Why is adding/subtracting allowed in the limit definition of derivatives when you're taking a derivative in a matrix Lie group?

Question

A matrix Lie group $G$ is a subgroup of $\operatorname{GL}(n, \mathbb{R})$ or $\operatorname{GL}(n, \mathbb{C})$ under matrix multiplication that also forms a compatible smooth submanifold structure.

For the sake of concreteness let us consider the group $\textrm{SO}(n)$. This is closed under matrix multiplication and it forms a specific kind of submanifold of $\operatorname{GL}(n, \mathbb{R})$. If we ever consider doing matrix addition in $\operatorname{SO}(n)$, that would be considered illegal since that would take us outside of the special orthogonal group.

However, when we are considering derivatives of paths in $\operatorname{SO}(n)$, the limit definition for the derivative involves componentwise addition/subtraction. Let $\gamma:(-1, 1)\rightarrow\operatorname{SO}(n)$ be such a path with $\gamma(0) = I$. Then the derivative at $t=0$ is \begin{align}\tag{1}\label{1} \gamma'(0) = \lim_{h\rightarrow 0} \frac{\gamma(h) - \gamma(0)}{h}. \end{align} Assuming $\gamma(0) = I$ (which is taken as granted here), this correctly gives us a skew-symmetric matrix (as it should), meaning we obtain an element of the Lie algebra $\mathfrak{so}(n)$.

However, \eqref{1} just doesn't make any sense to me. I thought we weren't allowed to add/subtract matrices since it is meaningless as far as the group $\operatorname{SO}(n)$ is concerned. Why is it all of a sudden okay in this context?

Answer 1

I want to start with a seemingly unrelated example and then circle back to OP's example.

Consider the unit sphere $S^{2}$ embedded in $\mathbb{R}^{3}$. This is a set of points $S^{2} = \{ (x, y, z) \,|\, x^{2}+y^{2}+z^{2}=1 \}$. With the structure we have so far defined, we can neither multiply nor add elements of $S^{2}$ with by each other in any sensible way. Adding coordinates componentwise will take you out of $S^{2}$ just like in the OP's example, so it's "not a valid operation" in a certain sense.

Actually, it's worth trying to elaborate what "not a valid operation" means here. By "not a valid operation," what is meant is that this operation (componentwise addition/subtraction of coordinates) explicitly relies on the particular embedding of $S^{2}$ in $\mathbb{R}^{3}$. Thus, coordinate addition/subtraction is meaningful from the extrinsic point of view but it is not meaningful from the intrinsic point of view. If we were to talk about $S^{2}$ in terms of intrinsic geometry, then addition of points of $S^{2}$ would be undefined.

Let $\gamma:(-1, 1)\rightarrow S^{2}$ be a differentiable path on $S^{2}$, and let's say it goes through the North pole $N = (0, 0, 1)$ at $t=0$. Despite the fact that we've said that addition is banned, the derivative of the path at $t=0$ is $$ \gamma'(0) = \lim_{h\rightarrow 0} \frac{\gamma(h)-\gamma(0)}{h} $$ which ends up using addition/subtraction. Here is the justification:

1. From the extrinsic point of view, the difference $\gamma(h)-\gamma(0)$ represents a vector from the North pole $\gamma(0) = N$ to $\gamma(h)$.
2. As we take the quotient shown above and send $h\rightarrow 0$, the vector tends more and more to become the velocity tangent vector to $\gamma$ at $t=0$. So in the limit, you should visualize a vector tangent to the sphere at the North pole pointing in the direction of the path.

The set of all such tangent vectors (set of all $\gamma'(0)$'s for smooth paths having $\gamma(0) = N$) forms the tangent plane to $S^{2}$ at $N$. We visualize this as a horizontal plane touching $S^{2}$ at $N$. So we have the following info:

1. Adding/subtracting points of $S^{2}$ embedded in $\mathbb{R}^{3}$ is explicitly something that makes sense only from the extrinsic point of view.
2. The tangent plane $P$ formed by the above description is also tied to the extrinsic point of view.
3. Despite this, the tangent plane $P$ as we have described it has a correspondence with the abstract tangent space $T_{N}S^{2}$, which is something that arises from the intrinsic point of view. There is a direct vector space isomorphism $P\cong T_{N}S^{2}$ between the two spaces in which $P$ is simply just a concrete realization of the more abstract $T_{N}S^{2}$.

So while adding/subtracting coordinates of $S^{2}$ on their own is meaningless, it ends up producing something meaningful in the sense described by points (1)-(3).

Going Back to Matrix Lie Groups

Now as the OP framed it, they are talking about matrix Lie groups as manifolds embedded in $\operatorname{GL}(n, \mathbb{R})$. This is part of the premise. This means the situation is no different than the one described with $S^{2}$ embedded in $\mathbb{R}^{3}$.

Addition/subtraction of $\operatorname{SO}(n)$-matrices doesn't have meaning from the intrinsic point of view of $\operatorname{SO}(n)$. However, as a manifold embedded in $\operatorname{GL}(n, \mathbb{R})$ (extrinsic point of view), we can think of the difference $A-B$ as like a vector with a tail at $B$ and head at $A$. If this doesn't make sense, then recognize that $\operatorname{GL}(n, \mathbb{R})\cong\mathbb{R}^{n^{2}}$ as vector spaces.

Given a smooth path $\gamma(t)$ with $\gamma(0) = I$, the limit for $\gamma'(0)$ produces a vector pointing tangent to the manifold of $\operatorname{SO}(n)$ at $I\in\operatorname{SO}(n)$. We can imagine forming a tangent space $P$ embedded in $\operatorname{GL}(n, \mathbb{R})$ exactly in the same manner as we did in the sphere case. Although we're relying on $P$ being embedded $\operatorname{GL}(n, \mathbb{R})$, its vector space structure corresponds to the abstract tangent space $T_{I}\operatorname{SO}(n) = \mathfrak{so}(n)$.

The point is, the entire premise of OP's scenario is that we're working from the extrinsic point of view, so there's nothing "illegal" happening when we are taking advantage of that fact. In principle, it is possible to work with all manifolds (including $S^{2}$ and the Lie group $\operatorname{SO}(n)$) in a fully intrinsic way, but that might require a bit of work. Luckily, the resulting tangent spaces (and tangent vectors) end up having a correspondence whether you derive them from the extrinsic viewpoint or the abstract intrinsic viewpoint.