Linearization of a tensor?
$begingroup$
Suppose that $(M,g)$ is a Riemannain manifold and $nabla$ denotes its Levi-Civita connection. Define a $(0,2)$ tensor field $pi$ by
$$pi (X,Y)=(nabla_X omega)(Y)-omega (X)omega (Y)+dfrac{1}{2}omega (U)g(X,Y)$$
where $omega$ is an 1-form on $(M,g)$ and $U$ is its equivalent vector field, in fact for all $Xin mathcal{X}(M)$
$$omega (X)=g(X,U).$$
Suppose $g(t)=g+ts$ and $omega (t)=omega +tdelta$, I want to linearize $|pi |^2$. By means of a local coordinate system on $M$ with local frame $partial_i$, set $pi (partial_i,partial_j)=pi_{ij}$. consequently,
$$
pi_{ij}=nabla_i omega_j -omega_i omega_j +dfrac{1}{2}g_{ij} |omega|^2
$$
where, $omega(partial_i)=omega_i$. We have
$$|pi |^2=g^{ij}pi_{ij}=mathrm{div}(omega) +(dfrac{n}{2}-1)|omega |^2 .$$
Consequently,
$$
(|pi (t) |^2 )'(0)=<d omega +(dfrac{n}{2}-1)omegaotimes omega,-s>+<g ,ddelta +(n-2)omegaotimes delta>.
$$
Is my result correct?
proof-verification differential-geometry
$endgroup$
add a comment |
$begingroup$
Suppose that $(M,g)$ is a Riemannain manifold and $nabla$ denotes its Levi-Civita connection. Define a $(0,2)$ tensor field $pi$ by
$$pi (X,Y)=(nabla_X omega)(Y)-omega (X)omega (Y)+dfrac{1}{2}omega (U)g(X,Y)$$
where $omega$ is an 1-form on $(M,g)$ and $U$ is its equivalent vector field, in fact for all $Xin mathcal{X}(M)$
$$omega (X)=g(X,U).$$
Suppose $g(t)=g+ts$ and $omega (t)=omega +tdelta$, I want to linearize $|pi |^2$. By means of a local coordinate system on $M$ with local frame $partial_i$, set $pi (partial_i,partial_j)=pi_{ij}$. consequently,
$$
pi_{ij}=nabla_i omega_j -omega_i omega_j +dfrac{1}{2}g_{ij} |omega|^2
$$
where, $omega(partial_i)=omega_i$. We have
$$|pi |^2=g^{ij}pi_{ij}=mathrm{div}(omega) +(dfrac{n}{2}-1)|omega |^2 .$$
Consequently,
$$
(|pi (t) |^2 )'(0)=<d omega +(dfrac{n}{2}-1)omegaotimes omega,-s>+<g ,ddelta +(n-2)omegaotimes delta>.
$$
Is my result correct?
proof-verification differential-geometry
$endgroup$
add a comment |
$begingroup$
Suppose that $(M,g)$ is a Riemannain manifold and $nabla$ denotes its Levi-Civita connection. Define a $(0,2)$ tensor field $pi$ by
$$pi (X,Y)=(nabla_X omega)(Y)-omega (X)omega (Y)+dfrac{1}{2}omega (U)g(X,Y)$$
where $omega$ is an 1-form on $(M,g)$ and $U$ is its equivalent vector field, in fact for all $Xin mathcal{X}(M)$
$$omega (X)=g(X,U).$$
Suppose $g(t)=g+ts$ and $omega (t)=omega +tdelta$, I want to linearize $|pi |^2$. By means of a local coordinate system on $M$ with local frame $partial_i$, set $pi (partial_i,partial_j)=pi_{ij}$. consequently,
$$
pi_{ij}=nabla_i omega_j -omega_i omega_j +dfrac{1}{2}g_{ij} |omega|^2
$$
where, $omega(partial_i)=omega_i$. We have
$$|pi |^2=g^{ij}pi_{ij}=mathrm{div}(omega) +(dfrac{n}{2}-1)|omega |^2 .$$
Consequently,
$$
(|pi (t) |^2 )'(0)=<d omega +(dfrac{n}{2}-1)omegaotimes omega,-s>+<g ,ddelta +(n-2)omegaotimes delta>.
$$
Is my result correct?
proof-verification differential-geometry
$endgroup$
Suppose that $(M,g)$ is a Riemannain manifold and $nabla$ denotes its Levi-Civita connection. Define a $(0,2)$ tensor field $pi$ by
$$pi (X,Y)=(nabla_X omega)(Y)-omega (X)omega (Y)+dfrac{1}{2}omega (U)g(X,Y)$$
where $omega$ is an 1-form on $(M,g)$ and $U$ is its equivalent vector field, in fact for all $Xin mathcal{X}(M)$
$$omega (X)=g(X,U).$$
Suppose $g(t)=g+ts$ and $omega (t)=omega +tdelta$, I want to linearize $|pi |^2$. By means of a local coordinate system on $M$ with local frame $partial_i$, set $pi (partial_i,partial_j)=pi_{ij}$. consequently,
$$
pi_{ij}=nabla_i omega_j -omega_i omega_j +dfrac{1}{2}g_{ij} |omega|^2
$$
where, $omega(partial_i)=omega_i$. We have
$$|pi |^2=g^{ij}pi_{ij}=mathrm{div}(omega) +(dfrac{n}{2}-1)|omega |^2 .$$
Consequently,
$$
(|pi (t) |^2 )'(0)=<d omega +(dfrac{n}{2}-1)omegaotimes omega,-s>+<g ,ddelta +(n-2)omegaotimes delta>.
$$
Is my result correct?
proof-verification differential-geometry
proof-verification differential-geometry
edited Jan 6 at 17:20
Ramand
asked Jan 6 at 12:19
RamandRamand
492210
492210
add a comment |
add a comment |
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