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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.

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 Brian Hall's An Elementary Introduction to Groups and Representations 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.

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 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.

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.

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 Jian, 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 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.

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 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.