Mixing
Existence of Orthonormal Basis of a Metric in a Manifold and Uniqueness of Signature of a Metric
Clash Royale CLAN TAG#URR8PPP
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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 5 hours ago
Sou
2,6142720
asked 7 hours ago
Jerry
364111
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up vote
1
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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 5 hours ago
Sou
2,6142720
asked 7 hours ago
Jerry
364111
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 5 hours ago
Sou
2,6142720
asked 7 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 5 hours ago
Sou
2,6142720
asked 7 hours ago
Jerry
364111
edited 5 hours ago
Sou
2,6142720
edited 5 hours ago
Sou
2,6142720
edited 5 hours ago
Sou
2,6142720
2,6142720
asked 7 hours ago
Jerry
364111
asked 7 hours ago
Jerry
364111
asked 7 hours ago
Jerry
364111
364111
add a comment |Â
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 5 hours ago
Batominovski
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I think Gram-Schimidt process still applies though, only with a bit modification.
Suppose we have a semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the couple $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form.
$textbfChoosing a local frame$
Choose arbitrary point $p in M$ and let $v_1,dots,v_n$ be an orthonormal basis for $T_pM$ (guaranteed by this lemma. Having this, we can find a smooth local frame $(X_1,dots,X_n)$ on some neighbourhood $Usubseteq M$ of $p$ such that $X_i|_p=v_i$ for all $i =1,dots,n$ (See John Lee's Introduction to Smooth Manifolds 2nd Ed, Proposition 8.11).
$textbfNormalizing$
We normalize them as follows: Since $g(X_1,X_1): U to BbbR$ is a smooth function with $g(X_1,X_1)(p) = g_p(X_1|_p,X_1|_p) =g_p(v_1,v_1)= pm 1$, then by continuity we have a smaller neighbourhood $U_1 subseteq U$ of $p$ such that $g(X_1,X_1)> 0 $ or $<0$ in this region, depends on the sign $g_p(v_1,=v_1)pm 1$. Because of this $|X_1| = sqrt$ is a non-vanishing smooth vector field on $U_1$, then we can normalize $X_1$, by setting
$$
E_1 = fracX_1.
$$
Similarly, the vector $V_2 = X_2 - varepsilon_1 g(X_2,E_1)E_1$ is a smooth vector field with $g(V_2,V_2)(p) = g_p(V_2|_p,V_2|_p) = g_p(v_2,v_2) = pm 1$. So $g(V_2,V_2) $ has a neighbourhood $U_2 subseteq U$ of $p$ such that it has only positive or negative values. So we have normalize smooth vector field
$$
E_2 = fracX_2 - varepsilon_1 g(X_2,E_1)E_1, quad varepsilon_1=g(E_1,E_1)=pm1
$$
with $g(E_2,E_1) = 0$.
Working inductively, we have $n$-tuple of smooth vector fields $(E_1,dots,E_n)$ defined on $W = U_1 cap cdots cap U_n$ given by
$$
E_j = fracX_j - sum_i=1^j-1 colorredvarepsilon_i , g(X_j,V_i)E_i
$$
with $varepsilon_i=g(E_i,E_i)=pm1$. By construction $g(E_i,E_j) = 0$ for $i neq j$ and $g(E_i,E_i) = pm 1$, and $textspan(X_1,dots,X_n) = textspan(E_,dots,E_n)$.
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. We have to show that $k=textdim W$. Therefore this number 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 5 hours ago
Sou
2,6142720
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
4 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
4 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
4 hours ago
I just leave it that way.
– Sou
4 hours ago
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2 Answers
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2 Answers
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active
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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 5 hours ago
Batominovski
22.2k22675
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up vote
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 5 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 5 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 5 hours ago
Batominovski
22.2k22675
edited 5 hours ago
answered 5 hours ago
Batominovski
22.2k22675
answered 5 hours ago
Batominovski
22.2k22675
answered 5 hours ago
Batominovski
22.2k22675
22.2k22675
add a comment |Â
add a comment |Â
up vote
0
down vote
I think Gram-Schimidt process still applies though, only with a bit modification.
Suppose we have a semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the couple $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form.
$textbfChoosing a local frame$
Choose arbitrary point $p in M$ and let $v_1,dots,v_n$ be an orthonormal basis for $T_pM$ (guaranteed by this lemma. Having this, we can find a smooth local frame $(X_1,dots,X_n)$ on some neighbourhood $Usubseteq M$ of $p$ such that $X_i|_p=v_i$ for all $i =1,dots,n$ (See John Lee's Introduction to Smooth Manifolds 2nd Ed, Proposition 8.11).
