# Hessian of the inverse exponential map on a Riemannian manifold

Let $(M,g)$ be a Riemannian manifold. Then define

$f: T^*M \times M \to \mathbb{R}$

$f(x,\xi, y) = \langle exp_x^{-1} y, \xi \rangle$

where $exp_{\cdot}\cdot$ is the the exponential map and it's inverse is defined on a neighborhood of the diagonal and $\langle,\rangle$ is the natural pairing between tangent and cotangent vectors.

I'd like to compute the Hessian of $f$ in the x and y variables.

So far I've found that for $\partial^2_{xx} f(x_0,\xi_0,y_0)$ I can put $x$ in normal coordinates at $y_0$ and then $exp_x^{-1} y_0$ is simply $-x$ so

$\partial^2_{xx} f(x_0,\xi_0,y_0) = 0.$

Similarly if $y$ is normal coordinates at $x_0$ I have

$\partial^2_{yy} f(x_0,\xi_0,y_0) = 0.$

But how can I compute the mixed derivatives $\partial^2_{xy}$. I am guessing this will involve Jacobi fields? Thanks.

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Your formulas so far don't look right to me, mostly because since you are treating $\xi_0$ as a constant, it should appear in the formula for the Hessian. Also, if $M$ is $\mathbb{R}^n$ with the standard flat metric, then $f(x,\xi,y) = \langle y-x,\xi\rangle$, whose Hessian is zero. – Deane Yang Mar 9 '13 at 22:42
Thanks Deane! I think they are correct now. As for $\xi$, I am doing the computation in some neighborhood $U$ where $T^*U$ is trivialized. I'm not differentiating in $\xi$ so it should be okay no? – Eric Mar 10 '13 at 0:32
I think you also need to figure out carefully what exactly the Hessian of a function on a Riemannian manifold is and how to compute it. It's not correct to call $x$ and $y$ "variables". They are points on a Riemannian manifold. They can be viewed as "variables" or "vectors" only if the manifold is $\mathbb{R}^n$. In particular, I have no idea what you mean by $\partial^2_{yy}$ and $\partial^2_{xx}$. – Deane Yang Mar 10 '13 at 3:01
And once you've got the definitions straight, then indeed the Hessian can be calculated using Jacobi fields. – Deane Yang Mar 10 '13 at 4:19