What is exponential map in differential geometry 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.
 A: 
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"


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:

