Real projective space and transersality problem

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I am presented with the following problem:

Given $pi: S^3 to mathbbRP^3$, the quotient map to the real projective space.

For any $t in [0,1]$, we define $S_3=(x,y,z,t): x^2+y^2+z^2+t^2=1 subset S^3$

And we define $Lambda: (x,0,0,w): x^2+y^2=1subset S^3$.

Prove that for any $t in [0,1]$: $pi(S_t)$ is a dimension $2$ sub-manifold of $mathbbRP^3$, that intersects $pi(Lambda)$ transversally.

Ok.. so, proving that $pi(S_t)$ is a sooth manifold for $t>0$, is pretty trivial, considering that $S_t$ is a smooth manifold, and $S_t$ is homeomorphic to $pi(S_t)$.

And when $t=0$, just pick local charts that sends an small enough open set back to $S_t$, to specified, strategically chosen open sets, then sending them by the charts of $S_t$.

For the ladder part, I thought of this, since $Y:=pi(Lambda)$ has dimension $1$ and $Z:=pi(S_t)$ dimension $2$, you have that given $xin Ycap Z$,

Since $dim(T_xYcap T_xZ)$ can be at most $1$:

$dimT_xY + dimT_xZ-dim(T_xYcap T_xZ)$ is either $3$ or $2$. (If it is $3$ I am done)

My idea is to prove that $T_xYcap T_xZ$ cannot have dimension $1$, so if $(T_xYcap T_xZ) subset T_xY$ has dimension $1$, you must have $(T_xYcap T_xZ) = T_xY$, hence $T_xY subset T_xZ$.

Now, all I got to do is find an $v in T_xY$ that it is not in $T_xZ$, but my problem is seeing $T_xY$ and $T_xZ$. I tried proving that $pi$ is differentiable submersion but I am not sure if this is true.

Any help would be appreciated.

differential-topology transversality

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asked Aug 3 at 21:38

Bajo Fondo

376213

This question has an open bounty worth +50
reputation from Bajo Fondo ending ending at 2018-08-12 23:33:55Z">in 6 days.

This question has not received enough attention.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, any (smooth) covering map is always a submersion. In fact, it's a local diffeomorphism.
  – Ted Shifrin
  Aug 3 at 22:40
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

I am presented with the following problem:

Given $pi: S^3 to mathbbRP^3$, the quotient map to the real projective space.

For any $t in [0,1]$, we define $S_3=(x,y,z,t): x^2+y^2+z^2+t^2=1 subset S^3$

And we define $Lambda: (x,0,0,w): x^2+y^2=1subset S^3$.

Prove that for any $t in [0,1]$: $pi(S_t)$ is a dimension $2$ sub-manifold of $mathbbRP^3$, that intersects $pi(Lambda)$ transversally.

Ok.. so, proving that $pi(S_t)$ is a sooth manifold for $t>0$, is pretty trivial, considering that $S_t$ is a smooth manifold, and $S_t$ is homeomorphic to $pi(S_t)$.

And when $t=0$, just pick local charts that sends an small enough open set back to $S_t$, to specified, strategically chosen open sets, then sending them by the charts of $S_t$.

For the ladder part, I thought of this, since $Y:=pi(Lambda)$ has dimension $1$ and $Z:=pi(S_t)$ dimension $2$, you have that given $xin Ycap Z$,

Since $dim(T_xYcap T_xZ)$ can be at most $1$:

$dimT_xY + dimT_xZ-dim(T_xYcap T_xZ)$ is either $3$ or $2$. (If it is $3$ I am done)

My idea is to prove that $T_xYcap T_xZ$ cannot have dimension $1$, so if $(T_xYcap T_xZ) subset T_xY$ has dimension $1$, you must have $(T_xYcap T_xZ) = T_xY$, hence $T_xY subset T_xZ$.

Now, all I got to do is find an $v in T_xY$ that it is not in $T_xZ$, but my problem is seeing $T_xY$ and $T_xZ$. I tried proving that $pi$ is differentiable submersion but I am not sure if this is true.

Any help would be appreciated.

differential-topology transversality

share|cite|improve this question

asked Aug 3 at 21:38

Bajo Fondo

376213

This question has an open bounty worth +50
reputation from Bajo Fondo ending ending at 2018-08-12 23:33:55Z">in 6 days.

This question has not received enough attention.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, any (smooth) covering map is always a submersion. In fact, it's a local diffeomorphism.
  – Ted Shifrin
  Aug 3 at 22:40
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I am presented with the following problem:

Given $pi: S^3 to mathbbRP^3$, the quotient map to the real projective space.

For any $t in [0,1]$, we define $S_3=(x,y,z,t): x^2+y^2+z^2+t^2=1 subset S^3$

And we define $Lambda: (x,0,0,w): x^2+y^2=1subset S^3$.

