No holomorphic injective mapping $f:D\to \Bbb C^n$ if $\operatorname{\dim} D>n$

Question

Take an open subset $D\subset\Bbb C^m$.

I suspect there is no holomorphic injective mapping $f:D\to \Bbb C^n$ if $m>n$.

Is this true?

Being $f$ injective holomorphic then it must be regular (that is its differential has maximum rank at each point), so the image is an open set in $\Bbb C^n$, then, intuitively, being the target too tight because of the smaller dimension, there is no room for the image of an injective mapping. But how can we prove it?

Answer 1

Invariance of domain is one way to go, but I think that's too much machinery. An alternative proof can be given using the constant-rank theorem (which is equivalent to the inverse function theorem and also to the implicit function theorem, and these are certainly available to students after a multivariable calculus course). We can actually prove this theorem:

Let $m,n$ be integers with $m>n$, let $D\subset \Bbb{R}^m$ be non-empty and open, and $f:D\to \Bbb{R}^n$ be a $C^1$ map. Then, $f$ is not injective.

To prove this, let $r$ be the maximum rank of $f$ on $D$, i.e \begin{align} r:=\max\{\text{rank}(Df_x)\,|\,\,\, x\in D\} \end{align} The maximum exists because this is a finite non-empty set of integers. Suppose $a\in D$ is a point where $r=\text{rank}(Df_a)$. Since the rank cannot decrease locally, and $r$ is the maximum rank, it follows there is an open set $A\subset D$ containing $a$, such that $f$ has constant rank $r$ on $A$. Therefore, by the constant rank theorem, there exist $C^1$ diffeomorphisms

• $\phi:A'\to \phi[A']\subset\Bbb{R}^m$, where $A'$ is some neighborhood of $a$ in $A$
• $\psi:B\to \psi[B]\subset\Bbb{R}^n$ where $B$ is a neighborhood of $f(a)$ in $\Bbb{R}^n$

such that $f(A')\subset B$ and the composition $\psi\circ (f|_{A'})\circ \phi^{-1}:\phi[A']\subset\Bbb{R}^m\to\Bbb{R}^n$ is equal to the rank $r$ linear transformation \begin{align} (x^1,\dots, x^m)\mapsto (x^1,\dots, x^r,0,\dots, 0) \end{align} Since $m>n\geq \min(m,n)\geq r$, it follows that $\psi\circ f\circ \phi^{-1}$ is not injective. Since $\psi,\phi$ are $C^1$ diffeomorphisms (in particular bijections), it follows that the restriction $f|_{A'}$ and hence $f$ are not injective.

Finally, in your case, holomorphic maps are certainly $C^1$, so by considering $\Bbb{C}^n\cong \Bbb{R}^{2n}$ etc it follows there is no holomorphic injective mapping $f:D\subset\Bbb{C}^m\to\Bbb{C}^n$ if $m>n$. (or you could repeat this entire argument, because the inverse function theorem, implicit function theorem, constant rank theorem all have natural analogues in the holomorphic case).