There are three questions:

  1. Please let me know your proof of the following theorem:

If $CP^3$ can be immersed in $R^8$ with an Euler class $W_{2}(\nu)$ for the normal bundle of $CP^3$ respect to $R^8$ then $$\int_{CP^3} \!W_{{2}} \left( \nu \right) c^2$$ is divisible by 3.

Where $c$ is the cohomological generator of $CP^3$.

  1. What is the exact value of the integral

$$\int_{CP^3} \!W_{{2}} \left( \nu \right) c^2$$

  1. It is possible to have an immersion of $CP^3$ in $R^8$?

Many thanks.


1 Answer 1


I may have miscalculated, but writing $e(\nu)$ for the Euler class of the normal bundle to such an immersion I find that $$\int_{\mathbb{CP}^3} e(\nu) c^2 = 2,$$ which appears to contradict 1).

To see this, write $\nu$ for the oriented 2-plane bundle occurring as the normal bundle of the immersion, so that $T\mathbb{CP}^3 \oplus \nu = \epsilon^8$. As the total Pontrjagin class is $p(T\mathbb{CP}^3) = (1-c^2)^4 = 1-4c^2$, we find that $p(\nu) = p(T\mathbb{CP}^3)^{-1} = 1+4c^2$, and hence $p_1(\nu) = (2c)^2$. For an oriented 2-plane bundle we have $p_1 = e^2$, and so $e(\nu)=2c$. Thus $$\int_{\mathbb{CP}^3} e(\nu) c^2 = \int_{\mathbb{CP}^3} 2 c^3 = 2.$$

  • $\begingroup$ The computation performed by @OscarRandal-Williams is correct and the result contradicts the question 1. Then the answer for the question 3 is: it is not possible to have an immersion of $CP^3$ in $R^8$. $\endgroup$ Nov 16, 2014 at 19:12

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