Harmonic polynomials on complex 2-space

Question

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 ${\Bbb C}^2$?

Answer 1

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 $|w|$.

Answer 2

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 $|w|$.