Principal curvatures of $\mathbb{R}^{n^2}$-embedded SO(n) It's well known that the sectional curvatures of a Lie group, endowed with a left-invariant metric have a nice closed-form formula $k(X,Y) = \frac{1}{4} \|[X Y]\|^2$. 
I'm wondering if the following (extrinsic) question has a simple answer: let me consider the natural embedding of $\mathrm{SO}(n)$ into $\mathbb{R}^{n^2}$, such that the induced Euclidean metric agrees with the left-invariant metric. 
Can one derive closed-form expressions for the principal curvatures of the Weingarten map (for a normal direction, as the co-dimension is $> 1$) ?  
 A: What you are asking about is the second fundamental form of the embedding.  Since this is a Lie group, it's enough to know what the second fundamental form is at the identity matrix $I_n=e$.  Since the tangent space of $\mathrm{SO}(n)$ at the identity is the space of $n$-by-$n$ skew-symmetric matrices and the normal space is the space of $n$-by-$n$ symmetric matrices, the second fundamental form, which is a quadratic map from the tangent space to the normal space, is simply given by the order-2 term in the exponential series that gives the embedding.  Thus, the quadratic map $\mathrm{I\!I}_e:T_e\mathrm{SO}(n)\to N_e$ (i.e., from the tangent space to the normal space), is given by 
$$
\mathrm{I\!I}_e(A) = \tfrac12 A^2.
$$
Added remark [28 Oct 2018]  I was asked how to prove this.  Here is a little extra detail on this computation: When one writes the vector space $M_n$ of $n$-by-$n$ symmetric matrices with real entries as the direct sum of the subspaces $A_n$ (the $n$-by-$n$ anti-symmetric matrices) and the subspace $S_n$ (the $n$-by-$n$ symmetric matrices), the exponential map $\exp: A_n\to \mathrm{SO}(n)\subset M_n$ splits into two parts as 
$$
\exp(a) = I_n + a + \tfrac12a^2 + \cdots = (a + \tfrac16a^3+\cdots)+(I_n+\tfrac12a^2+\cdots) = \sinh(a) + \cosh(a),
$$
where $\sinh:A_n\to A_n$ is a local diffeomorphism near $a = 0$, and $\cosh(a):A_n\to S_n$ is smooth. Thus, writing $b = \sinh(a)$, we have, for small $b\in A_n$,
$$
\exp(a) = b + \cosh(\sinh^{-1}b) = b + \sqrt{I_n+b^2}.
$$ 
Hence, near the identity $I_n\in\mathrm{SO}(n)$, the submanifold $\mathrm{SO}(n)$ can be regarded as a graph in $M_n = A_n\oplus S_n$ of the form
$$
\bigl(b, \sqrt{I_n+b^2}\,\bigr).
$$
Since one has the convergent Taylor series expansion $\sqrt{I_n+b^2} = I_n + \tfrac12b^2 + \cdots$, the result follows.
To compute the principal curvatures in a given direction $S\in N_e$, where $S$ is a symmetric $n$-by-$n$ matrix, you just need to take the inner product of $S$ with the above map and find the eigenvalues of the quadratic form 
$$
Q_S(A) = \tfrac12 \mathrm{tr}(SA^2) = \tfrac12 \mathrm{tr}(ASA).
$$
with respect to the natural quadratic form $Q(A) = -\mathrm{tr}(A^2)$.  By equivariance, it suffices to consider the case in which $S$ is diagonal, with eigenvalues (i.e., diagonal entries) $s_1,\ldots,s_n$.  Then one finds that the eigenvalues of $Q_S$ with respect to $Q(A)$ are of the form $\tfrac12(s_i{+}s_j)$ for $1\le i<j\le n$.
Note, to normalize $S$, i.e., to consider unit normals instead of arbitrary normal vectors, one should arrange that ${s_1}^2+\cdots+{s_n}^2 = 1$.
Added remark:  By the way, by the above computation, when $n\ge 3$, we have that $Q_S\not=0$ when $S\not=0$.  Thus, it follows that $\mathrm{SO}(n)$ does not lie in any proper affine subspace of the space of $n$-by-$n$ matrices, which gives an alternative proof of the answer to this question about the linear span of $\mathrm{SO}(n)$ in $n$-by-$n$ matrices.
