# De Rham cohomology of $T^*\mathbb{CP}^n$

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 $\mathbb{C}^{n+1}$)?

• 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. – Aaron Mar 22 '14 at 0:06
• What does $T^*$ means in this context? – Seth Mar 22 '14 at 0:22
• There is a very beautiful proof in Bott and Tu using some spectral sequence technique. The ring structure come along the proof. – user99914 Mar 22 '14 at 3:06
• @Seth Here $T^*$ denotes the cotangent bundle (which is a vector bundle associated to every smooth manifold such as $\mathbb{CP}^n$). – Daniel Robert-Nicoud Mar 23 '14 at 11:06

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.
• 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? – Daniel Robert-Nicoud Mar 23 '14 at 15:43
• @PeterCrooks Can you give a detailed proof of the non-exactness of $\omega^j$? – TheWanderer Feb 5 '15 at 11:52