Let $T^n$ be the $n$-dimensional torus and let $F$ be the set of all volume preserving continuous mappings $f:T^n\to T^n$. I would like to know if $F$ is connected in the sense that for any $f\in F$ there is a continuous mapping $h:[0,1] \to F$ such that $h(0)=I$ and $h(1)=f$ ($F$ is endowed with the sup-norm). If $F$ is not connected, is there any description for the connected component containing the identity $I$?
1 Answer
It's certainly not connected: All of these $F$ lift to maps $\bar F$ from $\mathbb R^d$ to itself. Each of these maps has the property that $\Phi_n(\bar F,x)=\bar F(x+\mathbf n)-F(x)\in \mathbb Z^d$. $\Phi_n$ is a continuous function both of $\bar F$ and $x$, and hence $\Phi_{\mathbf n}(h(t),0)$ is a constant function. In fact $\Phi_{\mathbf n}(h(t),0)=A\mathbf n$ for an integer-valued matrix.
(This is the degree of the map $F$). I suspect that the component of the identity is exactly the set of volume-preserving maps with the same degree, but can't prove this yet.
To see it's not everything, if $B$ is any matrix in $GL(n,\mathbb Z)$, the map $x\mapsto Bx\bmod 1$ is a volume-preserving homeomorphism of the torus with degree $B$.
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$\begingroup$ I think $SL(n,\mathbb{Z})$ is discrete and the matrix degree is a homotopy invariant. so the component of the identity is completly determined. $\endgroup$ Dec 22, 2013 at 22:20
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$\begingroup$ @AnthonyQuas: Why should the component of the identity be exactly the set of volume-preserving maps with the same degree? Already on the circle $S^1$ there are homeomorphisms isotopic to the identity that do not preserve the (1-dim.) volume. $\endgroup$ Dec 22, 2013 at 23:03
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$\begingroup$ @Wlodek: agreed about the 1D case. But: 1 dimension is a very rigid situation. I think you have a lot of flexibility in 2 dimensions. I don't claim a proof though. $\endgroup$ Dec 23, 2013 at 2:05
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$\begingroup$ @AnthonyQuas: Thanks for your answer. The mapping is volume preserving, so it has the same degree (and matrix $A$) as the identity. So I do not see yet any obstructions. $\endgroup$ Dec 28, 2013 at 12:11
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$\begingroup$ @John: The map $(x,y)\mapsto (2x+y,x+y)$ is area-preserving, but has a different degree (in the sense of my answer) than the identity: the maps are not homotopic and therefore my map cannot lie in the component of the identity. $\endgroup$ Dec 30, 2013 at 20:08