7
$\begingroup$

I take the following quote from Huybrecht's notes on hyperkähler manifolds and mirror symmetry:

Mirror symmetry in a first approximation predicts for any Calabi-Yau manifold (M,g) the existence of another Calabi-Yau manifold $(M^v,g^v)$ together with an isomorphism $M^{\text{cpl}}(M) \simeq M^{\text{khl}}(M^{v})$.

Here $M^{\text{cpl}}(M)$, and $M^{\text{khl}}(M^{v})$, refer to the moduli spaces of complex, resp. symplectic, structures on $M$, resp $M^v$.

Mirror symmetry is supposed to be much simpler for hyperkähler manifolds, as usually the mirror manifold $M^v$ as a real manifold is $M$ itself.

If one looks at quaternionic-Kähler manifolds, is there any analogous simplification that holds. Discussion of some simple examples (ie the Grassmannian $Gr(4,2)$) would be appreciated.

$\endgroup$

2 Answers 2

7
$\begingroup$

Let me summarize and supplement my remarks above. A quaternion-Kahler manifold $X$ is a Riemannian manifold with holonomy $Sp(n)Sp(1)$, its definition of course includes hyperkahler manifolds as a special case. For a hyperkahler manifold, mirror symmetry can sometimes be realized as a hyperkahler rotation, e.g. elliptic $K3$ surfaces. However, such an understanding fails in general, as was pointed out by Huybrechts in his lecture notes: http://arxiv.org/pdf/math/0210219.pdf. For the method to handle mirror symmetry in the general case, see the work of Gross-Siebert: http://arxiv.org/abs/math/0703822.

Now let's exclude hyperkahler manifolds by assuming the scalar curvature is nonzero. For simplicity, let's further assume that $X$ is symmetric, then it can be written as $X=G/H$ with $G$ a simple Lie group and $H=K\cdot SU(2)$ , $K$ is the centralizer of $SU(2)$ in $G$. In this case, as far as I konw, only the mirror symmetry for the case when $G=SU(n+2)$ and $H=S\big(U(n)\times U(2)\big)$ is systematically studied.

In general, for the homogeneous space $X=G/P$, where $G$ is a semisimple Lie group and $P$ a parabolic subgroup, its mirror is given by the Landau-Ginzburg model $(R,W)$, where $R\subset G^L/P^L$ is the Richardson variety, and $W:R\rightarrow\mathbb{C}$ is a holomorphic function called superpotential. Here $G^L$ and $P^L$ denote their Langlands dual. The main reason for using $X^\vee=R$ to partially compactify $(\mathbb{C}^\ast)^N$ is to get the following isomorphism between the quantum cohomology ring of $X$ and the Jacobi ring of $W$:

$QH^\ast(X)\cong Jac(W)$.

This isomorphism is generally expected to hold for mirror symmetry for Fano manifolds. For the case when $X$ is toric Fano, such an isomorphism is proved by Fukaya-Oh-Ohta-Ono: http://projecteuclid.org/euclid.dmj/1262271306. The general mirror construction for $X=G/P$ is done by Rietsch: http://arxiv.org/abs/math/0511124.

The isomorphism $QH^\ast(X)\cong Jac(W)$ should be compared with Kontsevich's conjecture, which asserts the following:

$QH^\ast(X)\cong HH^\ast\big(\mathcal{F}(X)\big)$,

where $\mathcal{F}(X)$ is the Fukaya category of $X$. The interesting thing is that when $X$ is Grassmannian (e.g. $X=Gr(2,n+2)$, in which case $X$ is quaternion-Kahler), $W:X^\vee\rightarrow\mathbb{C}$ has only isolated critical points, therefore it should provide a Lefschetz fibration on $X^\vee$, so Seidel's theory (http://www.ems-ph.org/books/book.php?proj_nr=12) can be applied and one may expect that all the geometric information of the Fukaya category $\mathcal{F}(X^\vee,W)$ is contained in the superpotential $W$. This suggests that $\mathcal{F}(X)$ and $\mathcal{F}(X^\vee,W)$ should be related to each other.

