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I have questions referred to the second fundamental form on riemannian manifolds and mean curvature:

Is there a notion of mean curvature for submanifolds with arbitrary codimension?

I couldn't find something in the net. But I have herad about the notion of second fundamental form (=II) for arbitrary submanifolds. Maybe one can define it as the trace of II? On the other hand, as I can see, the second fundamental form II depends on the choice of a normal vector $\nu$. Is this true?

For instance, I'm faced with the following situation:

We consider an isometric immersion $f:M\rightarrow \mathbb{R}^m$, s.t. M is a submanifold of dimension n. Then we take the function $\nu:M\rightarrow \mathbb{R}$, which assigns to every point $p\in M$ the trace of its second fundamental form. Is the choice of second fundamental form unique yet?

Regards

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1 Answer 1

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Yes, the second fundamental form does depend on the choice of outward normal. For a general codimension $1$ submanifold, choose a normal vector (i.e. $\nabla \tau$ where $\tau$ is the distance function to the boundary) and then define the second fundamental form

$$S(X, Y) = g(\nabla_X \nabla \tau, Y)$$

The mean curvature is then the trace of this form. If the submanifold has codimension $k$, then there will be $k$ linearly independent normal vectors to your submanifold, for each of these you may define a second fundamental form as above and then take the trace, to obtain a version of mean curvature in the direction of each particular normal.

If you like, you can then combine all of these mean curvatures into a "mean curvature vector." More specifically, if $\nu_1, \ldots, \nu_k$ are your choices of unit normals to your submanifold, and $H_i$ is the mean curvature w.r.t $\nu_i$ as defined above, you may consider the vector $$\hat H = \sum H_i \nu_i$$

Then it follows that $g(\hat H, \nu_k) = H_k$ so that in this sense the mean curvature normal has the property that $g(\hat H, X)$ is "the mean curvature in the direction $X$".

Regarding the second part of your question, you need to properly interpret the phrase "trace of the second fundamental form" in the context I have described above.

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thank you for your help! –  lightningsnail Mar 2 '12 at 14:02
    
You're welcome :) –  treble Mar 2 '12 at 15:47

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