Genericity of Nondegenerate Free Boundary CMC Embeddings

Abstract

Let $\Sigma ^n$ and $M^{n+1}$ be smooth manifolds with smooth boundary. In this paper, following the techniques developed by White (Indiana Univ Math J 40:161–200, 1991) and Biliotti–Javaloyes–Piccione (Indiana Univ Math J, 1797–1830, 2009), we prove that, given a compact manifold with boundary $\Sigma ^n$ and a manifold with boundary $M^{n+1}$, for a generic set of Riemannian metrics on M every free boundary CMC embedding $\phi :\Sigma \rightarrow M$ is non-degenerate.

Appendices

Appendix A

In this appendix, we will give some definitions and important results of functional analysis, which were necessary during the development of the whole paper.

Definition 6.1

If $f:N\rightarrow M$ is a smooth map and $S\subset M$ is an embedded submanifold, we say that f is transverse to S if, for every $p\in f^{-1}(S)$, $T_{f(p)}M=T_{f(p)}S+df_p(T_pN)$.

Definition 6.2

The definition of transversality between a map $F:\mathcal {X}\rightarrow \mathcal {Y}$ and $\mathcal {Z} \subset \mathcal {Y}$ a smooth submanifold, where $\mathcal {X}$ and $\mathcal {Y}$ are Banach manifolds, is that presented in the Definition 6.1 but with the additional assumption that $dF_{\mathfrak {x}_0}^{-1}(T_{F(\mathfrak {x}_0)}\mathcal {Z})$ is a complemented subspace of $T_{\mathfrak {x}_0}\mathcal {X}$, i.e., there is a subspace $\mathcal {V}\subset T_{\mathfrak {x}_0}\mathcal {X}$ such that $T_{\mathfrak {x}_0}\mathcal {X}=dF_{\mathfrak {x}_0}^{-1}(T_{F(\mathfrak {x}_0)}\mathcal {Z})\oplus \mathcal {V}$.

Definition 6.3

If A is a bounded operator in a Hilbert space H and $\{e_i:i\in I\}$ is a orthonormal bases for H, is defined the Hilbert–Schmidt norm as $||A||_{HS}^2=Tr(A^*A)=\sum _{i\in I}||Ae_i||_H^2$, where $||\cdot ||_H$ is the norm of H.

Definition 6.4

The Hölder space $C^{j,\alpha }(\Omega )$, where $\Omega$ is an open subset of some Euclidean space and $j\ge 0$ an integer, consists of those functions on $\Omega$ having continuous derivatives up to order j and such that the jth partial derivatives are Hölder continuous with exponent $\alpha$, where $0<\alpha \le 1$. A real-valued function f on n-dimensional Euclidean space is Hölder continuous, when there are nonnegative real constants c, such that

$$\begin{aligned}|f(x)-f(y)|\le c||x-y||^{\alpha }\end{aligned}$$

Definition 6.5

A linear continuous operator $T:E\rightarrow F$ between normed spaces is Fredholm if $\mathrm {Ker}\ T$ is finite dimensional and $\mathrm {Im}\ T$ is close and finite codimensional, the index of T is $\mathrm {ind}\ T=\mathrm {dim}\ \mathrm {Ker}\ T - \mathrm {dim}\ \mathrm {coker}\ T$. A Fredholm map is a $C^1$ map $f: M \rightarrow N$, M and N being differentiable Banach manifolds, such that for each $x\in M$, the derivative $df_x: T_x(M) \rightarrow T_{f(x)}(N)$ is a Fredholm operator. The index of f is defined to be the index of $df_x$ for some x. The definition does not depend on x, see [12].

Theorem 6.1

[13, The Inverse Mapping Theorem, 5.2] Let E, F Banach spaces, U an open subset of E, and Let $f:U\rightarrow F$ a $C^p$-morphism with $p\ge 1$. Assume that for some point $x_0\in U$ The derivative $f'(x_0):E \rightarrow F$ is a toplinear isomorphism. Them f is a local $C^p$-isomorphism at $x_0$.

