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In perfect elastostatics, the unknown is the displacement $x\mapsto y$, where $x\in\Omega\subset{\mathbb R}^3$ is the reference configuration, and $y\in{\mathbb R}^3$. It obeys to an 2nd-order PDEs. When we rewrite the PDE as a 1st-order system, the unknown becomes the deformation gradient $F(x):=\nabla y$.

A fundamental constraint is that the matter cannot interpenetrate. In particular, one must have $\det F>0$ everywhere. This is highly non-trivial, in the sense that the set $P$ of $3\times3$ real matrices with positive determinant has a non-trivial homotopy type. Thanks to the polar factorization, $P$ is homeomorphic to the product of ${\bf SPD}_3$ (a convex cone, something trivial) with ${\bf SO}_3$ whose fundamental group is ${\mathbb Z}_2$.

This raises the following questions. Given a domain $\Omega\subset{\mathbb R}^3$, how many connected components are there in $C(\partial\Omega;P)$ ? How can we determine if an $f\in C(\partial\Omega;P)$ can be lifted as an $F\in C(\Omega;P)$ ? How can we determine if an $f\in C(\partial\Omega;P)$ can be lifted as an $F\in C(\Omega;P)$ such that $F$ is homotopic to the identity ?

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What is $C(\partial \Omega; P)?$ Is it continuous functions with values in $P?$ –  Igor Rivin Sep 25 '11 at 15:17
    
In elastostatics I think you usually assume that the deformation is gradually applied, which means that $F$ is assumed homotopic to the identity. But, it you don't assume this, suppose that the body $\Omega$ is homeomorphic to a solid ball, then $C(\partial\Omega,P)$ is classified by $\pi_2(SO(3))$, which I am pretty sure(?) is 0, since $S^3$ is the double cover. In other cases of $\Omega$ I don't know. If this comment is all wrong, which is possible, somebody please remove it. –  Bob Terrell Sep 25 '11 at 15:44
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Yes, $\pi_2(SO(3))=0,$ and you are correct. –  Igor Rivin Sep 25 '11 at 18:45

2 Answers 2

Complementing Dmitri's answer: one can see $SO(3)\cong \mathbb{P^3}(\mathbb{R})$ as the 3-skeleton of $K(\mathbb{Z}/2,1)\cong \mathbb{P^\infty}(\mathbb{R})$. Assuming $\Omega$ and its boundary are non-pathological, they will have the homotopy type of 2-dimensional $CW$-complexes. If $X$ is a 2-dimensional $CW$-complex, one can deform any mapping $X\to K(\mathbb{Z}/2,1)$ into a mapping to the 2-skeleton and any homotopy between such mappings into a mapping to the 3-skeleton. So $H^1(X,\mathbb{Z}/2)\cong [X,K(\mathbb{Z}/2,1)]\cong [X,SO(3)]$.

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Algori, thanks a lot for the answer! –  Dmitri Sep 25 '11 at 16:24
    
Dmitri -- welcome! –  algori Sep 25 '11 at 16:31

The answer to this question can be formulated in the following way. All maps from $\Omega$ to $P$ are classified up to homotopy by the elements $\alpha\in H^1(\Omega,\mathbb Z_2)$, and maps from $\Omega$ to $P$ are classified by the elements $\beta\in H^1(\partial\Omega,\mathbb Z_2)$. Since we have the inclusion $i:\partial \Omega\to \Omega$ we get the map $i^*: H^1(\Omega,\mathbb Z_2)\to H^1(\partial\Omega,\mathbb Z_2)$. In order to be able to extend any map $\partial \Omega\to P$ to the map $\Omega\to P$ we just need to check that $i^*\alpha=\beta$.

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