The cylinder $\Bbb R\times S^1$ can be viewed as a complex manifold with a flat metric by viewing it has the quotient $\Bbb R\times\Bbb R/\Bbb Z$, where $\Bbb R\times\Bbb R=\Bbb C$. (In fact it makes the cylinder into a Kähler manifold.)
Problem: Show that there is no isometric holomorphic embedding $$\varphi:\Bbb R\times S^1\hookrightarrow \Bbb C^n,$$ for any $n$, where $\Bbb C^n$ has the standard Kähler structure.
Motivation: I was trying to answer another of my questions here, and after some research I found this mathoverflow post of Peter Kronheimer claiming the above. He gives the following reason.
Hint: Apply the maximum modulus principle to the derivative of $\varphi$.
I tried this approach without any success. Does anybody know how it works?
How about this? If $\phi$ is an isometric holomorphic embedding $\Bbb C/\Bbb Z\hookrightarrow\Bbb C^n$, it lifts to a $\Bbb Z$-invariant holomorphic map $\tilde\phi\colon\Bbb C\to\Bbb C^n$ pulling back the standard Kähler form on $\Bbb C^n$ to the standard Kähler form on $\Bbb C$, i.e., $$\tilde\phi^*\left(\sum_{j=1}^n dz^j\wedge d\bar z^j\right) = dz\wedge d\bar z.$$ This means that $\sum\limits_{j=1}^n \big|(\tilde\phi{}^j)'(z)\big|^2 = 1$ for all $z$. Applying Liouville's Theorem to $(\tilde\phi{}^j)'$, we infer that all $(\tilde\phi{}^j)'$ must be constant. But each $\tilde\phi{}^j$ is also $\Bbb Z$-invariant, so the only constant that can work is $0$. Oops.