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Is $\exp_{q}(v)$ a projection of point $q$ to some point $q'$ along the geodesic whose tangent (right?) at $q$ is the vector $v$? And so $\exp_{q}(v)$ is the projection of point $q$ to some point along the geodesic between $q$ and $q'$?

(Another post gives an explanation: Riemannian geometry: ...Why is it called 'Exponential' map? So now I'm wondering how we know where $q$ exactly falls on the geodesic after it travels for a unit amount of time. Does it uniquely depend on $p, v, M$ only, is it affected by any other parameters as well, or is it arbitrarily set to any point in the geodesic?)

The reason that it is called exponential map seems to be that the function satisfy that two images' multiplication $\exp_{q}(v_1)\exp_{q}(v_2)$ equals the image of the two independent variables' addition (to some degree)?

But that simply means a exponential map is sort of (inexact) homomorphism. Is there any other reasons for this naming?


(To make things clearer, what's said above is about exponential maps of manifolds, and what's said below is mainly about exponential maps of Lie groups. And I somehow 'apply' the theory of exponential maps of Lie group to exponential maps of Riemann manifold (for I thought they were 'consistent' with each other). What I tried to do by experimenting with these concepts and notations is not only to understand each of the two exponential maps, but to connect the two concepts, to make them consistent, or to find the relation or similarity between the two concepts. Now it seems I should try to look at the difference between the two concepts as well.)

It seems that, according to p.388 of Spivak's Diff Geom, $\exp_{q}(v_1)\exp_{q}(v_2)=\exp_{q}((v_1+v_2)+[v_1, v_2]+...)$, where $[\ ,\ ]$ is a bilinear function in Lie algebra (I don't know exactly what Lie algebra is, but I guess for tangent vectors $v_1, v_2$ it is (or can be) inner product, or perhaps more generally, a 2-tensor product (mapping two vectors to a number) (length) times a unit vector (direction)). It seems $[v_1, v_2]$ 'measures' the difference between $\exp_{q}(v_1)\exp_{q}(v_2)$ and $\exp_{q}(v_1+v_2)$ to the first order, so I guess it plays a role similar to one that first order derivative $/1!$ plays in function's expansion into power series.

With such comparison of $[v_1, v_2]$ and 2-tensor product, and of $[v_1, v_2]$ and first order derivatives, perhaps $\exp_{q}(v_1)\exp_{q}(v_2)=\exp_{q}((v_1+v_2)+[v_1, v_2]+ T_3\cdot e_3+T_4\cdot e_4+...)$, where $T_i$ is $i$-tensor product (length) times a unit vector $e_i$ (direction) and where $T_i$ is similar to $i$th derivatives$/i!$ and measures the difference to the $i$th order.

(According to the wiki articles https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory) mentioned in the answers to the above post, it seems $\exp_{q}(v))$ does have an power series expansion quite similar to that of $e^x$, and possibly $T_i\cdot e_i$ can, in some cases, written as an extension of $[\ , \ ]$, e.g. $[v_1,[v_1,v_2]]$ so that $T_i$ is $i$-tensor product but remains a function of two variables $v_1,v_2$.)

Besides, if so we have $\exp_{q}(tv_1)\exp_{q}(tv_2)=\exp_{q}(t(v_1+v_2)+t^2[v_1, v_2]+ t^3T_3\cdot e_3+t^4T_4\cdot e_4+...)$.

I'm not sure if my understanding is roughly correct.

