Explicit description of tangent spaces of $O(n)$ I would like someone to verify the calcuations that I'm making about the tangent spaces of $O(n)$, and help answer some general questions that I have from this special case. I don't know much about Lie theory at all, so I'm flying by the seat of my pants here :) 
So, we have the space $O(n) \equiv \{ X | X^TX = I \}$. It's not hard to show that the tangent space should satisfy the constraint: tangent space at $P \in O(n) \equiv T_P O(n) \equiv \{ Z | Z^T P + P^T Z = 0\}$.
Now, by performing the usual Lie algebra trick of "calculate tangent space at identity", we see that:
$$
T_I O(n) = \{ Z | Z^T + Z = 0 \}
$$
Next, to calculate the tangent space at some arbitrary point $P \in O(n)$, we consider the map:
$$
f: O(n) \rightarrow O(n) \qquad f(X) = PX
$$
Note that $f(I) = P$, and hence the differential of this map, $df$ will have the type $df: T_I O(n) \rightarrow T_P O(n)$. Now, all that's left to do is to calculate the differential.
To perform this, we take a curve:
$$
c: (-\epsilon, \epsilon) \rightarrow O(n) \qquad
c(t) = e^{Kt} \qquad
K^T = -K
$$
This has image in $O(n)$, since 
$$
c(t)^T c(t) = e^{K^T t}e^{Kt} = e^{t(K^T + K)} = e^{t\cdot 0} = I
$$
Hence, this is a valid curve.
Now, we compute $df$:
$$
df \equiv \frac{d}{dt} (f \circ c)(t) \vert_{t = 0} = \frac{d}{dt} P  e^{Kt} \vert_{t=0} = PKe^{K0} = PK
$$
Hence, the tangent space at the point $P$ is $T_PO(n) \equiv \{ PK | K^T + K = 0 \}$
I am uncomfortable with many things that "automatically work" in this proof:


*

*Is this proof correct? If not, where is it wrong?

*Why is the choice of the integral curve $c(t) = e^{Kt}$ the "correct" choice? How do I prove that the mapping of tangent spaces is indeed a bijection, and not some artefact of the curve parametrization I chose?

*What is the "general form" of this proof, for an arbitrary matrix Lie group $M$? Can I state that the tangent space at a point $P$ will be of the form $T_P M \equiv \{ PZ \mid Z \in T_I M \}$? If not, why is this not correct?

*Is there a slick proof of this fact?
 A: *

*Yes, this proof is correct.

*You could use any curve $c:(-\epsilon, \epsilon)\to O(n)$ that satisfies $c(0) = I$ and $c'(0)=K$. Your calculation of $"df"$ (actually, you should write $(df)_I(K)$) would be identical.

*For a matrix Lie group $M$, this is pretty much the general form of the proof. For a general Lie group (which does not necessarily have a matrix representation), the only difference would be that since $f(X) = L_PX$, where $L_P:G\to G$ denotes left multiplication by $P\in G$, you would have 
$$
(df)_I(K) = (dL_P)_I(K).
$$
This expression is commonly just written as $P\cdot K$ for $P\in G, K\in\mathfrak{g}$, however.

*Not really, as far as I can tell. Maybe you'd prefer this one: let $d:(-\epsilon,\epsilon)\to G$ be an arbitrary curve through $P\in G$. Then $t\mapsto P^{-1}\cdot d(t)$ is a curve passing through $I$, so its derivative $P^{-1}\cdot d'(0)\in T_IG = \mathfrak{g}$. Hence $d'(0)\in P\cdot\mathfrak{g}$, implying $T_PG = P\cdot\mathfrak{g}$. Incidentally, using $d(t)\cdot P^{-1}$ instead would yield $T_PG = \mathfrak{g}\cdot P$, which is also true. This proof is essentially the same as the one you gave, however.

