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Suppose that $U(x)$ is an element of the gauge group say $SU(2)$ and suppose $U(x)=1$ as $|\vec{x}|\to\infty$. Then, why does space have the topology of $S^3$?

This is done in Srednicki page 571. Note that I'm not asking how to prove that $SU(2)\cong S^3$. What I'm asking is how to prove that when $U(x)=1$ as $|\vec{x}|\to\infty$ the space $\mathbb{R}^3$ is compactified to $S^3$ space.

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    $\begingroup$ Write a general matrix with entries $a$, $b$, $c$, $d$. Note that these are complex numbers. Find out what constraints being on $SU(2)$ impose on $a$, $b$, $c$, $d$. You will find that they satisfy the equation of a sphere. $\endgroup$
    – OkThen
    Mar 8, 2017 at 2:05

2 Answers 2

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The question can be formulated more generally as:

Why is it that when we consider functions over $\mathbb{R}^n$ such that the limit as $|\vec{x}|\to\infty$ is the same in any direction then we can identify their domain with $S^n$?

Note that although the functions that we consider might be $\mathbb{R}^3\to SU(2)$, the statement is valid for any target space and any dimension $n$. The argument is the following:

First, notice that $\mathbb{R}^n$ is topologically the same as the open $n$-dimensional ball $B^n$. We can identify the functions in $\mathbb{R}^n$ (with existing limit in any direction as $|\vec{x}|\to\infty$) with the functions on the closed ball $\bar{B}^n$ by identifying $\mathbb{R}^n$ with $B^n$, and then assigning to every point in the boundary the limiting value of the function in that direction.

When a function has the same limit as $|\vec{x}|\to\infty$ in any direction, the corresponding function in $\bar{B}^n$ takes the same value in every point of the boundary. We can then identify all the points in the boundary, getting $S^n$ and a well-defined function over it. You can gain intuition about this identification by thinking about the case $n=2$. If we have the disk with all the boundary points glued together, we clearly get a $2$-sphere.

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  • $\begingroup$ Thank you for your answer. Could you give some reference for the second answer $\endgroup$ Mar 8, 2017 at 10:20
  • $\begingroup$ @amiltonmoreira I don't remember any specific reference for it. All the ideas should appear in any topology book (for example, Munkres). For $\mathbb{R}^n\cong B^n$ it is easy to find an homeomorphism between the two spaces. If you want to learn about this "identification of points" you should look for "quotient space". "One-point compactification" is another important related concept. $\endgroup$
    – coconut
    Mar 8, 2017 at 10:33
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I think i got it,correct me if i am wrong.

We consider stereographic projection from the North pole $p$. Since Stereographic

projection is a one-to-one correspondence between {$S^n−p$} and $R^n$ and since,

$U(x)=1$ as $|\vec{x}|\to\infty$ we can regard $U(\infty)$ as the image of the

point $p$ then, instead of having a map between $R^n$ and $SU(2)$ we can consider

a map between $S^n$ and $SU(2)$

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