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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 $> 1$) ?

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

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  • $\begingroup$ 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?] $\endgroup$
    – Andy Mack
    Oct 21 '18 at 18:14
  • $\begingroup$ @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. $\endgroup$ Oct 21 '18 at 23:07
  • $\begingroup$ 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? $\endgroup$
    – Andy Mack
    Feb 15 '19 at 18:22
  • $\begingroup$ @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. $\endgroup$ Feb 15 '19 at 20:41

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