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Consider a complex-valued harmonic polynomial $f$ on ${\Bbb C}^2$ and assume that $f(0)=0$. Suppose also that $f$ does not vanish on the unit sphere $S^3\subset{\Bbb C}^2\simeq{\Bbb R}^4$. Does it follow that $f$ is a function of $|z|$ and $|w|$ alone, where $z,w$ are complex coordinates in ${\Bbb C}^2$?

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I don't quite understand. A non-constant polynomial $f(z,w)\in\mathbb{C}[z,w]$ is never a function of $|z|$ and $|w|$ alone. Maybe you mean something different? –  Pietro Majer Dec 14 '12 at 11:01
    
Maybe he means a polynomial in z,w,$\bar z$ and $\bar w$. –  Michael Renardy Dec 14 '12 at 11:31
    
For example, the harmonic function $|z|^2-|w|^2$ is a function of $|z|$ and $|w|$ alone. By the way, in my question I should have probably added "after a linear change of the real variables". –  Alexander Dec 14 '12 at 11:50
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It looks like you have harmonic polynomials on $\mathbb{R}^4$. You can compute the dimensions of the spaces of homogeneous harmonic polynomials, and I guess the condition $f(0)=0$ alone would not suffice to cut out only the functions of $|z|$ and $|w|$. –  Dima Pasechnik Dec 14 '12 at 13:58
    
Do you have an example of such a polynomial that is a function of |z| and |w|? I note that $|z|^2-|w|^2$ does have zeros on the unit sphere. –  Michael Renardy Dec 14 '12 at 19:44
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Since it doesn't really involve the complex structure, can't we use the standard real coordinates on $\mathbb{C}^2=\mathbb{R}^4$ as $x_0,x_1,x_2,x_3$ and consider the harmonic polynomial $$ f = x_0 + i\ (3{x_0}^2-{x_1}^2-{x_2}^2-{x_3}^2)? $$ This doesn't vanish on the unit sphere and doesn't seem (even under linear changes of variables) to be of the kind you seem to want to avoid (though you didn't define it very carefully). If you want to re-express it in terms of $z, \bar z, w,\bar w$ where $z = x_0+ix_1$ and $w = x_2 + ix_3$, you can certainly do that.

Added comment based on further conditions in the comments below: The OP has written that he wants an example of a harmonic polynomial map $f:\mathbb{C}^2\to\mathbb{C}$ that has the following properties: First, $f(0)=0$; second, $f$ does not vanish on the unit $3$-sphere, and, third, $f$ has no purely anti-holomorphic terms. He would like to know whether, up to a unitary transformation, such an $f$ must be a polynomial in $|z|$ and $|w|$ (where $z$ and $w$ are the standard complex coordinates on $\mathbb{C}^2$).

Now, the answer is that this is not so. The best way to see this is to realize that this latter condition is more geometrically phrased as asking whether or not such an $f$ is invariant under some $2$-torus $T\subset\mathrm{U}(2)$. This is because all $2$-tori in $\mathrm{U}(2)$ are conjugate in $\mathrm{U}(2)$ to the standard $2$-torus consisting of the diagonal elements of $\mathrm{U}(2)$. Thus, to find a counterexample, it's enough to construct an $f$ satisfying the $3$ properties whose symmetry group does not contain a $2$-torus. To that end, start with the example that the OP himself provided: $$ f_0 = \bigl(|z|^4-4\ |z|^2|w|^2+|w|^4\bigr)+ i\ \bigl(|z|^2-|w|^2\bigr) $$ (which does have a $2$-torus of symmetries). Note that the unitary symmetry group of the real and imaginary parts separately of $f_0$ is the standard, diagonal $2$-torus in $\mathrm{U}(2)$. Now, select any real-valued, harmonic polynomial that has no purely anti-holomorphic terms and that is not a polynomial in $|z|$ and $|w|$, say $h= z\bar w + w\bar z$. Now, for some small (real) $\epsilon>0$, consider the polynomial $$ f = f_0 + \epsilon\ h. $$ Clearly, for small enough $\epsilon$, this $f$ will satisfy the three conditions. Moreover, Because the unitary symmetry group of this polynomial must be intersection of the unitary symmetry groups of its individual real and imaginary parts, it follows that the unitary symmetry group of $f$ is just the circle of scalar multiplication by $e^{i\theta}$. Consequently, there is no unitary change of variables that will carry $f$ into a polynomial in $|z|$ and $|w|$.

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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. –  Alexander Isaev Dec 15 '12 at 9:48
    
@Alexander Isaev: I think your comment to Michael just above probably should have been a response to his comment to your question and not to my answer. You might consider moving it. As for 'dependent only on $|z|$ and $|w|$ (after a linear change of variables)', I wasn't sure what you were trying for, but I was suspecting that you wanted to know whether examples of the kind you seek must necessarily have a $2$-parameter symmetry group (such as being invariant under multiplication by unit complex numbers in two different lines, which is what happens for polynomials in $|z|$ and $|w|$). –  Robert Bryant Dec 15 '12 at 13:59
    
@Alexander Isaev: By the way, the linear symmetry group of the polynomial $f$ that I wrote down is a copy of $\mathrm{O}(3)$, acting as the orthogonal transformations in the variables $x_1,x_2,x_3$. This symmetry group does not contain an abelian $2$-parameter symmetry group, so there is no linear change of variables that would bring it to being a polynomial in $|z|$ and $|w|$. –  Robert Bryant Dec 15 '12 at 14:38
    
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 $z,w$). –  Alexander Isaev Dec 16 '12 at 12:52
    
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. –  Alexander Isaev Dec 16 '12 at 22:25
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Complex polynomial $f(z,w)=z+iw$ is harmonic, vanishes at the origin, and never equals $0$ on the unit sphere. However it is not a function of $|z|$ and $|w|$.

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$z=1/\sqrt{2}$, $w=i/\sqrt{2}$. –  Michael Renardy Dec 14 '12 at 15:35
    
Yes, Michael is right, $z+iw$ does vanish on the sphere. Certainly, my question must be understood up to a linear change of the real coordinates. Otherwise, one can take a function of $|z|,|w|$ on-vanishing on the sphere and then simply swap, say, the real part of $z$ with the imaginary part of $w$. The resulting function will not be a function of $|z|,|w|$ any more. So, my question needs to be understood modulo such things. –  Alexander Dec 14 '12 at 16:44
    
Dima, the main condition is the condition of non-vanishing on the sphere. That rules out a lot of homogeneous polynomials. –  Alexander Dec 14 '12 at 16:49
    
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. –  Alexander Isaev Dec 14 '12 at 18:32
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