A hand-wavy answer to the question in the title could be "3D rotations".

But I would like to build a solid and intuitive geometric understanding of the $SO(3)$ manifold together with the Lie group and the Lie algebra of 3D rotations.

I read this article on Lie algebras for 2D and 3D rotations, but I am confused when it comes 3D rotations. I would like to build a similar intuition for $SO(3)$ as I have formed for $SO(2)$.

2D Rotations

The article says the following about $SO(2)$:

$SO(2)$ is a 1D manifold living in a 2D Euclidean space, e.g. moving around a circle.

I seem to understand this intuitively; with 2D rotations, there is just 1 degree of freedom and in my mind the $SO(2)$ manifold looks exactly like the unit circle in $\mathbb{R}^2$.

Each point $p \in \mathbb{R}^2$ on the unit circle has its $X$ and $Y$ coordinate. The $X$ coordinate is the cosine and the $Y$ coordinate is the sine of the angle between the $X$ axis of the coordinate system and the vector $p$.

Given an arbitrary point $p$ on the unit circle, I can easily compute the angle $\alpha$ it corresponds to by e.g. $\alpha = cos^{-1}(p_x)$, where $p_x$ is the $X$ coordinate of the point $p$.

3D Rotations

This is where I am confused. The article says the following about $SO(3)$:

There are only 3 degrees of freedom in describing a rotation. But this object doesn’t live in 3D space. It is a 3D manifold, embedded in a 4-D Euclidean space.

If the $SO(3)$ manifold lives in 4D space, then each point on its surface seems to be an $r \in \mathbb{R}^4$. But what does each of the 4 components $r_x, r_y, r_z, r_w$ of $r$ represent (in the 2D case, it was sine and cosine of the angle)?

I also found this image. I assume that if the $SO(3)$ manifold is embedded in 4D space, then it cannot be visualized as a 3D sphere.

enter image description here

Moreover, the tangent space in the image is "just a 2D plane". If my understanding is correct, then the purpose of the tangent space (Lie algebra) is mainly to allow expressing angle parameters manipulation in a (linear) vector space (convenient for optimization problems) instead of a movement on the manifold (difficult for optimization problems).

Given that 3D rotations have 3 degrees of freedom, it would make more sense to me if the tangent space was 3D volume (with 3 DoF) instead of 2D plane (with 2 DoF). Is the image just an analogue for the $SO(3)$ manifold embedded in 4D?

I think I am missing something obvious. What is your intuitive understanding of the $SO(3)$ manifold and how do you visualize it?

  • $\begingroup$ Sometimes I visualize a rotation as a circular arc; more precisely, an equivalence class of directed circular arcs on the unit sphere. Two arcs are equivalent if they're part of the same circle and have the same length and direction; you can slide one arc along the circle to get the other. But this doesn't help much in visualizing the set of all 3D rotations. For that, you might try quaternions and stereographic projection. $\endgroup$
    – mr_e_man
    Jan 28, 2022 at 1:07
  • $\begingroup$ An element of the Lie algebra is best thought of as a bivector. $\endgroup$
    – mr_e_man
    Jan 28, 2022 at 1:11
  • $\begingroup$ @mr_e_man Quaternions also came to my mind, especially since they have 4 components (similar to the points on the $SO(3)$ manifold) and they also have unit length, which made me question whether the points on the $SO(3)$ manifold aren't actually quaternions. But then I remembered that quaternions are just one of the several ways how 3D rotations can be represented, so this might be just a coincidence. I'll definitely learn about bivectors though, thanks for introducing them to me. $\endgroup$
    – jordi
    Jan 28, 2022 at 1:52
  • 2
    $\begingroup$ Sadly, the article you are reading is misleading: The manifold $SO(3)$ does not embed in the 4-dimensional space (my answer here). The author of the article confused $SO(3)$ with a closely related group, called $SU(2)$, which does embed in the 4D space (as the group of unit quaternions). $\endgroup$ Jan 28, 2022 at 4:59

1 Answer 1


Abstract group

Oftentimes, when one describes a group like $SO(3)$, they are referring to the abstract group, which is just a set of elements with a prescribed composition operation. An abstract group does not "live in" any larger space, and the "points" need not have any internal structure whatsoever. We can then endow this set with the structure of a smooth manifold and talk about Lie groups, Lie algebras, etc. This approach is theoretically useful, but when working with a specific group, it's often best to explicitly construct the group (or, more precisely, construct an isomorphic group) out of more familiar objects. There are many ways of doing this; I'll call these models of the group. Questions like "what are the points" or "what is the ambient space" don't make sense when applied to the abstract group, rather they are questions about a particular model of the group, and different models will yield different answers.

