# Bundle over $\mathbb CP^\infty$

Good time of day. I have the following question

Let $$\eta$$ -tautological bundle over $$\mathbb CP^\infty$$. I don't understand why there is no such complex vector bundle $$\xi$$ over $$\mathbb CP^\infty$$ that bundle $$\eta \bigoplus \xi$$ is trivial.

I'm not sure about my attempt. I try to use Pontryagin classes for solving this task. If these classes are non-vanishing then this bundle will be non-trivial.

We know that $$2p(E \bigoplus F)=2p(E)\smile p(F)$$

and it's famous fact that if $$\theta$$ is oriented real bundle of rank $$2k$$ then $$p_k(\theta)=e(\theta)^2$$, where $$e(\theta)^2$$ is the square of Euler class.

I don't know how to continue and compute this

• Given that these bundles are complex vector bundles, I would suggest using Chern classes instead. Are you familiar with them? Mar 21 at 12:02
• @Michael Albanese. Yes, I'm familiar with them a bit. As I understand these Chern classes should be also non-vanishing for non-triviality of this bundle $\eta \bigoplus \xi$? Am I correct? Mar 21 at 12:12
• That's not necessarily true. It is possible to have a non-trivial bundle with trivial Chern classes. Instead, try contradiction. Suppose such a $\zeta$ exists. What can you say about its Chern classes? Mar 21 at 12:20
• So $c_1(\eta\oplus\zeta) = 0$. What does this tell you about $c_1(\eta)$ and $c_1(\zeta)$? Mar 21 at 12:33
• You can express the Chern classes of $\zeta$ in terms of the Chern classes of $\eta$. For the first Chern class, we have $0 = c_1(\eta\oplus\zeta) = c_1(\eta) + c_1(\zeta)$, so $c_1(\zeta) = -c_1(\eta)$. What do you get for $c_2(\zeta)$? Mar 21 at 15:44

Michael Albanese the following method based on Chern classes to me.

Suppose that a complex vector bundle $$\zeta$$ exists such that $$\eta \oplus \zeta$$ is trivial. We have known that the first Chern classes for trivial bundles vanish, i.e. $$c_1(\eta\oplus\zeta) = 0$$.

It follows that $$c_1(\eta\oplus\zeta) = c_1(\eta)+c_1(\zeta)=0$$, so $$c_1(\zeta)=-c_1(\eta)$$.

We have expressed the first Chern class of $$\zeta$$ in terms of the Chern classes of $$\eta$$. We will continue to do the same for higher Chern classes.

Since we the tautological bundle over $$\mathbb CP^\infty$$ is a complex line bundle we have $$c_2(\eta) = 0$$. Also we have $$c_2(\eta\oplus\zeta) = 0$$ since we assumed that bundle $$\eta\oplus\zeta$$ is trivial. Therefore

$$0 = c_2(\eta\oplus\zeta) = c_2(\eta) + c_1(\eta)c_1(\zeta) + c_2(\zeta) = 0 + c_1(\eta)(-c_1(\eta)) + c_2(\zeta),$$

so $$c_2(\zeta) = c_1(\eta)^2$$.

Likewise $$c_3(\eta\oplus\zeta)=c_3(\eta)+c_1(\eta)c_2(\zeta)+c_1(\zeta)c_2(\eta)+c_3(\zeta)=0$$, and we see that $$c_3(\zeta)=-c_1(\eta)^3$$.

Also $$c_4(\zeta)=c_4(\eta)+c_2(\eta)c_2(\zeta)+ c_1(\eta)c_3(\zeta)+c_3(\eta)c_1(\zeta)+c_4(\zeta)=0$$, so $$c_4(\zeta)=c_1(\eta)^4$$.

One can see that the $$k^{\text{th}}$$ Chern class satisfies $$c_k(\zeta)=(-1)^k c_1(\eta)^k$$.

Combining the famous fact from the theory of fiber bundles that for a vector bundle of rank $$k$$, all Chern classes for $$i > k$$ are vanishing and the results of our computations showing that $$c_i(\zeta) = (-1)^ic_1(\eta)^i \neq 0$$ we have obtained a contradiction.

In summary, one can see from the triviality of the bundle $$\eta \oplus \zeta$$ that $$c(\eta\oplus\zeta) = 1 \implies c_i(\zeta) = (-1)^ic_1(\eta)^i \neq 0$$ for every $$i$$. This is impossible since the Chern classes of $$\zeta$$ should vanish above its rank.

• I hope you don't mind, I edited your answer to clean it up a bit. Just a couple of comments. First, you can prove that $c_k(\zeta) = (-1)^kc_1(\eta)^k$ by induction. Second, the fact that $c_i(\zeta) \neq 0$ requires one to know that $c_1(\eta)^i \neq 0$. This is the case because $c_1(\eta) \in H^2(\mathbb{CP}^{\infty}; \mathbb{Z})$ is non-zero and $H^*(\mathbb{CP}^{\infty}; \mathbb{Z}) \cong \mathbb{Z}[\alpha]$ where $\deg\alpha = 2$. Mar 22 at 12:00