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Mathematical expression for map from $[0,1]$ to $S^2$

+ 4 like - 3 dislike
189 views

A topological space is called arcwise connected if, for any points $x,y\in X$, there exists a continuous map $f: [0,1]\rightarrow X$ such that $f(0)=x$ and $f(1)=y$. Although it is intuitively understandable but how does such a map mathematically look like for $S^2$?

According to this definition is there way to show that $SU(2)$ is connected but $O(3)$ is not? As I continuously change the group parameters (up to their allowed ranges) can I show in the first case that I can reach all points on the $SU(2)$ manifold and in case of $O(3)$ I cannot exhaust all points? Only this can prove the nature of connectedness, in this definition, Right?

This post imported from StackExchange Mathematics at 2014-10-05 10:04 (UTC), posted by SE-user Roopam Sinha
asked Oct 5, 2014 in Closed Questions by Roopam Sinha (25 points) [ no revision ]
closed Jun 22, 2015 as per community consensus

Undergrad level question, voted to close, but one way to show that a Lie group is connected and construct explicit connecting curves is using the exponential map, so that every point x can be written (nonuniquely) as exp(A) for A in the Lie algebra. The Lie algebra is a real vector space, you can connect any two points using straight lines, so x and y connect by tx + (1-t)y in the vector space, and then exponentiating exp(tA + (1-t)B) interpolates exp(A) and exp(B).

SO(3) and SU(2) are both connected. O(3) is disconnected because there is a determinant for the rotation matrix which can be either +1 or -1. This is a complete answer, but I made it a comment because I voted to close this question.

Also, this is an early duplicate of the much clearer and well answered: http://physicsoverflow.org/24226/connectedness-of-%24o-3-%24-group-manifold , which is also technically undergrad level, but who cares, it's a well phrased question with clear answers.

2 Answers

+ 1 like - 0 dislike

(This answer was written to address the original version of the question, before the second paragraph was added.)

As an example, on $S^2$ it's always possible to define a coordinate patch

$$\psi:S^2\rightarrow R^3$$

using typical spherical coordinates $(r,\theta,\phi)$, defined such that $\psi(x)=(1,0,0)$ and $\psi(y)=(1,0,\phi_y)$. The coordinate patch can be defined everywhere on $S^2$ except in some neighborhoods of the poles. Like any coordinate patch, $\psi$ is injective (although it isn't surjective), so $\psi^{-1}$ is defined on $\psi$'s image. Then the simplest possible definition for the map $f$ using those coordinates would be

$$f(\lambda)=\psi^{-1}(1,0,\lambda \phi_y)\ .$$

There are of course many other possible ways to define $f$, which take the image of $f$ along different paths on $S^2$ between $x$ and $y$.

This post imported from StackExchange Mathematics at 2014-10-05 10:04 (UTC), posted by SE-user Red Act
answered Oct 5, 2014 by Red Act (10 points) [ no revision ]
+ 0 like - 2 dislike

$f:[0,1]\to S^2$ given by $f(x)=(\sin x\cos x,\cos^2 x,\sin x)\in S^2$, I think it is very easy to check now taking any two arbitrary point from sphere and connect by a path like $tx+(1-t)y;x,y\in S^2$

This post imported from StackExchange Mathematics at 2014-10-05 10:04 (UTC), posted by SE-user Une Femme Douce
answered Oct 5, 2014 by Une Femme Douce (-10 points) [ no revision ]

This is nonsense-- you just mapped a line to a particular curve in the spehre. The correct version of this argument uses the exponential map, or on the sphere, the manifold exponential map, not this, which is meaningless.





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