Return to Answer

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 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.

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.

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 $0$. 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.

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 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.

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. 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 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.

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 $0$. 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.