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Consider the following SDE on $\mathbf R^d$: \begin{equation}\tag{*} dX_t^i = -\frac{d-1}{2}X_t^i dt + \sum_{j=1}^d(\delta^{ij}-X_t^iX_t^j)dW_t^j, \quad i=1,2,...,d, \end{equation} where $W = (W^1,W^2,...,W^d)$ is a $d$-dimensional standard Brownian motion. The paper 'Kinetic Brownian motion on Riemannian manifolds', Subsection 2.2 has the following claim,

Claim: The $\mathbf R^d$-valued process $X=(X^1,X^2,...,X^d)$ lies on the sphere $\mathbf S^{d-1}$.

But why?

I try to solve this problem in two options. But neither works.

OPTION 1. I rewrite the SDEs (*) to the Stratonovich form. To this purpose, let $\sigma_j^i(x) = \delta^{ij} - x^ix^j$. Then $\partial_k\sigma_j^i = -(\delta^{ik}x^j+\delta^{jk}x^i)$. Thus, \begin{equation} \begin{split} dX_t^i &= -\frac{d-1}{2}X_t^i dt + \sum_{j=1}^d \sigma^i_j(X_t)dW_t^j \\ &= -\frac{d-1}{2}X_t^i dt + \sum_{j=1}^d \sigma^i_j(X_t)\circ dW_t^j - \frac{1}{2} \sum_{j,k=1}^d \sigma^k_j\partial_k\sigma_j^i(X_t) dt\\ &= \left(1-\sum_{j=1}^d(X_t^j)^2\right) X_t^i dt + \sum_{j=1}^d(\delta^{ij}-X_t^iX_t^j)\circ dW_t^j. \end{split} \end{equation}

Let $Z_t=(t,W_t)$ and \begin{align} V_0(x) &= \left(1-\sum_{j=1}^d(x^j)^2\right) x^i \frac{\partial}{\partial x^i}, \\ V_j(x) &= (\delta^{ij}-x^i x^j)\frac{\partial}{\partial x^i}. \end{align} Then we can write $$dX_t = \sum_{j=0}^d V_j(X_t) \circ dZ_t^j.$$ If the vector fields $V_j, j=0,1,...,d$ are all tangent to $\mathbf S^{d-1}$, then the claim follows by Proposition 1.2.8 of Hsu's book 'Stochastic Analysis On Manifolds'. But \begin{align} V_0 \cdot x &= \sum_i V_0^i x^i = \left(1-\sum_j(x^j)^2\right) \left( \sum_i (x^i)^2 \right), \\ V_j \cdot x &= \sum_i V_j^i x^i = \left(1-\sum_i(x^i)^2\right) x^j, \end{align} they are not zero. Hence we cannot assert $V_j, j=0,1,...,d$ to be tangent to $\mathbf S^{d-1}$.

OPTION 2. For the function $r(x) = \|x\| = \sqrt{\sum_{i=1}^d (x^i)^2}$, we have $\partial_i r = \frac{x^i}{r}$, $\partial_i \partial_j r = \frac{\delta^{ij}}{r} - \frac{x^ix^j}{r^3}$. Let $R_t = \|X_t\|$. Using Ito's formula, \begin{equation} \begin{split} dR_t &= \sum_i \partial_i r(X_t) dX_t^i + \frac{1}{2}\sum_{i,j} \partial_i \partial_j r(X_t) d\langle X^i, X^j\rangle_t \\ &= -\frac{d-1}{2}R_t dt + \sum_{j=1}^d\left(\frac{1}{R_t}-R_t\right)X^j_tdW_t^j + \frac{d-1}{2R_t} dt \\ &= \frac{d-1}{2}\left(\frac{1}{R_t}-R_t\right) dt + \sum_{j=1}^d\left(\frac{1}{R_t}-R_t\right)X^j_tdW_t^j. \end{split} \end{equation} We cannot get a zero for $dR_t$, hence we cannot assert $R_t$ to be constant.

Can anyone figure out how to prove the claim? Or point out the mistake in my arguments? TIA...

PS: This is a crosspost from math.stackexchange.

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    $\begingroup$ Your $V_0$ and $V_j$ above (just before OPTION 2), seem to be zero when restricted to the unit sphere, so what's the problem? $\endgroup$
    – José Figueroa-O'Farrill
    May 3, 2019 at 16:40
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    $\begingroup$ Under OPTION 2, note that $R_t = 1$ is a.s. a fixed point of the SDE for $R_t$. $\endgroup$
    – Nawaf Bou-Rabee
    May 3, 2019 at 17:04
  • $\begingroup$ @JoséFigueroa-O'Farrill Thank you for your comments. When restricted to the unit sphere, $V_0$ vanishes and each $V_j$ is tangent to the unit sphere. $\endgroup$
    – Dreamer
    May 4, 2019 at 2:20

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