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David Roberts ♦
David Roberts ♦
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What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv J. Eur. Math. Soc. 19 (2017), no. 5, pp. 1507–1544, doi:140310.4171/JEMS/698, arXiv:1403.2089.2089

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXivdoi:math/040960510.3934/jgm.2013.5.281, arXiv:math/0409605.

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344, doi:10.3934/jgm.2013.5.xx, arXiv:1209.6576, (pdf).

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in$\mathbb{R}^d$, Journal of Differential equationsEquations 258 (Mar 2015), no. 6, 2010-2053, doi:10.1016/j.jde.2014.11.021, arXiv:1411.4081.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ has many more open neighborhoods, and for these nothing follows.

What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv:1403.2089

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXiv:math/0409605

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344 (pdf)

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in Journal of Differential equations.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ has many more open neighborhoods, and for these nothing follows.

What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. J. Eur. Math. Soc. 19 (2017), no. 5, pp. 1507–1544, doi:10.4171/JEMS/698, arXiv:1403.2089.

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. doi:10.3934/jgm.2013.5.281, arXiv:math/0409605.

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344, doi:10.3934/jgm.2013.5.xx, arXiv:1209.6576, (pdf).

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on $\mathbb{R}^d$, Journal of Differential Equations 258 (Mar 2015), no. 6, 2010-2053, doi:10.1016/j.jde.2014.11.021, arXiv:1411.4081.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ has many more open neighborhoods, and for these nothing follows.

typo corrected
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Peter Michor
Peter Michor
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  • 2
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  • 112

What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv:1403.2089

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXiv:math/0409605

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344 (pdf)

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in Journal of Differential equations.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ and havehas many more open neighborhoods, and for these nothing follows.

What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv:1403.2089

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXiv:math/0409605

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344 (pdf)

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in Journal of Differential equations.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ and have many more open neighborhoods, and for these nothing follows.

What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv:1403.2089

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXiv:math/0409605

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344 (pdf)

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in Journal of Differential equations.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ has many more open neighborhoods, and for these nothing follows.

deleted 34 characters in body
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Peter Michor
Peter Michor
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What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv:1403.2089

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXiv:math/0409605

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344 (pdf)

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in Journal of Differential equations.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ and have many more open neighborhoods (e.g., all $H^k$-open, for $k>s$), and for these nothing follows.

What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv:1403.2089

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXiv:math/0409605

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344 (pdf)

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in Journal of Differential equations.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ and have many more open neighborhoods (e.g., all $H^k$-open, for $k>s$), and for these nothing follows.

What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators is contractible to 1 (slide each generator to 1 along the curve).

EDIT

Some references:

  • (1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv:1403.2089

  • (2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXiv:math/0409605

  • (3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344 (pdf)

  • (4) M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in Journal of Differential equations.

In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) $$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$ Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$. Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally $\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$.

Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large but finite $k$, and we can apply the result with the $H^k$-metric.

But $Diff_c(\mathbb R^n)$ and have many more open neighborhoods, and for these nothing follows.

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Peter Michor
Peter Michor
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Peter Michor
Peter Michor
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