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As explained here, the infinitesimal generator of a 1D Brownian motion is $\frac{1}{2}\Delta$. As discussed here, for the Brownian motion on circle we can write

$$Y_1=\cos(B) \\ Y_2= \sin(B)$$

and its Ito formula is

$$dY_1=-\frac12 Y_1 \, dt-Y_2 \, dB, \\ dY_2=-\frac12 Y_2 \, dt+Y_1 \, dB,$$

Then, to find its generator (as discussed here) we can write

$$\mathcal{A}=\frac12\left(-y_1\partial_{y_1}-y_2\partial_{y_2}+y_2^2\partial_{y_1}^2+y_1^2\partial_{y_2}^2 \right)=\frac12\partial_\theta^ 2.$$

I wonder what would be the three steps above for the Brownian motion on $S_2$ sphere? Unlike above where we hade only ^one Brownian^, i.e., $B$, should we start with

$$Y=( \cos(B_1)\sin(B_2), \sin(B_1)\sin(B_2), \cos(B_1))$$

where $B_1$ and $B_2$ are two independent Brownians?

  • Are the statements above correct?
  • What is the intuitive explanation of the ^two Brownians^?
  • How to calculate the SDE and also the infinitesimal generator from $Y$?

Thanks in advance.

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3 Answers 3

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The generator of Brownian motion on $S^2$ (with the round metric) is $\frac12\Delta$, where $\Delta$ is the Laplacian on $S^2$, in spherical coordinates $$\Delta = \frac{1}{\sin\theta}\partial_{\theta}(\sin\theta\cdot\partial_{\theta})+\frac{1}{\sin^2\theta}\partial_{\varphi}^2.$$ Let $$ X_t=\sin\theta_t\cos\varphi_t,\\ Y_t=\sin\theta_t\sin\varphi_t,\\ Z_t=\cos\theta_t.$$ Now you suggest that if $B_t^{(i)}$, $i=1,2$ are two independent Brownian motions, then $$ d\theta_t = dB_t^{(1)},\\ d\varphi_t = dB_t^{(2)},\tag1$$ defines a Brownian motion on $S^2$. We have $$dZ_t=d\cos\theta_t=-\sin\theta_t\cdot dB_t^{(1)}-\frac12\cos\theta_t\cdot dt.$$ But since $\Delta\cos(\theta)=-2\cos\theta$, we have $$d\cos\theta_t-\frac12\Delta\cos\theta_t=-\frac{3}{2}\cos\theta_t\cdot dt-\sin\theta_t\cdot dB_t^{(1)},$$ Thus $\cos\theta_t-\int_0^t\frac12\Delta\cos\theta_sds$ is not a local martingale. Therefore $\frac12\Delta$ is not the generator of your process (1), and therefore (1) does not define a Brownian motion on $S^2$.


There are many different ways to construct Brownian motion on the sphere. One of them works in Stratonovich form and reads $$ d\mathbf{X}_t = \mathbf{X}_t\otimes d\mathbf{B}_t, \tag2$$ where $\otimes$ denotes a Stratonovich cross product and $\mathbf{B}_t$ is a 3d Brownian motion. In other words, $$ dX_t = Y_t\circ dB^{(3)}_t - Z_t\circ dB^{(2)}_t,\\ dY_t = Z_t\circ dB^{(1)}_t - X_t\circ dB^{(3)}_t, \\ dZ_t = X_t\circ dB^{(2)}_t - Y_t\circ dB^{(1)}_t.$$ First of all we can check that by the Stratonovich chain rule $$ d(X_t^2+Y_t^2+Z_t^2) = 2(X_t\circ dX_t+Y_t\circ dY_t+Z_t\circ dZ_t) = ... = 0,$$ hence $(X_t,Y_t,Z_t)$ is on $S^2$ for all $t\geq 0$ iff $(X_0,Y_0,Z_0)$ is on $S^2$. Then by the Stratonovich chain rule we obtain $$ d\mathbf{X}_t = \frac{\partial \mathbf{x}}{\partial\theta}\circ d\theta_t + \frac{\partial \mathbf{x}}{\partial\varphi}\circ d\varphi_t,$$ By expanding everything and matching with the expressions above, we can solve for $d\theta_t$ and $d\varphi_t$ in terms of the Brownian motions: \begin{align} d\theta_t&=\sin\varphi_t\circ dB_t^{(1)}-\cos\varphi_t\circ dB_t^{(2)},\\ d\varphi_t&=\cot\theta_t\left(\cos\varphi_t\circ dB_t^{(1)}+\sin\varphi_t\circ dB_t^{(2)}\right)-dB_t^{(3)}. \end{align} We get this back into Itô form (please check), which leads to a drift term in $\theta_t$ \begin{align} d\theta_t&=\frac12\cot\theta_t dt+\sin\varphi_t dB_t^{(1)}-\cos\varphi_t dB_t^{(2)},\\ d\varphi_t&=\cot\theta_t\left(\cos\varphi_t dB_t^{(1)}+\sin\varphi_t dB_t^{(2)}\right)-dB_t^{(3)}. \end{align} So now take a $C^2$ function $f(\theta_t,\varphi_t)$ and use Itô's lemma to check that $$df(\theta_t,\varphi_t) = \frac12\Delta f(\theta_t,\varphi_t) dt + ... dB_t^{(1)}+ ... dB_t^{(2)}+ ... dB_t^{(3)}$$ (only the $dt$ term has to be calculated). This shows that the generator of the process (2) is indeed $\frac12\Delta$.

