Return to Question

Let $X$ be a complex space.

We say that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

We say that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-distance $d_X$ on $X$ is a genuine distance.

Since holomorphic maps are distance-decreasing withe respect to the Kobayashi pseudo-distance, and since $d_\mathbb C\equiv 0$, we immediately see that any Kobayshi hyperbolic complex space is also Brody hyperbolic: if $f\colon\mathbb C\to X$ is any holomorphic map then $$ d_X(f(z),f(w))\le d_\mathbb C(z,w)=0, $$ so that $f$ must be constant, $d_X$ being a true distance.

Now, by Bordy's reparametrization lemma, if $X$ is moreover compact, then we also have that Brody hyperbolicity implies Kobayashi hyperbolicity. But in the non-compact case it is well known that Kobayashi hyperbolicity is actually stronger. A standard example is given by the domain $X$ in $\mathbb C^2$ described as follows: $$ X=\{(z,w)\in\mathbb C^2\mid|z|<1,|zw|<1\}\setminus\{(0,w)\mid|w|\ge 1\}. $$ This domain $X$ can be quite straightforwardly seen to be not Kobayashi hyperbolic (the origin has zero distance from points of the form $(0,b)$) but nevertheless there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

Here is my question:

Question. Do you know any example of a complex affine variety $X$ which is Brody hyperbolic but not Kobayashi hyperbolic? Or do you at least know a Stein example?

If such an example exists, then any compactification $\overline X$ of $X$ is not Kobayashi hyperbolic and so must contain non-constant holomorphic images of the complex plane. But then, all these images must cut the boundary $\overline X\setminus X$ of $\overline X$ in at least two points...

Thanks in advance!

Let $X$ be a complex space.

We say that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

We say that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-distance $d_X$ on $X$ is a genuine distance.

Since holomorphic maps are distance-decreasing withe respect to the Kobayashi pseudo-distance, and since $d_\mathbb C\equiv 0$, we immediately see that any Kobayshi hyperbolic complex space is also Brody hyperbolic: if $f\colon\mathbb C\to X$ is any holomorphic map then $$ d_X(f(z),f(w))\le d_\mathbb C(z,w)=0, $$ so that $f$ must be constant, $d_X$ being a true distance.

Now, by Bordy's reparametrization lemma, if $X$ is moreover compact, then we also have that Brody hyperbolicity implies Kobayashi hyperbolicity. But in the non-compact case it is well known that Kobayashi hyperbolicity is actually stronger. A standard example is given by the domain $X$ in $\mathbb C^2$ described as follows: $$ X=\{(z,w)\in\mathbb C^2\mid|z|<1,|zw|<1\}\setminus\{(0,w)\mid|w|\ge 1\}. $$ This domain $X$ can be quite straightforwardly seen to be not Kobayashi hyperbolic (the origin has zero distance from points of the form $(0,b)$) but nevertheless there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

Another exemple of algebraic nature (a reference of which can be found in S. Lang's book "Introduction to complex hyperbolic spaces" as pointed out to me by W. Cherry) is the following. Take four lines in general position in $\mathbb P^2$ and let $\ell_1, \ell_2,\ell_3$ the diagonal lines of this configuration. Now, choose three distinct points $P_1,P_2,P_3$, such that $P_i\in\ell_i$ but $P_i$ is not on the starting four lines. Then, $$ W:=\mathbb P^2\setminus\{\text{the four lines}\}\setminus\{P_1,P_2,P_3\} $$ is Brody hyperbolic but not Kobayashi hyperbolic. Of course, $W$ is quasi-projective, but not affine.

Here is my question (which is indeed already asked by S. Lang right after the above example):

Question 1. Is there an example of a complex affine variety $X$ which is Brody hyperbolic but not Kobayashi hyperbolic? Or do you at least know a Stein example?

If such an example exists, then any compactification $\overline X$ of $X$ is not Kobayashi hyperbolic and so must contain non-constant holomorphic images of the complex plane. But then, all these images must cut the boundary $\overline X\setminus X$ of $\overline X$ in at least two points...

Possibly, a natural place to look at, might be the complement of an ample divisor in an abelian variety: by a result of G.-E. Dethloff and S. S.-Y. Lu this complement is always Brody hyperbolic. Moreover, it is of course affine. So, Question 1 would be answered in the affirmative if one were able to answer positively to the following.

Question 2. Is there an example of a pair $(A,D)$, where $A$ is an abelian variety and $D\subset A$ an ample divisor, such that $A\setminus D$ is not Kobayashi hyperbolic?

Thanks in advance!

Let $X$ be a complex space.

We say that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

We say that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-distance $d_X$ on $X$ is a genuine distance.

Since holomorphic maps are distance-decreasing withe respect to the Kobayashi pseudo-distance, and since $d_\mathbb C\equiv 0$, we immediately see that any Kobayshi hyperbolic complex space is also Brody hyperbolic: if $f\colon\mathbb C\to X$ is any holomorphic map then $$ d_X(f(z),f(w))\le d_\mathbb C(z,w)=0, $$ so that $f$ must be constant, $d_X$ being a true distance.

Now, by Bordy's reparametrization lemma, if $X$ is moreover compact, then we also have that Brody hyperbolicity implies Kobayashi hyperbolicity. But in the non-compact case it is well known that Kobayashi hyperbolicity is actually stronger. A standard example is given by the domain $X$ in $\mathbb C^2$ described as follows: $$ X=\{(z,w)\in\mathbb C^2\mid|z|<1,|zw|<1\}\setminus\{(0,w)\mid|w|\ge 1\}. $$ This domain $X$ can be quite straightforwardly seen to be not Kobayashi hyperbolic (the origin has zero distance from points of the form $(0,b)$) but nevertheless there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

Here is my question:

Question. Do you know any example of a complex affine variety $X$ which is Brody hyperbolic but not Kobayashi hyperbolic?

