How Does Nakahara Compactify I=[0,1] to S^1 in De Rham Cohomology Proof?

[Silviu]

Hello! In Nakahara's Geometry, Topology and Physics in chapter 6.4.5 second edition, he proves at a point that on a simply connected manifold, the first de Rham cohomology group is trivial. In the proof he defines $\alpha : I=[0,1] \to M$, homotopic to a point. Now, by the rules of integration on a manifold, we have $\int_{\alpha(I)} \omega = \int_I \alpha^* \omega$, where the * represents the pullback. However in the book he writes $\int_{\alpha(I)} \omega = \int_{S^1} \alpha^* \omega$, and he mentions that $I=[0,1]$ in the LHS is compactified to $S^1$. Can someone explain this to me? All the proof is based on the fact that $S^1$ has no boundary, so can someone explain to me how can you go from I to $S^1$? Thank you!

 

[WWGD]

EDIT But this seems confusing: $I$ is compact to start with, and so is $\alpha(I)$ , as the continuous image of $I=[0,1]$ ( So why is he compactifying something that is already compact *). Maybe I am missing something? Does he define $\alpha$ and/or give other/additional conditions on it? Any chance you can scan and post the page where he wrote this part?

* IIRC there are some issues with compactification in that in some spaces the compactification of an already compact space may not be homeomorphic to the original, though I doubt this is the case here..

 

[lavinia]

$α(0)=α(1)$ so $α$ may be viewed as a map from the circle.

 

[Silviu]

$α(0)=α(1)$ so $α$ may be viewed as a map from the circle.

Yes, but he doesn't replace $\alpha$ with $S^1$. He replaces I.

 

[lavinia]

Yes, but he doesn't replace $\alpha$ with $S^1$. He replaces I.

Right.

 

[Silviu]

Right.

Wait I am confused. $\alpha$ is a closed loop, so I can see why would it be homeomorphic to a circle. But I is a closed interval, how can it be homeomorphic to a circle? I think I am missing something.

 

[lavinia]

Wait I am confused. $\alpha$ is a closed loop, so I can see why would it be homeomorphic to a circle. But I is a closed interval, how can it be homeomorphic to a circle? I think I am missing something.

OK. I see your question now.

A loop is defined as a continuous mapping from the closed interval $I$ into a topological space $α: I→M$ such that $α(0) = α(1)$. If one is integrating 1 forms then $α$ must be piecewise smooth. In general $α$ is not a homeomorphism because its image may cross over itself. For instance its image may be a figure eight.

Since $α(0)$ equals $α(1)$ $α$ can be factored through the circle. $I→S^1→M$.

 

[Silviu]

OK. I see your question now.

A loop is defined as a continuous mapping from the closed interval $I$ into a topological space $α: I→M$ such that $α(0) = α(1)$. If one is integrating 1 forms the $α$ must be piece wisesmooth. In general $α$ is not a homeomorphism because its image may cross over itself. For instance its image may be a figure eight.

Since $α(0)$ equals $α(1)$ $α$ can be factored through the circle. $I→S^1→M$.

Oh I see so you basically change the conditions such that $\alpha(x)=\alpha(x+2\pi)$. Thank you! One more question: Why does he defines alpha on I in the first place and not on $S^1$, from the beginning?

 

[lavinia]

Oh I see so you basically change the conditions such that $\alpha(x)=\alpha(x+2\pi)$. Thank you! One more question: Why does he defines alpha on I in the first place and not on $S^1$, from the beginning?

Not sure. But in general one thinks about paths in a space. A path may have different end points. A line integral is just the integral of a 1 form over a piecewise smooth path.

 

[WWGD]

OK. I see your question now.

A loop is defined as a continuous mapping from the closed interval $I$ into a topological space $α: I→M$ such that $α(0) = α(1)$. If one is integrating 1 forms then $α$ must be piecewise smooth. In general $α$ is not a homeomorphism because its image may cross over itself. For instance its image may be a figure eight.

Since $α(0)$ equals $α(1)$ $α$ can be factored through the circle. $I→S^1→M$.

Actually, $\alpha$ is assumed to be contractible, according to the OP. Supposedly author is compactifying the unit interval, which is compact to start with (??) so $\alpha$ cannot be too wild and contractible.

 

[lavinia]

Actually, $\alpha$ is assumed to be contractible, according to the OP. Supposedly author is compactifying the unit interval, which is compact to start with (??) so $\alpha$ cannot be too wild and contractible.

$α$ is "not too wild" because it is smooth/ piecewise smooth. If it were only continuous then it could be wild even if it is null homotopic. For instance, there are continuous closed loops on the sphere that are space filling.

 

[WWGD]

$α$ is "not too wild" because it is smooth/ piecewise smooth. If it were only continuous then it could be wild even if it is null homotopic. For instance, there are continuous closed loops on the sphere that are space filling.

Never mind, I said something dumb: Since M is simply-connected, no matter how wild, the curve can be homotoped to a point within the space.