Skip to main content
Alexander Isaev's user avatar
Alexander Isaev's user avatar
Alexander Isaev's user avatar
Alexander Isaev
  • Member for 13 years, 7 months
awarded
awarded
comment
Harmonic polynomials on complex 2-space
Robert Bryant: Thank you for your latest example. Indeed, you are right, and a simple perturbation of $f_0$ is a counterexample to what I asked. I have to think more about what conditions to impose. The three conditions that I stated come from the geometry of totally real embeddings of $S^3$ in ${\Bbb C}^3$. The fact that these conditions do not guarantee dependence on just $|z|$, $|w|$ means that such embeddings can have a more interesting structure than I previously thought. Once again, thank you very much for your help.
comment
Harmonic polynomials on complex 2-space
Robert Bryant: Once again, thank you for your example and remarks. My objective is to find a reasonable sufficient condition for a complex-valued harmonic polynomial $f$ on ${\Bbb C}^2$ to be a function of $|z|$, $|w|$. I need that to study totally real embeddings of the 3-sphere into ${\Bbb C}^3$. I am now looking at the following three conditions: (i) $f(0)=0$, (ii) $f$ does not vanish on $S^3$, (iii) $f$ contains no purely anti-holomorphic terms. I hope that under these three assumptions $f$ is a function of $|z|$, $|w|$ (modulo a unitary change of $z,w$).
comment
Harmonic polynomials on complex 2-space
Michael, one example of such a polynomial is |z|4+|w|4−4|z|2|w|2+i(|z|2−|w|2).
awarded
comment
Harmonic polynomials on complex 2-space
Dear Robert, maybe you are right, and one cannot make a linear change to bring your example into the form dependent only on $|z|$, $|w|$. Maybe I should think more carefully about what exactly I want. For example, the polynomials I am dealing with do not have purely anti-holomorphic terms (when written in some complex coordinates), whereas if your example is rewritten through of complex variables, it will always contain anti-holomorphic terms (or so it seems). Thank you for your example anyway.
comment
Harmonic polynomials on complex 2-space
Alexandre Eremenko, a non-constant entire function (for example, $z+iw$) cannot have a non-empty compact zero set, so if a holomorphic polynomial vanishes at the origin, it also vanishes somewhere on the unit sphere. The question that I asked above is an analogue of this statement for harmonic polynomials. In a sense, harmonic polynomials that are not functions of $|z|$, $|w|$ (after a suitable linear change of the real coordinates) should behave like holomorphic functions: their zero sets should be non-compact. I hope this explains my motivation for the question.
comment
Harmonic functions on the plane
In fact, I think I know how to adapt your proof to this more general situation. One has to consider the region where $-kA<v<nA$ for suitable $k$, $n$.
comment
Harmonic functions on the plane
Actually, I am interested in showing that the tube domain in ${\Bbb C}^2$ with base given by $y>x^3$ is Kobayashi-hyperbolic. If the question that I have asked had a positive answer, it would mean that the domain is Brody-hyperbolic, which is a somewhat weaker statement.
comment
Harmonic functions on the plane
Thank you very much for your solution. Your argument proves that there exists no non-constant harmonic map from the plane into the domain in the plane lying above the cubic parabola $y>x^3$. Consider now the region defined as follows: $y>tx^3$ for $x\ge 0$ and $y>sx^3$ for $x<0$, where $s,t>0$. If one applies your argument to this more general domain, it seems that the answer depends on $s$ and $t$. However, I believe that there is no non-constant map into this domain for any $s$ and $t$. Can your proof be adopted to the case of general $s$ and $t$? Alex Isaev.
accepted
asked
Loading…
awarded
comment
Classical invariant theory: absolute rational invariants and $GL(2)$-orbits
I have thought about what you said, and I now agree. I am wondering if you let me mention this fact in one of my papers. Can I refer to "personal communications" with you? If you do not mind my doing that, what is your name? Thank you again for your answer.
comment
Classical invariant theory: absolute rational invariants and $GL(2)$-orbits
Thank you for that, but does $GL_2({\Bbb C})$ really act with finite stabilizers? For example, for the quadratic form $xy$ all the maps $x\mapsto c x$, $y\mapsto 1/c y$ are in the stabilizer.
awarded
Loading…