# Connection 1-forms of a Riemannian metric and the norm of the Hessian and ( seemingly ) two different definitions of Hessian and its norm

In the paper "On Quasiconformal Harmonic Maps " (link here) by L. F. Tam and T.Y.H. Wan, Pacific Journal of Mathematics, vol 182, no 2, 1998, in section 1, they define the Hessian of a function $f :H^m \to H^n$ w.r.t the Hyperbolic metric in section 1, "Estimates on Douady-Earle extension" by using the connection forms as follows: (for me, I am happy with $m=n=2$, so I am just re-writing everything in $2$-dimensions). Let $\theta_i, \theta_{ij}$ respectively denote the local orthonormal coframe on $H^2$ (or $D^2$) and its corresponding matrix of connection forms, $1\le i,j,\le 2$. For a map $\Phi: D^2 \to D^2$, one defines the energy density of $\Phi$ by $e(\Phi)= \sum_{i,\alpha} {(f_{i}^{\alpha}})^2$, where $f_{i}^{\alpha}$'s are given by $F^*(\theta_\alpha)= \sum_{i}f_{i}^{\alpha}\theta_i$. Then according to this paper, the Hessian of $\Phi$ is denoted by $f_{ij}^{\alpha}$, is defined by :

$\sum_{j} f_{ij}^{\alpha}. \theta_j = df_i^{\alpha} - \sum_j{f_j^\alpha}.\theta_{ij} + \sum_{\beta}f_{i}^{\beta}F^*(\omega_ {\beta} {\alpha} )$

and they write the norm of the Hessian w.r.t. this orthonormal coframe as :$|Hess(f)|^2=\sum_{ij\alpha} (f_{ij}^{\alpha})^2$.

I guess I am a little confused by this definition. So my questions are :

1) Previously I knew ( from Riemmanian Geometry textbook of Peter Petersen or John Lee P.54 ) that $Hess(\Phi)=\nabla^2(\Phi)$ is a $(2,0)$ tensor defined by $Hess(\Phi)(X,Y)= g( \nabla_X(\nabla\Phi),Y)$, where $g$ denotes the hyperbolic metric on the open unit disk $D^2$.

Is the new definition of the paper the same as the old definition ? And what should be the norm according to the old definition ?

2) The proposition 1.2, (iii) states that under certain condition, the norm of the Hessian ( as defined before ) of $\Phi$ is $< \epsilon$ .Does it mean that we can say the standard partial derivatives of $\Phi$, i.e. $\Phi_{xx}, \Phi{xy}, \Phi{yy}$ have norms less than $< 4\epsilon ( 1- |z^2| )^2$ at the point $z$ under the same condition ( that condition is not necessary for my question here ) , given that we are working with $D$ with the standard Poincare metric on it ,i.e., $g= 4\frac{|dz|^2}{( 1- |z^2| )^2}$?

Thanks so much !

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I'm sorry to be unhelpful, but a rather important skill to develop as a differential geometer is figuring out why apparently different definitions and formulas are either really the same or are actually different. There is nothing particularly complicated about the definition of the Hessian of a scalar function, so this is a good learning exercise. The recommended approach is to first figure out what notation and definitions you like best and then slowly learn how to convert anything someone else writes into your preferred notation and definition. –  Deane Yang Sep 4 '11 at 17:19
There is only one possible co-ordinate-free definition of the Hessian as a tensor. –  Deane Yang Sep 4 '11 at 17:21
I took the liberty of adding a link to the paper. –  Joseph O'Rourke Sep 4 '11 at 17:41
Oops. I didn't read the question carefully enough. This is about the Hessian of a map between two Riemannian manifolds. That's a lot trickier to define properly. You can't just use the definition of the Hessian of a real-valued function (as mentioned in 1)) and just use it for a map. Defining the Hessian of a map takes real effort. –  Deane Yang Sep 4 '11 at 18:55
First, make sure you know how to define the differential of $f$ and note that it defines a map between vector bundles that lie over different manifolds (the domain and range of $f$). Then figure out how to define the covariant derivative of a vector bundle map, where both bundles have connections. The Hessian is then just the covariant derivative of the differential of $f$. –  Deane Yang Sep 4 '11 at 18:57

Let $f: M \to N$ be a smooth map between two riemannian manifolds $(M,h)$ and $(N,g)$.
You can view $df$ as a one-form on $M$ with values in the pullback bundle $f^{-1}TN$. In other words, $df \in \Omega^1(M; f^{-1}TN)$ or if you prefer a smooth section of the vector bundle: $T^*M \otimes f^{-1}TN$ over $M$.
Both of these bundles have natural connections: on $T^*M$ you have the Levi-Civita connection associated to the metric $h$ on $M$, whereas on $f^{-1}TN$ you have the pullback of the Levi-Civita connection on $TN$ associated to the metric $g$ on $N$. The Hessian of $f$ is then simply $\nabla df$ where $\nabla$ is the product connection on $T^*M \otimes f^{-1}TN$.