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

Thank you for your help

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    $\begingroup$ Given that these bundles are complex vector bundles, I would suggest using Chern classes instead. Are you familiar with them? $\endgroup$ Mar 21 at 12:02
  • $\begingroup$ @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? $\endgroup$
    – Victory
    Mar 21 at 12:12
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    $\begingroup$ 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? $\endgroup$ Mar 21 at 12:20
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    $\begingroup$ So $c_1(\eta\oplus\zeta) = 0$. What does this tell you about $c_1(\eta)$ and $c_1(\zeta)$? $\endgroup$ Mar 21 at 12:33
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    $\begingroup$ 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 $c_2(\zeta)$? $\endgroup$ Mar 21 at 15:44

1 Answer 1


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.

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    $\begingroup$ 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 $\deg\alpha = 2$. $\endgroup$ Mar 22 at 12:00

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