Index of a Morse function via the Hessian tensor For a smooth function    $f:M\to \mathbb{R}$ one usually defines the degeneracy and index of a critical point    $p\in M$ in terms of the eigenvalues of the Hessian matrix    $(\partial^2 f/\partial x_i\partial x_j)$.  
On the other hand, if we have a Riemannian metric    $g$ we can define the Hessian tensor    $H(f,g)=\nabla df$.  From what I understand, one can recover the Hessian matrix from this tensor and define the index and degeneracy of a critical point as above.
My question is: can we cut out the middle man, i.e., is there a natural way to define the index and degeneracy of a critical point from the Hessian tensor without using the Hessian matrix?
 A: Let $f : M\to \mathbb R$ be a smooth function, where $M$ is a smooth (infinite-dimensional, if you like) manifold.  Pick a point $m\in M$.  Then the value $f(m) = f^{(0)} \in \mathbb R$ is well-defined, of course, and also $f^{(1)}(m) = {\rm d}f_m$ is well-defined as a linear map ${\rm T}_mM \to \mathbb R$, i.e. as an element in ${\rm T}^*_m M$.  In general, the higher derivatives are not well-defined, even if you choose a metric, although they are if you choose coordinates.   What is well-defined is the "$k$-jet" of the function for each $k$; the $k$-jet of $f$ takes values in a vector bundle over $M$ that's locally isomorphic to $\bigoplus_{j=0}^k (\operatorname{Sym}^j( {\rm T}_m M))^*$, but choosing such an isomorphism essentially requires choosing local coordinates.
(Actually, more is true: the $k$-jet bundle surjects onto the $(k-1)$-jet bundle, so that there is a well-defined $\infty$-jet bundle; these surjections are compatible with the local trivializations, so that a trivialization identifies the $\infty$-jet bundle with $(\operatorname{Sym}^\bullet({\rm T}_mM))^*$.  Naively you would define the Taylor series as living here.  The system of surjections defines a filtration on the $\infty$-jet bundle, and $(\operatorname{Sym}^\bullet({\rm T}_mM))^*$ is the corresponding associated graded bundle.)
But there is one situation in which you can define a symmetric tensor $f^{(k)}(m) : ({\rm T}_mM)^{\otimes k} \to \mathbb R$.  Namely, if all $f^{(1)}(m), \dots, f^{(k-1)}(m)$ vanish, then $f^{(k)}(m)$ is well-defined.  This is because under changes of coordinates, the coordinate-dependent term $f^{(k)}(m) \in (\operatorname{Sym}^k({\rm T}_m M))^*$ changes by some term depending linearly on $\{f^{(1)}(m), \dots, f^{(k-1)}(m)\}$ (and polynomially on the derivatives of the change-of-coordinates function).
This is the situation for the Hessian.  If $m$ is a critical point for $f$, then $f^{(1)}(m) = {\rm d}f_m = 0$, and so the second derivative is well-defined as a symmetric map $H = f^{(2)}(m) : ({\rm T}_mM)^{\otimes 2} \to \mathbb R$.
In any case, if $V$ is any vector space, and $\phi: V^{\otimes 2} \to \mathbb R$ any symmetric linear map, then the Morse index of $\phi$ is defined as the dimension (if it is finite, or $\infty$ otherwise) of any maximal subspace $W\subseteq V$ so that $\phi\rvert_{W\otimes W}$ is negative-definite.  It is a straightforward lemma that this number (or $\infty$) does not depend on which maximal subspace you use.  Of course, you can also define a "Morse coindex" as the Morse index of $-\phi$, and finally one can write down the degeneracy of $f$ at $m$ as the dimensional of the kernel of $\phi$ thought of as a map $\phi : V \to V^*$.
Point being, there are no other choices necessary to go: smooth function $f: M\to \mathbb R$ and choice $m\in M$ of critical point $\Rightarrow$ Hessian of $f^{(2)}(m) : ({\rm T}_mM)^{\otimes 2} \to \mathbb R$ $\Rightarrow$ Morse index of $f^{(2)}(m)$.  At non-critical points, you cannot do the first step.

Finally, let me comment on the use of a metric.  For applications of Morse theory to the topology of manifolds, picking a metric is important, as it lets you convert the differential one-form ${\rm d}f$ into a vector field and thus consider the corresponding flow lines, etc.  From a topological point of view, what particular metric you use doesn't matter much: the space of metrics is contractible.  But it is important; just not for the definition of Morse index.
Recall that a metric $g$ determines for each $m\in M$ a linear map $\operatorname{Sym}^2 {\rm T}_m M \to \mathbb R$, and hence a linear map $g_m: {\rm T}_m M \to {\rm T}^*_m M$.  Being a metric means that this linear map is an injection (and hence an iso in finite-dimensions).  So you should try to write down an inverse $g^{-1}_m: {\rm T}_m^* M \to {\rm T}_m M$.  You can do this in finite-dimensions; in infinite dimensions the inverse is not everywhere-defined.  Then you can define $\nabla f_m = {\rm d}f_m \, g^{-1}_m$ and, if you want, the Hessian to be the matrix $f^{(2)}_m g^{-1}_m$.  Since $g$ is positive-definite, so is $g^{-1}$, and so the dimensions of the kernel and positive and negative eigenspaces of $f^{(2)}_m g^{-1}_m$ are the same as for $f^{(2)}$.
A: What you call the Hessian matrix is defined using a choice of local co-ordinates near the critical point. The Hessian matrix at the critical point consists of the components of the Hessian tensor written with respect to the local-ordinates. In other words, at any critical point the two don't just have the same index; they are the same thing. 
A: The Hessian tensor (defined at a critical point of a function) is a symmetric bilinear form on the tangent space of the manifold at that point. For a symmetric bilinear form on a finite-dimensional vector space (in characteristic different from $2$) there is always a basis such that the corresponding symmetric matrix is diagonal. The diagonal entries are not well-defined independent of diagonalizing basis, but if your field is $\mathbb R$ then the number of positive (or negative) entries is well-defined. You can intrinsically define the index of a nondegenerate critical point as the maximal dimension of a tangent vector subspace on which the Hessian is negative definite.
