$SU(2)$ and the three sphere - MathOverflow most recent 30 from http://mathoverflow.net2013-05-20T04:21:24Zhttp://mathoverflow.net/feeds/question/11821http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/11821/su2-and-the-three-sphere$SU(2)$ and the three sphereAston Smythe2010-01-15T04:52:08Z2010-02-01T10:45:57Z
<p>Can anyone give me an explicit isomorphism between $SU(2)$ and the three sphere?</p>
<p>What about for higher spheres? This question <a href="http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotient" rel="nofollow">link text</a> seems to indicate that there exists a homeomorphism from $SU(n)/SU(n-1)$ to the $(2n-1)$-sphere.</p>
http://mathoverflow.net/questions/11821/su2-and-the-three-sphere/11823#11823Answer by MTS for $SU(2)$ and the three sphereMTS2010-01-15T05:08:27Z2010-01-15T05:08:27Z<p>Elements of $SU(2)$ look like this:
<code>$$ x = \begin{pmatrix} a & - \overline{b} \\ b & \overline{a} \end{pmatrix},$$</code>
where $|a|^2 + |b|^2 = 1$. This follows easily from $x^* = x^{-1}$. So you map that matrix to the point $(a,b)$ in $\mathbb{C}^2$, and this is your diffeomorphism.</p>
http://mathoverflow.net/questions/11821/su2-and-the-three-sphere/11824#11824Answer by Anirbit for $SU(2)$ and the three sphereAnirbit2010-01-15T05:10:31Z2010-01-15T08:55:57Z<p>Just write $SU(2)$ in some local coordinates (some of the standard systems are the double-polar system (a.k.a Hyperspherical coordinates) or the single angle coordinates (a.k.a Hopf coordinates) and then one sees that the special unitary condition forces the $4$ parameters you will need to satisfy the $3$-sphere equation. Hence one gets an explicit local map to the $3$-sphere by mapping that matrix to the ordered tuple of $4$ numbers. Smoothness is assured since the functions are all polynomials. </p>
<p>These coordinate systems make what MTS said below explicit.</p>
<p>After all one is very likely to want to do some geometry on $SU(2)$ now that one knows it is $S^3$ and these coordinate systems are naturally adapted to do them. Like computing vielbiens on $SU(2)$ in these systems look very natural and make the symmetries of the spherical structure underneath very clear. </p>
<p>A related extra stuff: </p>
<p>One can look up a very nice analysis of this in the first chapter of Gregory Naber's book "Geometry, Topology and Gauge Fields" Volume 1.</p>
<p>In that section he will do what gets called the Heegard Decomposition of $S^3$ using very simple high-school maths! </p>
<p>Basically Naber will rationalize why these coordinate systems are in some sense natural. </p>
<p>You can also look up these analysis in these two write-ups I had done during my undergrad , <a href="http://www.cmi.ac.in/~anirbit/S3%20Part-I" rel="nofollow">this</a> and <a href="http://www.cmi.ac.in/~anirbit/Anirbit%27s%20Projective%20spaces%20to%20Spheres%20talk." rel="nofollow">this one</a>. </p>
http://mathoverflow.net/questions/11821/su2-and-the-three-sphere/11881#11881Answer by Deane Yang for $SU(2)$ and the three sphereDeane Yang2010-01-15T16:43:32Z2010-01-15T16:43:32Z<p>Regarding the higher dimensional question:</p>
<p>Please try to figure this out yourself. Just think about the map from SU(n) to $C^n$, mapping each matrix to the first column. What is the image and what is the preimage of each point in the image? Hint: confirm that it suffices to figure out the preimage of (1,0,...,0) and all other preimages are essentially the same.</p>
http://mathoverflow.net/questions/11821/su2-and-the-three-sphere/13654#13654Answer by Misha Verbitsky for $SU(2)$ and the three sphereMisha Verbitsky2010-02-01T10:45:57Z2010-02-01T10:45:57Z<p>$SU(2)$ is a group of unitary quaternions $U(1,H)$, which are of form $a + bI + cJ + dK$, with
$a^2+b^2+c^2+d^2=1$. This is clearly $S^3$. The action of unitary quaternions
from the right preserves complex structure acting from the left (or vice versa), this
gives a map from $U(1,H)$ to $U(2)$. It also preserves the complex
volume, because quaternionic structure can be used to define the symplectic form.</p>