# Lie group Homomorphism $SU(2) \to SO(3)$

The Lie algebra of $$\mathfrak{so(3)}$$ and $$\mathfrak{su(2)}$$ are respectively

$$[L_i,L_j] = i\epsilon_{ij}^{\;\;k}L_k$$ $$[\frac{\sigma_i}{2},\frac{\sigma_j}{2}] = i\epsilon_{ij}^{\;\;k}\frac{\sigma_k}{2}$$

And of course, there is an isomorphism between these two algebras, $$\Lambda : \mathfrak{su(2)} \rightarrow \mathfrak{so(3)}$$ such that $$\Lambda(\sigma_i/2) =L_i$$

Now is it possible, using $$\Lambda$$, to construct a group homomorphism between $$SU(2)$$ and $$SO(3)$$?

I was checking up on Lie group homomorphism, and in Wikipedia, there is a beautiful image

In this image's language, how are $$\phi$$ and $$\phi_*$$ related to each other (just like the algebra and group elements are).

Note : I know there is a one-to-two homomorphism between these two groups which can be directly found using the group elements. I am not looking for this.

EDIT 1 : In $$SL(2,\mathbb{R})$$ the generators, say $$X_1,X_2,X_3$$, they obey the following commutation rules :

$$[X_1,X_2] = 2X_2$$ $$[X_1,X_3] = -2X_3$$ $$[X_2,X_3] = X_1$$

And in the case of $$SO(3)$$ with a different basis, $$L_{\pm} = L_1 \pm i L_2$$ and $$L_z = L_3$$ with the commutators being,

$$[L_z,L_{\pm}]= \pm L_{\pm}$$ $$[L_+,L_-]= 2 L_z$$

This algebra is very similar to the algebra of the previous one, so why is that we can't define a map ?

EDIT 2:

Can the group homomorphism between these two groups be written like this (Something like what I expected) : $$R = \exp(\sum_k i t_k L_k) = \exp\left(\sum_k i t_k \frac{\sigma_k}{2}\right) = \exp\left(\sum_k i t_k \frac{1}{2}ln(U_k)\right)$$

Now this seems like the map $$\phi$$,

$$R = \phi(U) = \exp\bigg(\sum_k i t_k \frac{1}{2}ln(U_k)\bigg)$$

• In this language, $\phi_\star$ is essentially the differential of $\phi$ (at identity), sometimes also called the pushforward (this is what the star in the index means). en.wikipedia.org/wiki/Pushforward_(differential) This question is probably more appropriate at math.SE. – Heidar Apr 1 '14 at 6:11
• Comment to the question (v4): Note that the two generators $L_{\pm}$ do not belong to the Lie algebra $so(3)$. Rather they belong to the complexification $so(3,\mathbb{C})$. Note that the complexifications of the real Lie algebras $su(2)\cong so(3)$ and $sl(2,\mathbb{R})\cong so(2,1)$ are all the same, namely $sl(2,\mathbb{C})\cong so(3,\mathbb{C})$. – Qmechanic Apr 2 '14 at 16:31
• @Qmechanic : Oh thanks a lot, so a these two things have a similarity ? – user35952 Apr 2 '14 at 16:38
• @Qmechanic : So is there also a general rule that, the groups with real/complex parameters have their corresponding Lie Algebra LVS defined over a real/complex field respectively ? – user35952 Apr 2 '14 at 16:57
• $\uparrow$ Yes. – Qmechanic Apr 2 '14 at 17:08

First notice that the generators are $-i\sigma_k/2$ and $-iL_k$, since the groups are real Lie groups and thus the structure tensor must be real.

The answer to your question is positive. In principle it is enough to take the exponential of the Lie algebra isomorphism and a surjective Lie group homomorphism arises this way $\phi : SU(2)\to SO(3)$: $$\phi\left(\exp\left\{-\sum_k t^k i\sigma_k/2\right\}\right) =\exp\left\{-\sum_k t^k iL_k\right\}\:.$$ The point is that one should be sure that the argument in the left-hand side covers the whole group. For the considered case, this is true because $SU(2)$ is compact.

If you instead consider no compact Lie groups, like $SL(2,\mathbb C)$, the exponential does not cover the group. However it is possible to prove that products of exponential do. In that case a product of two exponentials is sufficient, in practice decomposing an element of $SL(2,\mathbb C)$ by means of the polar decomposition, mathematically speaking, or as a (unique) product of a rotation and a boost physically speaking.

