Variation of compact support and minimal surfaces

I was reading the classical book of Colding and Minicozzi II "A Course in Minimal Surfaces". In page 6, they say the following

Let $$F:\Sigma \times (-\varepsilon,\varepsilon) \to M$$ be a variation of $$\Sigma$$ with compact support and fixed boundary. That is, $$F=Id$$ outside a compact set, $$F(x,0)=x,$$ and for all $$x \in \partial \Sigma$$, $$F(x,t)=x.$$

Here, $$\Sigma$$ is a $$k-$$submanifold of $$M$$, and $$M$$ is a compact (Riemannian) manifold. Well here we have three conditions:

1. $$F=Id$$ outside a compact. What I understand of this condition is that we only make variations to some compact set that it's inside of $$\Sigma$$, and then outside of the compact, $$\Sigma$$ don't change.
2. $$F(x,0)=x$$, that is, at time zero $$\Sigma$$ doesn't change,
3. $$F(x,t)=x$$, $$\forall x \in \partial \Sigma$$. This tell us that we don't move the boundary of $$\Sigma$$.

My question here is about the condition 1. and 3. When they say that $$F=Id$$ outside the compact set (here I asumming that the compact set is in $$\Sigma$$), I understand that $$F(x,t)=x$$ for all $$x \in \Sigma$$ outside the compact set. So in particular, this will imply the condition 3. At least if we take a compact set which don't intersect $$\partial \Sigma$$. So the third condition it's only to cover this last example, or appears for other reasons? Is it my reasoning right?.

Thanks!

As you say if that compact set contains all of $$\partial \Sigma$$, then 1. implies 3.
It is reason to state 1. and 3. separately. 3. is normally called the Dirichlet boundary condition: one fixes the boundary in $$\mathbb R^3$$ and consider all surfaces with that boundary.
Indeed in their book, $$M, \Sigma$$ can be non-compact (For example, they study minimal surfaces in $$\mathbb R^n$$, when $$\Sigma$$ has no boundary it has to be non-compact).
• I understand that 3. is some kind of boundary condition. My problem is other, but I think that now I understand. For example, if we take our compact set as $\Sigma$, then condition 1. doesn't imply that tha boundary of $\Sigma$ is fixed by the variation. So, as you say, the condition 3 is imposed separately from 1. Thanks! Jun 26, 2019 at 3:07