Lower bound for Jacobian of matrix exponential map near origin What is a lower bound for the Jacobian of the exponential map from the skew-symmetric matrices to the orthogonal matrices near the origin?
 A: For simplicity, I work with $2n \times 2n$ matrices, the odd by odd case is similar.
Summary: If $B$ is a skew symmetric matrix with eigenvalues $\pm i \theta_1$, $\pm i \theta_2$, ..., $\pm i \theta_n$ then the Jacobian matrix of the exponential near $B$ has eigenvalues 
$$\frac{1-e^{i(\mp \theta_j \mp \theta_k)}}{i(\pm \theta_j \pm \theta_k)}, \quad 1 \leq j < \leq n$$
as well has having the eigenvalue $1$ with multiplicity $n$. (This is $4 \binom{n}{2}+n=\binom{2n}{2}$ eigenvalues in total.) 
The determinant of the Jacobian matrix is thus
$$\prod_{1 \leq j < k \leq n} \frac{1-e^{i(\mp \theta_j \mp \theta_k)}}{i(\pm \theta_j \pm \theta_k)} = \prod_{1 \leq j < k \leq n} \frac{16 \sin^2 \frac{\theta_j-\theta_k}{2} \sin^2 \frac{\theta_j+\theta_k}{2}}{(\theta_j^2-\theta_k^2)^2}.$$
Take whatever sort of bound you have on $B$ and turn it into a lower bound on the above quantity. Notice that, if $\theta_j + \theta_k$ gets as large as $2 \pi$, the above quantity is zero, so there is no nontrivial lower bound in that case.

A practical note on minimizing the above quantity: the log of the above is the sum of many terms of the form $f(\phi) := \log \sin(\phi) - \log \phi$, where $\phi$ is a linear function. By the double derivative test, $f$ is concave. So the sum of many terms of the form $f(\mbox{linear function})$ will be concave. This means that, on any convex region, the minimum will occur somewhere on the boundary.

Notation:  We write $\mathfrak{so}$ for the vector space of Skew symmetric matrices. We fix $B$ and $\theta_j$ as above.
Explanation: Let me first point out why the question makes sense. The orthogonal matrices are a manifold, not a vector space, so one might be tempted to wonder whether it even makes sense to speak of a Jacobian; let alone to speak of the eigenvalues of the Jacobian matrix. 
There are two ways to fix this, a naive way, and a sophisticated way, and they both give the same answer. The naive way is to point out that the orthogonal matrices are contained in the $n \times n$ matrices. So we certainly have a $\binom{2n}{2} \times (2n)^2$ matrix, giving the Jacobian matrix of the exponential map as a map from skew-symmetric matrices to all matrices. This matrix is not square; its image is the tangent plane at $e^B$ to the space of orthogonal matrices. Explicitly, that tangent plane is $e^B \mathfrak{so}$. We can rotate that tangent plane by the orthogonal matrix $e^{-B}$, giving us a map from $\mathfrak{so}$ to itself; it is now sensible to discuss the eigenvalues of that map.
The sophisticated way is say that the Jacobian matrix is a map from $\mathfrak{so}$ to the tangent space of $SO$ at $e^B$. But that tangent space is canonically identified with the tangent space of $SO$ at the identity, and the latter tangent space is $\mathfrak{so}$.
Either way, we are being asked to consider the following map from $\mathfrak{so}$ to itself:
$$E \mapsto e^{-B} \lim_{t \to 0} (e^{B+tE} - e^B)/t. \quad (*)$$
Let $A$ be the map $X \mapsto [B,X]$ from $\mathfrak{so}$ to itself. By the Baker-Campbell-Hausdorff formula, $(*)$ is
$$\frac{1-e^{-A}}{A} E$$
where $(1-e^{-A})/A$ must be understood as the power series $1-A/2+A^2/6-\cdots$. If written out as matrices, $A$ would be a $\binom{2n}{2} \times \binom{2n}{2}$ matrix and $E$ would be a vector of length $\binom{2n}{2}$.
Now, if the eigenvalues of $B$ are $\pm i \theta_j$, as above, then the eigenvalues of $A$ are $0$ with multiplicity $n$ and $i(\pm \theta_j \pm \theta_k)$. (Because the root system of type $D_n$ is $\{ \pm e_j \pm e_k \}$, or because it is an easy computation.) If $\alpha_1$, $\alpha_2$, ..., $\alpha_N$ are the eigenvalues of $A$, then the eigenvalues of $(1-e^{-A})/A$ are $(1-e^{-\alpha_j})/\alpha_j$, where $(1-e^{-0})/0$ is interpreted as $1$.
Putting this all together, we get the above formulas.
