Difference between SO(3) and so(3)

[priyansh]

Hi,

What is the difference between lie group SO(3) and lie algebra so(3)? I just got the idea in terms of terminology that one represents the group and the other stands for algebra. But can anyone please provide details as to what exactly is the difference?

Regards,
Priyanshu

 

[shoehorn]

SO(3) is a Lie group; so(3) is the Lie algebra of SO(3). SO(3) is a manifold; so(3) is a vector space with a bilinear, antisymmetric bracket that satisfies the Jacobi identity.

That really is all there is to it.

 

[Fredrik]

In terms of differential geometry, so(3) is the tangent space at the identity element of SO(3), with a Lie bracket $(X,Y)\mapsto[X,Y]$ defined by setting [X,Y] equal to the commutator at the identity, of the left-invariant (or right-invariant) vector fields corresponding to X and Y. (Those are the fields defined in the quote below. The two Lie algebras defined this way are isomorphic, so it doesn't matter if we use left multiplication or right multiplication to define the Lie bracket).

In terms of matrices, SO(3) is the set of 3×3 orthogonal matrices with determinant 1, and so(3) is the set of matrices X such that exp(tX) is in SO(3) for all real numbers t. (This is a trick to find a Lie algebra that's isomorphic to the group's Lie algebra. It works for matrix Lie groups). So it's the set of all antisymmetric traceless 3×3 matrices. (Mathematicians use the term "skew symmetric" or just "skew" instead of "antisymmetric". It means X^T=-X. "Traceless" means Tr X=0, where Tr X is the sum of the diagonal elements).

Suppose that M and N are manifolds and $\phi:M\rightarrow N$ is a diffeomorphism. Then we can define a function

$$\phi(p)_*:T_pM\rightarrow T_{\phi(p)}N$$

for each $p\in M$.

A Lie group is both a group and a manifold. We can use any member g of a group G to construct two diffeomorphisms $\rho_g$ and $\lambda_g$ that map G onto itself:

$$\rho_g(h)=hg$$
$$\lambda_g(h)=gh$$

The Lie algebra associated with the Lie Group is defined as the tangent space at the identity element, with a Lie bracket that will be defined below. Let's use the notation $\mathfrak{g}=T_eG$

We can use either right or left multiplication to map the Lie algebra onto the tangent space at any other point g:

$$\rho_g_*:\mathfrak{g}\rightarrow T_gG$$
$$\lambda_g_*:\mathfrak{g}\rightarrow T_gG$$

Let's simplify the notation a bit:

$$\rho_g_*(L)=Lg$$
$$\lambda_g_*(L)=gL$$

We can use these maps to construct two vector fields $X_L^\rho$ and $X_L^\lambda$ for each vector L in the Lie algebra:

$$X_L^\rho|_g=Lg$$
$$X_L^\lambda|_g=gL$$

Either of these two vector fields can be used to define a Lie bracket on the Lie Algebra:

$$[K,L]=[X_K^\rho,X_L^\rho]_e$$
$$[K,L]=[X_K^\lambda,X_L^\lambda]_e$$

 

[priyansh]

Thanks Fredrick and shoehorn.

 

[tmccullough]

I imagine your question is ultimately rooted in quantum mechanics reading. The other posts definitely are correct, but there is a missing piece.

Given the Lie group, you get the Lie algebra as described, but Lie algebras aren't unique. Different Lie groups share the same Lie algebra, but to do so they are topologically "close". They are locally isomorphic - they share the same universal cover (algebraic topology concept).

The Lie algebra gets you to the unique simply connected version of the Lie group, which for $SO(3)$ happens to be $SU(2).$ This is the reason that $SU(2)$ is used as the rotation group for three dimensional space.

 

[priyansh]

Thanks for the addition tmccullough. I came across it while reading some proofs in computer vision.

 

[lavinia]

btw: SO(3) is topologically the real projective space of lines in Euclidean three space. It is diffeomorphic to the tangent circle bundle of the 2 sphere. With this picture you can also visualize the Lie algebra structure of the tangent space at the identity - an instructive exercise.