Wednesday 17 December 2014

Is this intuition behind product manifolds correct?



I've been studying differential geometry on Spivak's books and recently I proved that the cartesian product of manifolds is another manifold. Right, however, what's the intuition behind this? I've tried to develop some intuition, but I'm not sure it's correct.




For instance, we can say that the cylinder $C$ is the product of the circle and the real line, in other words $C = S^1 \times \mathbb{R}$. My intuition says that we do define this way because identifying a point on a cylinder is equivalent to identifying a point on a circle and then the position of this point on a line. In other words, describing one point on a cylinder should be equivalent to describing one point on a circle plus one point on a line.



Another example, the thorus given by $T = S^1 \times S^1$, again what's the intuition behind it? My idea is: given that we can describe the torus as the result of rotating a circle around some axes we would describe a point on the torus to be one point on the circle being rotated plus one point on the circle that represents the trajectory of that particular point during the rotation.



So in general, my intuition is that given $r$ manifolds $M_1, \dots, M_r$ their product is a manifold that each point can be described by one point on each manifold in a particular way depending on the case we are considering.



Is this intuition correct? Is this the way we should think about product manifolds?



Thanks in advance for your help.


Answer




For the product manifold there is a different explanation.



Take the example $S^1\times\mathbb R$. This means that at each point of the circle $S^1$ you attach the whole real line!! So in this way, going round the circle and attaching everywhere a real line you get a cylinder.



For the other example : $\mathbb T^2=S^1\times S^1$. At each point of a circle attach another circle (orthogonal to the 1st circle). Then, if for all points of the circle you attach an orthogonal circle, you are going to get the torus!



It's pretty fascinating how intuition works here!


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...