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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
asked Aug 3 at 21:38
Bajo Fondo
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reputation from Bajo Fondo ending ending at 2018-08-12 23:33:55Z">in 6 days.
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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
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 |Â
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
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
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
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 |Â
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 |Â
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Yes, any (smooth) covering map is always a submersion. In fact, it's a local diffeomorphism.
– Ted Shifrin
Aug 3 at 22:40