If $\iota: E \to \Bbb C^k$ is a holomorphic bundle map, and $p_i$ is projection of $\Bbb C^k \to \Bbb C$ onto the $i$th factor, then $p_i \iota$ is a map $E \to \Bbb C$. Aka, a section of the dual bundle.
And since everything we did was holomorphic, that should be holomorphic.
On the other hand, this idea inspires me. Let $s_i$ be the sections of $E$. Define $E^* \to \Bbb C^k$ by $\varphi \mapsto (\varphi(s_1), \varphi(s_2), \dots, \varphi(s_k))$.
In your chosen holomorphic trivialization of the bundle over $U_i$, there's no reason to believe $s_i(v)$ doesn't go to infinity as the basepoint goes towards the boundary.
My construction didn't actually involve any Hermitian inner product. I just took the trivializations outside the complement of $s_k = 0$ and multiplied them together. I thought $s_i$ were holomorphic sections?
It's probably futile to talk about a failed construction though, so feel free to stop conversing if you want.
@Balarka Your trivialization don't preserve the "size" of s_i. There's absolutely no reason that in these trivializations, the s_i die near "infinity".
Lots of languages have no 'h' — in French, it's never aspirate, common part of diphthongs, yes, otherwise virtually unpronounced. None in Russian. Yes in German.
I arrived at this expression earlier while adding a function that has a pole of order $k$ at $z_0$ and a function that has an essential singularity at $z_0$
@TedShifrin If $f$ has a pole at $z_0$ and $g$ has an essential singularity at $z_0$, would that mean $f \cdot g$ has an essential singularity at $z_0$?
I look at the product of the Laurent series and would guess that is the case
So can I say $$\sum_{j=-k}^{\infty} a_j(z-z_0)^j \cdot \sum_{j=-\infty}^{\infty} b_j(z-z_0)^j = \sum_{j=-\infty}^{\infty} c_j(z-z_0)^{j}$$ where $c_j$ is the Cauchy product between $f$ and $g$?
@TedShifrin Ugh, can I just say that multiplying any term from the Laurent series of $f$ with $g$ produces a function with an essential singularity at $z_0$?
I think I learnt more about manifolds talking to people than reading a specific textbook (although G&P is the only book I tried to seriously read which is about smooth manifolds, and I like it a lot).
Analytic means the function is locally bounded, pole means $|f(z)|\to\infty$ nearby, essential singularity means the values of $f$ are dense in $\Bbb C$ nearby.
@TedShifrin I don't know what dense means. But I know that essential singularity means the modolus of the function is unbounded but does not approach infinity
I've still got some old questions about the Albanese torus, but screw that I'll talk to someone else about that, or you if they can't answer my questions
But seriously , when you study mathematics do you just solve problems and exercises only? Because I have trouble concentrating when I read math unless I'm already familiarized with some concepts in the text.
Let $L$ be a (holo.) line bundle globally generated by global sections $s_1,\dots,s_n$. Then show that the Hermitian structure induced on $L^*$ by dualizing coincides with that induced by restricting the standard Hermitian structure on $\mathcal O^{\oplus n}$. That it embeds into $\mathcal O^{\oplus n}$ is something I've already asked Mike and Balarka about.
Mike eventually solved it. So I get that part. So I just need to understand why the two Hermitian structures agree.
Right, @Lozansky, and clearly if you multiply that by a function with an essential singularity you still have an essential singularity, because the modulus of that is locally bounded.
On the other hand, this idea inspires me. Let $s_i$ be the sections of $E$. Define $E^* \to \Bbb C^k$ by $\varphi \mapsto (\varphi(s_1), \varphi(s_2), \dots, \varphi(s_k))$.
Without it embedding the entire exercise doesn't even make sense :\
In any case, I assume the Hermitian structure induced on the dual by dualizing is given by taking $(x,z^*)$ and $(x,w^*)$ and just using $\langle z,w\rangle$ (we already have a Hermitian structure on $L$)
Well okay so $h(z)$ has an essential singularity at $z_0$. Then we can add any number of poles at $z_0$ and still say the function has an essential singularity at $z_0$
Any submanifold, @Balarka. I mentioned that ages ago.
But we don't need any of that.
So once you map injectively to the trivial bundle, use the hermitian inner product on $\Bbb C^n$ to induce one on the image of your bundle, hence on sections of the bundle.
In any case, I assume the Hermitian structure induced on the dual by dualizing is given by taking $(x,z^*)$ and $(x,w^*)$ and just using $\langle z,w\rangle$ (we already have a Hermitian structure on $L$)
Well, think about how you do the hermitian inner product on $\mathscr L$ in a trivialization and how it transforms according to the transition functions?
I thought it should be analogous to this since I view Hermitian structures on vector bundles as a generalization of Hermitian structures on manifolds (i.e. on tangent bundles)
That's how you make sure something works in a well-defined way. If you have a global intrinsic definition (which you never told me), you can still check.
Okay, so you wanted me to tell you how a Hermitian structure acts in a trivialization. So let $x\in X$ and let $\psi$ be a trivialization of $L\cong U\times \Bbb C$ around $x$. Then $h_x(\psi_x^{-1}-,\psi^{-1}_x-)$ is a Hermitian scalar product on $\Bbb C$. By definition it should vary smoothly with $x$.
So let me modify the notation a bit: We have triv's $\psi_1,\psi_2$ and a transition function $\psi_{21}$ from $U_1$ to $U_2$. Then I guess we must have $h_x(\psi_{2,x}^{-1}-,\psi_{2,x}^{-1}-)=h_x((\psi_{21,x}\psi_{1,x})^{-1}-,(\psi_{{21,x}}\psi_{1,x})^{-1}-)$
please help me improve this terrible notation haha
