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 $P(E)$.
[Math] Chern Classes of a Trivial Bundle
characteristic-classesvector-bundles
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I made a number of relevant comments to this question on tigu's previous question. Here they are again, lightly edited.
No, but this there is a true statement like this. When you describe $V_i$ as the locus where some sections are dependent, there are a lot of subtleties in that description. It is only literally correct if there are "enough" holomorphic sections and if those sections are "generic enough". Otherwise, you have to start talking about keeping tracks of multiplicities and signs.
The details of how you deal with the technicalities will be different depending on whether you are looking at algebraic/holomorphic sections (in this case, your best source is Fulton's "Intersection Theory") or whether you are looking at smooth sections (in this case, Milnor's "Characteristic Classes" has a good reputation, thought I haven't read it). I'll discuss a single example.
Look at $\mathbb{P}^2$ and let $U:= \mathcal{O}(1) \oplus \mathcal{O}(−1)$. The Chern class is $1−h^2$, where $h$ generates $H^2(\mathbb{P}^2, \mathbb{Z})$. The point is that $\mathcal{O}(−1)$ has no nonzero holomorphic sections. So all the holomorphic sections of $U$ are of the form $(0, \sigma)$ for $\sigma$ a section of $\mathcal{O}(1)$. In particular, any two holomorphic sections of $U$ are proportional everywhere on $\mathbb{P}^2$.
If we perturb the above sections so that they are not holomorphic then, I think, there will be are two curves $C$ and $C'$ on which they become dependent. One of these curve should be weighted positively and one negatively, and their degrees cancel in $H^2(\mathbb{P}^2)$. It would be fun to work out an example of this. UPDATE: I worked this out below and, at least for the two sections I tried, I got a single contractible sphere rather than two curves.
Similarly, if you look at one holomorphic section of $U$, it will vanish along an entire line in $\mathbb{P}^2$. If you perturb this section to be smooth, then it will vanish along a codimension $2$ subvariety, with multiplicity $−1$. You might want to read this question for a further understanding of how negative numbers show up as intersection multiplicities. That $-1$ explains that $-h^2$.
The true statement I allude to above should be something like "If $U$ has enough holomorphic sections, and $c_1(U)=0$, then $U$ is trivial." I do not know what the right definition of "enough" is; I would guess "globally generated".
Okay, let's actually work out the examples discussed above. We will describe a point of $\mathbb{P}^2$ using homogenous coordinates $(x:y:z)$. The fiber of $\mathcal{O}(-1)$ over $(x:y:z)$ is the line in $\mathbb{C}^3$ spanned by $(x,y,z)$. The fiber of $\mathcal{O}(1)$ over $(x:y:z)$ is the dual of that, in other words, $\mathrm{Hom}(\mathrm{Span}_{\mathbb{C}} (x,y,z), \mathbb{C})$. It is easy to give global holomorphic sections of $\mathcal{O}(1)$: Just take an element of $\mathrm{Hom}(\mathbb{C}^3, \mathbb{C})$ and restrict it to every fiber of $\mathcal{O}(-1)$. Let $\sigma_1$ be the section obtained by restricting $(x,y,z) \mapsto x$ and let $\sigma_2$ and $\sigma_3$ use the other two coordinate functions.
There are no holomorphic sections of $\mathcal{O}(-1)$, but there are lots of smooth sections. Let $$\tau_1(x:y:z) = \frac{1}{|x|^2+|y|^2+|z|^2} ( x \overline{x}, y \overline{x}, z \overline{x}).$$ Define $\tau_2$ and $\tau_3$ similarly.
