I am a beginner in differential geometry and would appreciate some pointers on how to answer the following question.

Let $M$ be a closed orientable Riemannian manifold with $\{e^1,...,e^n\}$ an oriented orthonormal basis of the cotangent bundle $T^* M$. My question is: how would one go about showing whether the operator $$c(e^k) : \Omega^*(M)\rightarrow\Omega^*(M),\,\,\omega\mapsto e^k\wedge\omega - i_{e^k}\omega$$ is bounded or not with respect to the usual $L^2$ inner product on $\Omega^*(M)$? Here $i_v$ denotes interior multiplication by $v$.


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    $\begingroup$ I think it's true $ c(e_i) $ is an isometry isn't it? ie. inside of $ T^*M $, $$ || c(e_i) \omega || == || \omega || $$ $\endgroup$ – user226970 Aug 1 '15 at 0:52
  • $\begingroup$ Is this somehow obvious? $\endgroup$ – ougoah Aug 1 '15 at 10:01

It's straightforward to see for a 1-form: take an orthonormal frame $e_1,\ldots,e_n$ and then $$|e_1\wedge\omega|^2=4\sum_{i<j}\Big((e_1\wedge\omega)(e_i,e_j)\Big)^2=4\sum_{j\geq 2}\Big((e_1\wedge\omega)(e_1,e_j)\Big)^2=\sum_{j\geq 2}\Big(\omega(e_j)\Big)^2$$ and $$|\iota_{e_1}\omega|^2=\Big(\omega(e_1)\Big)^2$$ so $|c(e_i)\omega|^2=|e_1\wedge\omega|^2+|\iota_{e_1}\omega|^2=|\omega|^2.$ The only extra difficulty for a $k$-form is in keeping track of all the normalization constants in defining norms and wedge products and such.

  • $\begingroup$ Thanks! I think i understand all the steps except for the first equality. (I'm probably missing a definition - could you please elaborate?) $\endgroup$ – ougoah Aug 2 '15 at 0:18
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    $\begingroup$ $e_1\wedge\omega$ is a 2-form and the norm of a 2-form $\eta_{ij}$ is defined as $|\eta|^2=2g^{ij}g^{kl}\eta_{ik}\eta_{jl}$. In an orthonormal frame, $g^{ij}=\delta^{ij}$ and so this becomes $|\eta|^2=2\sum_{ij}\eta_{ij}^2.$ If $i=j$ then $\eta_{ij}=0$ due to $\eta$ being a 2-form, and so $|\eta|^2=4\sum_{i<j}\eta(e_i,e_j).$ $\endgroup$ – youler Aug 2 '15 at 1:26
  • $\begingroup$ (edited my answer to allow for $n>2$) $\endgroup$ – youler Aug 2 '15 at 1:33

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