Prove that for any $t in [0,1]$: $pi(S_t)$ is a dimension $2$ sub-manifold of $mathbbRP^3$, that intersects $pi(Lambda)$ transversally.

Ok.. so, proving that $pi(S_t)$ is a sooth manifold for $t>0$, is pretty trivial, considering that $S_t$ is a smooth manifold, and $S_t$ is homeomorphic to $pi(S_t)$.

And when $t=0$, just pick local charts that sends an small enough open set back to $S_t$, to specified, strategically chosen open sets, then sending them by the charts of $S_t$.

For the ladder part, I thought of this, since $Y:=pi(Lambda)$ has dimension $1$ and $Z:=pi(S_t)$ dimension $2$, you have that given $xin Ycap Z$,

Since $dim(T_xYcap T_xZ)$ can be at most $1$:

$dimT_xY + dimT_xZ-dim(T_xYcap T_xZ)$ is either $3$ or $2$. (If it is $3$ I am done)

My idea is to prove that $T_xYcap T_xZ$ cannot have dimension $1$, so if $(T_xYcap T_xZ) subset T_xY$ has dimension $1$, you must have $(T_xYcap T_xZ) = T_xY$, hence $T_xY subset T_xZ$.

Now, all I got to do is find an $v in T_xY$ that it is not in $T_xZ$, but my problem is seeing $T_xY$ and $T_xZ$. I tried proving that $pi$ is differentiable submersion but I am not sure if this is true.

Any help would be appreciated.

differential-topology transversality

share|cite|improve this question

asked Aug 3 at 21:38

Bajo Fondo

376213

I am presented with the following problem:

Given $pi: S^3 to mathbbRP^3$, the quotient map to the real projective space.

For any $t in [0,1]$, we define $S_3=(x,y,z,t): x^2+y^2+z^2+t^2=1 subset S^3$

And we define $Lambda: (x,0,0,w): x^2+y^2=1subset S^3$.

Prove that for any $t in [0,1]$: $pi(S_t)$ is a dimension $2$ sub-manifold of $mathbbRP^3$, that intersects $pi(Lambda)$ transversally.

Ok.. so, proving that $pi(S_t)$ is a sooth manifold for $t>0$, is pretty trivial, considering that $S_t$ is a smooth manifold, and $S_t$ is homeomorphic to $pi(S_t)$.

And when $t=0$, just pick local charts that sends an small enough open set back to $S_t$, to specified, strategically chosen open sets, then sending them by the charts of $S_t$.

For the ladder part, I thought of this, since $Y:=pi(Lambda)$ has dimension $1$ and $Z:=pi(S_t)$ dimension $2$, you have that given $xin Ycap Z$,

Since $dim(T_xYcap T_xZ)$ can be at most $1$:

$dimT_xY + dimT_xZ-dim(T_xYcap T_xZ)$ is either $3$ or $2$. (If it is $3$ I am done)

My idea is to prove that $T_xYcap T_xZ$ cannot have dimension $1$, so if $(T_xYcap T_xZ) subset T_xY$ has dimension $1$, you must have $(T_xYcap T_xZ) = T_xY$, hence $T_xY subset T_xZ$.

Now, all I got to do is find an $v in T_xY$ that it is not in $T_xZ$, but my problem is seeing $T_xY$ and $T_xZ$. I tried proving that $pi$ is differentiable submersion but I am not sure if this is true.

Any help would be appreciated.

differential-topology transversality

share|cite|improve this question

asked Aug 3 at 21:38

Bajo Fondo

376213

share|cite|improve this question

share|cite|improve this question

asked Aug 3 at 21:38

Bajo Fondo

376213

asked Aug 3 at 21:38

Bajo Fondo

376213

asked Aug 3 at 21:38

Bajo Fondo

376213

376213

This question has an open bounty worth +50
reputation from Bajo Fondo ending ending at 2018-08-12 23:33:55Z">in 6 days.

This question has not received enough attention.

This question has an open bounty worth +50
reputation from Bajo Fondo ending ending at 2018-08-12 23:33:55Z">in 6 days.

This question has not received enough attention.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, any (smooth) covering map is always a submersion. In fact, it's a local diffeomorphism.
  – Ted Shifrin
  Aug 3 at 22:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, any (smooth) covering map is always a submersion. In fact, it's a local diffeomorphism.
  – Ted Shifrin
  Aug 3 at 22:40
  

  
  
  
  

Yes, any (smooth) covering map is always a submersion. In fact, it's a local diffeomorphism.
– Ted Shifrin
Aug 3 at 22:40

Yes, any (smooth) covering map is always a submersion. In fact, it's a local diffeomorphism.
– Ted Shifrin
Aug 3 at 22:40

add a comment | 

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