Title: Can Calabi-Yau manifolds have nonabelian discrete symmetry groups?
Tags: gr.group-theory;dg.differential-geometry;complex-geometry;calabi-yau
Question: A particle physicist asked me the above question.  Let me try to make it more precise.  Suppose $M$ is a 3-dimensional Calabi-Yau manifold: that is, a compact K&auml;hler manifold of complex dimension 3 whose holonomy group is contained in $\mathrm{SU}(3)$.  Let $\mathrm{Aut}(M)$ be its group of holomomorphic metric-preserving diffeomorphisms.  What can this group be like?  In particular:
1) which nonabelian discrete groups can $\mathrm{Aut}(M)$ contain?
or if that's unmanageable:
2) which nonabelian discrete groups can appear as the group of connected components of $\mathrm{Aut}(M)$?
I believe he is particularly curious as to whether we can get $\mathrm{PSL}(2,7)$ as the answer to either of these questions.
Comment: Probably CY was intended -- CY 3-folds are what string theorists love.

Comment: Why there is "Calabi-Yau" in the title and only "Kähler" in the body of the question? Is $M$ supposed to be Ricci flat? Trivial fundamental group?

Comment: If you don't insist on compactness, it seems to me that the total space $M$ of the rank 2 bundle $T_{C}^{*}\oplus \mathcal{O}_C$, where $C$ is the Klein genus 3 quartic curve of automorphism group $PSL(2,7)$, is CY and its $Aut(M)$ contains $PSL(2,7)$.

Comment: I suggest the following, which I will carry out in detail later if I have time.  $PSL(2, 7)$ has an irreducible $3$-dim'l representation $V$ (as well as $\bar V$), I claim it suffices to find a large trivial summand of $\text{Sym}^8(V\oplus 1)$, which probably exists (for example I think $\text{Sym}^{2k}(V)$ has a trivial summand).  Indeed, $PSL(2, 7)$ acts on $\mathbb{P}(V\oplus 1)$ in the obvious way; take a generic octic invariant under the group action.  Then the double cover of $\mathbb{P}(V\oplus 1)$ branched over this octic should have an action of $PSL(2, 7)$... (cont.)

Comment: We want to find an invariant octic which is smooth, whence the double cover will be a smooth CY $3$-fold with an action of $PSL(2, 7)$.  If the trivial summand of $\text{Sym}^8(V\oplus 1)$ is big enough (and "sufficiently generic") this will follow by Bertini; we must actually check that the linear system $|\Gamma(\mathbb{P}^3, \mathcal{O}(8))^{PSL(2, 7)}|$ is base-point free, or produce a smooth invariant octic.  I think one can see by "pure thought" that the trivial summand of $\text{Sym}^8(V\oplus 1)$ is at least $5$-dim'l, so it would take a miracle for this not to work...

Comment: (And just to answer the question in the title, the CY $3$-fold defined by $\sum x_i^5=0$ certainly has discrete non-Abelian symmetry group.)

Comment: I got distracted and forgot to add a definition of Calabi-Yau.  I fixed the question.  That's an interesting noncompact example, but I think my friend needs something compact!


Here is another answer: Assuming you actually meant "Kähler" and not "Calabi-Yau":
In the book Fundamental Groups of Compact Kähler Manifolds by Amorós et al., on page 6 (example 1.11) it is asserted that every finite group can occur as the fundamental group of a compact Kähler manifold, and the result is attributed to Serre (J.P. Serre, Sur la topologie des variétés algébriques en caractéristique p).
Hence, by taking covers, it can also occur as a subgroup of holomorphic isometries of a compact Kähler manifold.
Comment for this answer: I meant Calabi-Yau, since my physicist friend was curious about discrete symmetries of string theories.  But I left the crucial Calabi-Yau condition out of my more detailed statement!  I'll fix that now.

Here is another answer: If you work one dimension down, at the level of K3 surfaces, there's a very pretty classification of finite group actions preserving the holomorphic form due to Mukai.  In that classification, the simple group of order 168 is extremal.  Oguiso and Zhang have a nice article on the properties of K3 surfaces which admit such an action.  To be specific, you can get a K3 surface with this group action (the "Klein-Mukai surface") by taking a fourfold cover of the plane branched over Klein's quartic: $x^3 y + y^3 z + z^3 x + w^4=0$.  It would be fun to look at Calabi-Yau threefolds with a geometric relationship to the Klein-Mukai surface.

Here is another answer: The main theorem of Fine and Panov's 'The diversity of symplectic Calabi-Yau six-manifolds'   (http://arxiv.org/abs/1108.5944) implies that, in particular, every finitely presented group arises as the fundamental group of a Calabi--Yau. As in the Kaehler case, what you want then follows by passing to the universal cover.
Edit: As abx points out, Fine and Panov construct symplectic Calabi--Yaus, which are not necessarily complex Calabi--Yaus as required by the question.  But I'll leave this answer up, as I suspect that this kind of construction may be what's needed.  Both Fine and Panov are sometimes active on MO, so perhaps one of them will answer.
Comment for this answer: Good point! (Distinguishing different types of Calabi--Yau is beyond my pay grade).  I seem to remember that twistor-type constructions can be used to produce compact six-dimensional complex manifolds with arbitrary (finitely presented) fundamental group, so it seems plausible to me that Fine and Panov's techniques might be able to answer this question, but I'm no expert in this area.
Comment for this answer: (Six-real-dimensional complex manifolds, that is.)
Comment for this answer: In the paper you quote Calabi-Yau is taken in the symplectic sense; it is not clear to me that these examples admit a complex structure.

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