# Connection between SU(2) and SO(3)

1. Aug 18, 2014

### HomogenousCow

I am somewhat confused with the connection between the two groups.
In the text I'm reading (An Introduction to Tensors and Group theory for physicists N. Jeevanjee), there is a chapter quite early on (in the group theory part) which outlines a homomorphism from SU(2) to SO(3), however I find this homomorphism very odd.
Why is it that we start with SU(2) on traceless anti-Hermitian matrices? There seems to be some connection here with the Lie Algebra of SU(2) however I can't put my finger on it.

2. Aug 20, 2014

### Fredrik

Staff Emeritus
An element of SO(3) is a linear operator on $\mathbb R^3$. The set of traceless hermitian (not anti-hermitian) 2×2 matrices is a 3-dimensional vector space over $\mathbb R$, so it's isomorphic to $\mathbb R^3$. The Pauli spin matrices are a basis for this space, since an arbitrary x in that space can be written as
$$x=\sum_{i=1}^3 x_i\sigma_i =\begin{pmatrix}x_3 & x_1-ix_2\\ x_1+ix_2 & -x_3\end{pmatrix}.$$ The map $(x_1,x_2,x_3)\mapsto x$ is an isomorphism. Because of this isomorphism, every linear transformation on this space induces a linear transformation on $\mathbb R^3$.

We'd like to find a linear transformation that preserves the self-adjointness, the tracelessness, and also the determinant, because $\det x=-|x|^2$. Transformations of the form $x\mapsto uxv$ are linear. If $v=u^{-1}$, the determinant and the trace are preserved. If $v=u^*$, the self-adjointness is preserved. If u is unitary, the two conditions $v=u^{-1}$ and $v=u^*$ are equivalent.

Hm, at the moment I don't see why we choose u to have determinant 1.

3. Aug 20, 2014

### Matterwave

The Lie Algebra of SU(2) is exactly the set of all traceless anti-Hermitian matrices.

@Frederik: if you allow determinant to be not 1, you get the group of unitary matrices U(2) which is actually one dimension higher than SU(2) (U(2) has dimension 4 not 3) and so there is no 2 to 1 covering of SO(3) (that I'm aware of) in that case.

Last edited: Aug 20, 2014
4. Aug 20, 2014

### Fredrik

Staff Emeritus
OK, I see. The other conditions I mentioned ensure that the codomain (of the map $x\mapsto uxv^*$) isn't larger than the domain, and that these maps correspond to linear operators on $\mathbb R^3$ that preserve the Euclidean norm. But the point of the condition det u=1 is to ensure that there are only two u's corresponding to each of those linear operators on $\mathbb R^3$ instead of infinitely many.

One detail is still bugging me. I said "linear operators on $\mathbb R^3$ that preserve the Euclidean norm". Those are $O(3)$ transformations. I see nothing here that ensures that we're dealing with SO(3) instead of O(3). None of the conditions discussed so far, including the choice between U(2) and SU(2) seems to have any effect on the sign of the determinant of the O(3) transformation corresponding to a given u in U(2) or SU(2).

We should be able to define the rotation corresponding to a given u in the following way. We have $$\sum_j x'_j\sigma_j =uxu^*=\sum_i x_iu\sigma_i u^*=\sum_ix_i\sum_j(u\sigma_i u^*)_j\sigma_j,$$ where the $(u\sigma_iu^*)_j$ is the jth component of $u\sigma_iu^*$ in the spin matrix basis. We get $x'_j=\sum_i(u\sigma_iu^*)_j x_i$, so we can define the O(3) transformation R(u) by $R(u)_{ji}=(u\sigma_iu^*)_j$.

What exactly is forcing the determinant of R(u) to be 1?

5. Aug 20, 2014

### Ben Niehoff

The homomorphism $\varphi : SU(2) \to SO(3)$ preserves local topology (i.e. is smooth), and $SU(2)$ only has one connected component. Therefore the image of $\varphi$ cannot be larger than the identity component of $O(3)$.

6. Aug 21, 2014

### HomogenousCow

Well I mean the argument begins with considering U(2) as transformations acting on traceless anti-hermitian matricies, does that mean the homomoprhism can only be established this way? Can we not just consider U(2) on arbitrary vector spaces? (which are applicable of course)

7. Aug 21, 2014

### Fredrik

Staff Emeritus
I checked what the book is saying. (It's on page 106 if anyone else wants to look. Unfortunately page 107 doesn't show up in the preview at Google Books). He's using anti-hermitian matrices instead of hermitian matrices as I'm used to. Apparently that works just as well, and it works for exactly the same reason. So my comments in post #2 still apply. In particular, this is a 3-dimensional vector space over $\mathbb R$ that is easily seen to be isomorphic to $\mathbb R^3$. This is essential when you try to show that the SU(2) matrices correspond to rotations.

8. Aug 21, 2014

### HomogenousCow

What I find strange is that we have to "make" U(2) act on a specific vector space to show the connection while the same can be done without any mention of a vector space using the lie algebra homomorphisms.

9. Sep 2, 2014

### Incnis Mrsi

Complex 2 × 2 matrices are also one of presentations of quaternions, where SU(2) corresponds to unit quaternions (also known as versors or the group Sp(1)). Representation of quaternions as Euclidean rotations, namely v ↦ q−1vq for a quaternion q, is a well-known thing.

In theory of Lie groups, it is called “adjoint representation” and can be specified independently of quaternionic algebra (note that tangent space of SU(2) is real three-dimensional).

The topological fact about SO(3) that permits for SU(2) → SO(3) covering (i.e. a map that is locally a diffeomorphism) is that SO(3) is not simply connected.