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clean up statement -- although it's maybe a little vague
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Ryan Budney
Ryan Budney
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The answer is no.

Perhaps the simplest counter-example is for vector bundles over $0$-manifolds. They all immerse in $\mathbb R^n$ where $n$ is the dimension of the bundle. This is a considerably better number than $2n-a(n)$.

You might say, "but what about for non-discrete manifolds?" Such manifolds do immerse in $\mathbb R^{2n-a(n)}$ (according to Cohen). The issue is whether or not you can do better for this restricted subclass of manifolds.

Consider for example the line bundles over surfaces. Note that they all immerse in $\mathbb R^4$, which is better than what theThe immersion conjecture would say they all immerse in $\mathbb R^{2\cdot 3 - 1} = \mathbb R^5$ but I think we can argue they immerse in $\mathbb R^4$.

One argument for this is to immerse the surface in $\mathbb R^3$Said another way, and take the normal bundle. You then modify thewe are asking what line bundles are sub-bundles of normal bundlebundles of immersed surfaces in $\mathbb R^4$ to achieve. For orientable surfaces you can quickly generate any line-bundleisomorphism type. Line bundles over surfaces are basically just mod For non-2orientable surfaces I believe you can, as well. You immerse the surface in $\mathbb R^3$ and use the $1$-cocycles, sococycle as a way to guide how you define them by choosing curve systems where"flip" the normal orientation flipsbundle, when thought of as inside the larger -- vaguely speaking$\mathbb R^4$.

The answer is no.

Perhaps the simplest counter-example is for vector bundles over $0$-manifolds. They all immerse in $\mathbb R^n$ where $n$ is the dimension of the bundle. This is a considerably better number than $2n-a(n)$.

You might say, "but what about for non-discrete manifolds?" Such manifolds do immerse in $\mathbb R^{2n-a(n)}$ (according to Cohen). The issue is whether or not you can do better for this restricted subclass of manifolds.

Consider for example the line bundles over surfaces. Note that they all immerse in $\mathbb R^4$, which is better than what the immersion conjecture would say $\mathbb R^{2\cdot 3 - 1} = \mathbb R^5$.

One argument for this is to immerse the surface in $\mathbb R^3$, and take the normal bundle. You then modify the normal bundle in $\mathbb R^4$ to achieve any line-bundle type. Line bundles over surfaces are basically just mod-2 $1$-cocycles, so you define them by choosing curve systems where the normal orientation flips -- vaguely speaking.

The answer is no.

Perhaps the simplest counter-example is for vector bundles over $0$-manifolds. They all immerse in $\mathbb R^n$ where $n$ is the dimension of the bundle. This is a considerably better number than $2n-a(n)$.

You might say, "but what about for non-discrete manifolds?" Such manifolds do immerse in $\mathbb R^{2n-a(n)}$ (according to Cohen). The issue is whether or not you can do better for this restricted subclass of manifolds.

Consider for example the line bundles over surfaces. The immersion conjecture would say they all immerse in $\mathbb R^{2\cdot 3 - 1} = \mathbb R^5$ but I think we can argue they immerse in $\mathbb R^4$.

Said another way, we are asking what line bundles are sub-bundles of normal bundles of immersed surfaces in $\mathbb R^4$. For orientable surfaces you can quickly generate any isomorphism type. For non-orientable surfaces I believe you can, as well. You immerse the surface in $\mathbb R^3$ and use the $1$-cocycle as a way to guide how you "flip" the bundle, when thought of as inside the larger $\mathbb R^4$.

Source Link
Ryan Budney
Ryan Budney
  • 44.4k
  • 2
  • 139
  • 245

The answer is no.

Perhaps the simplest counter-example is for vector bundles over $0$-manifolds. They all immerse in $\mathbb R^n$ where $n$ is the dimension of the bundle. This is a considerably better number than $2n-a(n)$.

You might say, "but what about for non-discrete manifolds?" Such manifolds do immerse in $\mathbb R^{2n-a(n)}$ (according to Cohen). The issue is whether or not you can do better for this restricted subclass of manifolds.

Consider for example the line bundles over surfaces. Note that they all immerse in $\mathbb R^4$, which is better than what the immersion conjecture would say $\mathbb R^{2\cdot 3 - 1} = \mathbb R^5$.

One argument for this is to immerse the surface in $\mathbb R^3$, and take the normal bundle. You then modify the normal bundle in $\mathbb R^4$ to achieve any line-bundle type. Line bundles over surfaces are basically just mod-2 $1$-cocycles, so you define them by choosing curve systems where the normal orientation flips -- vaguely speaking.