$textbfNormalizing$
We normalize them as follows: Since $g(X_1,X_1): U to BbbR$ is a smooth function with $g(X_1,X_1)(p) = g_p(X_1|_p,X_1|_p) =g_p(v_1,v_1)= pm 1$, then by continuity we have a smaller neighbourhood $U_1 subseteq U$ of $p$ such that $g(X_1,X_1)> 0 $ or $<0$ in this region, depends on the sign $g_p(v_1,=v_1)pm 1$. Because of this $|X_1| = sqrt$ is a non-vanishing smooth vector field on $U_1$, then we can normalize $X_1$, by setting
$$
E_1 = fracX_1.
$$
Similarly, the vector $V_2 = X_2 - varepsilon_1 g(X_2,E_1)E_1$ is a smooth vector field with $g(V_2,V_2)(p) = g_p(V_2|_p,V_2|_p) = g_p(v_2,v_2) = pm 1$. So $g(V_2,V_2) $ has a neighbourhood $U_2 subseteq U$ of $p$ such that it has only positive or negative values. So we have normalize smooth vector field
$$
E_2 = fracX_2 - varepsilon_1 g(X_2,E_1)E_1, quad varepsilon_1=g(E_1,E_1)=pm1
$$
with $g(E_2,E_1) = 0$.
Working inductively, we have $n$-tuple of smooth vector fields $(E_1,dots,E_n)$ defined on $W = U_1 cap cdots cap U_n$ given by
$$
E_j = fracX_j - sum_i=1^j-1 colorredvarepsilon_i , g(X_j,V_i)E_i
$$
with $varepsilon_i=g(E_i,E_i)=pm1$. By construction $g(E_i,E_j) = 0$ for $i neq j$ and $g(E_i,E_i) = pm 1$, and $textspan(X_1,dots,X_n) = textspan(E_,dots,E_n)$.
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. We have to show that $k=textdim W$. Therefore this number 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 5 hours ago
Sou
2,6142720
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
4 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
4 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
4 hours ago
I just leave it that way.
– Sou
4 hours ago
add a comment |Â
up vote
0
down vote
I think Gram-Schimidt process still applies though, only with a bit modification.
Suppose we have a semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the couple $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form.
$textbfChoosing a local frame$
Choose arbitrary point $p in M$ and let $v_1,dots,v_n$ be an orthonormal basis for $T_pM$ (guaranteed by this lemma. Having this, we can find a smooth local frame $(X_1,dots,X_n)$ on some neighbourhood $Usubseteq M$ of $p$ such that $X_i|_p=v_i$ for all $i =1,dots,n$ (See John Lee's Introduction to Smooth Manifolds 2nd Ed, Proposition 8.11).
$textbfNormalizing$
We normalize them as follows: Since $g(X_1,X_1): U to BbbR$ is a smooth function with $g(X_1,X_1)(p) = g_p(X_1|_p,X_1|_p) =g_p(v_1,v_1)= pm 1$, then by continuity we have a smaller neighbourhood $U_1 subseteq U$ of $p$ such that $g(X_1,X_1)> 0 $ or $<0$ in this region, depends on the sign $g_p(v_1,=v_1)pm 1$. Because of this $|X_1| = sqrt$ is a non-vanishing smooth vector field on $U_1$, then we can normalize $X_1$, by setting
$$
E_1 = fracX_1.
$$
Similarly, the vector $V_2 = X_2 - varepsilon_1 g(X_2,E_1)E_1$ is a smooth vector field with $g(V_2,V_2)(p) = g_p(V_2|_p,V_2|_p) = g_p(v_2,v_2) = pm 1$. So $g(V_2,V_2) $ has a neighbourhood $U_2 subseteq U$ of $p$ such that it has only positive or negative values. So we have normalize smooth vector field
$$
E_2 = fracX_2 - varepsilon_1 g(X_2,E_1)E_1, quad varepsilon_1=g(E_1,E_1)=pm1
$$
with $g(E_2,E_1) = 0$.
Working inductively, we have $n$-tuple of smooth vector fields $(E_1,dots,E_n)$ defined on $W = U_1 cap cdots cap U_n$ given by
$$
E_j = fracX_j - sum_i=1^j-1 colorredvarepsilon_i , g(X_j,V_i)E_i
$$
with $varepsilon_i=g(E_i,E_i)=pm1$. By construction $g(E_i,E_j) = 0$ for $i neq j$ and $g(E_i,E_i) = pm 1$, and $textspan(X_1,dots,X_n) = textspan(E_,dots,E_n)$.
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. We have to show that $k=textdim W$. Therefore this number 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 5 hours ago
Sou
2,6142720
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
4 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
4 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
4 hours ago
I just leave it that way.
– Sou
4 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think Gram-Schimidt process still applies though, only with a bit modification.
Suppose we have a semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the couple $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form.
$textbfChoosing a local frame$
Choose arbitrary point $p in M$ and let $v_1,dots,v_n$ be an orthonormal basis for $T_pM$ (guaranteed by this lemma. Having this, we can find a smooth local frame $(X_1,dots,X_n)$ on some neighbourhood $Usubseteq M$ of $p$ such that $X_i|_p=v_i$ for all $i =1,dots,n$ (See John Lee's Introduction to Smooth Manifolds 2nd Ed, Proposition 8.11).