If such an example exists, then any compactification $\overline X$ of $X$ is not Kobayashi hyperbolic and so must contain non-constant holomorphic images of the complex plane. But then, all these images must cut the boundary $\overline X\setminus X$ of $\overline X$ in at least two points...

Thanks in advance!

Let $X$ be a complex space.

We say that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

We say that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-distance $d_X$ on $X$ is a genuine distance.

Since holomorphic maps are distance-decreasing withe respect to the Kobayashi pseudo-distance, and since $d_\mathbb C\equiv 0$, we immediately see that any Kobayshi hyperbolic complex space is also Brody hyperbolic: if $f\colon\mathbb C\to X$ is any holomorphic map then $$ d_X(f(z),f(w))\le d_\mathbb C(z,w)=0, $$ so that $f$ must be constant, $d_X$ being a true distance.

Now, by Bordy's reparametrization lemma, if $X$ is moreover compact, then we also have that Brody hyperbolicity implies Kobayashi hyperbolicity. But in the non-compact case it is well known that Kobayashi hyperbolicity is actually stronger. A standard example is given by the domain $X$ in $\mathbb C^2$ described as follows: $$ X=\{(z,w)\in\mathbb C^2\mid|z|<1,|zw|<1\}\setminus\{(0,w)\mid|w|\ge 1\}. $$ This domain $X$ can be quite straightforwardly seen to be not Kobayashi hyperbolic (the origin has zero distance from points of the form $(0,b)$) but nevertheless there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

Here is my question:

Question. Do you know any example of a complex affine variety $X$ which is Brody hyperbolic but not Kobayashi hyperbolic? Or do you at least know a Stein example?

If such an example exists, then any compactification $\overline X$ of $X$ is not Kobayashi hyperbolic and so must contain non-constant holomorphic images of the complex plane. But then, all these images must cut the boundary $\overline X\setminus X$ of $\overline X$ in at least two points...

Thanks in advance!

Let $X$ be a complex space.

We say that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

We say that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-distance $d_X$ on $X$ is a genuine distance.

Since holomorphic maps are distance-decreasing withe respect to the Kobayashi pseudo-distance, and since $d_\mathbb C\equiv 0$, we immediately see that any Kobayshi hyperbolic complex space is also Brody hyperbolic: if $f\colon\mathbb C\to X$ is any holomorphic map then $$ d_X(f(z),f(w))\le d_\mathbb C(z,w)=0, $$ so that $f$ must be constant, $d_X$ being a true distance.

Now, by Bordy's reparametrization lemma, if $X$ is moreover compact, then we also have that Brody hyperbolicity implies Kobayashi hyperbolicity. But in the non-compact case it is well known that Kobayashi hyperbolicity is actually stronger. A standard example is given by the domain $X$ in $\mathbb C^2$ described as follows: $$ X=\{(z,w)\in\mathbb C^2\mid|z|<1,|zw|<1\}\setminus\{(0,w)\mid|w|\ge 1\}. $$ This domain $X$ can be quite straightforwardly seen to be not Kobayashi hyperbolic but nevertheless there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

Here is my question:

Question. Do you know any example of a complex affine variety $X$ which is Brody hyperbolic but not Kobayashi hyperbolic?

If such an example exists, then any compactification $\overline X$ of $X$ is not Kobayashi hyperbolic and so must contain non-constant holomorphic images of the complex plane. But then, all these images must cut the boundary $\overline X\setminus X$ of $\overline X$ in at least two points...

Thanks in advance!

Let $X$ be a complex space.

We say that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

We say that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-distance $d_X$ on $X$ is a genuine distance.

Since holomorphic maps are distance-decreasing withe respect to the Kobayashi pseudo-distance, and since $d_\mathbb C\equiv 0$, we immediately see that any Kobayshi hyperbolic complex space is also Brody hyperbolic: if $f\colon\mathbb C\to X$ is any holomorphic map then $$ d_X(f(z),f(w))\le d_\mathbb C(z,w)=0, $$ so that $f$ must be constant, $d_X$ being a true distance.

Now, by Bordy's reparametrization lemma, if $X$ is moreover compact, then we also have that Brody hyperbolicity implies Kobayashi hyperbolicity. But in the non-compact case it is well known that Kobayashi hyperbolicity is actually stronger. A standard example is given by the domain $X$ in $\mathbb C^2$ described as follows: $$ X=\{(z,w)\in\mathbb C^2\mid|z|<1,|zw|<1\}\setminus\{(0,w)\mid|w|\ge 1\}. $$ This domain $X$ can be quite straightforwardly seen to be not Kobayashi hyperbolic (the origin has zero distance from points of the form $(0,b)$) but nevertheless there is no non-constant holomorphic map $f\colon\mathbb C\to X$.

Here is my question:

Question. Do you know any example of a complex affine variety $X$ which is Brody hyperbolic but not Kobayashi hyperbolic?

If such an example exists, then any compactification $\overline X$ of $X$ is not Kobayashi hyperbolic and so must contain non-constant holomorphic images of the complex plane. But then, all these images must cut the boundary $\overline X\setminus X$ of $\overline X$ in at least two points...

Thanks in advance!