Defining Representation

The classical way of constructing a model of a group is to write it as a set of matrices (or, equivalently, as a set of linear maps $\mathbb{R}^n\to\mathbb{R}^n$; this is called a (linear) representation. $SO(3)$ is often defined this way, as the set of $3\times 3$ matrices $A$ satisfying $A^TA=I$ and $\det(A)>0$. In this model, we can view $SO(3)$ as a subset of $\mathbb{R}^9$, and it is in fact a smooth 3-dimensional submanifold, and the induced manifold structure on $SO(3)$ is exactly the Lie group structure. In this model, the points are $3\times 3$ matrices, and the ambient space is $\mathbb{R}^9$. If we define the Lie algebra $\mathfrak{so}(3)$ as the tangent space at the identity, we can identify it with a subspace of $\mathbb{R}^9$, and this subspace corresponds to the antisymmetric matrices, with the lie bracket given by the matrix commutator.

Quaternion Model

We can also construct a model of $SO(3)$ out of the quaternions $\mathbb{H}\cong\mathbb{R}^4$. The set of unit quaternions form a Lie group under quaternion multiplication, and this group is ismorphic to $SU(2)$, a double cover of $SO(3)$. Identifying each unit quaternion $q$ with its opposite $-q$, we obtain a group isomorphic to $SO(3)$. This allows one to localy view $SO(3)$ as equivalent to the 3-sphere in $\mathbb{R}^4$, but it is not actually an embedding onto the $3$-sphere, since there is the added step of identifying opposite points.

There are many other models of $SO(3)$, of course, but I find these two to be the most useful. When trying to build some intuition for an abstract group, it's useful to familiarize oneself with multiple ways of constructing it, and, perhaps more importantly, with the correspondences between these constructions.

  • $\begingroup$ I think I have a better understanding now thanks to your answer and @MoisheKohan's comment, thank you both. May I ask what does this image depict? Is it just a simplified illustration to help understand what is meant by the tangent space and exponential map in general? $\endgroup$
    – jordi
    Jan 28, 2022 at 13:17
  • $\begingroup$ @jordi It seems to be an illustration of the Riemannian exponential map on the $2$-sphere. This isn't the same as the Lie exponential map of $SO(3)$, but the exponential map on the quaternion model behaves in an analogous way. $\endgroup$
    – Kajelad
    Jan 28, 2022 at 18:01
  • $\begingroup$ @Kajelad: It may be worth noting that $SO(3)$ is diffeomorphic to the unit tangent bundle of $S^2$, so there's actually a very close connection between $SO(3)$ and the Riemannian exponential map on the $2$-sphere. $\endgroup$ Jan 29, 2022 at 2:10
  • $\begingroup$ @JasonDeVito The manifolds are related, but I don't see how the exponential map on $SO(3)$ relates to the exponential map on $S^2$ in any simple way. Rather than great circles, the one-parameter subgroups of $SO(3)$ correspond to curves of constant speed on $S^2$ with a unit vector field such that the velocity and the unit vector field both rotate at the same constant rate. There's some interesting geometry there, but it's beyond the explanatory power of the diagram. $\endgroup$
    – Kajelad
    Jan 29, 2022 at 3:46
  • $\begingroup$ @jordi: The author of the image attempted to draw the group $SU(2)$ as unit quaternions in the 4-space, with its tangent space at $1$ and the exponential map. However, they could not draw the 3d sphere in the 4d space and decided to reduce the dimension by projecting everything to the 3-d subspace. Does this help? $\endgroup$ Jan 29, 2022 at 8:18

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