On the Affine Gauss Maps of Submanifolds of Euclidean Space

Abstract

It is well known that the space of oriented lines of Euclidean space has a natural symplectic structure. Moreover, given an immersed, oriented hypersurface $\mathcal{S}$ the set of oriented lines that cross $\mathcal{S}$ orthogonally is a Lagrangian submanifold. Conversely, if $\overline{\mathcal{S}}$ an n-dimensional family of oriented lines is Lagrangian, there exists, locally, a 1-parameter family of immersed, oriented, parallel hypersurfaces $\mathcal{S}_t$ whose tangent spaces cross orthogonally the lines of $\overline{\mathcal{S}}.$ The purpose of this paper is to generalize these facts to higher dimension: to any point x of a submanifold $\mathcal{S}$ of ${\mathbb {R}^{}} ^m$ of dimension n and co-dimension $k=m-n,$ we may associate the affine k-space normal to $\mathcal{S}$ at x. Conversely, given an n-dimensional family $\overline{\mathcal{S}}$ of affine k-spaces of ${\mathbb {R}^{}} ^m$, we provide certain conditions granting the local existence of a family of n-dimensional submanifolds $\mathcal{S}$ which cross orthogonally the affine k-spaces of $\overline{\mathcal{S}}$. We also define a curvature tensor for a general family of affine spaces of ${\mathbb {R}^{}} ^m$ which generalizes the curvature of a submanifold, and, in the case of a 2-dimensional family of 2-planes in ${\mathbb {R}^{}} ^4$, show that it satisfies a generalized Gauss–Bonnet formula.

Appendices

Appendix A: The Curvature of the Tautological Bundles

We prove here Theorem 4. We only deal with the case of the tautological bundle $\tau _T,$ since the proof for the bundle $\tau _N$ is very similar. We consider $u,v\in \Gamma (T{\mathcal {Q}}_o)$ and $\xi \in \Gamma (\tau _T)$ such that, at $p_o,$

$$\begin{aligned} dv(u)-du(v)=0~~~~\text{ and }~~~~\nabla ^{T}\xi =0. \end{aligned}$$

(41)

Since, in a neighborhood of $p_o,$

$$\begin{aligned} \nabla _u\xi =d\xi (u)-{d\xi (u)}^N=d\xi (u)-u(\xi ), \end{aligned}$$

(42)

where in this last expression $u\in T_{p}{\mathcal {Q}}_o$ is regarded as an element of $L({\tau _T}_p,{\tau _N}_p)$, we get

$$\begin{aligned} R^{T}(u,v)\xi= & {} \nabla ^{T}_u\nabla ^{T}_v\xi -\nabla ^{T}_v\nabla ^{T}_u\xi \nonumber \\= & {} \nabla ^{T}_u(d\xi (v)-v(\xi ))-\nabla _v(d\xi (u)-u(\xi ))\nonumber \\= & {} \big ( d(u(\xi ))(v)-d(v(\xi ))(u)\big )^T, \end{aligned}$$

(43)

where we used that $d\circ d\ \xi =0.$ The superscript T means that we take the component of the vector belonging to $p_o.$ We will need the following

Lemma 5

For all $u\in T_{p}{\mathcal {Q}}_o$ and $\xi \in p,$

$$\begin{aligned} u(\xi )=-\frac{\epsilon _n}{2}\big (\xi \cdot p\cdot u+u\cdot p\cdot \xi \big ) \end{aligned}$$

where in the left hand side u is considered as a linear map $p\rightarrow p^\perp$ and in the right hand side $\xi ,p,u$ are viewed as elements of the Clifford algebra $Cl({\mathbb {R}^{}} ^m)$; the dot ”$\cdot$” stands for the Clifford product in $Cl({\mathbb {R}^{}} ^m).$

