Can anyone give me an explicit isomorphism between $SU(2)$ and the three sphere?
What about for higher spheres? This question link text seems to indicate that there exists a homeomorphism from $SU(n)/SU(n-1)$ to the $(2n-1)$-sphere.
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4$\begingroup$ A simple answer -- $U(2)$ acts on $S^3$ since it acts on $\mathbb C^2$ by linear isometries. So $SU(2)$ also acts on $S^3$. That is the isomorphism, as it's a faithful representation. $\endgroup$– Ryan BudneyJan 15, 2010 at 7:03
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7$\begingroup$ I'm sorry, but I don't think that this is appropriate for MO, since the first 100 google results will give you the isomorphism. so -1 $\endgroup$– SpinorbundleJan 15, 2010 at 9:51
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$\begingroup$ @Ryan: You need to know that that action of $SU(2)$ on $S^3$ is transitive: $SU(2)$ also acts faithfully and unitarily on $\mathbb C^{23}$, yet it is not homeomorphic to $S^{22}$ :) $\endgroup$– Mariano Suárez-ÁlvarezJan 15, 2010 at 16:06
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$\begingroup$ Yes, of course, I just mean homeomorphisn (or diffeomorphism if you want to get differential). $\endgroup$– Aston SmytheJan 15, 2010 at 17:41
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5$\begingroup$ Not only is this easily found on google, it is also in any book on Lie groups. Yes, it is possible to learn things without using the internet. ;-) $\endgroup$– Kevin H. LinJan 16, 2010 at 0:27
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4 Answers
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Elements of $SU(2)$ look like this: $$ x = \begin{pmatrix} a & - \overline{b} \\ b & \overline{a} \end{pmatrix},$$ 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.
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$\begingroup$ Of course you also need that $\mathrm{det}x = 1$ to get that form for the matrix. $\endgroup$– MTSJan 15, 2010 at 5:09
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2$\begingroup$ It's not so much that you also need det(x) = 1, as that this is exactly the same as saying |a|<sup>2</sup> + |b|<sup>2</sup> = 1. $\endgroup$ Dec 24, 2012 at 17:27
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$SU(2)$ 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.
answered Feb 1, 2010 at 10:45
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Regarding the higher dimensional question:
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.
answered Jan 15, 2010 at 16:43
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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.
These coordinate systems make what MTS said below explicit.
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.
A related extra stuff:
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.
In that section he will do what gets called the Heegard Decomposition of $S^3$ using very simple high-school maths!
Basically Naber will rationalize why these coordinate systems are in some sense natural.
You can also look up these analysis in these two write-ups I had done during my undergrad , this and this one.
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