Chern Classes of a Trivial Bundle Could someone explain to me why the chern classes of a trivial bundle are zero? 
(I'm studying it from Bott & Tu book) To be more specific I can't understand why, given the vector bundle $E$ on $M$, it must be $C_1(S^*_E)^n=0$. $S^∗_E$ is the dual of the tautological subbundle of the pullback of $E$ on $P(E)$.
 A: At least rationally you can also see this easily since any trivial bundle admits a flat connection. But then by Chern Weil theory you see that any chern class has to be zero, since it can be computed in terms of the curvature which is $0$ in our case.
A: This is a trivial consequence of the naturality (or functoriality) of the Chern classes, which should be clear no matter which definition of the Chern classes you are using.
Fix a space $X$.  Let $P$ be a one-point space, and let $E \rightarrow P$ be the trivial $n$-dimensional complex vector bundle.  There is a unique map $f : X \rightarrow P$, and it is easy to see that the trivial $n$-dimensional complex vector bundle over $X$ is exactly the pullback $f^{\ast}(E) \rightarrow X$.  All the Chern classes of $E \rightarrow P$ have to be trivial since the cohomology groups of $P$ are trivial.  Thus by the naturality of Chern classes we have
$$c_i(f^{\ast}(E)) = f^{\ast}(c_i(E)) = 0.$$
A: I assume you're looking for a non-technical answer, one that will somehow tell you the intuative reason for why the Chern classes of a trivial bundle are zero without going into serious calculations. I can't promise you this will be that answer, but we'll damn well do our best. For the record, what I'll say makes sense in the algebraic and analytic categories. I think it does as well in the smooth one, but I wouldn't bet my life on it.
For the setup, I'll take my manifold $X$ to be compact and smooth, and I'll take a smooth vector bundle $E \to X$ over it. Now, what are the Chern classes of the bundle $E$ really?
Suppose we have a section $\sigma$ of the bundle $E$ and suppose also that the section $\sigma$ is not the zero section, but that it still has some zeros. Then we can consider the subvariety $(\sigma)_0$ of points of $X$ where the section $\sigma$ is zero. This subvariety defines a cohomology class on $X$; it's easiest to define it by duality, that is by simply integrating differential forms that represent different classes over the subvariety $(\sigma)_0$.
This cohomology class is the first Chern class of the vector bundle $E$. Thus the first Chern class measures, in some sense, how "often" a general section of $E$ is zero.
To get a feel for the second Chern class, take two sections $\sigma_1$ and $\sigma_2$ of $E$. Locally these are just vectors in a vector space, so they can be colinear or not. Let $(\sigma_{12})$ be the subvariety of $X$ which consists of the points where the sections $\sigma_1$ and $\sigma_2$ are colinear. This subvariety again defines a cohomology class, which is exactly the second Chern class of $E$.
The same happens for the higher Chern classes. Intuitively, the $k^{th}$ Chern class of a vector bundle measures, in some sense, how "likely" it is that $k$ generic sections of the bundle are linearly dependent.
I'm told this is more or less the original historical definition of Chern classes. This point of view is exposed in great detail in a forthcoming book of Harris and Morrison called "Intersection theory". Until that book comes out, Harris talks about this a great deal in his lectures available here: http://mate.dm.uba.ar/~visita16/ELGA-2011/version/v1/images-en.shtml
Now, this is relevant to our interests. Indeed, suppose the vector bundle $E$ is trivial, or $E = \mathbb R^r$. Then it is somehow clear that we have an incredible amount of room in which to fit our sections, as there are no global obstructions coming from the structure of the bundle itself. In vague but suggestive terms, we can take a global frame, or basis, of our vector bundle. If we need to find $k$ linearly independent sections of the bundle, we just take the $k$ first vectors of that basis. These are linearly independent everywhere, thus the subvariety which defines the $k^{th}$ Chern class is empty, so the $k^{th}$ Chern class of the bundle is zero.
This was all incredibly imprecise. I do suggest you look at Harris' lectures to get a better idea of what is going on.
A: This is a late answer to the question. I just wanted to explain how Bott & Tu deduced it in the book. In the chapter they computed the cohomology of $P(V)$ and showed that $H^*(P(V))=\frac{\mathbb{R[x]}}{(x^n)}$.
They write:

If $E$ is the trivial bundle $M \times V$ over $M$ then $P(E)=M \times P(V)$ so $x^n=0$.

In there construction of the chern class they are using the Leray–Hirsch theorem to represent $H^*(P(E))$ as a free module over $H^*(M)$ with basis $\{1, \dots, x^{n-1}\}$. But in this case we can use Kunneth formula for cohomology to deduce:
$$H^n(P(E)) \cong \left(\sum_{i + j = n} H^i(M) \otimes H^j(P(V))\right)$$
From this and the knowledge of the cohomology ring of $P(V)$ we deduce that $x^n=0$.
Edit: The answers to the question in the comment gos as follows.
At page 270 in there book the construct the following exact sequence
$$0 \to S \to \pi^{-1}E \to Q \to 0$$
Where $\pi^{-1}E $ is the pullback bundle over $P(E)$. If $E$ was the trivial bundle over $M$ then $\pi^{-1}E $ will be the trivial bundle over $P(E)$. Now we ask what will be $c_1(S^*) \in H^2(P(E))$?
If $\rho_2$ is the projection $M \times P(V)\to P(V)$. Then we can show that $\rho_2^{-1}\tilde{S} \cong S$, where $\tilde{S}$ is the universal sub-bundle of $P(V)$. Thus we have $c_1(S^*)=-\rho_2^*(c_1(\tilde{S}))$. Thus
we now have
$$x^n = \left(c_1(S^*)\right)^n=-\rho_2^*(c_1(\tilde{S})^n)=-\rho_2^*(0)=0.$$
