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I suspect this is true for some class of analytic manifolds (Riemann surfaces maybe), but my knowledge in differential geometry is very poor, so I could not conclude it. For complex manifolds, is it possible to find $h$ hermitian metric such that $Isom_{h}(X) \cong Aut(X)$ (where $Isom_{h}(X)$ means the isometries of $h$ that preserves orientation)? What are the obstructions? Any ideas on how to measure this? Is this true at least for Riemann surfaces?

Thanks in advance.

EDIT: As a friend of mine has just suggested, what about the complexification $Isom_{h}(X)^{\mathbb{C}}$? Will $Isom_{h}(X)^{\mathbb{C}} \cong Aut(X)$ for some special class of complex manifolds?

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  • $\begingroup$ By $\mathrm{Isom}_h(X)$, do you mean only the holomorphic isometries of $h$, as Liviu wants it to mean? Assuming this, I think that an appropriate hypothesis on $X$ might be that the isotropy group $\mathrm{Aut}(X,p)\subset \mathrm{Aut}(X)$ should be compact for each $p\in X$. This is clearly necessary and may even be sufficient. You don't actually need that $\mathrm{Aut}(X)$ itself be compact, as the example of the punctured complex plane shows. Note that this hypothesis can hold at some points and not others; for example, consider $\mathbb{CP}^2$ blown up at three non-collinear points. $\endgroup$
    – Robert Bryant
    Apr 24, 2014 at 18:07
  • $\begingroup$ @RobertBryant Yes, holomorphic isometries. I forgot to put that they must preserve orientation. $\endgroup$
    – user40276
    Apr 24, 2014 at 21:15
  • $\begingroup$ Well, orientation, by itself, is not enough. You can have orientation preserving isometries (in higher dimensions) that are still not complex linear. $\endgroup$
    – Robert Bryant
    Apr 24, 2014 at 23:12
  • $\begingroup$ @RobertBryant thanks for the observation, I have never heard about this, do you know some example? $\endgroup$
    – user40276
    Apr 25, 2014 at 1:00
  • $\begingroup$ Sure, start with an elliptic curve $C$ with an anti-holomorphic isometry $\tau:C\to C$. Of course, $\tau$ does not preserve orientation, but $\tau\times\tau:C\times C\to C\times C$ does preserve orientation and is not holomorphic. About your edit for the complexified isometry group: That won't always be possible either; just look at the Poincaré disk. (Of course, it will sometimes be possible, but it won't always.) $\endgroup$
    – Robert Bryant
    Apr 25, 2014 at 8:21

3 Answers 3

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The group of isometries of a compact Riemann manifold is a compact Lie group. Thus if $X$ is compact and $G={\rm Aut}\;(X)$ is not compact, then you cannot choose such a metric. On the other hand, if $G$ is compact you can produce such a metric by averaging with respect to an invariant probability measure on $G$.

Edit. As Robert Bryant indicated in his comment I have to be more careful. If $(X,h)$ is a complex hermitian manifold, then $\DeclareMathOperator{\Isom}{Isom}$ by $\Isom_h(X)$ I understand the group of biholomorphic maps $F: X\to X$ such that $F^*h=h$.

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    $\begingroup$ You do have to be a tad more careful, though: It's conceivable (and, in fact, there are examples) of $\mathrm{Aut}(X)$-invariant hermitian metrics on $X$ for which the isometry group is larger than $\mathrm{Aut}(X)$, so you do have to make sure that your metric doesn't have any 'extra' isomorphisms if you are going to get equality, which the OP wanted. $\endgroup$
    – Robert Bryant
    Apr 24, 2014 at 16:44
  • $\begingroup$ @ Robert You are absolutely correct. I tacitly assumed that in the holomorphic category an isometry is a biholomorphic map. $\endgroup$
    – Liviu Nicolaescu
    Apr 24, 2014 at 17:40
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No. Check the Riemann sphere or any elliptic curve, or a punctured complex plane, or a complex plane. Otherwise yes (on Riemann surfaces), by the uniformization theorem.

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  • $\begingroup$ More examples: Grassmannians, projective spaces, complex Euclidean spaces. The automorphism group of projective space is the projectivized special linear group, while the isometry group is the projectivized unitary group. $\endgroup$
    – Ben McKay
    Apr 24, 2014 at 16:29
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    $\begingroup$ The biholomorphism group of $\mathbb{C}^2$ contains beasts like $(z,w) \mapsto (z,w+f(z))$ for any entire holomorphic function $f(z)$: far from preserving any metric, since the stabilizer of a point is infinite dimensional and not compact. $\endgroup$
    – Ben McKay
    Apr 24, 2014 at 17:23
  • $\begingroup$ What about the complexification of $Isom_h (X)$? In the first case, it will hold. $\endgroup$
    – user40276
    Apr 24, 2014 at 21:22
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    $\begingroup$ Try $\mathbb{C}^2$; the complexification of the isometry group is the complex affine group, finite dimensional, while the biholomorphism group is infinite dimensional. $\endgroup$
    – Ben McKay
    Apr 25, 2014 at 7:54
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Yes, this is so, sometimes. The prime example in the unit disc $|z|<1$ in the complex plane. This is a $1$-dimensional complex analytic manifold. Its automorphism group is a $3$-parametric group which is isomorphic to $PSL_2(R)$. In the unit disc there is a Riemannian metric, which is called the Poincare metric and for which the isometries are exactly the automorphisms.

Now by the uniformization theorem, almost all Riemann surfaces have the unit disc as the universal cover. It follows that for almost all Riemann surfaces there is such a metric, that your equality holds. It is called the hyperbolic metric. (Exceptional Riemann surfaces are the sphere, the plane the punctured plane and the complex $1$-parametric family of tori.)

Some of this extends to complex manifolds in higher dimension. The analog of the hyperbolic metric is called the Kobayashi metric, and it is believed that "most" complex manifolds have it, though this is difficult to prove in sprecific cases. Moreover, Kobayashi metric is frequently not a Riemannian (Hermirtian) metric but only a Finsler metric. Much about this can be found in the books of S. Kobayashi.

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  • $\begingroup$ Thanks for the answer, but it is not clear for me that the uniformization implies the assertion. How do you know the isometries of the quotient of the disc by a discrete subgroup of the automorphisms, maybe you are lifting something to the universal covering. $\endgroup$
    – user40276
    Apr 25, 2014 at 0:56
  • $\begingroup$ Uniformizationb+Schwarz Lemma combined imply that every automorphism is an isometry. $\endgroup$
    – Alexandre Eremenko
    Apr 25, 2014 at 5:10

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