# 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 $$> 1$$) ?

• Weingarten map is defined when the codimension is 1 which is not the case here. Oct 21 '18 at 14:49
• Ah sure, but you can define it for a particular normal direction, though. I guess I was a bit sloppy with the phrasing. Oct 21 '18 at 14:54

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.

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.

• Thanks, that's very helpful! I'm a relative newcomer, so possibly a silly question: how does one see that the principle curvatures are eigenvalues with respect to the quadratic form $Q(A)$? [In the hypersurface case, the second fundamental form can be viewed as just a map $T_eSO(n) \to \mathbb{R}$, so they're just regular eigenvalues -- how does this port to the co-dimension $>1$ case?] Oct 21 '18 at 18:14
• @AndrewRugger: Well, in the hypersurface case, the definition of principal curvatures is the eigenvalues of the second fundamental form with respect to the first fundamental form. I suppose that I should have pointed out that the natural quadratic form $Q$ is the first fundamental form. Oct 21 '18 at 23:07
• Actually, another thing I can't quite parse: why can we assume S is diagonal? Trace is invariant under similarities, but S is "in the middle" of the above inner product. If we evaluate it for $U S U^T$, it isn't clear to me it is preserved? Feb 15 '19 at 18:22
• @AndyMack: The point is that we have the identity $$\mathrm{tr}(ASA) =\mathrm{tr}(UAU^T\ \ USU^T\ \ UAU^T)$$ for $S$ symmetric, $A$ anti-symmetric, and $U$ orthogonal, while all the inner products involved are invariant under the action $A\mapsto UAU^T$ and $S\mapsto USU^T$. Meanwhile, one can always choose $U$ so that $USU^T$ is diagonal. Feb 15 '19 at 20:41