Proof of Lemma 5

We write

$$\begin{aligned} u=\sum _{i=1}^ne_1\wedge \cdots \wedge u(e_i)\wedge \cdots \wedge e_n=\sum _{i=1}^ne_1\cdot \cdots \cdot u(e_i)\cdot \cdots \cdot e_n, \end{aligned}$$

and we compute

$$\begin{aligned} \xi \cdot p\cdot u= & {} \sum _i\xi \cdot e_1\cdot \cdots \cdot e_n \cdot e_1\cdot \cdots \cdot u(e_i)\cdot \cdots \cdot e_n\\= & {} \sum _i\xi \cdot e_1\cdot \cdots \cdot e_n \cdot e_1\cdot \cdots \cdot e_i\cdot \cdots \cdot e_n\cdot (-e_i)\cdot u(e_i)\\= & {} \sum _i\epsilon _n\ \xi \cdot e_i\cdot u(e_i) \end{aligned}$$

since $(e_1\cdot \cdots \cdot e_n)^2=-\epsilon _n.$ Similarly, we compute

$$\begin{aligned} u\cdot p\cdot \xi =\sum _i\epsilon _n\ u(e_i)\cdot e_i\cdot \xi , \end{aligned}$$

and thus get

$$\begin{aligned} \xi \cdot p\cdot u+u\cdot p\cdot \xi= & {} \epsilon _n\sum _i\left( \xi \cdot e_i\cdot u(e_i)+u(e_i)\cdot e_i\cdot \xi \right) \\= & {} \epsilon _n\sum _i\left( \xi _ie_i\cdot e_i\cdot u(e_i)+u(e_i)\cdot e_i\cdot \xi _ie_i\right) \\= & {} -2 \epsilon _n u(\xi ), \end{aligned}$$

which is the required formula. $\square$

Using Lemma 5, we get

$$\begin{aligned} d(u(\xi ))(v)= & {} -\frac{\epsilon _n}{2}\Big (d\xi (v)\cdot p_o\cdot u+\xi \cdot v\cdot u+\xi \cdot p_o\cdot du(v)\\&\quad +\, \, du(v)\cdot p_o\cdot \xi +u\cdot v\cdot \xi +u\cdot p_o\cdot d\xi (v)\Big ). \end{aligned}$$

Moreover, by (42) and since $\nabla \xi =0$ at $p_o,$ we have

$$\begin{aligned} d\xi (v)=v(\xi )=-\frac{\epsilon _n}{2}\big (\xi \cdot p_o\cdot v+v\cdot p_o\cdot \xi \big ), \end{aligned}$$

and we thus obtain an expression of $d(u(\xi ))(v)$ in terms of the Clifford products of $u,v,\xi ,p_o$ and du. Switching u and v we get a similar formula for $d(u(\xi ))(v).$ Plugging these two formulas in (43) and using the first identity in (41) we may then easily get

$$\begin{aligned} R(u,v)\xi =\epsilon _n\big ([\xi ,[u,v]]\big )^T \end{aligned}$$

(using moreover that $u\cdot p_o=-p_o\cdot u,$$v\cdot p_o=-p_o\cdot v$ and $p_o\cdot p_o=-\epsilon _n$); this gives the result since $[\xi ,[u,v]]$ is in fact a vector belonging to $p_o$: indeed, it is easy to check that [u, v] belongs to $\Lambda ^2p_o\oplus \Lambda ^2p_o^\perp$ if u and v are tangent to ${\mathcal {Q}}_o$ at $p_o,$ and then that $[\xi ,[u,v]]$ belongs to $p_o$ if $\xi$ belongs to $p_o.$

Finally, we provide another useful formula for the curvature of the tautological bundle $\tau _N\rightarrow {\mathcal {Q}}_o$:

Lemma 6

Let $\omega _o\in \Omega ^2({\mathcal {Q}}_o,End(\tau _N))$ be the curvature of $\tau _N\rightarrow {\mathcal {Q}}_o.$ Then, for $u,v\in T_{p}{\mathcal {Q}}_o,$

$$\begin{aligned} \omega _o(u,v)=u\circ v^*-v\circ u^*, \end{aligned}$$

where $u,v:p\rightarrow p^\perp$ are regarded as linear maps and $u^*,v^*:p^{\perp }\rightarrow p$ are their adjoint.

