In coordinates, the Laplace-Beltrami operator on a Riemannian manifold $(M,g)$ [can be written as](https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Gradient.2C_divergence.2C_Laplace.E2.80.93Beltrami_operator):
$$
\Delta\_g = g^{ij}\partial\_{ij} - g^{jk}\Gamma^\ell\_{jk}\partial\_\ell
$$
The second term:
$$
\mu^\ell = - g^{jk}\Gamma^\ell\_{jk}
$$
can be viewed as the "convection term" in the Riemannian heat equation or the drift term of the (Ito) stochastic differential equation defining Brownian motion on $(M,g)$.


**Question: What is the geometric meaning of this $\mu$?**


I would really like some intuition as to how the geometry generates this term (i.e. how to interpret it geometrically). Any other insights into intuitively understanding this term would be appreciated as well (e.g. other places where it appears).


(Note: this is a refinement of [this question](https://math.stackexchange.com/questions/2195289/why-does-brownian-motion-have-drift-on-riemannian-manifolds)).



Intuition for the Drift Term of the Laplace-Beltrami Operator
You really should think of $\Delta$ as an $L^2$ self-adjoint, elliptic operator in it's own right irrespective of coordinates. $-\Delta$ has positive spectrum, with countable eigenvalues accumulating at $\infty$ etc. It's the average of the Hessian $\nabla^2 f$ in the sense that contracting the $(0, 2)$ tensor $\nabla^2 f$ with the metric $g$ produces a $(1, 1)$ tensor that you can trace (which is an averaging operation). In this sense there is no drift term.


On the other hand, there is a way think of the extra term $\mu$ coming from the connection. For a smooth function $f : M \to \mathbb{R}$, from the smooth structure alone, you can always form the differential $Df \in \Gamma^{\infty}(T^{\ast} M)$ as smooth section of the cotangent bundle. Without a connection, you can think of $Df : TM \to \mathbb{R}$ as a map between smooth manifolds, differentiate again, and this gives
$$
D^2 f : TTM \to \mathbb{R}
$$


I used $D$ for the differential so as not to confuse this with the exterior derivative $d$ which would have $d^2 = 0$.


Now if you have a [connection](https://en.wikipedia.org/wiki/Ehresmann_connection), you can split $TTM$ as
$$
TTM \simeq VTM \oplus HTM
$$
into [vertical and horizontal sub-bundles](https://en.wikipedia.org/wiki/Vertical_and_horizontal_bundles). In this particular case both $VTM$ and $HTM$ are isomorphic to $TM$. $VTM$ is the kernel of the map $d\pi : TTM \to TM$ where $\pi: TM \to M$ is the bundle projection and this is isomorphic to $TM$. The connection allows you to "split" the map $d\pi : TTM \to TM$ obtaining an injective bundle morphism $TM \to TTM$ complementary to $VTM \simeq TM$ whose image I'll denoted $HTM$. The vertical bundle consists of elements of $TTM$ that are tangent to the fibres of $TM$ while the horizontal bundles consists of elements of $TTM$ that are tangent to the base $M$.


To help clarify, for a general vector bundle $\pi : E \to M$, the kernel $VE$ of $d\pi : TE \to TM$ is isomorphic to $E$ and a connection gives a splitting $TM \to TE$ whose image is denoted $HE$ and such that $TE \simeq VE \oplus HE \simeq E \oplus TM$.


Now, what this all has to do with "drift" is that under the identification $TTM \simeq TM \oplus TM$, the $g^{ij} \partial\_{ij}$ term comes from $VE$ - it's the term tangent to the fibre and corresponds to $\partial\_{ij} f = D^2 f$. The other term - the one with the connection coefficients $\Gamma$ corresponds to the part that is tangent to the base $M$.


In other words, the $g^{ij} \partial\_{ij}$ part arises from differentiating while moving along the fibres of $TM$ and the $\Gamma$ part arises from moving along the base $M$.


**Thus the drift is measuring the change in the fibres of the bundle $TM$ as measured by the connection as the basepoint $x \in M$ varies.**


