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David Treumann
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This Coste-Shiota result contradicts what I thought I knew about Thom's career. Isn't the point that there are examples where you just can't improve "topological stability" to "smooth stability"? (Semi-algebraic implies smooth, right?) Probably I don't know what "data in Thom's first isotopy lemma" meanssee Todd's comment for a simple explanation.)

Here is a famous example of Whitney's that I am probably misremembering, or never understood in the first place. Take in R^3 the four hypersurfaces x = 0, y = 0, x = y, and x = zy. Let's restrict our attention to where 0 < z < 1. Stratify so that the union of these hypersurfaces is the 2-skeleton and the line x = y = 0 is the 1-skeleton. Consider the map $(x,y,z) \mapsto z$. (Maybe we'd prefer a proper map, so require $x^2 + y^2 \leq 1$ as well and stratify the boundary in the obvious way.) The fibers of this map are R^2's (or disks) stratified by four lines through the origin. One of those lines moves around as z changes.

This map is a submersion on each stratum, so by Thom's isotopy lemma the fibers are all homeomorphic in a stratum-preserving way.

But claim: But any such trivialization must fail to be C^1 along z = 0. For, since any C^1 map from one fiber to another has to preserve the cross ratio of those four lines.

What's going on?

This Coste-Shiota result contradicts what I thought I knew about Thom's career. Isn't the point that there are examples where you just can't improve "topological stability" to "smooth stability"? (Semi-algebraic implies smooth, right?) Probably I don't know what "data in Thom's first isotopy lemma" means.

Here is a famous example of Whitney's that I am probably misremembering, or never understood in the first place. Take in R^3 the four hypersurfaces x = 0, y = 0, x = y, and x = zy. Let's restrict our attention to where 0 < z < 1. Stratify so that the union of these hypersurfaces is the 2-skeleton and the line x = y = 0 is the 1-skeleton. Consider the map $(x,y,z) \mapsto z$. (Maybe we'd prefer a proper map, so require $x^2 + y^2 \leq 1$ as well and stratify the boundary in the obvious way.) The fibers of this map are R^2's (or disks) stratified by four lines through the origin. One of those lines moves around as z changes.

This map is a submersion on each stratum, so by Thom's isotopy lemma the fibers are all homeomorphic in a stratum-preserving way.

But claim: any such trivialization must fail to be C^1 along z = 0. For any C^1 map from one fiber to another has to preserve the cross ratio of those four lines.

What's going on?

This Coste-Shiota result contradicts what I thought I knew about Thom's career. Isn't the point that there are examples where you just can't improve "topological stability" to "smooth stability"? (see Todd's comment for a simple explanation.)

Here is a famous example of Whitney's. Take in R^3 the four hypersurfaces x = 0, y = 0, x = y, and x = zy. Let's restrict our attention to where 0 < z < 1. Stratify so that the union of these hypersurfaces is the 2-skeleton and the line x = y = 0 is the 1-skeleton. Consider the map $(x,y,z) \mapsto z$. (Maybe we'd prefer a proper map, so require $x^2 + y^2 \leq 1$ as well and stratify the boundary in the obvious way.) The fibers of this map are R^2's (or disks) stratified by four lines through the origin. One of those lines moves around as z changes.

This map is a submersion on each stratum, so by Thom's isotopy lemma the fibers are all homeomorphic in a stratum-preserving way. But any such trivialization must fail to be C^1 along z = 0, since any C^1 map from one fiber to another has to preserve the cross ratio of those four lines.

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David Treumann
  • 4.9k
  • 26
  • 36

This Coste-Shiota result contradicts what I thought I knew about Thom's career. Isn't the point that there are examples where you just can't improve "topological stability" to "smooth stability"? (Semi-algebraic implies smooth, right?) Probably I don't know what "data in Thom's first isotopy lemma" means.

Here is a famous example of Whitney's that I am probably misremembering, or never understood in the first place. Take in R^3 the four hypersurfaces x = 0, y = 0, x = y, and x = zy. Let's restrict our attention to where 0 < z < 1. Stratify so that the union of these hypersurfaces is the 2-skeleton and the line x = y = 0 is the 1-skeleton. Consider the map $(x,y,z) \mapsto z$. (Maybe we'd prefer a proper map, so require $x^2 + y^2 \leq 1$ as well and stratify the boundary in the obvious way.) The fibers of this map are R^2's (or disks) stratified by four lines through the origin. One of those lines moves around as z changes.

This map is a submersion on each stratum, so by Thom's isotopy lemma the fibers are all homeomorphic in a stratum-preserving way.

But claim: any such trivialization must fail to be C^1 along z = 0. For any C^1 map from one fiber to another has to preserve the cross ratio of those four lines.

What's going on?