• Thanks !! But I wish to know what will be the functional form of this $\phi$. I am pretty new to this subject, can you please be more explicit on that part ? – user35952 Apr 1 '14 at 17:22
• Well, everything is already explicit in my answer concerning $\phi$. An element of $SU(2)$ can always written down as an exponential as I wrote (that exponential can be computed and the formula can be found in every book on QM). The fundamental point is that the element is completely fixed by the coefficients $t_k$. These coefficients appear in the right hand side determining the rotation (again written in terms of an exponential) associated with the element of $SU(2)$ in the left hand side. – Valter Moretti Apr 2 '14 at 6:35
• @V.Morreti : Thanks !! Since you have pointed about SL(2,R). I also checked that there is an isomorphism between the algebras sl(2,R) and so(3). Therefore, I am guessing there is a similar map for these groups. In that case what is the problem you're trying to address ? – user35952 Apr 2 '14 at 6:58
• No, the isomorphism is between sl(2,R) and o(3,1). There is a similar map, but now every element of $SL(2,\mathbb R)$ has to be decomposed as a product of two exponentials (using the polar decomposition procedure) and, correspondingly, a product of exponentials appears in the right-hand side corresponding to the standard decomposition of an element of $O(3,1)$ as a product of a $SO(3)$ rotation and a pure Lorentz transformation. – Valter Moretti Apr 2 '14 at 7:17
• It is the first time I see it. If you are sure that your map $SU(2) \ni U \mapsto R\in SO(3)$ preserves the composition of matrices and that its differential sends $i\sigma_k/2$ to $iL_k$, as it is continuous, it must coincide to the map I wrote using exponentials, since there is exactly one Lie group homomorphism which extends a Lie algebra homomorphism. Also the expression I provided for the covering homomorphism is two-one. This is because, for instance, $e^{i (t_3 +2\pi) L_3} = e^{i t_3 L_3}$ but $e^{i(t_3 + 2\pi) \sigma_3/2}= - e^{it_3 \sigma_3/2}$. – Valter Moretti Apr 4 '14 at 6:33

So I take it you are clearly aware that the big A Adjoint representation is the homomorphism you're after in this case, so you're seeking a more general method.

Also, I'm assuming you know that the homomorphism of Lie algebras can only lift to a group homomorphism if the homomorphism's domain is simply connected, in which case there is a unique group homomorphism with the given algebra homomorphism as its Lie map. In this case, we're in the clear because $SU(2)$ is simply connected. Page 73 through 76 of:

can then help you. Knapp gives you two methods of systematically constructing the simply connected Lie group: the first leaves you with differential equations for the left / right invariant vector fields, the second I believe is the same as V. Moretti's Answer.

A final "method" is to used Ado's theorem, which assures us that we can always realise a Lie algebra as a matrix Lie algebra; there is even an explicit software algorithm for this:

W. A. De Graaf, "Constructing Faithful Matrix Representations of Lie Algebras"

but if you can understand this algorithm, you are doing better than I (this paper has so far defeated me). Once you have a matrix algebra, you can used the matrix exponential to construct a neighbourhood of the identity, indeed the whole group if the latter is compact; as in V. Moretti's Answer the Lie algebra does not exponentiate to the whole group for noncompact groups (as far as I am aware, the problem of exactly what in a noncompact Lie group can be realised as an exponential of a Lie algebra element is to some extent still an open problem).

So, once you have the Lie group, you can in principle construct the universal cover with homotopy classes and carve out the discrete centre $\mathcal{Z}_d$ of the universal cover. Your original group will have as its fundamental group the quotient group of $\mathcal{Z}_d$ and one of its (normal) subgroups.

• De Graaf is a colleague of mine! Some doors next to mine. – Valter Moretti Apr 1 '14 at 10:53
• @V.Moretti Tell him I still tackle his paper from time to time but still leave off badly bruised! No slight intended on his technical writing abilities: an actual implementation of an Ado theorem algorithm is not an easy thing to convey, I'm likely not the sharpest audience and he's (de Graaf) has certainly done some very beautiful work. – WetSavannaAnimal Apr 1 '14 at 12:08
• OK, I will tell him all that you wrote. Bye – Valter Moretti Apr 1 '14 at 12:16
• the links to your notes appear to be dead. Did you put them somewhere else? – glS Aug 19 at 17:43