Free group actions on complex projective spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T03:31:24Z http://mathoverflow.net/feeds/question/82500 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82500/free-group-actions-on-complex-projective-spaces Free group actions on complex projective spaces COhrt 2011-12-02T18:35:51Z 2011-12-04T16:33:09Z <p>Hi,</p> <p>is there a free group action of the cyclic group $\mathbb{Z}/n\mathbb{Z}$ on the infinite dimensional projective space $\mathbb{CP}^\infty$ for every $n\in \mathbb{N}$? And if there is one, how does it work?</p> <p>Thanks</p> http://mathoverflow.net/questions/82500/free-group-actions-on-complex-projective-spaces/82539#82539 Answer by Alain Valette for Free group actions on complex projective spaces Alain Valette 2011-12-03T06:40:57Z 2011-12-03T07:39:27Z <p>Here is the spell-out of Will's comment, that there is no free linear action of $\mathbb{Z}/n$ on $\mathbb{CP}^N$, for $1\leq N\leq\infty$. View $\mathbb{CP}^N$ as the set of 1-dimensional linear subspaces in a complex Hilbert space $\cal{H}$ of dimension $N+1$. Any linear action $\pi$ of $\mathbb{Z}/n$ on $\cal{H}$ decomposes as a sum of isotypic components: fix a primitive $n$-th root of unit $\omega\in\mathbb{C}$; then $\cal{H}=\bigoplus_{i=0}^{n-1}\cal{H}_i$, where <code>$\cal{H}_i=\{v\in\cal{H}:\pi(k)v=\omega^{ik}v, \forall k\in\mathbb{Z}/n\}$</code>. Any one-dimensional subspace contained in some non-zero $\cal{H}_i$, will then be a fixed point for the corresponding action on $\mathbb{CP}^N$.</p> http://mathoverflow.net/questions/82500/free-group-actions-on-complex-projective-spaces/82583#82583 Answer by Vitali Kapovitch for Free group actions on complex projective spaces Vitali Kapovitch 2011-12-03T21:01:15Z 2011-12-04T16:33:09Z <p>I believe I can give a complete answer. First, let me collect several earlier comments by myself, Dylan Wilson and Alain Valette. When $N&lt;\infty$ then there can not be a free action of $\mathbb Z_n$ on $\mathbb{CP}^N$ when $n>2$. Indeed, if there were such an action then the square of the generator would act trivially on cohomology. It would therefore have a positive Lefschetz number and hence a fixed point. (This also means that for $n>2$ no free $\mathbb{Z}_n$ action on $\mathbb{CP}^\infty$ can leave any $\mathbb{CP}^N$ with $N&lt;\infty$ invariant).</p> <p>When $n=2$ then such an action is possible if and only if $N$ is odd. This action easily generalizes to a $\mathbb Z_2$ action on $\mathbb{CP}^\infty$ (with any definition of $\mathbb{CP}^\infty$ ) with the generator acting on $\mathbb{S}^\infty$ by $(z_1,z_2,z_3,z_4,\ldots)\mapsto (-\bar z_2,\bar z_1, -\bar z_4,\bar z_3,\ldots)$. This action normalizes the diagonal $S^1$ action and thus descends to an action on $\mathbb{CP}^\infty$ which is easily seen to be free.</p> <p>Before proceeding further let's discuss the fact that we have two competing definitions of $\mathbb{CP}^\infty$. They are homotopy equivalent to each other but that's not enough for this problem. Indeed, if we are only interested in the question up to homotopy equivalence then the answer is trivially "yes" by Borel construction.</p> <p>The first definition (what Alain calls topologist's definition) is that $\mathbb{CP}^\infty$ is the direct limit of $\mathbb{CP}^k$s under canonical inclusions $\mathbb{CP}^1\hookrightarrow \mathbb{CP}^2\hookrightarrow \mathbb{CP}^3\hookrightarrow\ldots$. The other (analyst's) definition is $\mathbb{CP}^\infty_H:=\mathbb{S}_H^\infty/\mathbb S^1$ where $\mathbb{S}^\infty_H$ (here $H$ stands for Hilbert) is the unit sphere in $l_2$.</p> <p>With the second definition we have that $\mathbb{CP}_H^\infty$ and $\mathbb{CP}_H^\infty\times \mathbb{S}_H^\infty$ are homotopy equivalent (since $\mathbb{S}_H^\infty$ is contractible) and hence are homeomorphic since any 2 homotopy equivalent $l_2$-manifolds are homeomorphic (see <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=2049453" rel="nofollow">section k-11 here)</a>.</p> <p>Moreover, I believe the same works for the first (topological) definition of $\mathbb{CP}^\infty$ as well. In that case $\mathbb{CP}^\infty$ is not modeled on $l_2$ but rather on $\mathbb C^\infty$ (the direct limit of $\mathbb C^k$). However, if I understand the definitions correctly, for such spaces we again have that homotopy equivalence of $\mathbb{CP}^\infty$ and $\mathbb{CP}^\infty\times \mathbb S^\infty$ implies homeomorphism (same reference as before, <a href="http://e-books.enigma.uran.ru/book_pdf/enc/k/k_11.pdf" rel="nofollow">see here for the specific chapter on infinite dimensional manifolds</a> and relevant definitions) since the spaces involved are $\mathcal C$-absorbing.</p> <p>To summarize, with whatever definition of $\mathbb{CP}^\infty$ free actions of $\mathbb Z_n$ on it exist because of Borel construction together with the fact that in infinite dimension the Borel construction does not change homeomorphism type for relevant classes of infinite dimensional manifolds.</p> <p>Lastly, let me add that as observed by Alain, with either definition such action can not be $\mathbb C$-linear for any $n$ (my example for $n=2$ is of course $\mathbb R$-linear only). See his answer for more details.</p>