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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?


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Linear or continuous? (or some other criterion?) Linear, there are none for any projective space, so I doubt there are any here. – Will Sawin Dec 2 '11 at 19:29
continuous would be enough. Thanks! – COhrt Dec 2 '11 at 19:31
One thing we know is that the action of the generator is homotopic to a map that induces $\pm 1$ on $H^2$. This is stronger than what we can usually say since we happen to know that $[CP^\infty, CP^\infty] = H^2CP^\infty$. I don't see how this can help, though, since Lefschetz fails badly for non-compact spaces... – Dylan Wilson Dec 3 '11 at 3:35
As an EML space, a $\mathbb Z/n$ action on $\mathbb CP^{\infty}$ is a $\mathbb $Z/n$ action on every cohomology group $H^2(X,\mathbb Z)$. Also, it permutes the complex line bundles on every space. But neither of these facts seems to help. – Will Sawin Dec 3 '11 at 7:03
When $n=2$ I believe the following is a free $\mathbb Z_2$ action. Look at the action on $\mathbb C^\infty$ with the generator acting by $(z_1,z_2,z_3,z_4,\ldots)\mapsto (-\bar z_2,\bar z_1,-\bar z_4,\bar z_3,\ldots)$. The induced action on $\mathbb{CP}^\infty$ is free. Of course, nothing similar can possibly work for $n>2$ because if the action leaves any $\mathbb{CP}^N$ with $N<\infty$ invariant then the square of the generator acts trivially on cohomology and hence has a nonzero Lefschetz number. – Vitali Kapovitch Dec 3 '11 at 16:29
up vote 9 down vote accepted

I believe I can give a complete answer. First, let me collect several earlier comments by myself, Dylan Wilson and Alain Valette. When $N<\infty$ 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 $N<\infty$ invariant).

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.

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.

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 $l_2$.

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 section k-11 here).

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, see here for the specific chapter on infinite dimensional manifolds and relevant definitions) since the spaces involved are $\mathcal C$-absorbing.

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.

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.

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+1 for a definitive, coherent, and nicely written answer. And I didn't know about that result for infinite-dimensional manifolds- very neat! – Dylan Wilson Dec 3 '11 at 23:07
$\mathbb R^\infty$ is smoothly isomorphic to $\mathbb S^\infty$. Who knew? – Tom Goodwillie Dec 4 '11 at 4:42
Thank you for this very well-written answer! – COhrt Dec 4 '11 at 7:42
Vitaly -- your first link requires a password and the second link seems broken (error 404). Could you perhaps give an alternative reference? – algori Dec 4 '11 at 16:16
@algori sorry about the second link. I just fixed it. please try now. – Vitali Kapovitch Dec 4 '11 at 16:34

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 $\cal{H}_i=\{v\in\cal{H}:\pi(k)v=\omega^{ik}v, \forall k\in\mathbb{Z}/n\}$. Any one-dimensional subspace contained in some non-zero $\cal{H}_i$, will then be a fixed point for the corresponding action on $\mathbb{CP}^N$.

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I don't think this works without some extra assumptions for $N=\infty$ because $\mathbb{CP}^\infty$ (at least the canonically constructed one) is smaller than the set of complex lines in a Hilbert space. – Vitali Kapovitch Dec 3 '11 at 16:20
Also, you may as well prove the stronger statement that there is no such continuous action, via the Lefschetz fixed point theorem... the cohomology is concentrated in even degrees, so the Lefschetz number has to be nonzero. But I agree with Vitali, I don't think either fo these arguments can be extended to the $CP^\infty$ case – Dylan Wilson Dec 3 '11 at 18:54
@Vitali: Your comment made me smile, because this is a recurring discussion between analysts and topologists (my definition of $\mathbb{CP}^\infty$ being the analysts'one - we are aware that we do not get a CW-complex!). The homotopy equivalence between the analyst's model and the topologist's model is proved e.g. by Guido Mislin in the appendix of my book "Introduction to the Baum-Connes conjecture", ETHZ Lecture Notes, Birkhauser 2000. – Alain Valette Dec 3 '11 at 19:18
@Alain I'm aware of the homotopy equivalence of the two models but I always considered the canonical model of $\mathbb{CP}^\infty$ to be the direct limit of $\mathbb{CP}^n$. And in this case a particular model makes a big difference because if we are only concerned with the homotopy type then there is always a free action of $\mathbb Z_n$ via Borel construction. In fact with your definition there is such an action on the original space because then $\mathbb{CP}^\infty$ and $\mathbb{CP}^\infty\times \mathbb S^\infty$ are homeomorphic since they are homotopy equivalent $l_2$-manifolds. – Vitali Kapovitch Dec 3 '11 at 19:41
@Vitali: Thanks a lot for your comment, which taught me things I was completely unaware of. But coming back to non-existence of linear free actions on $\mathbb{CP}^\infty$ (now with your model), I think that the same argument as for Hilbert space works, since the projections onto isotypic components can be defined purely algebraically, as elements of the complex group ring of $\mathbb{Z}/n$, and therefore make sense in any $\mathbb{Z}/n$-module... – Alain Valette Dec 3 '11 at 20:38

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