Proof

We first note that the adjoint map $u^*:p^{\perp }\rightarrow p$ is explicitly given in terms of the Clifford product by the formula

$$\begin{aligned} u^*(\xi )=-\frac{\epsilon _n}{2}\left( \xi \cdot p\cdot u+u\cdot p\cdot \xi \right) , \, \, \, \forall \xi \in p^{\perp }; \end{aligned}$$

This is the same formula than the formula for $u:p\rightarrow p^{\perp }$ given in Lemma 5. A straightforward computation then gives

$$\begin{aligned} (u\circ v^*-v\circ u^*)(\xi )= & {} \frac{1}{4}\left\{ (\xi \cdot p\cdot v+v\cdot p\cdot \xi )\cdot p\cdot u+ u\cdot p\cdot (\xi \cdot p\cdot v+v\cdot p\cdot \xi )\right. \\&\left. -(\xi \cdot p\cdot u+u\cdot p\cdot \xi )\cdot p\cdot v-v\cdot p\cdot (\xi \cdot p\cdot u+u\cdot p\cdot \xi )\right\} \end{aligned}$$

which simplifies to

$$\begin{aligned} (u\circ v^*-v\circ u^*)(\xi )= & {} \frac{1}{4}\left\{ \xi \cdot p\cdot v\cdot p\cdot u+ u\cdot p\cdot v\cdot p\cdot \xi -\xi \cdot p\cdot u\cdot p\cdot v\right. \\&\left. -v\cdot p\cdot u\cdot p\cdot \xi \right\} . \end{aligned}$$

Now, we have $p^2=-\epsilon _n$ and

$$\begin{aligned} p\cdot u+u\cdot p= 0\\ p\cdot v+v\cdot p = 0 \end{aligned}$$

$\forall u,v\in T_p{\mathcal {Q}}_o$, and we get

$$\begin{aligned} (u\circ v^*-v\circ u^*)(\xi )= & {} \frac{\epsilon _n}{4}\left\{ \xi \cdot (u\cdot v-v\cdot u)-(u\cdot v-v\cdot u)\cdot \xi \right\} \\= & {} \epsilon _n[\xi ,[u,v]]. \end{aligned}$$

This is the expression of the curvature of the bundle $\tau _N\rightarrow {\mathcal {Q}}_o$ given in Theorem 4. $\square$

We immediately deduce the following

Corollary 2

For a smooth map $\overline{\varphi }_o:\mathcal{M}\rightarrow {\mathcal {Q}}_o,$ we have

$$\begin{aligned} \overline{\varphi }_o^*\omega _o(X,Y)=d\overline{\varphi }_o(X)\circ d\overline{\varphi }_o(Y)^*-d\overline{\varphi }_o(Y)\circ d\overline{\varphi }_o(X)^*, \end{aligned}$$

$\forall X,Y\in T\mathcal{M},$ where $\overline{\varphi }_o^*\omega _o(X,Y)$ belonging to $\Lambda ^2\overline{\varphi }_o^{\perp }$ is regarded as a map $\overline{\varphi }_o^{\perp }\rightarrow \overline{\varphi }_o^{\perp }.$

Appendix B: The Abstract Shape Operator B Identifies to the Differential of the Gauss Map

The aim is to link the tensor B to the differential of the Gauss map. The next lemma first shows that B is the pull-back of a natural tensor on the tautological bundles on the Grassmannian ${\mathcal {Q}}_o$:

Lemma 7

Let us define

$$\begin{aligned} B': \tau _N\rightarrow & {} T^*{\mathcal {Q}}_o\otimes \tau _T\\ \xi\mapsto & {} Y\mapsto -(d\xi (Y))^T, \end{aligned}$$

where, in the right hand side, $\xi$ is exfvtended to a local section of $\tau _N\rightarrow {\mathcal {Q}}_o.$ We have:

• (i) $B'$ is a tensor; precisely,

  $$\begin{aligned}B'(\xi )(Y)=Y^*(\xi ),\end{aligned}$$

  for all $\xi \in {\tau _N}_{p_o}\simeq p_o^{\perp }$ and $Y\in T_{p_o}{\mathcal {Q}}_o\simeq L(p_o,p_o^{\perp }),$ where Y is considered as a linear map $p_o\rightarrow p_o^{\perp }$ and $Y^*: p_o^{\perp }\rightarrow p_o$ is its adjoint.

