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Jukka Kohonen
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The famous hopf theorem says that a smooth map from a oriented closed dimension $p$ manifold to $S^{p}$ is homotopic if and only if $f$ and $g$ have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':

  • Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .

  • Theorem B: If two Pontryagin manifold induced by $f$ and $g$ are frame cobordant, the $f$ and $g$ are homotopic (smooth).

  • Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping $f$.

First, it is well-know that if $f$ and $g$ is smooth homotopic, then they have the same brower degree.

Second, we need to prove that if $f$ and $g$ have the same degree, then they are homotopic. From above three theorems, we only have to prove that $f$ and g have the frame cobordant Pontryagin manifold. Since $dim M=p=dim$$\operatorname{dim} M=p=$ dimension of $p$-sphere, so the corresponding frame are of $0$ dim, i.e. discrete points in $M$, so if we define $sgn(x)=1$$\operatorname{sgn}(x)=1$ or $-1$ for $x$ in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

The famous hopf theorem says that a smooth map from a oriented closed dimension $p$ manifold to $S^{p}$ is homotopic if and only if $f$ and $g$ have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':

  • Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .

  • Theorem B: If two Pontryagin manifold induced by $f$ and $g$ are frame cobordant, the $f$ and $g$ are homotopic (smooth).

  • Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping $f$.

First, it is well-know that if $f$ and $g$ is smooth homotopic, then they have the same brower degree.

Second, we need to prove that if $f$ and $g$ have the same degree, then they are homotopic. From above three theorems, we only have to prove that $f$ and g have the frame cobordant Pontryagin manifold. Since $dim M=p=dim$ of $p$-sphere, so the corresponding frame are of $0$ dim, i.e. discrete points in $M$, so if we define $sgn(x)=1$ or $-1$ for $x$ in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

The famous hopf theorem says that a smooth map from a oriented closed dimension $p$ manifold to $S^{p}$ is homotopic if and only if $f$ and $g$ have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':

  • Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .

  • Theorem B: If two Pontryagin manifold induced by $f$ and $g$ are frame cobordant, the $f$ and $g$ are homotopic (smooth).

  • Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping $f$.

First, it is well-know that if $f$ and $g$ is smooth homotopic, then they have the same brower degree.

Second, we need to prove that if $f$ and $g$ have the same degree, then they are homotopic. From above three theorems, we only have to prove that $f$ and g have the frame cobordant Pontryagin manifold. Since $\operatorname{dim} M=p=$ dimension of $p$-sphere, so the corresponding frame are of $0$ dim, i.e. discrete points in $M$, so if we define $\operatorname{sgn}(x)=1$ or $-1$ for $x$ in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

Added $ symbols wherever appropriate
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The famous hopf theorem says that a smooth map from a oriented closed dimension p$p$ manifold to S^{p}$S^{p}$ is homotopic if and only if f$f$ and g$g$ have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':

  • Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .

  • Theorem B: If two Pontryagin manifold induced by f$f$ and g$g$ are frame cobordant, the f$f$ and g$g$ are homotopic (smooth).

  • Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping f$f$.

First, it is well-know that if f$f$ and g$g$ is smooth homotopic, then they have the same brower degree.

Second, we need to prove that if f$f$ and g$g$ have the same degree, then they are homotopic. From above three theorems, we only have to prove that f$f$ and g have the frame cobordant Pontryagin manifold. Since dim M=p=dim$dim M=p=dim$ of p$p$-sphere, so the corresponding frame are of 0$0$ dim, i.e. discrete points in M$M$, so if we define sgn(x)=1$sgn(x)=1$ or -1$-1$ for x$x$ in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

The famous hopf theorem says that a smooth map from a oriented closed dimension p manifold to S^{p} is homotopic if and only if f and g have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':

  • Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .

  • Theorem B: If two Pontryagin manifold induced by f and g are frame cobordant, the f and g are homotopic (smooth).

  • Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping f.

First, it is well-know that if f and g is smooth homotopic, then they have the same brower degree.

