Linearization of a tensor?












1












$begingroup$


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?











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$endgroup$

















    1












    $begingroup$


    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?











    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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?











      share|cite|improve this question











      $endgroup$




      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






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      edited Jan 6 at 17:20







      Ramand

















      asked Jan 6 at 12:19









      RamandRamand

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