Chern class of complex vector bundles

Let $\xi$ be an $n$-dimensional complex vector bundle. It is claimed that the Chern class of $\xi$ is
$$c(\xi)=(1+x_1)\cdots (1+x_n),$$ $|x_k|=2$, $c_j(\xi)$ is the $j$-th symmetric polynomial of $x_1,\cdots,x_n$. Is this claim true?

In the case $\xi=L_1\oplus\cdots\oplus L_n$, Whitney sum of complex line bundles, I have obtained the claim. How about general case?

• Would the splitting principle help? – Rolf Hoyer Apr 9 '15 at 4:06

This is not true.

For example, consider $$T\mathbb{CP}^2$$ which has total Chern class $$c(T\mathbb{CP}^2) = 1 + 3x + 3x^2$$ where $$x = c_1(\mathcal{O}(1)) \in H^2(\mathbb{CP}^2; \mathbb{Z})$$ is a generator. Suppose $$c(T\mathbb{CP}^2) = (1 + x_1)(1 + x_2)$$ for some $$x_1, x_2 \in H^2(\mathbb{CP}^2; \mathbb{Z})$$, then $$x_1 = kx$$ and $$x_2 = lx$$ for some integers $$k$$ and $$l$$. Then $$1 + 3x + 3x^2 = (1 + kx)(1 + lx) = 1 + (k + l)x + klx^2.$$ This is impossible: if $$k + l = 3$$, one of $$k$$ and $$l$$ is even, but then $$kl$$ would be even, so it can't be equal to $$3$$.

This shows that $$T\mathbb{CP}^2$$ is not isomorphic to the direct sum of line bundles. Note however that if the total Chern class did factor, it does not necessarily mean the bundle is isomorphic to a direct sum; see this MathOverflow question.

What is true however is the splitting principle:

Let $$\xi$$ be a rank $$n$$ vector bundle over $$X$$. There exists a space $$Y$$ (the total space of the flag bundle of $$\xi$$) and a map $$p : Y \to X$$ such that:

• the graded ring homomorphism $$p^* : H^*(X, \mathbb{Z}) \to H^*(Y, \mathbb{Z})$$ is injective, and
• $$p^*\xi = L_1\oplus\dots\oplus L_n$$ where $$L_1, \dots, L_n$$ are complex line bundles on $$Y$$.

Note that

$$p^*c(\xi) = c(p^*\xi) = c(L_1\oplus\dots\oplus L_n) = c(L_1)\dots c(L_n) = (1 + x_1)\dots(1+x_n)$$

where $$x_i = c_1(L_i)$$.