1
$\begingroup$

Let $M$ be a smooth orientable manifold with volume form $\Omega$. Fix two pints $x,y \in M$. Put $A$=all volume preserving diffeomorphism of M which maps $x$ to $y$. Define $B$=All linear volume preserving maps from $T_{x}M$ to $T_{y} M$, with respect to $\Omega_{x}$ and $\Omega_{y}$, respectively. We assume that $A\neq \text{The empty set}.$

My question:$\;$ Is the following map surjective? :

$$\phi:A\rightarrow B\\\phi(f)=Df_{x}$$

We can consider the same question by replacing the volume form by a riemanian metric (or some other structures) and revise $A$ and $B$ to isometries and linear isometries, respectively.

The motivation for the second part is "the spheres". However these are not reliable example because they are homogenous, isometrically.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

Let $M$ be an $n$-manifold endowed with a nonvanishing $n$-form $\Omega$, let $\mathrm{Diff}(M,\Omega)$ denote the group of $\Omega$-preserving diffeomorphisms of $M$, and, for $x\in M$, let $\mathrm{Diff}(M,\Omega,x)$ denote the subgroup that fixes $x$.

Your question, then, reduces to "Is the homomorphism $D:\mathrm{Diff}(M,\Omega,x)\to \mathrm{SL}(T_xM)$ defined by $D(f) = f'(x):T_xM\to T_xM$ surjective?"

The answer is 'yes'. The reason is that the image of $D$ has to be a Lie subgroup of $\mathrm{SL}(T_xM)$, and a simple local construction (see the remark at the end) shows that it must be all of $\mathrm{SL}(T_xM)$.

As for your more general question, this has been considered at length in the literature, beginning with the work of Élie Cartan on what are now called 'pseudo-groups' and continuing with a very extensive development in the 1950s and 1960s by Chern, Kuranishi, Singer, Sternberg, Guillemin, and many others. The basic question is this: "If an $n$-manifold is endowed with a $G$-structure $B\subset \mathcal{F}(M)$ (where $G\subset\mathrm{GL}(n,\mathbb{R})$ is a subgroup and $\mathcal{F}(M)$ is the (co-)frame bundle of $M$, when does $\mathrm{Diff}(M,B)$, the group of diffeomorphisms of $M$ that preserve $B$, act transitively on $M$ and when does it act transitively on $B$?" This latter transitivity is a very strict condition, and it can hold 'locally' without holding globally.

For example, when the $G$-structure is a Riemannian metric, $\mathrm{Diff}(M,B)$ acts transitively on $M$ iff $M$ is homogeneous as a Riemannian manifold, but $\mathrm{Diff}(M,B)$ acts transitively on $B$ iff $M$ has constant sectional curvature and is globally symmetric.

On the other hand, if the $G$-structure is a complex structure on $M$, all of these $G$-structures are locally equivalent, so they are locally homogeneous to all orders. Now, when $M=\mathbb{CP}^2$, the biholomorphisms act transitively on the complex frame bundle, but when $M=\mathbb{CP}^1\times \mathbb{CP}^1$, the biholomorphisms act transitively on $M$ but not on its complex frame bundle.

As a final example, when the $G$-structure is a symplectic structure on $M$ (and $M$ is connected), the symplectomorphisms do act transitively on both $M$ and $B$.

Thus, the answer to your question depends very much on which $G$-structure you want to study.

Remark added at the request of the OP: Showing that the image of the homomorphism $D$ contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$ can be done in a number of ways, but here is one that is relatively 'low-tech': Note that the result is obvious when $n=1$, so assume that $n>1$ from now on. Then, one can choose coordinates $u=(u^i)$ centered on $x\in M$ and defined on a neighborhood $U$ of $x$ in which $$ \Omega = \mathrm{d}u^1\wedge\mathrm{d}u^2\wedge\cdots\wedge\mathrm{d}u^n = \mathrm{d}u $$ There is an $r>0$, so that $u(U)\subset\mathbb{R}^n$ contains the open ball $B_r(0)$ of radius $r$ about $0\in\mathbb{R}^n$. Let $G_0 = \mathrm{Diff}_0\bigl(B_r(0),\mathrm{d}u,0\bigr)$ denote the group of diffeomorphisms $\phi:B_r(0)\to B_r(0)$ that satisfy $\phi(0)=0$, $\phi^*(\mathrm{d}u) = \mathrm{d}u$, and for which there exists a compact set $K_\phi\subset B_r(0)$ such that $\phi$ is the identity outside the compact set $K_\phi$. Clearly, it will be enough to show that the homomorphism $D:G_0\to \mathrm{SL}\bigl(T_0B_r(0)\bigr) = \mathrm{SL}(\mathbb{R}^n)$ is surjective.

To do this, consider the subset $C\subset \mathrm{Hom}_0(\mathbb{R}^n,\mathbb{R}^n)={\frak{sl}}(n,\mathbb{R})$ consisting of the operators that are skew-symmetric with respect to some positive definite inner product on $\mathbb{R}^n$. Let $L\in C$ be such an operator, and let $\langle,\rangle$ on $\mathbb{R}^n$ be a positive definite inner product with respect to which $L$ is skew-symmetric. In particular, the $1$-parameter subgroup $\mathrm{e}^{tL}$ preserves $\langle,\rangle$ and hence the volume form $\mathrm{d}u$. Now, let $\epsilon>0$ be so small that the set of vectors $y\in\mathbb{R}^n$ that satisfy $\langle y,y\rangle \le \epsilon$ is a compact subset of $B_r(0)$. Let $h:\mathbb{R}\to\mathbb{R}$ be a smooth function that is identically $1$ when $0\le t \le \epsilon/2$ and vanishes identically for $t\ge\epsilon$. Now consider the $1$-parameter family of smooth maps $$ f_t(y) = \mathrm{e}^{t\ h(\langle y,y\rangle) L}\ y. $$ This family preserves the level sets of $Q(y) = \langle y,y\rangle$, which are spheres, and isometrically rotates each one, so it preserves the volume form $\mathrm{d}u$. It clearly lies in $G_0$. When $Q(y)\ge\epsilon$, $f_t(y) = y$, and, when $Q(y)\le \epsilon/2$, one has $f_t(y) = \mathrm{e}^{tL}y$. Note that $f_t$ lies in $G_0$. Since $$ D(f_t) = \mathrm{e}^{tL}, $$ it follows that the image of $D$ contains the subset $\mathrm{e}^{C}\subset \mathrm{SL}(n,\mathbb{R})$. In particular, the image of $D$, which is a subgroup of $\mathrm{SL}(n,\mathbb{R})$, contains all of the compact subgroups of $\mathrm{SL}(n,\mathbb{R})$. Thus, it follows that the image of $D$ is all of $\mathrm{SL}(n,\mathbb{R})$, which is what needed to be proved.

$\endgroup$
3
  • $\begingroup$ Prof. Bryant, Thank you very much for your answer. May i ask you to explain the reason of your last statement "There is a neighborhood of Id. which is contained in the image of "D". $\endgroup$ Apr 28, 2014 at 13:34
  • $\begingroup$ @AliTaghavi: OK. See above. $\endgroup$ Apr 28, 2014 at 20:15
  • $\begingroup$ Thank you very much for your interesting answer and very deep information about G-structures. $\endgroup$ Apr 30, 2014 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.