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Consider the Lie algebra $so(n)$ equipped with the metric $\langle e_i \wedge e_j, e_k \wedge e_l \rangle = \delta_{i,k} \delta_{j,l}$. Similarly equip the tangent space at other points of $SO(n)$ by left translation. My question is, is the exponential map $\exp: so(n) \to SO(n)$ 1-Lipschitz? I can show it's Lipschitz using the following result from Wikipedia:

$$ \| e^{X+Y} - e^Y \| \le e^{\|X\|} e^{\|Y\|} \|X\|$$ for any matrix norm $\|\bullet \|$. But this yields an optimal Lipschitz constant of 2, by rescaling the Hilbert Schmidt norm. I really need the constant to be $1$. Maybe it's not even true?

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up vote 15 down vote accepted

This is also true for the Riemannian metric described in the original question. I'm sure there is an elementary argument to prove it - it should certainly follow from the formula for the differential of the matrix exponential for example but here is a quick argument which involves no computations at all (assuming you know basic Riemannian geometry).

Since $SO(n)$ is compact and the metric in question is bi-invariant, the Riemannian exponential map coincides with the matrix exponential and sectional curvature is nonnegative. By Rauch comparison this right away implies that exponential map is $1$-Lipschitz on a neighborhood of $0$. To get the same statement on the whole $so(n)$ one can observe that $SO(n)$ is a symmetric space which means that the curvature tensor is parallel and Jacobi equations along any radial geodesic giving differential of the exponential map essentially reduce to the trivial form $J(t)=f(t)X(t)$ where $X$ is a unit parallel vector field and $f(t)$ satisfies $f''+kf=0, f(0)=0, f'(0)=1$ with $k\ge 0$. This of course is easily solved and gives $f(t)=\frac{1}{\sqrt{k}}\sin (\sqrt k t)$ (and $f(t)=t$ when $k=0$) which certainly satisfies $|f(t)|\le |t|$ which means that the exponential map is $1$-Lipschitz.

John Jiang, To be clear, the argument about the 1-Lipschitz part near $0$ in the above can be skipped because the Jacobi field computation reproduces it not only locally but globally. It doesn't work on noncompact Lie groups because most of them don't have bi-invariant metrics. Also, there is a nice closed formula for the differential of the Lie group exponential which I was referring to above but was too lazy to look up when I wrote my first response. You can find it in most books on Lie groups. See here for example for a completely elementary proof of it.

The formula is this

$d\exp_X(Y)=L_{\exp(X)}(\int_0^1Ad_{\exp(-tX)}Ydt)$. For matrix groups this becomes $d\exp_X(Y)=\exp(X)\int_0^1 \exp(-tX)Y\exp(tX)dt$. It easily implies the fact that you want since $X$ is skew-symmetric and $\exp(tX)$ is orthogonal.

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Hi Prof. Kapovitch, Thank you for your detailed solution. I tried to prove the map is 1-Lipschitz using matrix exponentiation but I get terms of the form $X^n Y + X^{n-1}Y X + \ldots + Y X^n$ for $d \exp_X (Y)$, when $X$ and $Y$ do not commute, which I cannot bound easily. I am not sure why the local 1-Lipschitz argument is needed. But it must be since otherwise the same argument using Jacobi field could be pushed to noncompact Lie groups. –  John Jiang Oct 8 '11 at 1:02
    
Thanks for the great reference! –  John Jiang Oct 8 '11 at 16:42
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This is a really nice answer. I'm very happy that you've joined MO. –  Deane Yang Oct 8 '11 at 17:05
    
My approach was to use the Daleckii-Krein theorem to write the Frechet derivative $De_X(Y)$ as a Hadamard product of two matrices, and then bound every unitarily invariant norm appropriately. But you already updated your answer using the nice formula above, saving me the trouble of typing an anwer :-) –  Suvrit Oct 8 '11 at 17:19
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For the operator norm, this map has Lipschitz constant equal to $1$. See Exercise 106 on my list of supplementary exercises to my book on Matrices (Springer-Verlag GTM 216). The link to this list is http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf

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thank you, very valuable link! –  Pietro Majer Oct 7 '11 at 9:14
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