What does it mean to say "the cohomology of $\mathbb{C}P^{n-1}$ is of rank $n$"? From my knowledge, when considering the rank you would consider the individual ranks of each $H^{k}(\mathbb{C}P^{n-1}; G)$ component, and so I am confused what it means by simply saying cohomology.
Asked
Active
Viewed 33 times
3
-
Some authors like to look at (co)homology as a graded object. So, for example, statements like "the homology of the sphere $S^n$ is free of rank $2$ generated in degrees $0,n$" is a completely valid statement (and it remains valid when $n=0$). In your case, $H(\Bbb CP^n)$ is a free abelian group (it is also a ring, but that's irrelevant) generated in degrees $0,2,\cdots,2n$. – Pedro Mar 28 '16 at 00:34
1 Answers
4
The integral cohomology ring of $\mathbb{CP}^{n-1}$ is $H^*(\mathbb{CP}^{n-1}; \mathbb{Z}) \cong \mathbb{Z}[\alpha]/\langle\alpha^n\rangle$ where $|\alpha| = 2$. Note that $\{1, \alpha, \dots, \alpha^{n-1}\}$ is a basis for this $\mathbb{Z}$-module, so it is a free $\mathbb{Z}$-module of rank $n$.
Michael Albanese
- 99,526