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Given two riemannian manifolds $M$ and $N$ and a smooth map $f$ : $M$ $\rightarrow$ $N$, we define the energy density of $f$ as the smooth function $e(f)$ : $M$ $\rightarrow$ $\mathbb{R}$ given by $e(f)$ = $\frac{1}{2}$$|df|^2$, where $|df|^2$ is the Hilbert-Schmidt norm of the differential $df$ of $f$.

I've got a couple of questions about the definition above:

I believe that the differential in this definition is the global one, I mean, the differential $df$ defined on the tangent bundle of $M$ to the t.b. of $N$ ($df$ : $TM$ $\rightarrow$ $TN$), am I right? Beside that, I'd like to know what is the H-S norm of the differential $df$. The notes that I'm reading say that, if $(e_1,...,e_n)$ is a local orthonormal frame on an open set $U$ $\subset$ $M$, then $e(f)$ = $\frac{1}{2}$$\langle(f_{*})e_i,(f_{*})e_i\rangle$. I didn't understand how the global differential acts on frames, since the domain of $df$ is $TM$ (so it would have to be something like $df$$(p,v)$).

Thanks for your time.

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You probably need a Riemannian structure on your manifolds. Am I right? – Giuseppe Negro Feb 15 '14 at 18:07
Yes, sorry, I forgot to mention that. I'll edit the text. – Br09 Feb 15 '14 at 18:09
The differential $df$ acts pointwise, taking $T_pM\to T_{f(p)}N.$ – Neal Feb 16 '14 at 2:40
up vote 2 down vote accepted

If $T:V\to W$ is a linear transformation between two euclidean vector spaces (vector spaces with a positive definite inner product) then you define its norm square by taking the sum of squares of the entries of its matrix wrt some othonormal bases in $V,W$. You can check that this definition is independent of the bases chosen. An equivalent definition: $\text{tr}(T^*T)$, where the adjoint $T^*:W\to V$ is defined via the identity $\langle w,Tv\rangle= \langle T^*w,v\rangle$. Now apply this to $df:T_xM\to T_{f(x)}N$.

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