# Is there a notion of mean curvature for general submanifolds?

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

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 $X$".