Let $X$ be a complex manifold and $g$ a hermitian metric on $X$. Consider the Riemannian exponential $\exp_p: T_p X \to X$.
If $\exp_p$ is holomorphic for every $p \in X$, then $(\exp_p)^{-1}$, suitably restricted, provide holomorphic normal coordinates near $p$, with respect to which the metric osculates to order 2 to the standard metric at the origin. This shows that $g$ is a Kähler metric.

However, Kähler is not sufficient to ensure that $\exp_p$ is holomorphic: take $X$ a curve of genus $g \geq 2$. If $\exp_p:T_pX \to X$ is holomorphic, then it lifts to a holomorphic map from $T_pX$ to the universal cover $\widetilde{X} = \Delta$, giving a holomorphic map $T_pX \simeq \mathbb{C} \to \Delta$, which must be constant by Liouville's theorem. In fact, one can see that $\exp$ cannot be holomorphic if $X$ is Kobayashi hyperbolic.

This leaves the question: What are the hermitian manifolds/metrics whose exponential map is holomorphic?

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    $\begingroup$ In the $1$ complex dimension case this forces the metric to be locally isomorphic to the standard metric. The exponential map preserves length in the radial direction, so if it is conformal then it must also preserve length in the other direction. $\endgroup$ Jun 16 '11 at 3:52

NB: I've had a little time to think about this and can now improve my answer, in particular, removing the real-analytic assumption, which, as I suspected, was not necessary. Here is the improved answer:

If the metric $g$ is Kähler, then having the exponential map from a point $p\in M$ be holomorphic makes it flat in a neighborhood of $p$.

Suppose that $\exp_p:T_pM\to M$ is holomorphic near $0_p\in T_pM$ (where we use the natural holomorphic structure on the complex vector space $T_pM$). Let $z:T_pM\to\mathbb{C}^n$ be a complex linear isometry, so that the hermitian metric on $T_pM$ is just $|z|^2$ in the usual sense. Let $Z$ be the holomorphic 'radial' vector field on $\mathbb{C}^n$, whose real part is the standard radial vector field on $\mathbb{C}^n$.

Then $$ {\exp_p}^*g = g_{i\bar j}(z)\ dz^i\ d\overline{z}^j $$ for some functions $g_{i\bar j}$ on a neighborhood of $0\in\mathbb{C}^n$. Since $g$ is Kähler, there is a function $f$ defined on a neighborhood of $0\in\mathbb{C}^n$ such that $$ g_{i\bar j} = \frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}. $$

Now, the condition that $z$ furnish Gauss normal coordinates for ${\exp_p}^*g$ is easily seen to be that
$$ \mathcal{L}_Z\bigl(\bar\partial f\bigr) = \bar\partial\bigl(|z|^2\bigr). $$ In particular, $ \bar\partial\bigl(\mathcal{L}_Z(f - |z|^2)\bigr) = 0$, so $\mathcal{L}_Z(f - |z|^2) = h$ for some holomorphic function $h$ on a neighborhood of $0$. This $h$ must vanish at $0$, so it is easy, by adding the real part of the appropriate holomorphic function to $f$ (which won't change $g$) to arrange that $h\equiv0$ and, moreover, that $f(0) = 0$. But this now implies that the real-valued function $f-|z|^2$ vanishes at the origin and also is constant along the radial vector field. Thus, $f = |z|^2$, and the metric $g$ is flat in these coordinates.


I do not like complex numbers and can make a mistake easily...

Let $L_p$ be a complex line in a tangent space $T_pX$. It is easy to see that $\exp_p$ gives an isometric embedding $L_p\hookrightarrow X$ which is also star-shaped with center at $p$; set $L=\exp_p(L_p)$

Take any other point $q\in L$, and let $L_q\subset T_q$ be the tangent subspace to $L$. Note that the maps $\exp_p$ and $\exp_q$ coinside (up to a shift) on the geodesic $(pq)$. [Here I use that if two holomorphic maps coincide on the real line then they coincide in the complex plane.] It follows that $L=\exp_q(L_q)$. Therefore $L$ is totally geodesic.

In other words: For any complex sectional direction in $X$, there is a tangent totally geodesic surface which is isometric to complex plane.

In particular, the sectional curvature in all complex sectional directions is zero and therefore the curvature of $X$ is identically zero; the later stated in Kobayashi--Nomizu, Foundations of differential geometry, Volume 2 IX, Prop. 7.1. (thanks to RdN).

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    $\begingroup$ I don't quite understand your arguement for exp of a complex line to be totally geodesic. But if so there is a lemma in vol II of Kobayashi Nomizu (Prop. 7.1) that says if two tensors with the symmetry and J invariance of a Kahler curvature tensor agree on complex lines they must be equal. It's clear curves have to be flat by Gauss lemma and totally geodesic would mean the second fundamental vanishes, so we can measure R of the ambient space from such curves. Thus the Prop. says total curvature must vanish. $\endgroup$
    – RdN
    Jun 25 '11 at 16:43
  • $\begingroup$ @RdN, Thank you, I will include it in the answer $\endgroup$ Jun 28 '11 at 10:30

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