Addendum: By setting \begin{align} dB_t^{\theta}&=\sin\varphi_tdB^{(1)}_t-\cos\varphi_tdB^{(2)}_t,\\ dB^{\phi}_t&=\cos\theta_t(\cos\varphi_tdB^{(1)}_t+\sin\varphi_tdB^{(2)}_t)-\sin\theta_tdB^{(3)}_t \end{align} and checking that these are two independent Brownian motions, one can rewrite the process as \begin{align} d\theta_t&=\frac12\cot\theta_t dt+dB_t^{\theta},\\ d\varphi_t&=\frac{1}{\sin\theta_t}dB^{\phi}_t. \end{align}

There is a third Brownian motion $dB^{N}_t=\mathbf{X}_t\cdot d\mathbf{B}_t$ that is normal to the sphere and therefore gets cancelled out.

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    $\begingroup$ You can read off the generator from the SDEs in spherical coordinates, or compute the dt term in Ito’s lemma and compare it as I did. The overall logic is: we define what a BM is in terms of the generator (this is something we don’t derive, we have to start somewhere). Then we check that the SDEs you suggested in your question don’t have that generator by using Ito’s lemma on one specific function. Then we guess a different set of SDEs and check that they have the right generator by checking that the dt term in Ito’s lemma comes out as expected for every test function. $\endgroup$
    – S.Surace
    Jun 21, 2020 at 5:26
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    $\begingroup$ A more direct approach is to rearrange the spherical Laplacian in terms of first order and second order terms and then read off the SDE coefficients for $\theta$ and $\varphi$ based on the general formula for the generator of a diffusion process. $\endgroup$
    – S.Surace
    Jun 21, 2020 at 5:38
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    $\begingroup$ If we don‘t know what BM is supposed to be, we don‘t have any means to check whether a system of SDEs, expressed in some arbitrary set of local coordinates, is BM. How do you define BM on Riemannian manifolds? You could also take as a (perhaps less elementary) definition that BM is a martingale wrt. to the Levi-Civita connection with Riemannian quadratic variation equal to $t$. But you have to start somewhere and know what you are looking for before you can check against that in local coordinates. Different authors define Brownian motion differently. $\endgroup$
    – S.Surace
    Jun 22, 2020 at 6:45
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    $\begingroup$ See math.kyoto-u.ac.jp/probability/sympo/Kyushu.pdf $\endgroup$
    – S.Surace
    Jun 22, 2020 at 6:46
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    $\begingroup$ You may like section 1.4 of Hsu's lecture notes linked above, which discusses the construction of BM by embedding in a similar way as this answer. $\endgroup$
    – S.Surace
    Jun 22, 2020 at 17:34
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As an alternative approach for the derivation of the advertised conclusion (in my opinion, this is a more elementary approach). You can take a look at Example 8.5.8 in Oksendal's classical textbook on SDEs, in which the author used a random time change argument to drive the SDE satisfied by a Brownian motion on the unit sphere in $\mathbb{R}^n$ (with $n \geq 3$). This SDE is written in the cartesian coordinates. However, when we consider the special case $n = 3$, a simple Ito-type computations enable us to transform the SDE in cartesian coordinate to the SDE in spherical coordinate $(\theta,\phi)$. I attached the detailed computations below, I hope you can follow and understand.

enter image description here

enter image description here

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  • $\begingroup$ why the downvote? I feel this is a detailed answer $\endgroup$
    – Fei Cao
    Jul 27, 2021 at 5:41
  • $\begingroup$ Very helpful answer. Can you elaborate more on the relation of the details you provided with the proof of example 8.5.8 of Oksendal? There he starts with applying a function to BM. Also there is important time change in the proof. I cannot however see information in this regard in your answer. Can you please extend and update your answer? $\endgroup$
    – Kevin
    Dec 1, 2023 at 15:51
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Brownian motion on the unit sphere will be given by

$$X_t=(\sin(\theta_t)\cos(\phi_t),\sin(\theta_t)\sin(\phi_t),\cos(\theta_t))$$

where $(\theta_t,\phi_t)$ solves the system of SDEs given by

$$d\theta_t=dB_t^{\theta}+\frac{1}{2}\cot(\theta_t)dt, \qquad d\phi_t=\frac{1}{\sin(\theta_t)}dB_t^{\phi}.$$ Here $B^{\theta}$ and $B^{\phi}$ are independent Brownian motions.

See Why does Brownian motion have drift on Riemannian Manifolds? for the general case.

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    $\begingroup$ This very nice representation can be obtained from the construction in my answer by setting $dB_t^{\theta}=\sin\varphi_tdB^{(1)}_t-\cos\varphi_tdB^{(2)}_t$ and $dB^{\phi}_t=\cos\theta_t(\cos\varphi_tdB^{(1)}_t+\sin\varphi_tdB^{(2)}_t)-\sin\theta_tdB^{(3)}_t$ and checking that these are two independent Brownian motions. The third direction basically gets cancelled because it is orthogonal to the sphere. $\endgroup$
    – S.Surace
    Jun 19, 2020 at 16:26

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