• (ii) B is the pull-back of $B'$ by the Gauss map:

  $$\begin{aligned}B={\overline{\varphi }_o}^*B'.\end{aligned}$$

Proof of (i)

Let $p_o\in {\mathcal {Q}}_o$ be an oriented n-plane, and $\xi \in p_o^\perp$ a vector normal to $p_o,$ extended to a local section of $\tau _N\rightarrow {\mathcal {Q}}_o.$ Let us consider $* p_o,$ the multi-vector belonging to $\Lambda ^k{\mathbb {R}^{}} ^m$ which represents the linear space $p_o^\perp ,$ with its natural orientation. We first observe that

$$\begin{aligned} (d\xi (Y))^T=i_{*p_o}\left( *p_o\wedge d\xi (Y)\right) , \, \, \, \forall Y\in T_{p_o}{\mathcal {Q}}_o. \end{aligned}$$

Since $*p_o\wedge \xi \equiv 0$ on ${\mathcal {Q}}_o,$ we have

$$\begin{aligned} *p_o\wedge d\xi (Y)=-(*Y)\wedge \xi , \end{aligned}$$

and thus

$$\begin{aligned} -(d\xi (Y))^T=i_{*p_o}\left( (*Y)\wedge \xi \right) . \end{aligned}$$

We finally observe that this expression is $Y^*(\xi ),$ where $Y^*$ is the adjoint of the map represented by Y: let $Y_{ij}$ be the scalar such that

$$\begin{aligned} Y(u_j)=\sum _{i=1}^kY_{ij}\, u_{n+i},\, \, \, \, \, \, \forall j, \, 1 \le j \le n, \end{aligned}$$

where $(u_1,\ldots ,u_n)$ and $(u_{n+1},\ldots ,u_{n+k})$ are positively oriented, orthonormal bases of $p_o$ and $p_o^{\perp }$ respectively. We have

$$\begin{aligned} Y= & {} \sum _{j=1}^n u_1\wedge \cdots \wedge u_{j-1}\wedge Y(u_j)\wedge u_{j+1}\wedge \cdots \wedge u_n\\= & {} \sum _{i=1}^k\sum _{j=1}^nY_{ij}\, u_1\wedge \cdots \wedge u_{j-1}\wedge u_{n+i}\wedge u_{j+1}\wedge \cdots \wedge u_n \end{aligned}$$

and

$$\begin{aligned} *Y= & {} \sum _{i=1}^k\sum _{j=1}^n Y_{ij}\ *\left( u_1\wedge \cdots \wedge u_{j-1}\wedge u_{n+i}\wedge u_{j+1}\wedge \cdots \wedge u_n\right) \\= & {} -\sum _{i=1}^k\sum _{j=1}^n Y_{ij}\ u_{n+1}\wedge \cdots \wedge u_{n+i-1}\wedge u_{j}\wedge u_{n+i+1}\wedge \cdots \wedge u_{n+k}. \end{aligned}$$

Since $*p_o=u_{n+1}\wedge \cdots \wedge u_{n+k},$ we get

$$\begin{aligned} i_{*p_o}\left( (*Y)\wedge u_{n+i}\right) =\sum _{j=1}^n Y_{ij}\, u_j, \, \, \, \, \, \, \forall j, \, 1 \le j \le k, \end{aligned}$$

and the claim follows. $\square$

Proof of (ii)

Assume that $\xi \in \Gamma (\tau _N);$ then $\xi \circ \overline{\varphi }_o\in \Gamma (E_N),$ and

$$\begin{aligned} B(\xi \circ \overline{\varphi }_o)(X)= & {} (d(\xi \circ \overline{\varphi }_o)(X))^T \\= & {} \big (d\xi (d\overline{\varphi }_o(X)) \big )^T\\= & {} B'(\xi )(d\overline{\varphi }_o(X)) \\= & {} \overline{\varphi }_o^*B'(\xi \circ \overline{\varphi }_o)(X), \quad \quad \forall X\in \Gamma (T\mathcal{M}). \end{aligned}$$

$\square$

Using Lemma 7, we deduce that if $\xi$ belongs to the fibre of ${\overline{\varphi }_o}^*\tau _N$ at the point $x_o\in \mathcal{M},$ then

$$\begin{aligned} B(\xi )(X)=d {\overline{\varphi }_o}_{x_o}(X)^*(\xi ), \quad \forall X\in T_{x_o}\mathcal{M}, \end{aligned}$$

where $d {\overline{\varphi }_o}_{x_o}(X)^*$ is regarded as a map $\overline{\varphi }_o(x_o)^{\perp }\rightarrow \overline{\varphi }_o(x_o)$; this naturally identifies B with $d{\overline{\varphi }}_o.$