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.

3 replaced original argument with a better one

Well

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

If the metric $g$ is Kählerand real-analytic, then having the exponential map from even one a point $p$ p\in M$be holomorphic makes it flat . (Probably the assumption of real-analytic is not necessary, but I haven't thought about that, and the following argument that uses it is very easy.) It's enough to show that a Kähler metric on in a neighborhood of$0\in\mathbb{C}^n$with the property p$.

Suppose that the linear holomorphic coordinates are Gauss normal coordinates $\exp_p:T_pM\to M$ is necessarily flat. To see this, suppose that holomorphic near $z^i$ are linear coordinates 0_p\in T_pM$(where we use the natural holomorphic structure on the complex vector space$\mathbb{C}^n$such T_pM$). Let $z:T_pM\to\mathbb{C}^n$ be a complex linear isometry, so that the given hermitian metric on $g$ agrees with 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 one at radial vector field on$0$. \mathbb{C}^n$.

let $f$ be ${\exp_p}^*g = g_{i\bar j}(z)\ dz^i\ d\overline{z}^jfor some functions$g_{i\bar j}$on a neighborhood of$0\in\mathbb{C}^n$. Since$g$is Kählerpotential for , there is a function$g$, so f$ defined on a neighborhood of $0\in\mathbb{C}^n$ such thatg=\frac{\partial^2f}{\partial g_{i\bar j} = \frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}\ dz^i d\overline{z}^j.$$The Now, the condition that the z^i be z furnish Gauss normal coordinates centered on for 0 {\exp_p}^*g is just easily seen to be that z^i\ \overline{z}^j\ \frac{\partial^2f}{\partial z^imathcal{L}_Z\bigl(\bar\partial f\bigr) = \bar\partial\bigl(|z|^2\bigr).In particular,  \partial\overline{z}^j}= bar\partial\bigl(\mathcal{L}_Z(f - |z^1|^2 + \cdots + z|^2)\bigr) = 0, so \mathcal{L}_Z(f - |z^n|^2.z|^2) = h for some holomorphic function h on a neighborhood of However, now assuming that 0. This f h must vanish at 0, so it is real-analyticeasy, expand it in a power series in by adding the appropriate holomorphic function to z^i f (which won't change g) to arrange that h\equiv0 and, moreover, that \overline{z}^i. Then f(0) = 0. But this last identity clearly shows now implies that all of the terms in f must be zero except the pure real-valued function z-terms, f-|z|^2 vanishes at the pure \overline{z}-terms, origin and also is constant along the terms of bidegree (1,1), which must simply be |z^1|^2 + \cdots + |z^n|^2. radial vector field. Thus, g is f = |z|^2, and the standard metric g is flat in these coordinates. 2 improved some phrases Well, if the metric g is Kähler and real-analytic, then having the exponential map from even one point p be holomorphic makes it flat. (Probably the assumption of real-analytic is not necessary, but I haven't thought about that, and the following argument that uses it is very easy.) It's enough to show that a Kähler metric on a neighborhood of 0\in\mathbb{C}^n with the property that the linear holomorphic coordinates are Gauss normal coordinates is necessarily flat. To see this, suppose that z^i are linear coordinates on \mathbb{C}^n such that the given metric g agrees with the standard one at the identity. 0. Then let f be a Kähler potential for g, so that$$ g=\frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}\ dz^i d\overline{z}^j. $$The condition that the z^i be Gauss normal coordinates centered on 0 is just that$$ z^i\ \overline{z}^j\ \frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j} = |z^1|^2 + \cdots + |z^n|^2.  However, now assuming that $f$ is real-analytic, expand it in a power series in $z^i$ and $\overline{z}^i$. Then this last identity clearly shows that all of the terms in $f$ must be zero except the pure $z$-terms, the pure $\overline{z}$-terms, and the terms of bidegree $(1,1)$, which must simply be $|z^1|^2 + \cdots + |z^n|^2$. Thus, $g$ is the standard metric.

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