One of the versions of the classical Levi problem asks the following:

Let $X$ be a complex manifold. Is it true that $X$ is Stein iff $X$ admits a smooth exhaustion strictly plurisubharmonic function?

The answer was proved to be affirmative by [Grauert68], I believe.

When we allow $X$ to have singularities we come straight away to a problem of defining a strictly plurisubharmonic function. Narasimhan uses the following definition in his proof of the singular version of Levi problem (see [FN80]).

Let $X$ be a reduced complex analytic space. A function $\rho\colon X\to \mathbb R$ is called plurisubharmonic if for every point $x\in X$ there exist an embedding of a neighbourhood $U$ of $x$ into $\mathbb C^N$ such that $\rho|_U$ comes as the restriction of a plurisubharmonic function $\rho'\colon \mathbb C^N\to \mathbb R$ to $U$. A plurisubharmonic function is called strictly plurisubharmonic if its small enough local perturbations are also plurisubharmonic functions.

Please check my words as I'm not an expert in complex analysis.

As far as I understand, Narasimhan does not impose any smoothness assumptions on plurisubharmonic functions in his proof. I wonder if a definition of a "smooth" strictly plurisubharmonic function can be given in easier terms, in terms of the $\partial\overline{\partial}$-operator for instance.

It seems too naïve to think that the following definition of a strictly plurisubharmonic function could allow one to make a conclusion about Steinity.

Let $X$ be a reduced complex analytic space. A function $\rho\colon X\to\mathbb R$ is called smooth strictly plurisubharmonic in a naïve sense if the following holds

  1. The restriction of $\rho$ to the set of non-singular points of $X$ is smooth.
  2. The form $\sqrt{-1}\partial\overline{\partial}\rho$ is a strictly positive $(1,1)$-form i.e. its restriction to the tangent space of any (maybe singular) point is strictly positive.

Now assume we have an exhaustion strictly plurisubharmonic function on $X$ in the sense of this naïve definition. What are we able to conclude about $X$? Which further assumptions on $\rho$ or on $X$ should one impose to conclude that $X$ is Stein?

My motivation for the question comes from the following observation made in [HKLR87].

Let $X$ by a hyperkähler manifold equipped with a HKLR-compatible $U(1)$ action i.e. the one that rotates complex structures. More precisely, for every $\lambda\in U(1)$ $$ \lambda^*\omega_I = \omega_I \:\:\:\:\: \lambda^*\Omega_I = \lambda\Omega_I $$ where $\omega_I\in \Lambda^{1,1}_IX$ is the Kähler form and $\Omega_I\in\Lambda^{2,0}_I X$ is the holomorphic symplectic form. Let $\mu\colon X\to \mathbb R$ be a moment map for this action i.e. $$ d\rho = \iota_\varphi\omega_I $$ where $\varphi$ is the vector field tangent to the $U(1)$-action.

Choose a complex structure $J\in\mathbb H$ which anticommutes with $I$. Let $\overline{\partial}_J$ be the $\overline{\partial}$-operator on differential forms for the complex structure $J$. Then $$ -\sqrt{-1}\partial_J\overline{\partial}_J \rho = \omega_J $$ In particular, $-\rho$ is a smooth strictly plurisubharmonic function.

The problem with the observation above is that most of examples of HKLR-compatible $U(1)$-actions arise on hyperkähler quotients of finite or infinite dimensional affine spaces. They are quite rarely smooth. So if one wants to conculde anything about Steinity of $X_J$ one needs an understanding of the notion of plurisubharmonicity in the singular case.

(a little bit disappointing) postscriptum. I am able to prove that under mild assumptions hyperkähler quotients of affine spaces are affine varieties when regarded as complex varieties with the complex structure $J$. My proof uses another methods. So alas, these varieties do not give us any interesting examples of Stein non-affine varieties. However, it is very tempting for me to apply some singular Levi problem to conclude more or less straightforwardly that these varieties are Stein. The exhaustion condition is not too difficult to check.


[Grauert68] Grauert, H.: On Levi's problem and the imbedding of real-analytic manifolds. Ann. of Math. 68, 460-472 (1968)

[FN80] Fornaess, J.E.; Narasimhan R.: The Levi Problem on Complex Spaces with Singularities. Math. Ann. 248, 47-72 (1980)

[HKLR87] Hitchin, N.J.; Karlhede, A.; Lindström, U.; Rocek, M.: Hyperkähler metrics and supersymmetry. Commun. Math. Phys. 108 (1987), 535 -- 589


1 Answer 1


Privet, Anya.

Your reference [FN80] actually seems to contain an answer to this problem! They state (in particular, see the question 1.5 in the introduction) that the class of weakly psh functions, i.e. functions which are psh being restricted on any holomorphic disk coincides with the class of psh functions in the restriction sense.

I think it is implied by your condition (I do not precisely understand the assumptions, but clearly if $\partial \bar \partial f$ is well defined and positive then its restriction on any curve is nonnegative too).

  • $\begingroup$ Privet, dorogoy. I mean the Zariski tangent space (en.wikipedia.org/wiki/Zariski_tangent_space). You made a nice remark about curves. I believe that the arguments with disk embeddings allow to give a definition of psh as "$\partial\overline{\partial}\rho$ is non-negative on the smooth part" plus the continuity of $\rho$ -- and it's equivalent to the one of Narasimhan (correct me if i'm wrong). But I'm not at all sure if the same argument should work for strictly psh functions. $\endgroup$ May 10, 2020 at 19:34
  • $\begingroup$ As far as I remember, the definition of $\partial\overline{\partial}\rho$ exists but isn't really tame. I don't know any reference. $\endgroup$ May 10, 2020 at 19:37

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