It turns out that $\sigma_1 \oplus \tau_1$ is not sufficiently generic is, but $\sigma_1 \oplus \tau_2$, so let's look at that. The section $\sigma_1$ vanishes whenever $x=0$, a line in $\mathbb{P}^2$. The section $\tau_2$ vanishes when $y=0$. However, if you check closely, you'll see that $\tau_2$ is anti-holomorphic, not holomorphic, in a neighborhood of its vanishing locus. So that vanishing should count negatively, and $\tau_2$ vanishes with multiplicity $-1$ on the line $y=0$. The section $(\sigma_1, \tau_2)$ vanishes at the point $(0:0:1)$, where $x=y=0$, and does so with sign $1 \times (-1) = -1$. This shows that $c_2 = -h^2$.
Similarly, let's look at the pair of sections $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$. Assuming that $x$ and $y$ are nonzero, the ratio $\sigma_1/\sigma_2$ is $x/y$, while the ratio $\tau_3/\tau_2 = \overline{z}/\overline{y}$. So $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$ are proportional when $x/y = \overline{y}/\overline{z}$, in other words, when $x \overline{z} = |y|^2$. I didn't check carefully, but I believe that $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_1)$ are linearly dependent precisely when $x \overline{z} = |y|^2$, including in the degenerate cases where these coordinates vanish.
The set $\{ (x:y:z) \in \mathbb{P}^2 : x \overline{z} = |y|^2 \}$ is a sphere. I can parameterize it as $(e^{i \theta} (1+\sin \phi) : \cos \phi : e^{i \theta} (1-\sin \phi))$ where $\theta$ is the longitude coordinate, running through $\mathbb{R}/(2 \pi \mathbb{Z})$, and $\phi$ is the lattitude coordinate, living in $[-\pi/2, \pi/2]$. I claim that this sphere is contractible in $\mathbb{P}^2$. To see this, map $S^2 \times [0,1]$ to $\mathbb{P}^2$ by $(\theta, \phi, t) \mapsto (e^{i \theta} (1+\sin \phi) : (1-t) \cos \phi : e^{i \theta} (1-\sin \phi))$. At one end of the homotopy, we have the previously described embedding. At the other end, we have the line segment $\{ (1+r:0:1-r) : r \in [-1, 1] \}$. This line segment is obviously contractible.
So the space where $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$ are dependent has homology class $0$, showing that $c_1=0$.
From purely algebraic arguments:
- If $E \to X$ is a complex vector bundle of complex dimension $n$, then the top Chern class $c_n(E) \in H^{2n}(X; \mathbb R)$ is equal to the Euler class $e(E)$.
- Every $n$-dimensional complex manifold $X$ (such as $\mathbb {CP}^n$) is orientable. In algebraic terms, orientability can be seen as the top homology group $H_{2n}(X; \mathbb R)$ having rank one, and we can choose an orientation, i.e. a choice of generator $[X] \in H_{2n}(X; \mathbb R)$.
Now let $\mathbb {CP}^n = X$ and $x$ be the Euler class of the tautological line bundle over $\mathbb {CP}^n$. Then there is a nice argument on p22 Example 2.9.3 of these lecture notes on characteristic classes (from the algebraic topology point of view). Essentially, it boils down to showing that
$$ \langle e(T \mathbb {CP}^n), [\mathbb {CP}^n] \rangle =(n+1)(-1)^n \langle x^n, [\mathbb {CP}^n] \rangle$$
where $\langle,\rangle : H^*(\mathbb {CP}^n, \mathbb R) \otimes_{\mathbb R} H_*(\mathbb {CP}^n, \mathbb R) \to \mathbb R$ is the canonical non-degenerate pairing between homology and cohomology and $x^n$ is the cup product of $x$ with itself $n$ times.
We know from standard theory that for an orientable manifold $M$, the pairing $\langle e(TM), [M] \rangle = \chi(M)$ -- the Euler characteristic. The homology groups of $\mathbb {CP}^n$ have rank one in every even grading that is $\leq 2n$ and zero otherwise, so $\chi(M)= n+1$. Hence, $x\neq 0$.
Best Answer
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 $$c_i(f^{\ast}(E)) = f^{\ast}(c_i(E)) = 0.$$