Second, we need to prove that if f and g have the same degree, then they are homotopic. From above three theorems, we only have to prove that f and g have the frame cobordant Pontryagin manifold. Since dim M=p=dim of p-sphere, so the corresponding frame are of 0 dim, i.e. discrete points in M, so if we define sgn(x)=1 or -1 for x in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

The famous hopf theorem says that a smooth map from a oriented closed dimension $p$ manifold to $S^{p}$ is homotopic if and only if $f$ and $g$ have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':

  • Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .

  • Theorem B: If two Pontryagin manifold induced by $f$ and $g$ are frame cobordant, the $f$ and $g$ are homotopic (smooth).

  • Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping $f$.

First, it is well-know that if $f$ and $g$ is smooth homotopic, then they have the same brower degree.

Second, we need to prove that if $f$ and $g$ have the same degree, then they are homotopic. From above three theorems, we only have to prove that $f$ and g have the frame cobordant Pontryagin manifold. Since $dim M=p=dim$ of $p$-sphere, so the corresponding frame are of $0$ dim, i.e. discrete points in $M$, so if we define $sgn(x)=1$ or $-1$ for $x$ in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

Fixed punctuation, spelling, formatting.
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Charles Rezk
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theThe famous hopf theorem says that a smooth map from a oriented closed dimension p manifold to S^{p} is homotopic if and only if f and g have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point': Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold . Theorem B:any two Pontryagin manifold induced by f and g are frame cobordant ,the f and g are homotopic (smooth). Theorem C:any frame cobordism Pontryagin manifold are induced by some smooth mapping f. First

  • Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .

  • Theorem B: If two Pontryagin manifold induced by f and g are frame cobordant, the f and g are homotopic (smooth).

  • Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping f.

First,it it is well-know that if f and g is smooth homotopic  ,then then they have the same brower degree. Second

Second,we we need to prove that if f and g have the same degree, then they are homotopic  .from From above three theorems  ,we we only have to prove that f and g have the frame cobordant Pontryagin manifold.since Since dim M=p=dim of p-sphere,so so the corresponding frame are of 0 dim  ,i i.e. discrete points in M,so so if we define sgn(x)=1 or -1for1 for x in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain mfdmanifold have the same degree(=sum sgn(x)),but but i don't know how to prove that if they have the same degree  ,they they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

the famous hopf theorem says that a smooth map from a oriented closed dimension p manifold to S^{p} is homotopic if and only if f and g have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point': Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold . Theorem B:any two Pontryagin manifold induced by f and g are frame cobordant ,the f and g are homotopic (smooth). Theorem C:any frame cobordism Pontryagin manifold are induced by some smooth mapping f. First ,it is well-know that if f and g is smooth homotopic  ,then they have the same brower degree. Second ,we need to prove that if f and g have the same degree, then they are homotopic  .from above three theorems  ,we only have to prove that f and g have the frame cobordant Pontryagin manifold.since dim M=p=dim of p-sphere,so the corresponding frame are of 0 dim  ,i.e discrete points in M,so if we define sgn(x)=1 or -1for x in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain mfd have the same degree(=sum sgn(x)),but i don't know how to prove that if they have the same degree  ,they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

The famous hopf theorem says that a smooth map from a oriented closed dimension p manifold to S^{p} is homotopic if and only if f and g have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':

  • Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .

  • Theorem B: If two Pontryagin manifold induced by f and g are frame cobordant, the f and g are homotopic (smooth).

  • Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping f.

First, it is well-know that if f and g is smooth homotopic, then they have the same brower degree.

Second, we need to prove that if f and g have the same degree, then they are homotopic. From above three theorems, we only have to prove that f and g have the frame cobordant Pontryagin manifold. Since dim M=p=dim of p-sphere, so the corresponding frame are of 0 dim, i.e. discrete points in M, so if we define sgn(x)=1 or -1 for x in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?

Note : we have the notations and definitions as in Milnor's book.

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