Elementary and purely topological proof of the non-triviality of tautological complex line bundle

Question

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 $\mathbb{PC}^n$

Answer 1

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 $j(S^3)\subset E$.

Answer 2

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

Answer 3

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.