What is the integral cohomology of $\mathrm{PGL}_n(\mathbb{C})$ as a space?

The computation of the cohomology of $\mathrm{GL}_n(\mathbb{C})$ is one of the basic applications of the Serre spectral sequence, using the fiber bundle $\mathrm{GL}_{n-1}(\mathbb{C})\to \mathrm{GL}_n(\mathbb{C}) \to \mathbb{C}^{n} \setminus0$. Using the same idea we can get a fiber bundle $\mathrm{PGL}_{n-1}(\mathbb{C})\to \mathrm{PGL}_n(\mathbb{C}) \to \mathbb{C}P^{n-1}$, but now the base has enough cohomology for the differentials to not be obviously zero.

In fact, I believe the following shows that $H^2(\mathrm{PGL}_n(\mathbb{C}))$ has torsion $\mathbb{Z}/n\mathbb{Z}$ (so not all of the differentials above can be zero), as well as computing $H^*(\mathrm{PGL}_n(\mathbb{C}); \mathbb{Q})$. The important observation is that in the fiber bundle $\mathbb{C}^\times \to \mathrm{GL}_n(\mathbb{C}) \to \mathrm{PGL}_n(\mathbb{C})$, the induced map $\pi_1(\mathbb{C}^\times) \to \pi_1(\mathrm{GL}_n(\mathbb{C}))$ is given by $\mathbb{Z} \xrightarrow{\times n} \mathbb{Z}$. So from the long exact sequence in homotopy, $\pi_1(\mathrm{PGL}_n(\mathbb{C})) = \mathbb{Z}/n\mathbb{Z}$, which shows up in $H^2$. Further, with $\mathbb{Q}$ coefficients, the map $H^*(\mathrm{GL}_n(\mathbb{C}); \mathbb{Q}) \to H^*(\mathbb{C}^\times; \mathbb{Q})$ is surjective, so by Leray-Hirsch, $H^*(\mathrm{GL}_n(\mathbb{C}); \mathbb{Q}) = H^*(\mathrm{PGL}_n(\mathbb{C}); \mathbb{Q}) \otimes H^*(\mathbb{C}^\times; \mathbb{Q})$, which lets us compute $H^*(\mathrm{PGL}_n(\mathbb{C}); \mathbb{Q})$ – in particular, its Poincaré polynomial is given by $(1+t^3)(1+t^5) \dotsm (1+t^{2n-1})$.

But how does one compute the integral cohomology of $\mathrm{PGL}_n(\mathbb{C})$?

• The mod $p$ cohomology (and other things) is computed in the paper "The cohomology of quotients of classical groups" by Baum and Browder (Topology, 1965). Presumably, together with your computation of the rational cohomology, this is enough to reconstruct the integral cohomology. – JHF Jul 26 '17 at 15:24
• @JHF Is it obvious that $\mathrm{PGL}_n$ is the same as (or deformation retracts to) $\mathrm{PU}_n$? – ronno Jul 26 '17 at 16:33
• $U(n) \hookrightarrow GL_n(\mathbb{C})$ is a deformation retract. Consider the map of fiber sequences from $(U(1) \to U(n) \to PU(n))$ to $(GL_1(\mathbb{C}) \to GL_n(\mathbb{C}) \to PGL_n(\mathbb{C}))$. Apply $\pi_*$ and use the five lemma to conclude that $PU(n) \to PGL_n(\mathbb{C})$ is a weak homotopy equivalence. But these are CW complexes, so they are in fact homotopy equivalent. I think there are also more geometric arguments, e.g., $PU(n)$ is the maximal compact subgroup in $PGL_n(\mathbb{C})$, various matrix decompositions, etc. – JHF Jul 26 '17 at 18:41
• @JHF, yes, I now believe the Gram-Schmidt argument for deformation retracting $\mathrm{GL}_n$ to $\mathrm{U}(n)$ also works. If you submit that as an answer, I'd accept it. – ronno Jul 26 '17 at 21:43

By arguments outlined in the comments, the projective linear group $PGL_n(\mathbb{C})$ is homotopy equivalent to the projective unitary group $PU(n)$, so it suffices to compute the cohomology of $PU(n)$.
One way to compute this is to use the Serre spectral sequence for the fiber bundle $U(1) \to U(n) \to PU(n)$. As the question points out, it's not too hard to use this to deduce rational information. For mod $p$ cohomology, potentially nontrivial differentials in this spectral sequence make this a more involved calculation. Nonetheless, this has been worked out in
The idea is that there is another fiber bundle $U(n) \to PU(n) \to \mathbb{C}P^\infty$, obtained by extending the previous fiber bundle once by delooping. The transgressions in this spectral sequence can be analyzed using Chern classes: this principal $U(n)$-bundle is classified by the composite map $\mathbb{C}P^\infty = BU(1) \to BU(1)^{\times n} \to BU(n)$, whose Chern classes, and hence the transgressions, are readily computed. Working at the prime $p$, the authors find (corollary 4.2): $$H^*(PU(n); \mathbb{Z}/p) \cong \Lambda[x_1, x_2, \ldots, \hat{x_{p^r}}, \ldots, x_n] \otimes P[y]/(y^{p^r}),$$ where $p^r$ is the largest power of $p$ dividing $n$, $\deg x_i = 2i - 1$, and $\deg y = 2$.