Let $a = (a_1, …, a_n) \in \mathbb{N}_{>0}^n$, I want to show that the following is a manifold of $2n – 3$ (real) dimension.
$$V(a) = \left\{ (z_1, …, z_n) \in \mathbb{C}^n : \sum_{j=1}^{n}{|z_j|^2} = 1, \sum_{j=1}^{n}{z_j^{a_j}} = 0 \right\}$$
I know I can do this using the regular value theorem. Writing $z_i = x_j + iy_j$, the formula I'm looking at is:
$$F(x_1, …, x_n, y_1, …, y_n) = \left(\sum_{j=1}^{n}{x_j^2} + \sum_{j=1}^{n}{y_j^2}, \Re\left({\sum_{j=1}^{n}{(x_j + iy_j)^{a_j}}}\right), \Im\left({\sum_{j=1}^{n}{(x_j + iy_j)^{a_j}}}\right)\right)$$
So $V(a) = F^{-1}(1, 0, 0)$ which means that what I need to do is show that $F^{-1}(a) \cap crit(F) = \emptyset$ where $crit(F)$ is the set of critical points. My problem is that I am having trouble figuring out when $DF$ has rank less than $3$.
My attempt: apply binomial expansion to $(x_j + iy_j)^{a_j}$ so
$$\Re\left((x_j+iy_j)^{a_j}\right) = \sum_{k=0}^{\lfloor a_j/2 \rfloor}{(-1)^k\binom{a_j}{2k}(x_j)^{2k}(y_j)^{a_j-2k}}$$
$$\Im\left((x_j+iy_j)^{a_j}\right) = \sum_{k=0}^{\lfloor (a_j – 1)/2 \rfloor}{(-1)^k\binom{a_j}{2k+1}(x_j)^{2k+1}(y_j)^{a_j-2k-1}}$$
Letting $U(x_1, …, x_n) = \Re\left({\sum_{j=1}^{n}{(x_j + iy_j)^{a_j}}}\right)$, $V(x_1, …, x_n) = \Im\left({\sum_{j=1}^{n}{(x_j + iy_j)^{a_j}}}\right)$
But then:
\begin{align}
DF
& = \begin{pmatrix}
2x_1 & \cdots & 2x_n & 2y_1 & \cdots & 2y_n \\
\partial U/\partial x_1 & \cdots & \partial U/\partial x_n & \partial U/\partial y_1 & \cdots & \partial U/\partial y_n \\
\partial V/\partial x_1 & \cdots & \partial V/\partial x_n & \partial V/\partial y_1 & \cdots & \partial V/\partial y_n
\end{pmatrix} \\
& = \begin{pmatrix}
2x_1 & \cdots & 2x_n & 2y_1 & \cdots & 2y_n \\
\partial U/\partial x_1 & \cdots & \partial U/\partial x_n & \partial U/\partial y_1 & \cdots & \partial U/\partial y_n \\
\partial U/\partial y_1 & \cdots & \partial U/\partial y_n & -\partial U/\partial x_1 & \cdots & -\partial U/\partial x_n
\end{pmatrix}
\end{align}
where:
$$\frac{\partial U}{\partial x_j} = \frac{\partial V}{\partial y_j} = \sum_{k=0}^{\lfloor a_j/2 \rfloor}{(-1)^k\binom{a_j}{2k}(2k)(x_j)^{2k-1}(y_j)^{a_j-2k}}$$
$$\frac{\partial U}{\partial y_j} = -\frac{\partial V}{\partial x_j} = \sum_{k=0}^{\lfloor (a_j – 1)/2 \rfloor}{(-1)^k\binom{a_j}{2k+1}(a_j-2k-1)(x_j)^{2k+1}(y_j)^{a_j-2k-2}}$$
So, for the rank to be less than $3$, this means that, for all $j$:
$$2x_j = p \frac{\partial U}{\partial x_j} + q \frac{\partial U}{\partial y_j}$$
$$2y_j = p \frac{\partial U}{\partial y_j} – q \frac{\partial U}{\partial x_j}$$
But the formulas for the partial derivatives are so massive, I have no idea how to proceed. Any hints would be appreciated!
Note: I also considered using polar coordinates, but I spoke with my professor about it and we can't use polar coordinates since: $\{(r, \theta) : r \geq 0, \theta \in [0, 2\pi)\}$ is not a smooth manifold because: (a) it has boundary; and (b) when $r = 0$, we get the same point for all $\theta$.
Best Answer
By definition of $V(a)$, up to permuting coordinates we can assume that $z_1\not=0\not=z_2$ and that $$ z_1^{a_1}/z_2^{a_2}\not\in\mathbb R_{>0}.\qquad (\clubsuit) $$ Using local real coordinates $$ (\rho_1,\phi_1,\rho_2,\phi_2,x_3,y_3,\dots,x_n,y_n) $$ with $$ z_b=\begin{cases}\rho_b\exp(i\phi_b/a_b),& b=1,2, \\ x_b+iy_b, & b=3,\dots,n, \end{cases} $$ on $\mathbb C^n$, the Jacobian matrix of the three constraints $$ \sum_{j=1}^n\rho_j^2=1,\ \ \sum_{j=1}^n\rho_j^{a_j}\cos(\phi_j)=0,\ \ \sum_{j=1}^n\rho_j^{a_j}\sin(\phi_j)=0, $$ looks like $$ \begin{pmatrix} 2\rho_1 & 2\rho_2 & 0 & 0 &\cdots\\ a_1\rho_1^{a_1-1}\cos(\phi_1) & a_2\rho_2^{a_2-1}\cos(\phi_2) & -\rho_1^{a_1}\sin(\phi_1) & -\rho_2^{a_2}\sin(\phi_2) &\cdots\\ a_1\rho_1^{a_1-1}\sin(\phi_1) & a_2\rho_2^{a_2-1}\sin(\phi_2) & \rho_1^{a_1}\cos(\phi_1) & \rho_2^{a_2}\cos(\phi_2)&\cdots \end{pmatrix}\qquad(\spadesuit) $$ and it is enough to show that the first $3\times 4$ block is always rank $3$ on $V(a)$. By removing either the first or second column from this block and taking det we see that the only problem is when $\phi_1=\phi_2$, which is not possible by our assumption $(\clubsuit)$ on $z_1,z_2$, or when $\phi_1=\phi_2\pm\pi$, in which case it is enough to compute the minor obtained from this first $3\times 4$ block by removing either the third or fourth column and observe that it is never zero when $\phi_1=\phi_2\pm\pi$.