k$. Therefore the number $k$ is fixed by $g$, independent of basis.
answered 15 hours ago
Sou
2,6142720
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
14 hours ago
@Jerry Ah. I think this is not gonna work, since $|cdot|$ may be zero on some places on $U$. It is defined as $|cdot| = sqrt$. I will delete this answer.
– Sou
14 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
14 hours ago
I just leave it that way.
– Sou
14 hours ago
@Jerry I have corrected my answer. I think this okay now. I'm sorry i misread your question. If you mean orthonormal basis just for a tangent space, then it's done in lemma 24 of barrett o'neill's (as linked above). My answer is kind of overkill since it's about construction of local orthonormal frame.
– Sou
9 hours ago
add a comment |Â
up vote
0
down vote
Suppose we have a Semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the map $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form. We can find an orthonormal basis for $T_pM$ for any $p in M$ using induction as shown in this lemma (lemma 24 of Barrett O'neill's Semi-Riemannian Geometry, p.50).
For the second question, it's enough to do that in the level of vector space. Suppose $V$ is a $n$-dimensional real vector space endowed with a non-degenerate, symmetric bilinear form $g : V times V to BbbR$. Let $e_1,dots,e_n$ is an arbitrary orthonormal basis for $V$.
Let $k leq n$ be the number of negative values in $g(e_i,e_i)=pm1 : i=1,dots ,n$. The case $k=0$ is trivial. If $k>0$, then $V$ will have a subspaces which $g$ is negative definite, e.g. the span of one of the basis elements. Let $W$ be the subspace of maximal dimension on which $g$ is negative definite. If we can show that $k=textdim W$ then we're done since this number $k$ is independent of basis.
By rearranging the basis $e_1,dots,e_k,e_k+1,dots,e_n$, we have $g(e_i,e_i)=-1$ for $1leq i leq k$ and $g(e_j,e_j)=1$ for $k+1leq j leq n$. Since $g$ negatives definite on $X = textspan (e_1,dots,e_k)$ then $k=textdim X leq textdim W$. To show $k geq textdim W$, define a map $T : W to X$ as follow; for any $w = sum_i=1^n w^i e_i in W$
$$
T(w) = sum_i=1^k w^ie_i.
$$
You can check directly that $T$ is injective and therefore $textdim W = 0 + textdim Im T leq textdim X=k$. Therefore the number $k$ is fixed by $g$, independent of basis.
answered 15 hours ago
Sou
2,6142720
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
14 hours ago
@Jerry Ah. I think this is not gonna work, since $|cdot|$ may be zero on some places on $U$. It is defined as $|cdot| = sqrt$. I will delete this answer.
– Sou
14 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
14 hours ago
I just leave it that way.
– Sou
14 hours ago
@Jerry I have corrected my answer. I think this okay now. I'm sorry i misread your question. If you mean orthonormal basis just for a tangent space, then it's done in lemma 24 of barrett o'neill's (as linked above). My answer is kind of overkill since it's about construction of local orthonormal frame.
– Sou
9 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Suppose we have a Semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the map $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form. We can find an orthonormal basis for $T_pM$ for any $p in M$ using induction as shown in this lemma (lemma 24 of Barrett O'neill's Semi-Riemannian Geometry, p.50).
For the second question, it's enough to do that in the level of vector space. Suppose $V$ is a $n$-dimensional real vector space endowed with a non-degenerate, symmetric bilinear form $g : V times V to BbbR$. Let $e_1,dots,e_n$ is an arbitrary orthonormal basis for $V$.
Let $k leq n$ be the number of negative values in $g(e_i,e_i)=pm1 : i=1,dots ,n$. The case $k=0$ is trivial. If $k>0$, then $V$ will have a subspaces which $g$ is negative definite, e.g. the span of one of the basis elements. Let $W$ be the subspace of maximal dimension on which $g$ is negative definite. If we can show that $k=textdim W$ then we're done since this number $k$ is independent of basis.
By rearranging the basis $e_1,dots,e_k,e_k+1,dots,e_n$, we have $g(e_i,e_i)=-1$ for $1leq i leq k$ and $g(e_j,e_j)=1$ for $k+1leq j leq n$. Since $g$ negatives definite on $X = textspan (e_1,dots,e_k)$ then $k=textdim X leq textdim W$. To show $k geq textdim W$, define a map $T : W to X$ as follow; for any $w = sum_i=1^n w^i e_i in W$
$$
T(w) = sum_i=1^k w^ie_i.
$$
You can check directly that $T$ is injective and therefore $textdim W = 0 + textdim Im T leq textdim X=k$. Therefore the number $k$ is fixed by $g$, independent of basis.
answered 15 hours ago
Sou
2,6142720
Suppose we have a Semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the map $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form. We can find an orthonormal basis for $T_pM$ for any $p in M$ using induction as shown in this lemma (lemma 24 of Barrett O'neill's Semi-Riemannian Geometry, p.50).
