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Let the operator $L$ be given by $Lf(x):=\nabla\cdot (A\nabla f(x))$, where $f:\mathbb{R}^d\rightarrow \mathbb{R}$ belongs to a suitable class of functions $\mathcal{A}$. The carre du champ operator $\Gamma$ on $\mathcal{A}\times\mathcal{A}$ is then given by $\Gamma(f,f)(x)=(\nabla f(x))^TA(x)\nabla f(x)$. For suitable matrix $A$, $\Gamma(f,f)^{1/2}$ is nothing else then the length of the gradient $\nabla_Mf(x)$ on the Riemannian manifold $(\mathbb{R}^d,A^{-1})$, where $A^{-1}$ denotes the metric on $\mathbb{R}^d$ induced by $x\mapsto A(x)^{-1}$. On this manifold, the distance between two points is given by $d_M(x,y)=\inf_{\gamma\in\Omega(x,y)}\int_0^1\sqrt{\gamma'(t)^TA(\gamma(t))^{-1}\gamma'(t)}dt$, where the inf runs over all Lipschitz paths $\gamma$ such that $\gamma(0)=x$ and $\gamma(1)=y.$

The operator $L$ induces also another metric on $\mathbb{R}^d$. A curve $\gamma$ is called subunit for $L$ if $\vert \gamma'(t)\cdot\xi\vert^2\leq \xi^T A(\gamma(t))\xi$ for all $\xi\in\mathbb{R}^d$. Let $C_{sub}(x,y)$ denote the set of all subunit curves for $L$ that connect $x$ and $y$. We then write $d_L(x,y)=\inf_{\gamma\in C_{sub}(x,y)}\{b:\gamma(0)=x,\gamma(b)=y\}$. Note that the inf is taken over all subunit curves $\gamma:[0,b]\rightarrow \mathbb{R}^d$ which connect $x$ and $y$ at times $t=0$ and $t=b$ respectively. $d_L(x,y)$ is then the smallest such $b$.

Are the metrics $d_M$ and $d_L$ the same or at least equivalent?

I am not very versed in differential geometry so I might be missing something obvious. Bakry and Ledoux use the above ideas (first paragraph) to motivate the name carre du champ. They then use the intrinsic metric $d_L$ (albeit another definition equivalent to this one) associated to $L$ without making explicit any links betweens these ideas. Since this field is also called geometry of markov diffusion operators I am curious to find more motivation regarding the geometric content for these ideas.

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  • $\begingroup$ What is $b$ in the definition of $d_{L}$? $\endgroup$ Commented Jan 25, 2016 at 12:33
  • $\begingroup$ Thanks for your comment. I edited the main post for more clarity. $\endgroup$ Commented Jan 25, 2016 at 12:49

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Yes, $d_M=d_L$. I discuss this distance on my blog: The distance associated to Laplace-Beltrami operators

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  • $\begingroup$ Thanks for your answer. Would the proof go like this for "$\leq$" 1) Let $\gamma \in C_{sub}(x,y), \gamma:[0,T]\rightarrow \mathbb{R}^d$, then $d_M(x,y)\leq T \int_0^1 \sqrt{\gamma'(Tt)^TA(\gamma(Tt))^{-1}\gamma'(tT)}dt\leq T$ because of $(\ast)$, see below. This holds for all $\gamma\in C_{sub}(x,y)$, we see that $d_M(x,y)\leq d_L(x,y)$. $\endgroup$ Commented Feb 5, 2016 at 22:05
  • $\begingroup$ 2) "$\geq$" On the other hand, let $\gamma\in \Omega(x,y)$. We parametrise $\gamma$ to $\tilde{\gamma}$ which runs in unit speed with respect to the metric induced by $A^{-1}$, so that $T=\int_0^T\sqrt{\tilde\gamma(s)^TA^{-1}(\tilde{\gamma}(s))\tilde{\gamma}(s)}ds$ is the length of the curve. By $(\ast)$ $\tilde{\gamma}$ is subunit, so that $d_L(x,y)\leq T$ for all $T$ which denote the length of any curve in $\Omega(x,y)$. Because this holds for all such curves, we see that $d_L(x,y)\leq d_M(x,y)$. $\endgroup$ Commented Feb 5, 2016 at 22:06
  • $\begingroup$ $(\ast)$. $\vert \gamma'(t)^T\xi\vert^2\leq \xi^TA(\gamma(t))\xi \quad \forall \xi \Longleftrightarrow \gamma'(t)^TA(\gamma(t))^{-1}\gamma'(t)\leq 1$. In $(\ast)$, I think I understand $\rightarrow$ implication but not the one the other way round. So I am not even sure this holds in such generality. $\endgroup$ Commented Feb 5, 2016 at 22:06

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