The Question

Given Hessian manifold $M$, there is a natural Kahler structures on $TM$. Is it possible to write the Kahler potential of these in terms of the Hessian potential?


To elaborate a bit more, let me give some background information. There are two separate definitions for a Hessian manifold $(M,g)$ which initially appear distinct but turn out to be the same.

  1. There are local coordinates $\{x_i \}_{i=1}^n$ and a convex potential $\psi$ so that in the $x$-coordinates, $$g_{ij}= \frac{ \partial^2 \psi}{\partial x_i x_j} $$
  2. $(M,g)$ locally admits a dually flat structure. That is to say, it admits two flat connections $\nabla$ and $\nabla^*$ satisfying $$X(g(Y,Z)) = g(\nabla_X Y,Z) + g(Y, \nabla^*_X Z). $$

If we consider the tangent bundle of $TM$, we can use the flat connection $\nabla$ to induce an almost complex structure $J^\nabla$ and a Sasaki metric $g^\nabla$ on $TM$. It turns out that $(TM, g^\nabla, J^\nabla)$ is a Kahler manifold if and only if $(M,g)$ is Hessian. It's worth noting that we can dualize all of this and obtain a second Kahler structure on $TM$ using the dual connection, as well.

In this setting, I'm wondering if it's possible to use potential $\psi$ to write the Kahler potential $\Psi$ for the Sasaki metric and induced complex structure.

For more details on the Sasaki metric and almost complex structure on $TM$, the paper of Satoh has more information.

Satoh, Hiroyasu, Almost Hermitian structures on tangent bundles, Suh, Young Jin (ed.) et al., Proceedings of the 11th international workshop on differential geometry, Taegu, Korea, November 9--11, 2006. Taegu: Kyungpook National University. 105-118 (2007). ZBL1125.53022..

  • $\begingroup$ Isn't it enough to just substitute $g=\text{Hess}^{\nabla}\psi$? By Thm. 3.1 in the Satoh paper the Kähler form is given by the pullback of the symplectic form on the cotangent bundle of $M$ by the isomorphism induced by the metric. You therefore have $i/2\partial\bar\partial\Psi=\varphi^{*}_{\text{Hess}^{\nabla}\psi}\Omega^{*}$. $\endgroup$ – S.Surace Dec 12 '18 at 13:08
  • $\begingroup$ My understanding is that expression gives a simple expression for the Kahler form, but I don't see how to use that to solve for the Kahler potential. Am I missing something easy? $\endgroup$ – Gabe K Dec 12 '18 at 15:17
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    $\begingroup$ If you calculate the expression with the symplectic form and use up all the knowledge of the metric and connection on $M$, you should end up with an equation with two and three covariant derivatives on one side, and two Dolbeault differentials on the other. I don't know enough about complex forms to say whether this allows you to 'guess' the Kähler potential. $\endgroup$ – S.Surace Dec 13 '18 at 15:44
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    $\begingroup$ I would suggest to try to do the explicit calculation for a simple example, like $M=\mathbb{R}$ and $\psi(x)=\frac12x^2$ (with Euclidean metric and Levi-Civita connection which is flat and self-dual). Maybe it is too simple to capture any interesting geometry, but if there is a nice relation between the two potentials you should be able to calculate it in this case and then, together with the equation you get from my argument above, make an educated guess for the general case. $\endgroup$ – S.Surace Dec 13 '18 at 15:51
  • $\begingroup$ Thanks for the suggestions. I had worked it out in the flat case, in which the Kahler potential is $\Psi=\frac{1/2}(x^2 + v^2)$ (the answer gives a good explanation of this). This suggests that the potential should be $\psi(x)+ \psi_{ij}(x) v^iv^j$, at least to first order. However, that guess doesn't work in general and I can't even find a Taylor series for the Kahler potential in terms of the Hessian potential. $\endgroup$ – Gabe K Dec 13 '18 at 22:27

My answer deals only with the case when $\nabla$ is the Levi-Civita connection.

As you seem to know the bundle $TTM$ naturally decomposes as a direct sum of the vertical subbundle $V$ and the horizontal subbundle $H$. In the case when the metric is flat, $H$ is integrable (these conditions are in fact equivalent). It means that there is a natural local decomposition of the total space to the tangent bundle as $TU \cong U\times F_p$ where $U$ is a small enough open subspace and $F_p$ is the fiber of the tangent bundle over a point $p$ (this decomposition depends only on the choice of the point). $TU$ is in fact isomorphic to $U\times F_p$ not only as a manifold but as a riemannian manifold equipped with the Sasaki metric. This means that $\Psi_0 = \psi + \varphi$ where I denote by $\varphi$ the potential of the constant metric on $F_p$ and by $\Psi_0$ I mean the function whose Hessian is the Sasaki metric. I claim that it is in fact the Kähler potential.

Unfortunately I can't invent a natural proof that $\Psi_0$ is the Kähler potential so I'll do this in coordinates. As the metric is flat we can choose coordinates $x_1,...x_n$ in $U$ and coordinates $y_1,... y_n$ in $F_p$ such that the Sasaki metric is written as the standard Euclidean and the complex structure $I$ acts like this: $I\frac{\partial}{\partial x_i} = \frac{\partial}{\partial y_i}$ and $I\frac{\partial}{\partial y_i} = -\frac{\partial}{\partial x_i}$. Then it is a direct computation that $\Psi = \Psi_0 = \frac{1}{2}\sum (x_i^2 + y_i^2)$.

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It turns out the answer was simpler than I initially thought, so thought I would provide an answer in case others were interested. S.Surace's suggestion in the comments was exactly right, but I didn't understand how to use it when I first asked the question.

Given any set of coordinates $\{ x_i \}_{i =1}^n$ on $M$, we can define smooth functions $y_1,\ldots,y_n$ on the tangent bundle $TM$ as follows. We define $y_j(X)=X^j$ for a tangent vector $X = \sum_{i=1}^n X^i \frac{\partial}{\partial x^i}$. The collection of functions $\{ (x_i, y_i) \}_{i=1}^n$ then form local coordinates for $TM$.

Furthermore, if we take $\{ x_i \}_{i =1}^n$ to be the coordinates where the metric is the second derivative of a potential $\Psi$ (more precisely, we need an atlas of such coordinates with affine transition maps), we can define complex coordinates $ \{ z^i = x^i + \sqrt{-1} y^i \}_{i=1}^n$ on $TM$. It turns out that these coordinates are holomorphic with respect to the complex structure $J^\nabla$.

In these coordinates, the K\"ahler metric is simply $$ \omega_{i \bar j} = \frac{\sqrt{-1}}{2} \frac{\partial^2}{\partial z^i \partial \bar z^j} \Psi, $$ where we define $\Psi(z)=\Psi(x)$ (independent of the $y$ coordinates). Since the original potential was strongly convex, $\Psi$ (defined on TM) is strongly plurisubharmonic. As such, the K\"ahler potential is simply $\Psi$, extended to be constant with respect to the imaginary part.

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