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I need some hints about the proof of the non triviality of the tautological complex line bundle, in a pure topological manner.

Let $E$ be the t.c.l. bundle defined in this way $$ E= \{ (x,v) \in \mathbb{P}\mathbb{C}^n \times \mathbb{C}^{n+1} \mid v \in x \}$$ with the obvious projection map $\pi: E \to\mathbb{P}\mathbb{C}^n$, $ (x,v) \mapsto x$

Looking around, (Milnor-Stasheff's book "Characteristic Classes" page 152 prob. 13-E) the hint to prove the non-triviality of $E$ is to define an holomorphic section and then prove that must be constant and so the zero section (Liouville Theorem I think, but I admit I didn't spend too much time over this hint)

I'm my work, I didn't introduce any kind of differential structure over $E$ ($E$ is a topological space with the projection, the complex vector space-fibers and local triviality property) so, according to me, I'm not allowed to speak about holomorphic function and so I need another approach to prove the statement. I saw that in the real case a kind of approach is to use the Intermediate Value Theorem (to prove that there can't be a non-vanishing continuous section), but here in the complex case it's not valid.

In conclusion I don't have any ideas on how to deal with this proof with my work hypothesis.

ADDENDUM: There is a very short answer using Chern Classes as shown by Georges Elencwajg (thanks for the answer!). This kind of tool is very advanced and I feel like there must be some more elementary tricks to exploit. The original hint was to use Liouville's Theorem, but I've modified the text deleting all references to the differential structure over $\mathbb{PC}^n$

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    $\begingroup$ One elementary idea that comes to mind is that you only need to prove this for $n=1$, and it will automatically follow for all $n \geq 1$. (Because if you have a nonzero section in $E_n$, then you can restrict it to $\mathbb{PC}^1$ and get a nonzero section in $E_1$). So you have to think about just one fixed bundle instead of a whole family of them. $\endgroup$
    – Dan Shved
    Commented Apr 30, 2014 at 16:25
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    $\begingroup$ Another idea is that if you have a global nonzero section in $E_1$, you can divide it by the norm at each point, and this will give you a global section in $E' = \{(x, v) \in \mathbb{PC}^1 \times S^3 \mid v \in x\}$, where $S^3 \subset \mathbb{C}^2$ consists of all elements with unit norm. But $E'$ is simply the Hopf fibration $S^3 \to S^2$. So what remains to show is that the Hopf fibration doesn't have a global section, which isn't hard. $\endgroup$
    – Dan Shved
    Commented Apr 30, 2014 at 16:48
  • $\begingroup$ Nice comments @Dan:+1. I have written an elementary answer in that spirit, since Riccardo seems to find characteristic classes too sophisticated. $\endgroup$ Commented Apr 30, 2014 at 19:30

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Here is an elementary answer, using only the second homology group $H_2(-,\mathbb Z)$

As Dan Shved noticed in a comment, we may reduce to the case $n=1$.
Assume for contradiction that we have a never zero section $\sigma$ of $\pi:E\to \mathbb P^1_\mathbb C$ .
Let $p:S^3 \to \mathbb P^1_\mathbb C$ be the restriction of $\pi$ to the sphere $S^3\subset \mathbb C^2$ formed by the vectors $x\in \mathbb C^2$ of norm ||x||=1.
By dividing the hypothetical section $\sigma $ of $\pi$ by the norm we obtain a section $s:\mathbb P^1_\mathbb C \to S^3: x\mapsto s(x)=\frac {\sigma(x)}{||\sigma(x)||}$ of $p$.
But then the composition $$p\circ s=Id_{\mathbb P^1_\mathbb C}: \mathbb P^1_\mathbb C \to S^3 \to \mathbb P^1_\mathbb C$$ yields in homology the composition $$Id_{H_2( \mathbb P^1_\mathbb C ,\mathbb Z)}:H_2( \mathbb P^1_\mathbb C,\mathbb Z)=\mathbb Z\to H_2( S^3,\mathbb Z)=0\to H_2( \mathbb P^1_\mathbb C,\mathbb Z)=\mathbb Z$$ a clear absurdity.
This contradiction shows that the section $\sigma$ cannot exist and that the tautological bundle is thus not topologically trivial.

