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.

  • 2
    $\begingroup$ 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


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.

  • $\begingroup$ 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

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 $SU(2)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.