$textbfNormalizing$
We normalize them as follows: Since $g(X_1,X_1): U to BbbR$ is a smooth function with $g(X_1,X_1)(p) = g_p(X_1|_p,X_1|_p) =g_p(v_1,v_1)= pm 1$, then by continuity we have a smaller neighbourhood $U_1 subseteq U$ of $p$ such that $g(X_1,X_1)> 0 $ or $<0$ in this region, depends on the sign $g_p(v_1,=v_1)pm 1$. Because of this $|X_1| = sqrt$ is a non-vanishing smooth vector field on $U_1$, then we can normalize $X_1$, by setting
$$
E_1 = fracX_1.
$$
Similarly, the vector $V_2 = X_2 - varepsilon_1 g(X_2,E_1)E_1$ is a smooth vector field with $g(V_2,V_2)(p) = g_p(V_2|_p,V_2|_p) = g_p(v_2,v_2) = pm 1$. So $g(V_2,V_2) $ has a neighbourhood $U_2 subseteq U$ of $p$ such that it has only positive or negative values. So we have normalize smooth vector field
$$
E_2 = fracX_2 - varepsilon_1 g(X_2,E_1)E_1, quad varepsilon_1=g(E_1,E_1)=pm1
$$
with $g(E_2,E_1) = 0$.
Working inductively, we have $n$-tuple of smooth vector fields $(E_1,dots,E_n)$ defined on $W = U_1 cap cdots cap U_n$ given by
$$
E_j = fracX_j - sum_i=1^j-1 colorredvarepsilon_i , g(X_j,V_i)E_i
$$
with $varepsilon_i=g(E_i,E_i)=pm1$. By construction $g(E_i,E_j) = 0$ for $i neq j$ and $g(E_i,E_i) = pm 1$, and $textspan(X_1,dots,X_n) = textspan(E_,dots,E_n)$.
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. We have to show that $k=textdim W$. Therefore this number 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 5 hours ago
Sou
2,6142720
I think Gram-Schimidt process still applies though, only with a bit modification.
Suppose we have a semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p in M$ the couple $g_p : T_pM times T_pM to BbbR$ is a non-degenerate, symmetric, bilinear form.
$textbfChoosing a local frame$
Choose arbitrary point $p in M$ and let $v_1,dots,v_n$ be an orthonormal basis for $T_pM$ (guaranteed by this lemma. Having this, we can find a smooth local frame $(X_1,dots,X_n)$ on some neighbourhood $Usubseteq M$ of $p$ such that $X_i|_p=v_i$ for all $i =1,dots,n$ (See John Lee's Introduction to Smooth Manifolds 2nd Ed, Proposition 8.11).
$textbfNormalizing$
We normalize them as follows: Since $g(X_1,X_1): U to BbbR$ is a smooth function with $g(X_1,X_1)(p) = g_p(X_1|_p,X_1|_p) =g_p(v_1,v_1)= pm 1$, then by continuity we have a smaller neighbourhood $U_1 subseteq U$ of $p$ such that $g(X_1,X_1)> 0 $ or $<0$ in this region, depends on the sign $g_p(v_1,=v_1)pm 1$. Because of this $|X_1| = sqrt$ is a non-vanishing smooth vector field on $U_1$, then we can normalize $X_1$, by setting
$$
E_1 = fracX_1.
$$
Similarly, the vector $V_2 = X_2 - varepsilon_1 g(X_2,E_1)E_1$ is a smooth vector field with $g(V_2,V_2)(p) = g_p(V_2|_p,V_2|_p) = g_p(v_2,v_2) = pm 1$. So $g(V_2,V_2) $ has a neighbourhood $U_2 subseteq U$ of $p$ such that it has only positive or negative values. So we have normalize smooth vector field
$$
E_2 = fracX_2 - varepsilon_1 g(X_2,E_1)E_1, quad varepsilon_1=g(E_1,E_1)=pm1
$$
with $g(E_2,E_1) = 0$.
Working inductively, we have $n$-tuple of smooth vector fields $(E_1,dots,E_n)$ defined on $W = U_1 cap cdots cap U_n$ given by
$$
E_j = fracX_j - sum_i=1^j-1 colorredvarepsilon_i , g(X_j,V_i)E_i
$$
with $varepsilon_i=g(E_i,E_i)=pm1$. By construction $g(E_i,E_j) = 0$ for $i neq j$ and $g(E_i,E_i) = pm 1$, and $textspan(X_1,dots,X_n) = textspan(E_,dots,E_n)$.
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. We have to show that $k=textdim W$. Therefore this number 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 5 hours ago
Sou
2,6142720
edited 4 mins ago
answered 5 hours ago
Sou
2,6142720
answered 5 hours ago
Sou
2,6142720
answered 5 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
4 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
4 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
4 hours ago
I just leave it that way.
– Sou
4 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
4 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
4 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
4 hours ago
I just leave it that way.
– Sou
4 hours ago
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
4 hours ago
How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function.
– Jerry
4 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
4 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
4 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
4 hours ago
Mind leaving your answer to the second part here? That means only delete the first part but not the second.
– Jerry
4 hours ago
I just leave it that way.
– Sou
4 hours ago
I just leave it that way.
– Sou
4 hours ago
add a comment |Â
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