What is exponential map in differential geometry

---

Why it is called the exponential

The reason it's called the exponential is that in the case of matrix manifolds, the abstract version of $exp$ defined in terms of the manifold structure coincides with the "matrix exponential" $exp(M) equiv sum_{i=0}^infty M^n/n!$.

---

A concrete example, the unit circle

For example, let's consider the unit circle $M equiv { x in mathbb R^2 : |x| = 1 }$. This can be viewed as a Lie group $M = G = SO(2) = left{ begin{bmatrix} cos theta & sin theta \ -sin theta & cos theta end{bmatrix} : theta in mathbb R right}$.

---

The unit circle: Tangent space at the identity, the hard way

We can derive the lie algebra $mathfrak g$ of this Lie group $G$ of this "formally" by trying computing the tangent space of identity. To do this, we first need a useful definition of the tangent space. One possible definition is to use the definition of the space of curves $gamma_{alpha}: [-1, 1] rightarrow M$, where the curves are such that $gamma(0) = I$. Then the tangent space $T_I G$ is the collection of the curve derivatives $frac{d(gamma(t)) }{dt}|_0$. Let's calculate the tangent space of $G$ at the identity matrix $I$, $T_I G$:

$$ gamma_alpha(t) = begin{bmatrix} cos (alpha t) & sin (alpha t) \ -sin (alpha t) & cos (alpha t) end{bmatrix} $$

This is a legal curve because the image of $gamma$ is in $G$, and $gamma(0) = I$. We can differentiate this and compute $d/dt(gamma_alpha(t))|_0$ to get:

begin{align*} &frac{d/dt} gamma_alpha(t)|_0 = frac{d}{dt} begin{bmatrix} cos (alpha t) & sin (alpha t) \ -sin (alpha t) & cos (alpha t) end{bmatrix}|_0 \ &= begin{bmatrix} frac{d(cos (alpha t))}{dt}|_0 & frac{d(sin (alpha t))}{dt}|_0 \ frac{d(-sin (alpha t))}{dt}|_0 & frac{d(cos (alpha t))}{dt}|_0 end{bmatrix} \ &= begin{bmatrix} -tsin (alpha t)|_0 & tcos (alpha t)|_0 \ -tcos (alpha t)|_0 & -tsin (alpha t)|_0 end{bmatrix} \ &= begin{bmatrix} 0 & t cdot 1 \ -t cdot 1 & 0 end{bmatrix} \ = text{skew symmetric matrix} end{align*}

So we get that the tangent space at the identity $T_I G = { S text{ is $2times2$ matrix} : S + S^T = 0 }$

Now recall that the Lie algebra $mathfrak g$ of a Lie group $G$ is defined to be the tangent space at the identity. So we have that $mathfrak g = T_I G = text{$2times2$ skew symmetric matrices}$.

---

The unit circle: Tangent space at the identity by logarithmization

On the other hand, we can also compute the Lie algebra $mathfrak g$ / the tangent space at the identity $T_I G$ "completely informally", by "logarithmizing" the group. We know that the group of rotations $SO(2)$ consists of orthogonal matrices group, so every element $U in G$ satisfies $UU^T = I$. We can logarithmize this condition as follows:

$$ M = G = { U : U U^T = I } \ mathfrak g = log G = { log U : log (U U^T) = log I } \ mathfrak g = log G = { log U : log (U) + log(U^T) = 0 } \ mathfrak g = log G = { log U : log (U) + log(U)^T = 0 } \ mathfrak g = log G = { S : S + S^T = 0 } \ $$

We got the same result: $mathfrak g$ is the group of skew-symmetric matrices by following the physicist derivation of taking a $log$ of the group elements.

---

The unit circle: The exponential map

Now, it should be intuitively clear that if we got from $G$ to $mathfrak g$ using $log$, we ought to have an nverse $exp: mathfrak g rightarrow G$ which does the opposite. Indeed, this is exactly what it means to have an exponential map: we can go from elements of the Lie algebra $mathfrak g$ / the tangent space at the identity $T_I G$ to the Lie group $G$.

