# Holonomy of a Warped Product Metric

A warped product metric on the "cone" $$\tilde{M} = \mathbb{R}^{+} \times M$$ is $$\tilde{g} =dr^2 + r^2g_M$$ where $$g_M$$ is the metric on $$M$$.

If we know the holonomy group of the manifold $$(M,g_M)$$, what can we say about the holonomy group of $$(\tilde{M}, \tilde{g})$$?

Any references?

Edit: Here is a much more specific question:

In this paper: https://arxiv.org/pdf/math/0703231.pdf, lemma 2 states that the $$\tilde{g}$$ is Ricci-flat if and only if $$g_M$$ is Einstein with Einstein constant $$(n-1)$$. So in the case where $$M$$ is a $$G_2$$ manifold, it has Einstein constant 0 (on account of being Ricci flat). So that means the cone metric over a $$G_2$$ manifold is never Ricci-flat (and hence, for example, can't have holonomy Spin(7)).

Is that correct?

In general, just knowing the holonomy of $$M$$ will not tell you much about the holonomy of $$\tilde M$$. For example, if $$M$$ is isometric to the sphere of radius $$r>0$$ (and the dimension of $$M$$ is at least $$2$$), then the holonomy of $$M$$ is $$\mathrm{SO}(n)$$. When $$r\not=1$$, the holonomy of $$\tilde M$$ will be $$\mathrm{SO}(n{+}1)$$, but when $$r = 1$$, the holonomy of $$\tilde M$$ will be trivial, because $$\tilde M$$ will be isometric to $$\mathbb{R}^{n+1}$$.
As another example, when $$M^6$$ is strictly nearly K\"ahler, then the holonomy of $$M$$ will be $$\mathrm{SO}(6)$$. In most cases, the holonomy of $$\tilde M$$ will be $$\mathrm{SO}(7)$$, but, if the scalar curvature is the right constant (I forget the exact value), the holonomy of $$\tilde M$$ will be $$\mathrm{G}_2$$ or trivial.
In general, you need to know more about the metric on $$M$$ than just its holonomy in order to compute the holonomy of $$\tilde M$$. Usually, the metric on $$M^n$$ has to be quite special in order for the holonomy of $$\tilde M$$ to be a proper subgroup of $$\mathrm{SO}(n{+}1)$$.
Answer to the added specific question: Yes, that's correct, the cone on a $$\mathrm{G}_2$$-holonomy metric never has holonomy $$\mathrm{Spin}(7)$$. On the other hand, the cone on a strictly nearly $$\mathrm{G}_2$$-structure (one that has the right scalar curvature and is not the $$6$$-sphere) will have holonomy $$\mathrm{Spin}(7)$$
• Robert listed examples where the holonomy of the warped product is smaller than the generic case. The converse also happens. If the fibre is $T^n$, you can construct hyperbolic cusps as warped products, which have full holonomy $SO(n+1)$. – Sebastian Goette Feb 12 at 20:26