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I am interested in the degree of isotropy of a connected (by arc) manifold in general.

Is it true that every connected manifold M (of dimension n) is maximally isotropic in the sense that you can find a diffeomorphism from M to M that:

  • relates any point $P$ to any point $P'$
  • relates any linear direction $d$ passing through $P$ to any linear direction $d'$ passing through $P'$
  • relates any surface direction $S$ passing through $P$ and $d$, to any surface direction $S'$ passing through $P'$ and $d'$ -...
  • until directional elements of dimension $n-1$

This proposition seems true for me for manifold that are of class $C^1, \cdots, \text{or } C^\infty$. Indeed, a manifold is locally diffeomorphic to an euclidean ball, and we can find a diffeomorphism of the ball into itself that relates any point to any other point, any linear direction to any other linear direction, etc., and which becomes smoothly the identity at the frontier of the ball.

My questions are:

1) Is my previous reasoning correct? 2) Is it always true that a connected manifold is maximally isotropic if we are talking about an analytical $C^\omega$ manifold?

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    $\begingroup$ For $C^\infty$-smooth manifolds all your claims are correct. $\endgroup$
    – asv
    Sep 3, 2014 at 8:58

1 Answer 1

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The answer is yes. The paper

  • Peter W. Michor, Cornelia Vizman: n-transitivity of certain diffeomorphism groups. Acta Math. Univ. Comenianiae 63, 2 (1994),221--225(pdf)

shows that many diffeomorphism groups (real analytic, volume preserving, symplectic, ...) act $N$-transitive for any finite $N$ if $\dim(M)>2$. So it remains to show that for any $x_0$ and any two linearly independent collections of tangent vectors $X^i_1,\dots,X^i_k\in T_{x_0}M$ for $i=0,1$ and $k<\dim(M)$ there exists a diffeomorphism $\phi$ with $\phi(x_0)= x_0$ and $T_{x_0}\phi.X^0_i = X^1_i$ for each $i$. We can do this with the method of the paper, and I do it for a real analytic real diffeomorphism:

We use a complete real analytic Riemannian metric $g$ on $M$. Let $\xi_1,\dots\xi_{n^2-1}$ be a basis of the Lie algebra $\mathfrak{gl}(T_{x_0}M)$, let $X_1,\dots,X_n\in T_{x_0}M$ be an orthonormal basis. Choose $Y_i$ for $i=1,\dots,N$ as real analytic vector fields which satisfy $$|Y_i(x_0)|_g<\epsilon,\quad 1\le i\le n^2-1$$ $$\|\nabla Y_i(x_0) - \xi_i\|_g<\epsilon,\quad 1\le i\le n^2-1,$$ $$|Y_i(x_0)-X_{i-n^2+1}|_g<\epsilon,\quad n^2\le i\le n^2-1+n =: N,$$ $$|Y_i(x)|_g<2$$ for all $x\in M$, for suitable small $\epsilon>0$. Each vector field $Y_i$ is complete, since it is bounded with respect to $g$. The the mapping $$ f(t_1,\dots,t_{n^2-1}) = T_{x_0}(\text{Fl}^{Y_1}_{t_1}\circ\dots\circ \text{Fl}^{Y_N}_{t_N}) $$ from $\mathbb R^{n^2-1}$ to $GL(T_{x_0}M, TM)$ has invertible differential at 0, and thus contains an open set in its image. Therefore each orbit of the map $$\text{Diff}^\omega(M)\ni\phi\mapsto T_{x_0}(\phi)\in GL(T_{x_0}M,TM)$$ contains an open set and thus this mapping is surjective onto the open component containing the identity of $T_{x_0}M$.

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