From an SYZ point of view, there is no such thing as mirror symmetry for Fano manifolds. Because everytime we talk about mirror symmetry for a Fano manifold $X$ we regard it as a Calabi-Yau manifold $X_0$ together with some boundary divisor $D$, i.e. $X_0=X\setminus D$. The Landau-Ginzburg mirror manifold $X^\vee$ is constructed from $X$, and $D$ determines only the superpotential $W$. For the SYZ mirror construction for some of the Fano examples, see Auroux's fundamental work: http://arxiv.org/abs/0706.3207.

In fact, such a philosophy coming from SYZ mirror symmetry applies also to the case when $X=G/P$. There is a Richardson variety $R^\vee$ which is dual to $R$ inside $G/P$. The open part $R^\vee\subset X$ should be regarded to be SYZ mirror to $R\subset G^L/P^L$. But when we add back the boundary divisor $X\setminus R^\vee$, we need to use the superpotantial $W$ to correct $R$, so we end up with the Landau-Ginzbug model $(R,W)$. From such a point of view, the meaning of the following mirror dualities should be clear:

$R^\vee\leftrightarrow R$

$(R,W)\leftrightarrow G/P$

$G^L/P^L\leftrightarrow (R^\vee,W^\vee)$

Let me also mention that the philosophy that $T$-duality should be realized as the Langlands duality in some special cases dates back to the work of Hausel and Thaddeus on mirror symmetry for Hitchin moduli spaces: http://arxiv.org/pdf/math/0205236v1.pdf. As the quaternion-Kahler case (with non-zero Ricci curvature) should be regarded as parallel to the hyperkahler case, it's not strange that the same philosophy works here.

To get a deep understanding of mirror symmetry in the quaternion-Kahler case, we will need to study the geometric structures of a quaternion-Kahler manifold, and understand how the geometric structures exchanges when passing from $X$ to $X^\vee$. Unfortunately, personally I don't know much development in these directions. However, there is the paper of Leung (http://arxiv.org/pdf/math/0303153v1.pdf) which aims to use normed division algebras to unify the geometric structures coming from different holonomy groups. I personally believe this unified picture should be fundamental in the study of mirror symmetry for quaternion-Kahler case.

$\endgroup$
1
  • $\begingroup$ In fact as comment $QH^*(X)\cong H^*(\wedge^* TX^\vee)$ $\endgroup$
    – user21574
    Jul 18, 2017 at 14:13
2
$\begingroup$

Quaternionic Kaehler manifolds are not Calabi Yau unless they are hyperkaehler.

$\endgroup$
6
  • $\begingroup$ I don't think they are complex manifolds. The structure group acts in a quaternionic representation, not preserving any complex structure. The subgroup fixing a complex structure is the hyperkaehler structure group. $\endgroup$
    – Ben McKay
    Sep 24, 2014 at 18:46
  • $\begingroup$ Quaternionic projective spaces admit no symplectic structure (because the 2nd cohomology group vanishes), so quaternionic Kaehler but not Kaehler. $\endgroup$
    – Ben McKay
    Sep 24, 2014 at 18:50
  • $\begingroup$ I would be curious to see a mirror symmetry example for quaternionic Kaehler manifolds. There seems to be a mirror symmetry notion for G2 holonomy manifolds, and Spin(7) holonomy manifolds (as far as I vaguely understand) so I don't know any reason why there wouldn't be. $\endgroup$
    – Ben McKay
    Sep 24, 2014 at 18:52
  • 1
    $\begingroup$ $Gr(N+2,2)$ is quaternionic Kaehler, and also a flag manifold. As far as I understand, mirror symmetry has been worked out for all the flag manifolds . . . so this should provide an example. $\endgroup$ Sep 24, 2014 at 19:19
  • 3
    $\begingroup$ @User1298 The mirror of $G/P$ with $P$ a parabolic subgroup is a Landau-Ginzburg model $(R,W)$ with $R\subset G^L/P^L$, where $G^L$ means the Langlands dual of $G$ and $R$ is the Richardson variety. For example, when $G=SL(n,\mathbb{C})$, $R$ is always an algebraic torus. But for $G/P=Gr(2,4)$, $R$ is not $(\mathbb{C}^\ast)^3$, it should be some partial compactification of $(\mathbb{C}^\ast)^3$. Whenever $G/P$ is Grassmannian, $W:R\rightarrow\mathbb{C}$ has isolated critical points. $\endgroup$
    – YHBKJ
    Sep 25, 2014 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.