# What is the significance of using Warped Products?

I have recently started studying differential geometry. I came across this idea of Warped Product Manifolds and I was wondering how is it different from regular Cartesian products of manifolds? What special properties or special intuition this gives compared to regular cartesian product of manifolds? Any insights are welcome. Thank you in advance!

• You take the warped product of Riemannian manifolds. The underlying manifold is just the Cartesian product, but the metric is not the product metric. – Michael Albanese Jun 21 '20 at 14:44
• I heard a lot of things about this book by Bang-Yen Chen. But it is expensive. – Si Kucing Jun 21 '20 at 14:48
• @MichaelAlbanese So for two Riemannian manifolds, warped product influences the metric that will be induced. But the product of those two manifolds is just cartesian product? – AdaMStrange Jun 21 '20 at 14:54
• @Eumenes Yes I have checked this out. Unfortunately I cant afford to buy this book :( – AdaMStrange Jun 21 '20 at 14:55
• @AdaMStrange: Yes. – Michael Albanese Jun 21 '20 at 15:15

Given two riemannian manifolds $$(M,g_M)$$ and $$(N,g_N)$$, there is one natural way to construct a product manifold $$M\times N$$ but plenty natural ways to construct product riemannian manifolds $$(M\times N,g)$$. The most simple way is to define $$g = g_M \oplus g_N$$, but it is a very restricted way as in this construction, geodesics are the curves whose projection on $$M$$ and $$N$$ are geodesics. Wraped product is made by choosing a positive function $$f$$ on $$M$$ and saying that $$g_{m,n} = {g_M}_m \oplus f(m){g_N}_n$$: in this construction, moving along $$M$$ modify the metric in the $$N$$ component.
The more important example of a natural wraped product is while choosing a polar decomposition in an euclidean space. The diffeomorphism \begin{align} \varphi : \mathbb{R}_+^* \times \mathbb{S}^n & \longrightarrow \mathbb{R}^{n+1}\setminus\{0\} \\ (r,\theta) & \longmapsto r\theta \end{align} allows to identify $$\mathbb{R}^{n+1}\setminus\{0\}$$ as a product of two manifolds but to have an isometry, you have to put on the product manifold the metric $$\varphi^*g_{\text{eucl}}$$. If $$\overset{\circ}{g}$$ denotes the standard metric of the sphere of radius $$1$$, one can show that in the product manifold $$\mathbb{R}^*_+\times\mathbb{S}^n$$, the metric $$\varphi^*g_{\text{eucl}}$$ has the form \begin{align} \mathrm{d}r^2 \oplus r^2 \overset{\circ}{g} \end{align} and thus, we can identify the euclidean space with this wraped product: wraped product come naturally. It reflects that at $$r$$ fixed, you are not on a sphere with radius $$1$$ but with radius $$r$$ (whose metric is $$r^2$$ times those of the sphere of radius 1)
You can generalize this construction, in a general riemannian manifold $$(M,g)$$. With some curvature assumptions, there exists some submanifold $$Y$$ such that there is a natural trivial bundle $$Y\times \mathbb{R}^k$$ which is isometric to $$(M,g)$$ while given a wraped product metric, etc.