Theorem 6.2

[13, The Implicit Mapping Theorem, 5.9] Let U, V be open sets in Banach Spaces E, F respectively, and set $f:U\times V\rightarrow G$ be a $C^p$ mapping. Let $(a,b)\in U\times V$, and assume that $D_2f(a,b):\mathbf{F}\rightarrow G$ is a isomorphism. Let $f(a,b)=0$. Then there exist a continuous map $g:U_0\rightarrow V$ defined on an open neighborhood $U_0$ of a such that $g(a)=0$ and such that $f(x,g(x))=0$ for all $x\in U_0$. If $U_0$ is taken to be a sufficiently small ball, then g is uniquely determined, and is also of class $C^p$.

Theorem 6.3

(Local Form of the Submersions) Let X and Y be Banach spaces and let $f:X\rightarrow Y$ be a submersion in $x_0$, e.i, $df(x_0):T_{x_0}\rightarrow T_{f(x_0)}Y$ is surjective and $\text {Ker}(df(x_0))$ is complemented. Then, there are open sets $U\subset X$ and $V\subset \text {Ker}(df(x_0))$, with $x_0\in U$ and $0\in V$, and a diffeomorphism $\varphi : V\times W \rightarrow U$, $W\subset Y$ closed subspace, such that $f\circ \varphi (x,w)=w$, for all $(x,w)\in V\times W$

Appendix B

In this appendix, we prove some lemmas of linear algebra in spaces of infinite dimension that are necessary in the proof of the genericity of the Bumpy Metrics in Sect. 3

Lemma 7.1

Let V be a infinite dimensional vector space with inner product $\langle , \rangle$ and $W\subset V$ an finite dimensional subspace. Let us suppose that $V=W\oplus W^\bot$. Let $Z\subset V$ be a subspace, $Z\supset W^\bot$. Then

• 1. $(W^\bot )^\bot =W$.

• 2. $V=Z\oplus Z^\bot$

Proof

• 1. Clearly $W\subset (W^\bot )^\bot$. Now, if $w\in (W^\bot )^\bot$, we can write $w=w_1+w_2$, with $w_1\in W$ and $w_2\in W^\bot$. From the fact $W\subset (W^\bot )^\bot$, we have $w_1\in (W^\bot )^\bot$. Thus $w_2=w+w_1 \in (W^\bot )^\bot$ (recall that $(W^\bot )^\bot$ is a closed subspace of V). Whence $w_2\in W^\bot \cap (W^\bot )^\bot =\{0\}$, that is, $w=w_1\in W$. So, we conclude that $(W^\bot )^\bot \subset W$.

• 2. Note that the quotient spaces $Z/W^\bot$ and $V/W^\bot$ are both finite dimensional and $Z/W^\bot \subset V/W^\bot$. Given $v\in V$, there are unique $v_1\in W$ and $v_2\in W^\bot$ such that $v=v_1+v_2$. Therefore, the map $\pi _1:V/W^\bot \rightarrow W$, $\pi _1(v+W^\bot )=v_1$, is an isomorphism. Let $i:Z/W^\bot \hookrightarrow V/W^\bot$ be the inclusion map. Consider $L=\pi _1\circ i:Z/W^\bot \rightarrow W$ and let $\widetilde{Z}=\text {Im}L\subset W$. Thus, $\widetilde{Z}=\{ z-z_2 : z\in Z, z=z_1+z_2, z_1\in W,z_2\in W^\bot \}\subset Z$ and $W=\widetilde{Z}\oplus \widetilde{Z}^\bot$. Let $\{x_1,\ldots ,x_k \}\subset W$ be a basis for $\widetilde{Z}^\bot$. We claim that $x_1,\ldots ,x_k\in Z^\bot$. Let $z+W^\bot \in Z/W^\bot$, $z=z_1+z_2\in Z$, $z_1\in W$ and $z_2\in W^\bot$. We have

  $$\begin{aligned} \langle z, x_i\rangle&= \langle z, x_i\rangle + \langle W^\bot , x_i\rangle =\langle z+W^\bot , x_i\rangle \\&= \langle z_1+z_2+W^\bot , x_i\rangle = \langle z_1+W^\bot , x_i\rangle \\&= \langle z_1, x_i\rangle + \langle W^\bot , x_i\rangle =0, \end{aligned}$$

  since $z_1\in \widetilde{Z}$, $x_i\in \widetilde{Z}^\bot \subset W$, $i=1,\ldots ,k$.

