# definition of Hessian with respect to connection

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,

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Try writing everything out carefully in local co-ordinates. Do a change of co-ordinates and figure out how everything changes. – Deane Yang Feb 3 '12 at 23:47
Assume $\nabla_i$ are two connections for $i=1,2$. You can calculate $(\nabla_1 -\nabla_2)_XY$ and you will see it's a tensor. – John B Feb 4 '12 at 2:33
The key point is that the difference of the two Hessians (defined by two different connections) is a linear function of $df$. This follows from the fact that the difference of two connections is a tensor, but you might as well verify this by a direct computation. – Deane Yang Feb 4 '12 at 11:27
Keep in mind that the definition H(f,p)= \nabla_v(df) does depend on the connection at an arbitrary point. This is very useful because you may want a second derivative tensor defined everywhere, not only at critical points, and to accomplish that a connection is needed. At a critical point where df vanishes, the Hessian is indeed independent of the choice of connection. – Pait Feb 7 '12 at 13:52

(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.

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I agree that in the long run your approach is the right and much clearer way to do things. And if you can understand it when you see it for the first time, then that's even better. But, alas, I didn't understand these formulas and calculations when I first saw them and only did after I unwound the formulas into local co-ordinates and watched the Christoffel symbols appear and disappear. – Deane Yang Feb 4 '12 at 11:24
Fair enough. Different people surely learn in different ways! I was scared to death by Christoffel symbols, and this explanation was like what one must feel enjoying the light and fresh air after having been locked in a basement for a while, or something of the sort. – Vladimir Dotsenko Feb 4 '12 at 13:47
I agree with your latest comment, too! I was unable to appreciate properly the light and fresh air without being locked in the Christoffel prison first. – Deane Yang Feb 4 '12 at 15:23
I know for any connection $\nabla$, $\nabla_X(0)= 0$. How $A(X, 0)=O$ give an information for independency of connection? – zatilokum Feb 4 '12 at 22:33
Morally, think of a tensor as of a multilinear function (well, a linear operator, to be precise). But you've to be a bit more careful, since you are working with manifolds, and things depend on the point where you evaluate stuff etc. A safe way to deal with this sort of stuff is to talk about sections of vector bundles. E.g., in our case, $A$ is a $C^\infty(M)$-linear operator from $\Gamma(TM)\otimes_{C^\infty(M)}\Gamma(T^*M)$ to $\Gamma(T^*M)$, where $\Gamma(E)$ denotes global sections of a bundle $E$ (e.g. $\Gamma(TM)$ would be vector fields). But your point-wise intuition is correct, yes. – Vladimir Dotsenko Feb 5 '12 at 18:26

There is no real need for a connection in this situation.

The underlying fact is that if $s$ is a section of a vector bundle $E$ over $X$, and $s$ vanishes at a point $x_0$, then it has an intrinsic derivative $Ds(x_0):T_{x_0}X\to E_{x_0}$, defined by $Ds(x_0)\cdot v=\lim_{t\to 0} s(x_t)/t$, where $t\mapsto x_t$, $t\in]-1,1[$ is any curve in $X$ with velocity $v$ at $t=0$.

In your case $E$ is $T^*X$ and $s=df$.

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