I'm trying to understand an algorithm that tries to map points from a lie group to its lie algebra using the exponential map. The background is the representation of 3d coordinate transformations as a lie group.
I can imagine, how one can map points on a manifold (eg a sphere) to the tangent space (tangent plane) at some point x on the sphere. I can imagine to create the tangent space by computing derivatives at that point x. But I cannot imagine how the exponential of some matrix is giving me that tangent space? How can I envision the exponential of a matrix? Is there some picture or analogy I can use as intuition?
Update
Meanwhile there are some nice videos (1) (2) (3) on youtube by Joan Sola, explaining the intuition behind Lie theroy
A few things that might be helpful:
1) The exponential of a matrix is always invertible.
2) Diagonal matrices are easy to exponentiate: the result is a diagonal matrix whose entries are the exponentials of the original diagonal entries.
3) Fact 2 suggests trying to diagonalize the input matrix before exponentiating it. And in fact: $\mathrm{exp}(gXg^{-1}) = g\mathrm{exp}(X)g^{-1}$.
4) Working with concrete examples is helpful! $G = SO(2)$ is a great place to start.
Also, you can find a really nice investigation of these ideas in Artin's "Algebra".
Good luck!
*** Additional comments on intuition:
Let's focus on $G = SL_n(\mathbb{R})$ so we have something concrete in mind. In fact, let's choose $n = 2$ so that $G$ is the space of $2 \times 2$ real matrics whose determinant is $1$. Perhaps the first thing to note is that we can think of $G$ as living inside the space $M_2(\mathbb{R})$ of all $2 \times 2$ matrices (this includes the non-invertible ones too.) And we can clearly identify $M_2(\mathbb{R})$ with $\mathbb{R}^4$ by the map: $$A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \mapsto (a, b, c, d).$$ The good news here is that we can now think of tangent vectors in terms of ordinary Euclidean space (and elementary vector calculus.)
Now, since $\det(A) = ad - bc$, we can actually think of $G$ as nothing more than the surface sitting inside $\mathbb{R}^4$ defined by $xw - yz = 1$. Cool! This is not much different from thinking of the sphere being defined by the equation $x^2 + y^2 + z^2 + w^2 = 1$. So, if you can find tangent vectors on spheres, then you can find tangent vectors on Lie groups!
In our present setting, we have a very nice way to find tangent vectors: Derivatives of curves. Let $A$ be some point in $SL_2(\mathbb{R})$, which we'll think of as a point $(a,b,c,d) \in \mathbb{R}^4$ with $ad - bc = 1$. If we could find a curve $\gamma(t)$ lying entirely in $SL_2(\mathbb{R})$ with $\gamma(0) = A$, then $\gamma'(0)$ would be a tangent vector based at $A$. And then we could ask: What's the relationship between the tangent vector $\gamma'(0)$ and the value of the exponential map $\exp(\gamma'(0))$?
OK, so before I go any further, is this helpful information? Does it get you closer to an answer to your question?