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Then I claim that the holonomy group of this connection is all of $G$. … Thus, since $G$ is connected, the holonomy of $\alpha$ is all of $G$. …

In fact, one can find a solution $\alpha_1$ that has holonomy in the maximal torus and a solution $\alpha_2$ whose holonomy is all of $\mathrm{SO}(3)$: Let $$ F = \begin{pmatrix}0&\mathrm{d}x_1\wedge\ …

However, I claim that the holonomy of $\nabla$ is $\mathrm{SL}(2,\mathbb{R})$. To see this, note that the Lie algebra of the holonomy group must contain $\mathfrak{so}(2)$. … Nevertheless, the holonomy of $\nabla$ is $\mathrm{SL}(3,\mathbb{R})$, not $\mathrm{SO}(3)$. This follows from an argument very similar to the one given above above. …

The holonomy $H$ of $g$ cannot be contained in $\mathrm{SU}(2)$ because the canonical bundle of $S$ is not trivial (though its square is trivial). … Meanwhile, the identity component of $H$ has to be equal to $\mathrm{SU}(2)$ because this is the holonomy of the (simply-connected, non-product) K3 surface that is the double cover of $S$ endowed with …

Section 3: The author makes some general remarks about holonomy and parallel differential forms and states that any Riemannian $7$-manifold with holonomy in $G_2$ supports a (nonzero) parallel $3$-form … and a (nonzero) parallel $4$-form and that any Riemannian $8$-manifold with holonomy in $\mathrm{Spin}(7)$ supports a (nonzero) parallel $4$-form. …

In general, just knowing the holonomy of $M$ will not tell you much about the holonomy of $\tilde M$. … In general, you need to know more about the metric on $M$ than just its holonomy in order to compute the holonomy of $\tilde M$. …

Suppose that the metric on $M^n$ has irreducible holonomy, is simply connected (or, slightly more generally, that the restricted holonomy $H^0$ acts irreducibly), and that there exist two independent parallel … Since there is at least one parallel complex structure $I$, the holonomy group $H^0$ is a subgroup of $\mathrm{U}(n/2)$. …

Conversely, if this invariance holds, then $R$ is the curvature of an irreducible $n$-dimensional symmetric space with holonomy $G$. … Your other question about characterizing curvature operators of Riemannian manifolds with reduced holonomy is not as easy to answer. …

In case one has a reasonably explicit formula for the map $\exp:{\frak{m}}\to G$ and its inverse, one can work out an explicit formula for the holonomy in that case. … . $$ Using this and a little spherical trigonometry, one sees that, for a geodesic triangle with vertices $A$, $B$, and $C$ (no pair antipodal), then the holonomy of the geodesic path $A\to B\to C\to A …

Observe that, in a $\mathrm{Spin}(7)$-manifold, since $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$, the $\mathrm{Spin}(7)$ decomposition $\Lambda^4 = \Lambda^4_1\oplus \Lambda^4_7\oplus\Lambda^4_{27}\oplus …

Berger's classification of the possible irreducibly acting holonomy groups then implies that the metric must have restricted holonomy equal to $\mathrm{Sp}(n)$, since no proper subgroup of this group can … act irreducibly and be the holonomy. …

When $n=2$, Ricci-flatness of a connection implies that it is flat, so, in that case, yes, you get holonomy reduction locally. … Thus, by the Ambrose-Singer Holonomy Theorem, the holonomy of $\nabla$ is $\mathrm{SL}(n,\mathbb{R})$. …

You're really asking an algebra question about how the various representations of $\mathrm{Spin}(8)$ interact. There are lots of places where you can read about this, but here is a set of notes that …

For which of these are you asking about the holonomy? …