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Let $PU(n) = U(n)/U(1)$ be the projective unitary group and denote by $BPU(n)$ its classifying space. Consider the algebra $M_n(\mathbb{C})$ as an $n^2$-dimensional Hilbert space equipped with the inner product $\langle T, S \rangle = \mathrm{tr}(T^*S)$. Then we have a group homomorphism $$ PU(n) \to U(n^2)\ , $$ which sends $[u] \in PU(n)$ to the linear map $\mathrm{Ad}_u \colon M_n(\mathbb{C}) \to M_n(\mathbb{C}) \ ;\ T \mapsto uTu^*$. The corresponding map $BPU(n) \to BU(n^2)$ on classifying spaces induces $$ \mathrm{Ad}^* \colon \mathbb{Z}[c_1, \dots, c_{n^2}] \cong H^*(BU(n^2), \mathbb{Z}) \to H^*(BPU(n), \mathbb{Z})\ . $$ The generators of the polynomial ring are the Chern classes of the universal bundle. Since every bundle of matrix algebras has the unit section as a non-vanishing section, we have $\mathrm{Ad}^*(c_{n^2}) = 0$.

For which $n \in \mathbb{N}$ do we have $\mathrm{Ad}^*(c_{n^2-1}) \neq 0$ ?

(I just realized that the map could rightfully be called $\mathrm{BAd}^*$, but decided against it :-)

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You may as well look at what happens on maximal tori, since Chern classes are seen there. Let $T$ be the maximal torus of $PU(n)$ and $D$ the maximal torus of the unitary group. Let's write the cohomology rings of their classifying spaces as $\mathbb{Z}[t_i-t_j]/~$ and $Z[x_{i,j}]$ resp. Here the quotient is by the evident linear relations on the generators (you get a polynomial ring in $n-1$ generators I just wanted to avoid choosing a basis).

It's not hard to check that the induced map on cohomology sends $x_{i,j}$ to $t_i-t_j$. (In particular the diagonal guys go to zero). So the effect on Chern classes is the effect on symmetric polynomials- you're asking when those go to zero in the target.

You've observed that the top polynomial vanishes, which here follows from the fact that the top symmetric polynomial has a factor of, say, $x_{1,1}$ in it.

Unless I'm mistaken- none of the other terms vanish. You can see this by noting that the total Chern class is given by coefficients of the polynomial in $z$, $\prod (z-t_i - t_j)$. Substitute $t_1=1$ and $t_i=0$ otherwise. The resulting polynomial is like $z(z-1)^k$ which has vanishing constant term but non vanishing coefficients otherwise (binomial coefficients).

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  • $\begingroup$ (To compute the map on cohomology you need to know that inversion on the circle gives multiplication by -1 on group cohomology and multiplication sends a generator to the sum of the two natural generators, since the Chern class of a tensor product is the sum.) $\endgroup$
    – Dylan Wilson
    Commented Mar 23, 2016 at 23:33
  • $\begingroup$ How do these classes behave under $\mathrm{Ad}^*$? $\endgroup$
    – Sebastian Goette
    Commented Mar 25, 2016 at 12:11
  • $\begingroup$ What do you mean? Ad* takes x_ij to t_i-t_j and Chern classes go to the corresponding symmetric polynomials. $\endgroup$
    – Dylan Wilson
    Commented Mar 25, 2016 at 12:45
  • $\begingroup$ There is one thing I don't understand though. What is wrong with the following reasoning: The $(n^2-1)$th Chern class in $H^*(BU(n^2), \mathbb{Z})$ is mapped to the $(n^2-1)$th elementary symmetric polynomial. This is a sum of monomials with each one containing at least one "diagonal" element $x_{ii}$. Since the induced map $H^*(BU(n^2)) \to H^*(BPU(n))$ is a ring homomorphism, this should be mapped to $0$ by your argument. $\endgroup$
    – Ulrich Pennig
    Commented Mar 30, 2016 at 9:50
  • $\begingroup$ It's not the case that each monomial has such an element. For example, the first symmetric polynomial is just the sum of all the $x_{i,j}$. Most of those will map to something nonzero. $\endgroup$
    – Dylan Wilson
    Commented Mar 30, 2016 at 14:38

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