I saw in a talk that a surface gradient of $f:M \to \mathbb{R}$ where $M$ is a hypersurface in $\mathbb{R}^n$ defined as $$\nabla_M f = \nabla f - (\nabla f \cdot N)N$$ where $N$ is the unit normal vector on $M$ and $\nabla$ is the ordinary gradient.

I just started learning about the connection/covariant derivative on a manifold and am wondering about the link. Is the surface gradient as defined above just a choice of a particular connection? Does it have anything to do with the Levi-Civita connection?

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    $\begingroup$ This question has been asked before it seems: math.stackexchange.com/q/200223/26695 $\endgroup$ – begeistzwerst Nov 26 '12 at 16:27
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    $\begingroup$ Should this really be closed? The question @begeistzwerst links to does not contain the answers to everything AC21 has asked about. $\endgroup$ – soup Nov 26 '12 at 17:08
  • $\begingroup$ Actually I did not intend to close it. I only wanted to draw attention to the other question. I am not sure what's best practice in this situation. $\endgroup$ – begeistzwerst Nov 26 '12 at 17:44

This doesn't really have anything to do with connections. Notice here that the $\nabla$ does not refer to a connection but instead denotes the gradient of a function, which is a vector field, and which can be defined for a manifold $M$ with metric $g$ by $$ g(\nabla f, X) = df(X) $$ where $X$ is a vector field on $M$. Unfortunately the symbol $\nabla$ is also used for connections and in that context $\nabla f$ is actually equal to $df$ for any connection $\nabla$ (basically by definition).

The usual metric on $\mathbb R^n$ restricts to a metric on the hypersurface $M$. Then the hypersurface gradient of $f$ is just the gradient of $f$ with respect to this metric (it'd be a good exercise to run through the definitions and check this); notice that in forming it you're taking the entire gradient of $f$ and then subtracting off the normal part, getting something tangent to $M$.

  • $\begingroup$ What about regularity of $\nabla_Mf$ when $f\in W^{1,p}(M)$? $\endgroup$ – Asatur Khurshudyan Sep 23 '16 at 14:54

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