Hi,
I am reading the lecture notes on Morse Homology written by M.Hutchings, in that notes definition of Hessian is defined in coordinate free way: given any connection $ H(f,p)= \nabla\_v(df)$ where $v$ is the tangent vector at critical point $p$, and $df$ is differential of $f$. I need to show this definition does not depend on choice of connection. Hutchings says that the difference of two connection is a tensor and $df$ vanishes at p, so the above fact holds. I can not understand the meaning of "the difference of two connection is a tensor" and how this observation solves my problem. 


Thanks,



definition of Hessian with respect to connection
(This is an elaboration on the comment of MG. I know I benefited a lot as an undergraduate from being shown this sort of argument once instead of having been told to check things in local coordinates, so I thought I'd do the same for you.)


The fact that $\nabla$ is a connection means that for every function $f$ $$\nabla\_{fX}(Y)=f\nabla\_X(Y)$$ and $$\nabla\_X(fY)=f\nabla\_X(Y)+L\_X(f)Y.$$ In other words, $\nabla$ it is linear over the ring of functions in the argument $X$, but not linear in $Y$, and instead satisfies some sort of a Leibniz rule for differentiating products. That non-linearity in $Y$ is precisely an obstruction for being a tensor. (There is nothing to check here: a tensor on a manifold is a linear over the ring of functions construction depending on several vectors and covectors, or, more eloquently put, a section of a tensor construction applied to the tangent and cotangent bundle! Saying that a tensor is something that transforms correctly in every coordinate system is such a mean thing to say to innocent students.) 


What you instantly see is that the "correction term" $L\_X(f)Y$ does not depend on $\nabla$, so when you compute the difference of two connections, it will disappear, and hence that difference will be linear in both $X$ and $Y$. 


Finally, for two connections $\nabla^{(1)}$ and $\nabla^{(2)}$, the tensor $A(X,Y)=\nabla^{(1)}\_X(Y)-\nabla^{(2)}\_X(Y)$, being a linear operator in $Y$ for every fixed $X$, satisfies $A(X,0)=0$, hence the claim you want.


