# ''Are Hermitian metric pullbacks automatically via biholomorphisms?''

The awkward title is an attempt at approximating the following specific question: Let $(M^{2n}, J)$ be a complex manifold, suppose $g_0$ is a Riemannian metric $M$ compatible with $J$, and suppose $\phi_t$ is a one parameter family of diffeomorphisms of $M$ generated by a vector field $X$ such that $\phi_t^* g$ is compatible with $J$ for all $t$. The questions is: must $X$ be a (real part of a) holomorphic vector field in the sense that $L_X J = 0$?

Assume that $g_0$ is Kahler and that $X = \nabla f$ for some smooth function $f$. As the family $\phi_t^* g_0$ remains compatible with $J$, it follows that $L_{\frac{1}{2} \nabla f} g_0 = \nabla^2 f$ is of type $(1,1)$. Using the K\"ahler hypothesis one can show that $(\nabla^2 f)^{1,1}(\cdot, J \cdot) = \frac{1}{2} d d^c f$. On the other hand, since the associated K\"ahler form $\omega$ is closed, it follows from the Cartan formula that $L_{\frac{1}{2} \nabla f} \omega = \frac{1}{2} d d^c f$. Thus it follows that $L_{\frac{1}{2} \nabla f} \omega = (L_{\frac{1}{2} \nabla f} g)(\cdot, J \cdot) = L_{\frac{1}{2} \nabla f} \omega + g(L_{\frac{1}{2} \nabla f} J\cdot , \cdot)$, and hence $L_{\nabla f} J \equiv 0$, as required.

• your "proof" is incorrect, because you don't have $\omega_t=\phi_t^*\omega_0$ in general. Just write down the action of both sides of this "equality" on a pair of vectors, and you will see that they need not be equal in general. – YangMills May 29 '17 at 21:54
• See the comment below Bryant's answer – Thisquestionisreallyhard May 30 '17 at 5:10
• I had read it, but it is not useful. The vector field being gradient doesn't help. – YangMills May 30 '17 at 15:43
• See edits above. – Thisquestionisreallyhard May 31 '17 at 16:59

## 1 Answer

The answer is 'no' in specific cases, and it doesn't help that $g$ is assumed to be Kähler (i.e., you have a mistake in your argument in the second paragraph). For example, let $M^{2n}=\mathbb{C}^n$ and let $g$ be the standard Kähler metric, and assume $n>1$. Then $$g = \mathrm{d}z_1\circ\mathrm{d}\bar z_1 + \cdots + \mathrm{d}z_n\circ\mathrm{d}\bar z_n\,.$$ Then $g$ is just the standard flat metric on $\mathbb{R}^{2n} = \mathbb{C}^n$, and hence the symmetry group of $g$ that fixes the origin is $\mathrm{O}(2n)$. Let $\phi_t$ be the flow of an orthogonal vector field vanishing at the origin that is not holomorphic. These exist because $\mathrm{SO}(2n)$ is much larger than $\mathrm{U}(n)\subset\mathrm{SO}(2n)$. Then $\phi_t$ actually preserves $g$, so $\phi_t^*g=g$ is $J$-compatible for all $t$, but, in general $\phi_t^*J\not= J$, so the the vector field $X$ generating $\phi_t$ not the real part of a holomorphic vector field.

• Thanks! I realize that in the Kahler setting I only proved it if the vector field is further assumed gradient (which I think is still correct). If I've understood your example, it possibly suggests a more general statement: namely that in the setting I describe there exists a one-parameter family $\psi_t$ of isometries of $g_t = \phi_t^* g_0$ such that $\phi_t \circ \psi_t$ are biholomorphisms? – Thisquestionisreallyhard May 26 '17 at 19:22