Covering map SU(2) x SU(2) to SO(4)

Scott Morrison

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Mar 9, 2001, 9:40:53 PM3/9/01

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Does anyone know an explicit form for this map?
SU(2) is a double cover for SO(3), and the homomorphism goes like this.
Write (x,y,z) in R^3 as a 2x2 matrix M
( z x+iy )
( x-iy -z )
Then A in SU(2) acts on M by
M \mapsto A M A^*
This induces a map of R^3, which gives the image on A in SO(3).
I'm looking for something similar for SU(2) x SU(2) -> SO(4)

Thanks, Scott Morrison
scott_re...@maths.unsw.edu.au

Christian Ohn

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Mar 11, 2001, 11:07:21 AM3/11/01

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Scott Morrison wrote:
>
> I'm looking for something similar for SU(2) x SU(2) -> SO(4)

This is a pretty standard example in courses on Lie groups; see e.g. the
text by Graeme Segal in R. Carter, G. Segal, I. Macdonald, "Lectures on
Lie Groups and Lie Algebras", LMS Student Texts 32, Cambridge U.P.

Identify R^4 with the quaternions and SU(2) with the group of unit
quaternions. Then let SU(2)xSU(2) act on R^4 by

(q_1,q_2).v = q_1 v q_2^-1

This yields the desired morphism SU(2)xSU(2) --> SO(4).

Christian Ohn
email: christian d o t ohn a t univ h y p h e n reims a t fr

Nikos

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Mar 12, 2001, 2:02:50 AM3/12/01

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One way to see this map is to think of SU(2) as the groups of unit
quaternions via the identification of quaternions with complex
matrices of the form

a+ib c+id
-c+id a-ib.

Then let SU(2)xSU(2) act on the 4-dimensional real vector space of
quartenions by
(U,V)Q=VQU^{-1}.
Now since determinant = norm this action is metric preserving and
therefore determines a map into SO(quartenions).
By choosing bases etc one can get explicit formulas.

Take care
Nikos.

Clarence Wilkerson

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Mar 14, 2001, 1:20:09 AM3/14/01

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I'm not sure what explicit means for you, but here's one that does it for
me.

The center of SU(2) is Z/2, generated by the scalar -1 diagonal matrix.
Therefore the center of SU(2) x SU(2) is Z/2 x Z/2, with a preferred basis
coming from the two factors.

Then SO(4) is the quotient group (SU(2) x SU(2))/ diagonal ( Z/2 \subset Z/2 x
Z/2).

--
Clarence Wilkerson \ HomePage: http://www.math.purdue.edu/~wilker
Prof. of Math. \ Internet: wil...@math.purdue.edu
Dept. of Mathematics \ Messages: (765) 494-1903, FAX 494-0548
Purdue University, \
W. Lafayette, IN 47907-1395 \

j.e.mebius

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Mar 23, 2001, 11:21:53 AM3/23/01

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Delft - March 23rd 2001

Dear Scott Morrison,

In the 1840s or 1850s (I do not know by heart) Arthur Cayley represented
points in 4D space by quaternions and found that combined left- and
right-multiplication by unit quaternions L, R amounts to a rotation about
the origin. Changing the signs of both L and R does not change the
resulting 4D rotation.
Actually, =any= 4D rotation can be represented as P -> LPR in two ways,
differing only in signs of L and R. As far as I know, Cayley failed to
prove this.

In 1897 the Dutch mathematician Lambertus van Elfrinkhof published a
formula expressing 4D rotations in terms of left and right unit quaternion
factors. He did not contend nor prove that his formula gives the =general=
4D rotation.

It is all very classical stuff. Nowadays we formulate this 150 years-old
theorem as
"S^3 x S^3 is 2:1 homomorphic to SO(4) and so as a manifold S^3 x S^3 is
the double cover of SO(4)".

Please look at http://www.xs4all.nl/~plast/So4exabs.htm for a matrix-based
proof of this theorem and for the explicit formula you want to know.

Good luck. Best regards: Johan E. Mebius