# Norm of clifford operator on forms

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 $v$.

Thanks!

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

## 1 Answer

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.

• Thanks! I think i understand all the steps except for the first equality. (I'm probably missing a definition - could you please elaborate?) – ougoah Aug 2 '15 at 0:18
• $e_1\wedge\omega$ 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 $|\eta|^2=4\sum_{i<j}\eta(e_i,e_j).$ – youler Aug 2 '15 at 1:26
• (edited my answer to allow for $n>2$) – youler Aug 2 '15 at 1:33