Is the mirror of a hyperkaehler manifold always a hyperkaehler manifold?

What I know so far is as follows:

In this paper (https://arxiv.org/pdf/hep-th/9512195.pdf) by Verbitsky, it is claimed that every hyperkaehler manifold is mirror to itself (which would meant the answer to my question is 'yes').

On the other hand, from the answer to this question - Mirror symmetry for hyperkahler manifold , it seems that this is not always the case, i.e., a K3 surface can have a nontrivial mirror.


2 Answers 2


I think Verbisky proves a refined form of the Mirror Conjecture for hyperkaehler manifolds, not the conjecture in the strict form. This is explained at page 3 of the paper that you link.

In fact, the assumptions for the conjecture are known to hold only for a hyperkaehler manifold which is generic in its deformation class. More precisely, Verbitsky proves the following result, see page 26.

Theorem 5.4. Let $M$ be a compact holomorphically symplectic manifold, which is generic in its deformation class. Then the Mirror Conjecture holds for $M$, which is mirror dual to itself.

For instance, $\mathrm{Pic}(M)=0$ is enough, see the footnote in the same page.

  • 3
    $\begingroup$ Typo correction: the transliteration that Verbitsky uses in his papers is "Verbitsky" (of course it would be different in other languages than English, but this seems to be the transliteration he uses himself). $\endgroup$ Jun 1, 2017 at 9:21
  • 1
    $\begingroup$ As comment: In fact Verbisky gave an isomorphism between the A-model and B-model of variation of Frobenius algebras $\endgroup$
    – user21574
    Jun 1, 2017 at 9:36
  • $\begingroup$ Thank you for the answer. I think for clarity, it would be good if you could comment on my original question, i.e., whether or not we know that the mirror of a hyperkaehler manifold is always hyperkaehler. $\endgroup$
    – Mtheorist
    Jun 2, 2017 at 11:24

The following example from Hausel–Thaddeus say that mirror hyperKahler is HyperKahler. Lets explain it

Let $\mathcal M$ be the moduli space of stable $GL_n$-Higgs bundles, non-singular and hyperkahler and $\tilde {\mathcal M}$ moduli space of stable $SL_n$-Higgs bundles, non-singular and hyperkahler and ${\hat{\mathcal{M}}}:= \tilde{\mathcal M} /Γ $ is $PGL_n$-Higgs moduli space which is an orbifold($\Gamma\cong \mathbb Z_n^{2g}$)

Hausel–Thaddeus , proved the following theorem about Strominger-Yau-Zaslow conjecture for the hyperKahler mirror pair $(\hat{\mathcal M}, \tilde {\mathcal M})$

\begin{array} ^\tilde{\mathcal M} & \stackrel{}{\longrightarrow} & \hat{\mathcal M}\\ \downarrow{\tilde\chi} & & \downarrow{\hat\chi} \\ \mathcal A & \stackrel{\cong}{\longrightarrow} & \mathcal A \end{array}

The generic fibers $\tilde\chi^{-1}(a)$ and $\hat \chi^{-1}(a)$ are dual Abelian varieties. The pair of hyperkahler manifolds $(\tilde{\mathcal M} , J)$ and $(\hat{\mathcal M} , J)$ satisfy SYZ conjecture and mirror to each other


Mirror symmetry, Langlands duality, and the Hitchin system, Inventiones mathematicae, July 2003, Volume 153, Issue 1, pp 197–229

Hausel - Thaddeus interpreted SYZ conjecture in the context of Hitchin system and Langlands duality. Let briefly explain it

Let $\pi : E \to Σ$ a complex vector bundle of rank $r$ and degree $d$ equipped with a hermitian metric on Riemann surface $\Sigma$ . Take th moduli space $$\mathcal M(r, d) = \{(A, Φ) \text{ solving }(\star)\}/\mathcal G $$

(which is a finite-dimensional non-compact space carrying a natural hyper-Kähler metric)


$$F^0_A + [Φ ∧ Φ^∗] = 0 ,\; \; \bar ∂AΦ = 0\; \; (\star)$$

Here $A$ is a unitary connection on $E$ and $Φ ∈ Ω^{1,0}(End E)$ is a Higgs field. $F^0$ denotes the trace-free part of the curvature and $\mathcal G$ is the unitary gauge group.

$\mathcal M(r, d)$ is the total space of an integrable system(which can be interpreted by the non-abelian Hodge theory due to Corlette), the Hitchin fibration, together with Langlands duality between Lie groups provides a model for mirror symmetry in the Strominger-Yau and Zaslow conjecture.

  • $\begingroup$ Thank you for your answer. Perhaps you could comment on my original question, i.e., whether it is true in general that the mirror of a hyperkaehler manifold is hyperkaehler? $\endgroup$
    – Mtheorist
    Jun 2, 2017 at 11:28
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    $\begingroup$ Every Calabi–Yau manifold in complex two dimensions is a hyperkähler manifold, so your question when $dim_{\mathbb C}X=3$ is good question $\endgroup$
    – user21574
    Jun 2, 2017 at 12:11

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