Now we prove that $\text {Span}\{Z,x_1,\ldots ,x_k\}=V$. Let $v=v_1+v_2\in V$ be given, with $v_1\in W$ and $v_2\in W^\bot \subset Z$. There holds that $v_1= \widetilde{z}_1 + \sum _{i=1}^k a_ix_i$, $\widetilde{z}_1\in \widetilde{Z}$, where $a_i$, $i=1,\ldots ,k$, are scalars. Whence $v=(\widetilde{z}+v_2)+\sum _{i=1}^k a_ix_i$.

It remains to show that $\text {Span}\{x_1,\ldots ,x_k\}=Z^\bot$. In fact, it suffices to show that $\text {Span}\{x_1,\ldots ,x_k\}\supset Z^\bot$. Let

$$\begin{aligned}i:Z^\bot \hookrightarrow V=Z\oplus \text {Span}\{x_1,\ldots ,x_k\}\end{aligned}$$

be the inclusion map, and

$$\begin{aligned}\pi _2:Z\oplus \text {Span}\{x_1,\ldots ,x_k\}\rightarrow \text {Span}\{x_1,\ldots ,x_k\}\end{aligned}$$

the natural projection. We have that $\text {Ker}(\pi _2\circ i)=\{0\}$, so $\pi _2\circ i$ is injective, thus $\text {Dim}(Z^\bot )\le k$. $\square$

Lemma 7.2

Let V be a infinite-dimensional vector space with inner product $\langle , \rangle$ and $W\subset V$ an finite dimensional subspace. Suppose that $V=W\oplus W^\bot$. Let $Z\subset V$ be a subspace, $Z\supset W^\bot$, such that for all $w\in W{\setminus }\{0\}$, there exist $z\in Z$ with $\langle z,w\rangle \ne 0$. Then $Z=V$

Proof

We show that

$$\begin{aligned} (\forall w\in W{\setminus }\{0\})(\exists z\in Z)(\langle z,w\rangle \ne 0) \Leftrightarrow W\cap Z^\bot =\{0\}). \end{aligned}$$

(7.1)

First suppose that ($\forall w\in W{\setminus }\{0\})(\exists z\in Z)(\langle z,w\rangle \ne 0$. Let $w\in W\cap Z^\bot$ be given, then $\forall z \in Z$, $\langle z,w\rangle = 0$, so $w=0$, that is, $W\cap Z^\bot =\{0\}$.

Now suppose that $W\cap Z^\bot =\{0\}$ and let $w\in W{\setminus } \{0\}$, then $w\notin Z^\bot$, so there is $z\in Z$ such that $\langle z,w\rangle \ne 0$.

Since $Z\supset W^\bot$, by 1) of Lemma 7.1 we have $Z^\bot \subset (W^\bot )^\bot = W$. From the fact $W\cap Z^\bot =\{0\}$, we deduce that $Z^\bot =\{0\}$. Also by 2) of Lemma 7.1 we have $V=Z\oplus Z^\bot$. So, $V=Z$. $\square$

Next lemma were taken from [8, Lemma 2.2].

Lemma 7.3

Let $L : U \rightarrow V$ be a linear map between vector spaces, and let $S\subset V$ be a subspace of finite codimension. Then $L^{-1}(S)$ is finite co-dimensional in U, and

$$\begin{aligned} \text {Codim}_U (L^{-1}(S))=\text {Codim}_V (S)-\text {Codim}_V (\text {Im}(L)+S)\end{aligned}$$

.