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I've began to study the relationship between calibrations and holonomy, mainly through D.D. Joyce's Riemannian Holonomy Groups and Calibrated Geometry and partly through internet material.

Pretty much everyone explains this relationship by the holonomy principle: if $H=\text{Hol}_x$ and $\varphi_0$ is an $H$-invariant $k$-form in $T_pM$, then there is a parallel $k$-form $\varphi$ in $M$ with $\nabla\varphi=0$. In particular, this means $d\varphi=0$. Rescaling $\varphi_0$ if necessary, we get that $\varphi$ is a calibration.

So far, so good.

After this, people start saying something about special holonomy and invariably mention Berger's classification.

1) What does special mean in this context? I thought this was an informal adjective used by Joyce, but apparently everyone uses it and I haven't found a definition for it.

2) I understand Berger's list is interesting, for they deal with irreducible manifolds. But why don't they mention symmetric manifolds, which are not on the list, like $\mathbb{R}^n$, $\mathbb{S}^n,\mathbb{R}H^n$, compact Lie groups etc. They seem pretty interesting (and numerous) to me, so why not consider them?

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(1) A generic metric has (restricted) holonomy group $SO(n)$ (More proper statement: the set of holonomy $SO(n)$ metrics is comeagre in the space of all Riemannian metrics). Hence the adjective special is coined (as in the opposite of "generic") when we can reduce it to smaller subgroups. It definitely predates Joyce (certainly Harvey and Lawson used that in their seminal paper introducing calibrations in the early 1980s).

(2) Elie Cartan proved that for Riemannian symmetric spaces $G/H$, the restricted holonomy group is the identity component of the isotropy group $H$. So this is just a pure algebra problem as to which Lie group is a subgroup of another Lie group (or equivalently which Lie algebra is a subalgebra of another), hence not interesting (as in having little if not no geometry content).

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  • $\begingroup$ If I got it right, you're saying that determining $\text{Hol}_p$ of a symmetric manifold is essentially an algebraic problem, not a geometric one. But in the context of calibrations, our goal is to find calibrated submanifolds, right? By the method I described, we try that by looking for $\text{Hol}_p$-invariant forms. $\endgroup$
    – rmdmc89
    Commented Jun 18, 2019 at 23:00
  • $\begingroup$ My point is: if we already know $\text{Hol}_p$ (like we do in $\mathbb{S}^n$, for example), we have everything we need to find new calibrations, so being symmetric or not is irrelevant, right? $\endgroup$
    – rmdmc89
    Commented Jun 18, 2019 at 23:33
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    $\begingroup$ Yes, you have that. But finding calibration alone isn't the end-goal --- we want to study the submanifolds they calibrate in order to learn more about the manifold. The questions we ask in symmetric spaces (knowing it is symmetric) is, therefore, in a sense the opposite of what we do elsewhere. For nonsymmetric spaces we don't have this luxury of turning to algebra (well, unless you count algebraic geometry for things like Kahler or Calabi-Yau) so a lot remains in analysis of PDEs and suchlike. $\endgroup$
    – user10354138
    Commented Jun 19, 2019 at 0:37
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    $\begingroup$ A symmetric space $M$ is the quotient of two Lie groups $\operatorname{Isom}(M)/\operatorname{Isom}_p(M)$. $\endgroup$
    – user10354138
    Commented Jun 19, 2019 at 1:54
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    $\begingroup$ If you are on the same underlying manifold, yes. $\endgroup$
    – user10354138
    Commented Jun 19, 2019 at 12:44

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