Hello I'm a physics student, not a math student, who is trying to understand how to write down the metric of a manifold which is constructed via a cartesian product of two other manifolds I know the metrics for. For example S^1 X S^1 is a Torus and I know the metric for a Circle, and I even know the metric for a Torus, but I do not know how to derive the metric for a Torus from the Cartesian product of two circles. In the end I want to construct the metric for the S^1 X S^2. Any help will be appreciated thank you.
The Metric Tensor Of A Cartesian Product
differential-geometryphysics
Related Solutions
A point on a circle $x^2+y^2=r^2$ is defined by an angle $\theta $ via $x=r\cos\theta $ and $y=r\sin\theta $. And it wraps around as $\theta$ goes beyond $2\pi $. A point with angle just under $2\pi $ is close to a point with angle $0$.
A point on the embedded torus $\left (\sqrt {x^2+y^2}-R\right)^2+z^2=r^2 $ is defined by two angles, $\theta $ and $\phi $, via $x=(R+r\cos\theta)\cos\phi $, $y=(R+r\cos\theta)\sin\phi $, $z=r\sin\theta $. And you go around the torus (in two independent directions) as an angles goes beyond $2\pi $.
By the shape of a torus, two points on the torus are close in 3d space when the corresonding pairs of angles $(\theta,\phi) $ for those points are close (an angle just under $2\pi$ being close to $0$). That's why a torus is topologically (rough shape wise) like a product of circles.
The question is not entirely clear to me, but according to the clarification given in the comments, the answer is no.
I guess the source of the confusion lies in inheriting structure. To give an example:
- The set $\mathbb{S}^2 = \{ (x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 = 1 \}$ is just that - a set.
- Since we usually think of $\mathbb{R}^3$ as equipped with the standard metric, we may also think of $\mathbb{S}^2$ as a metric space (with the metric inherited from $\mathbb{R}^3$).
- Similarly, if $\mathbb{R}^3$ is viewed as a topological space, one can consider the induced topology on $\mathbb{S}^2$.
- The standard smooth structure on $\mathbb{R}^3$ induces a smooth structure on $\mathbb{S}^2$. In contrast to all previous cases, this is not automatic - one has to check first that $\mathbb{S}^2 \subseteq \mathbb{R}^3$ is indeed a smooth submanifold.
- Finally, $\mathbb{R}^3$ has a standard Riemannian metric, which induces a Riemannian metric on any smooth submanifold, in particular on $\mathbb{S}^2$.
Nowadays, we have separate notions of sets, metric spaces, topological spaces, smooth manifolds and Riemannian manifolds. But without this more abstract point of view, it's easy to mix all these together; after all, the underlying set $\mathbb{S}^2$ is the same in all the above examples.
One could also imagine similar (somewhat ill-defined) questions; for each of them the answer is again no:
- Given a set $A$, can I decide which functions $f \colon A \to \mathbb{R}$ are continuous and which are not?
- Given a topological space $A$, can I calculate the distance between two points $x,y \in A$?
- Given a metric space $A$ and a Lipschitz function $f \colon A \to \mathbb{R}$, can I define its differential? (actually, there are some weak workarounds here)
And please don't get me wrong - I'm not listing these to suggest that your question was silly, quite the opposite. I just think other examples might help.
Best Answer
Given two Riemannian Manifolds $(M,g)$ and $(N,h)$ then we define their product metric at the point $(p,q)\in M\times N$ as $$g \oplus h: (T_pM\times T_qN) \times (T_pM\times T_qN) \rightarrow \mathbb R$$ $$(g\oplus h)_{(p,q)}((v,w),(\hat v,\hat w)):=g_p(v,\hat v)+h_q(w,\hat w)$$