Edit
As an answer to Riccardo's comment let me clarify the following.
There is a slight abuse of language above: we have an embedding of the sphere $j:S^3\hookrightarrow E$ in which the pair $(v,w)$ with $||v||^2+||w||^2=1$ is sent to $j(v,w)=([v:w],(v,w))\in E$.
In my answer, when I write $S^3$ I actually mean its homeomorphic image $j(S^3)\subset E$.

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  • $\begingroup$ thanks for your effort, now I need some times to understand your answer and eventually accept it ! Thanks again $\endgroup$
    – Riccardo
    Commented Apr 30, 2014 at 19:33
  • $\begingroup$ Nice answer. If I was forced to provide a proof that Hopf's fibration isn't trivial, I would probably end up comparing $H_2(-,\mathbb{Z})$ of $S^3$ and $S^2 \times S^1$ too. $\endgroup$
    – Dan Shved
    Commented Apr 30, 2014 at 19:34
  • $\begingroup$ You are welcome, Riccardo! $\endgroup$ Commented Apr 30, 2014 at 19:34
  • $\begingroup$ Dear @Dan, yes I'm sure you could have given an equivalent (or better) answer. I took the liberty of posting a detailed answer because I feared that your excellent comments might be too crisp for a complete beginner (I don't allude to Riccardo, whose familiarity with topology I don't know) $\endgroup$ Commented Apr 30, 2014 at 19:45
  • $\begingroup$ I was reluctant to post a complete answer. My topology's a bit rusty due to the rare use, and I could have easily overcomplicated things if asked for details. So thanks for posting :) $\endgroup$
    – Dan Shved
    Commented Apr 30, 2014 at 20:40
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There is an elementary proof which mirrors the proof of the corresponding real statement given in Milnor & Stasheff (Theorem $2.1$).

Let $p : S^{2n+1} \to \mathbb{CP}^n$ be the quotient map and $\sigma : \mathbb{CP}^n \to E$ be a continuous section of $E$. Then $\sigma\circ p : S^{2n+1} \to E$ is given by $(\sigma\circ p)(v) = ([v], s(v))$ where $s : S^{2n+1} \to \mathbb{C}^{n+1}$ is such that $s(v) \in [v]$. So $s(v) = t(v)v$ for some continuous map $t : S^{2n+1} \to \mathbb{C}$.

Note that

$$(\sigma\circ p)(-v) = \sigma(p(-v)) = \sigma([-v]) = ([-v], s(-v)) = ([-v], t(-v)(-v)) = ([v], -t(-v)v)$$

but

$$(\sigma\circ p)(-v) = \sigma(p(-v)) = \sigma([-v]) = \sigma([v]) = ([v], t(v)v)$$

so $t(v) = -t(-v)$. By the Borsuk-Ulam Theorem (applied to $t|_{S^2}$), there is $v_0 \in S^{2n+1}$ such that $t(-v_0) = t(v_0)$. Therefore, $t(v_0) = -t(-v_0) = -t(v_0)$, so $t(v_0) = 0$ and hence

$$\sigma([v_0]) = ([v_0], t(v_0)v_0) = ([v_0], 0) = 0 \in E_{[v_0]}.$$

As $E$ does not admit a nowhere-zero section, $E$ is non-trivial.

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Use first Chern class $c_1$, a purely topological invariant ( Milnor-Stasheff, Theorem 14.4, page 160):
$$c_1(E)=-1\neq c_1(\mathbb P^n_\mathbb C\times\mathbb C)=0\in H^2(\mathbb P^n_\mathbb C,\mathbb Z)=\mathbb Z$$

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  • $\begingroup$ More propaganda for characteristic classes here and here $\endgroup$ Commented Apr 29, 2014 at 22:55
  • $\begingroup$ sorry for the late answer, I've read about Chern Classes, but I need some more elementary method (if possible): For real tautological line bundle the intermediate value theorem is the way, but in the complex case without a differential structure I don't know what to use. Even if your method is purely topologic, I need some more elementary disquisition, as I already said, if possible. $\endgroup$
    – Riccardo
    Commented Apr 30, 2014 at 15:34
  • $\begingroup$ Anyway, the up vote is well deserved, because the answer is correct, but I need elementary reasonings $\endgroup$
    – Riccardo
    Commented Apr 30, 2014 at 15:36

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