---

The unit circle: Computing the exponential map

Assume we have a $2 times 2$ skew-symmetric matrix $S$. We want to show that its exponential lies in $G$:

$$ exp(S) = exp left (begin{bmatrix} 0 & s \ -s & 0 end{bmatrix} right) = sum_{n=0}^infty S^n/n! $$

We can compute this by making the following observation:

begin{align*} S^2 = begin{bmatrix} 0 & s \ -s & 0 end{bmatrix} begin{bmatrix} 0 & s \ -s & 0 end{bmatrix} = begin{bmatrix} -s^2 & 0 \ 0 & -s^2 end{bmatrix} = -begin{bmatrix} s^2 & 0 \ 0 & s^2 end{bmatrix} end{align*}

We immediately generalize, to get $S^{2n} = -(1)^n begin{bmatrix} s^{2n} & 0 \ 0 & s^{2n} end{bmatrix}$

This gives us $S^{2n+1} = S^{2n}S$:

begin{align*} S^{2n+1} = S^{2n}S = (-1)^n begin{bmatrix} s^{2n} & 0 \ 0 & s^{2n} end{bmatrix} begin{bmatrix} 0 & s \ -s & 0 end{bmatrix} = (-1)^n begin{bmatrix} 0 & s^{2n+1} \ -s^{2n+1} & 0 end{bmatrix} end{align*}

We can now compute the exponential as:

begin{align*} &exp(S) = I + S + S^2 + S^3 + .. = \ &(I + S^2/2! + S^4/4! + cdots) + (S + S^3/3! + S^5/5! + cdots) \ &= begin{bmatrix} 1 - s^2/2! + s^4/4! + cdots & 0 \ 0 & 1 - s^2/2! + s^4/4! + cdots end{bmatrix} + begin{bmatrix} 0 & s - s^3/3! + s^5/5! + cdots \ s - s^3/3! + s^5/5! + cdots & 0 end{bmatrix} \ &= begin{bmatrix} cos(s) & sin(s) \ -sin(s) & cos(s) end{bmatrix} end{align*}

We get the result that we expect: We get a rotation matrix $exp(S) in SO(2)$. We can check that this $exp$ is indeed an inverse to $log$.

---

Why skew-symmetric?

What does it mean that the tangent space at the identity $T_I G$ of the group of rotations are the skew-symmetric matrices?

One explanation is to think of these as curl, where a curl is a sort of "infinitesimal rotation". See that a skew symmetric matrix $S equiv begin{bmatrix} a & b \ -b & a end{bmatrix}$

can be viewed as having two vectors $S_1 = (a, b)$ and $S_2 = (-b, a)$, which represents an infinitesimal rotation from $(a, b)$ to $(-b, a)$.

This is skew-symmetric because rotations in 2D have an orientation. Flipping the order of the vectors gives us the rotations in the opposite order: It takes clockwise to anti-clockwise and anti-clockwise to clockwise. If we wish to fancy, we can talk about this in terms of exterior algebra

See the picture which shows the skew-symmetric matrix $begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}$ and its transpose as "2D orientations"

---

The unit circle: What about the other tangent spaces?!

So far, I've only spoken about the lie algebra $mathfrak g$ / the tangent space at the identity $T_I G$. What about all of the other tangent spaces?

A very cool theorem of matrix Lie theory tells us that the tangent space at some point $P$, $T_P G$ is always going to be translates of $T_I G$. Formally, we have the equality:

$$T_P G = P T_I G = { P T : T in T_I G }$$

This lets us immediately know that whatever theory we have discussed "at the identity" can be easily translated to "any point" $P in G$, by simply multiplying with the point $P$.

---

Recap We saw the following equivalences:

• We have a Lie group $G$ with Lie algebra $mathfrak g$, which is the tangent space at the identity $T_I G$.
• For this, computing the Lie algebra by using the "curves" definition co-incides with simply invoking $log: G rightarrow mathfrak g$ on the definiton of the matrix group.
• Vice versa, the $exp$ (inverse of $log$) can be computed from the series definition, giving us a map $exp: mathfrak g rightarrow G$.
• These maps allow us to go from the "local behaviour" to the "global behaviour".
• We gained an intuition for the concrete case of $G = SO(2)$, $mathfrak g$ as skew-symmetric matrices, and why skew-symmetric matrices are the "infinitesimal rotations"
• We refer to the fact that if we know $mathfrak g$/$T_I G$, we automatically know all tangent spaces due to the group being a Lie group. Hence, knowing the lie algebra $mathfrak g$ is "as good as" knowing the tangent space structure everywhere.

A summary picture: