Let $${\mathcal {C}}^*_{\epsilon ,\mathbf{S}^2}$$ C ϵ , S 2 ∗ denote the cover time of the two dimensional sphere by a Wiener sausage of radius $$\epsilon $$ ϵ . We prove that $$\begin{aligned} \sqrt{{\mathcal {C}}^{*}_{\epsilon ,\mathbf{S}^2} } -\sqrt{\frac{2A_{\mathbf{S}^2}}{\pi }}\left( \log \epsilon ^{-1}-\frac{1}{4}\log \log \epsilon ^{-1}\right) \end{aligned}$$ C ϵ , S 2 ∗ - 2 A S 2 π log ϵ - 1 - 1 4 log log ϵ - 1 is tight, where $$A_{\mathbf{S}^2}=4\pi $$ A S 2 = 4 π denotes the Riemannian area of $$\mathbf{S}^2$$ S 2 .

中文翻译：

---

二维球体覆盖时间的紧密度

令 $${\mathcal {C}}^*_{\epsilon ,\mathbf{S}^2}$$ C ϵ , S 2 ∗ 表示半径为 $$ 的维纳香肠对二维球体的覆盖时间\epsilon $$ ϵ 。我们证明 $$\begin{aligned} \sqrt{{\mathcal {C}}^{*}_{\epsilon ,\mathbf{S}^2} } -\sqrt{\frac{2A_{\mathbf{ S}^2}}{\pi }}\left( \log \epsilon ^{-1}-\frac{1}{4}\log \log \epsilon ^{-1}\right) \end{对齐}$$ C ϵ , S 2 ∗ - 2 AS 2 π log ϵ - 1 - 1 4 log log ϵ - 1 是紧的，其中 $$A_{\mathbf{S}^2}=4\pi $$ AS 2 = 4 π 表示 $$\mathbf{S}^2$$ S 2 的黎曼面积。