For the second question, it's enough to do that in the level of vector space. Suppose $V$ is a $n$-dimensional real vector space endowed with a non-degenerate, symmetric bilinear form $g : V times V to BbbR$. Let $e_1,dots,e_n$ is an arbitrary orthonormal basis for $V$.
Let $k leq n$ be the number of negative values in $g(e_i,e_i)=pm1 : i=1,dots ,n$. The case $k=0$ is trivial. If $k>0$, then $V$ will have a subspaces which $g$ is negative definite, e.g. the span of one of the basis elements. Let $W$ be the subspace of maximal dimension on which $g$ is negative definite. If we can show that $k=textdim W$ then we're done since this number $k$ is independent of basis.
By rearranging the basis $e_1,dots,e_k,e_k+1,dots,e_n$, we have $g(e_i,e_i)=-1$ for $1leq i leq k$ and $g(e_j,e_j)=1$ for $k+1leq j leq n$. Since $g$ negatives definite on $X = textspan (e_1,dots,e_k)$ then $k=textdim X leq textdim W$. To show $k geq textdim W$, define a map $T : W to X$ as follow; for any $w = sum_i=1^n w^i e_i in W$
$$
T(w) = sum_i=1^k w^ie_i.
$$
You can check directly that $T$ is injective and therefore $textdim W = 0 + textdim Im T leq textdim X=k$. Therefore the number $k$ is fixed by $g$, independent of basis.
answered 15 hours ago
Sou
2,6142720
edited 38 mins ago
answered 15 hours ago
Sou
2,6142720
answered 15 hours ago
Sou
2,6142720
answered 15 hours ago
Sou
2,6142720
2,6142720
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
14 hours ago
@Jerry Ah. I think this is not gonna work, since $|cdot|$ may be zero on some places on $U$. It is defined as $|cdot| = sqrt$. I will delete this answer.
– Sou
14 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
14 hours ago
I just leave it that way.
– Sou
14 hours ago
@Jerry I have corrected my answer. I think this okay now. I'm sorry i misread your question. If you mean orthonormal basis just for a tangent space, then it's done in lemma 24 of barrett o'neill's (as linked above). My answer is kind of overkill since it's about construction of local orthonormal frame.
– Sou
9 hours ago
add a comment |Â
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
14 hours ago
@Jerry Ah. I think this is not gonna work, since $|cdot|$ may be zero on some places on $U$. It is defined as $|cdot| = sqrt$. I will delete this answer.
– Sou
14 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
14 hours ago
I just leave it that way.
– Sou
14 hours ago
@Jerry I have corrected my answer. I think this okay now. I'm sorry i misread your question. If you mean orthonormal basis just for a tangent space, then it's done in lemma 24 of barrett o'neill's (as linked above). My answer is kind of overkill since it's about construction of local orthonormal frame.
– Sou
9 hours ago
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
14 hours ago
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
14 hours ago
@Jerry Ah. I think this is not gonna work, since $|cdot|$ may be zero on some places on $U$. It is defined as $|cdot| = sqrt$. I will delete this answer.
– Sou
14 hours ago
@Jerry Ah. I think this is not gonna work, since $|cdot|$ may be zero on some places on $U$. It is defined as $|cdot| = sqrt$. I will delete this answer.
– Sou
14 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
14 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
14 hours ago
I just leave it that way.
– Sou
14 hours ago
I just leave it that way.
– Sou
14 hours ago
@Jerry I have corrected my answer. I think this okay now. I'm sorry i misread your question. If you mean orthonormal basis just for a tangent space, then it's done in lemma 24 of barrett o'neill's (as linked above). My answer is kind of overkill since it's about construction of local orthonormal frame.
– Sou
9 hours ago
@Jerry I have corrected my answer. I think this okay now. I'm sorry i misread your question. If you mean orthonormal basis just for a tangent space, then it's done in lemma 24 of barrett o'neill's (as linked above). My answer is kind of overkill since it's about construction of local orthonormal frame.
– Sou
9 hours ago
add a comment |Â
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Sylvester's theorem
– Tom Chalmer
8 hours ago