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    $\begingroup$ It's worth noting that there are two types of exponential maps typically used in differential geometry: one for Riemannian manifolds, which you refer to in your question, and one for Lie groups, which Spivac is referring to. The expression $\exp(u)\exp(v)=\exp(u+v+[u,v]+\dots)$ is known as the BCH formula and applies specifically to Lie groups. $\endgroup$ – Kajelad Jul 23 at 5:07
  • $\begingroup$ Trying to understand the second variety. Where can we find some typical geometrical examples of exponential maps for Lie groups? $\endgroup$ – Narasimham Jul 23 at 6:16
  • $\begingroup$ Just to clarify, what do you mean by $\exp_q$? Specifically, what are the domain the codomain? At the beginning you seem to be talking about a Riemannian exponential map $\exp_q:T_qM\to M$ where $M$ is a Riemannian manifold, but by the end you are instead talking about the map $\exp:\mathfrak{g}\to G$ where $G$ is a Lie group and $\mathfrak{g}$ is its Lie algebra. These maps have the same name and are very closely related, but they are not the same thing. $\endgroup$ – Kajelad Jul 23 at 6:38
  • $\begingroup$ @Narasimham Typical simple examples are the one demensional ones: $\exp:\mathbb{R}\to\mathbb{R}^+$ is the ordinary exponential function, but we can think of $\mathbb{R}^+$ as a Lie group under multiplication and $\mathbb{R}$ as an Abelian Lie algebra with $[x,y]=0$ $\forall x,y$. Fitting this into the more abstract, manifold based definitions/constructions can be a useful exercise. $\endgroup$ – Kajelad Jul 23 at 6:47
  • $\begingroup$ Yes, I do confuse the two concepts, or say their similarity in names confuses me a bit. Thanks for clarifying that. $\endgroup$ – Charlie Chang Jul 23 at 7:16
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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} -t\sin (\alpha t)|_0 & t\cos (\alpha t)|_0 \\ -t\cos (\alpha t)|_0 & -t\sin (\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 $2\times2$ 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{$2\times2$ 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"

diagram-skew-symmetric


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: summary-relationships

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  • $\begingroup$ I see $S^1$ is homeomorphism to rotational group $SO(2)$, and the Lie algebra is defined to be tangent space at (1,0) in $S^1$ (or at $I$ in $SO(2)$. Exponential maps from tangent space to the manifold, if put in matrix ‘representation’, since powers of a vector $v$ of tangent space (in matrix ‘representation’, i.e. an anti symmetric matrix $\lambda [0, 1; -1, 0]$, say $\lambda T$ ) alternates between $\lambda^n\cdot T$ or $\lambda^n\cdot I$, leading to that exponentials of the vector‘s matrix representation being combination of $\cos(v), \sin(v)$ which is (matrix repre of) a point in $S^1$. $\endgroup$ – Charlie Chang Jul 23 at 13:32
  • $\begingroup$ Since the matrices involved only have two independent components we can repeat the process similarly using complex number, (v is represented by $0+i\lambda$, identity of $S^1$ by $ 1+i\cdot0$) i.e. $\exp(v)=\exp(i\lambda)$ = power expansion = $cos(\lambda)+\sin(\lambda)$. (For both repre have two independents components, the calculations are almost identical.) However, this complex number ‘repre’ can’t be easily extended to slanting tangent space in 2-dim and higher dim. $\endgroup$ – Charlie Chang Jul 23 at 13:46
  • $\begingroup$ Should be ‘Exponential maps from tangent space to the manifold, if put in matrix ‘representation’, are called exponential, since powers of...’. Besides, I’m not sure why Lie algebra is defined this way, perhaps it’s because that makes tangent spaces of all Lie groups easily inferred from Lie algebra? $\endgroup$ – Charlie Chang Jul 23 at 13:53
  • $\begingroup$ Also, in this example $\exp(v_1)\exp(v_2)= \exp(v_1+v_2)$ and $[v_1, v_2]=AB-BA=0$, where A B are matrix repre of the two vectors. I’m not sure if these are always true for exponential maps of Riemann manifolds. $\endgroup$ – Charlie Chang Jul 23 at 14:04
  • $\begingroup$ @CharlieChang Indeed, this example $SO(2) \simeq U(1)$ so it's commutative. To the see the "larger scale behavior" wth non-commutativity, simply repeat the same story, replacing $SO(2)$ with $SO(3)$. All the explanations work out, but rotations in 3D do not commute; This gives the example where the lie group $G = SO(3)$ isn't commutative, while the lie algbera `$\mathfrak g$ is [thanks to being a vector space]. $\endgroup$ – Siddharth Bhat Jul 23 at 18:44

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