7
$\begingroup$

I am a bit rusty on my de Rham cohomology, and I'm hoping that someone here could help me.

I want to find the cohomology of $T^*\mathbb{CP}^n$ (seen as a real manifold). Now, this should be equal to the cohomology of $\mathbb{CP}^n$ since the two are homotopic (by homotopy of each fibre with a point), thus the problem reduces to the computation of $H^\bullet(\mathbb{CP}^n)$. How can I proceed to find it? Would something as the third possibility proposed in this answer work (by taking $G=\mathbb{C}^*$ acting on $\mathbb{C}^{n+1}$)?

$\endgroup$
  • 1
    $\begingroup$ The cohomology is $\mathbb R[x]/(x^{n+1})$ where $x$ is of degree $2$. It is easy to get the module structure by taking a cellular approach (there is one cell in each even dimension up to $2n$), but I can't remember a good way to see the ring structure. $\endgroup$ – Aaron Mar 22 '14 at 0:06
  • $\begingroup$ What does $T^*$ means in this context? $\endgroup$ – Seth Mar 22 '14 at 0:22
  • $\begingroup$ There is a very beautiful proof in Bott and Tu using some spectral sequence technique. The ring structure come along the proof. $\endgroup$ – user99914 Mar 22 '14 at 3:06
  • 1
    $\begingroup$ @Seth Here $T^*$ denotes the cotangent bundle (which is a vector bundle associated to every smooth manifold such as $\mathbb{CP}^n$). $\endgroup$ – Daniel Robert-Nicoud Mar 23 '14 at 11:06
4
$\begingroup$

Let me elaborate on @Aaron's comment. Let $\omega$ be the Fubini-Study symplectic form on $\mathbb{C}\mathbb{P}^n$. You can use Aaron's approach to conclude that $H^i(\mathbb{C}\mathbb{P}^n)$ is one-dimensional for even $i$ between $0$ and $2n$ and is trivial otherwise. To compute the ring structure, one can show $\omega^j$ is closed but not exact for $0\leq j\leq n$. Therefore, its class in cohomology generates $H^{2j}(\mathbb{C}\mathbb{P}^n)$ as a vector space for all such $j$. Therefore, the cohomology of $\mathbb{C}\mathbb{P}^n$ is generated by $[\omega]$ as an algebra. The only relation is that $[\omega]^{n+1}=0$. This gives you Aaron's answer.

$\endgroup$
  • $\begingroup$ Thanks. Do you have any reference where I can find the details? Also, for $T^*\mathbb{CP}^n$ I suppose that the cohomology is the same and the (representative) forms are trivial on the fibres? $\endgroup$ – Daniel Robert-Nicoud Mar 23 '14 at 15:43
  • $\begingroup$ @PeterCrooks Can you give a detailed proof of the non-exactness of $\omega^j$? $\endgroup$ – TheWanderer Feb 5 '15 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.